<중딩, 생명과학을 논문으로 만나다.>




 The article I'll be explaining is "Single master regulatory gene coordinates the evolution and development of butterfly color and iridescence". This article was shared by prof. Robert D. Reed and his team.

  Scientists used to think that multiple genes were responsible for the wing color and patterns of butterflies by interacting with each other. The article itself states that dozens of genes were implicated in wing pattern development as well. But researches showed that only a few genes owned the role of actual, effective causation of the difference in wing color and patterns. The optix gene was one of them. However, the actual developmental function of optix was unclear. This article shows that the optix gene is responsible for the determination of chromatic coloration by repression of melanins and promotion of ommochromesiridescence and wing scale structure in nymphalid butterflies (Also known as "four - footed butterflies")The team used the CRISPR - cas9 technology to drive targeted deletion of the optix gene in four species of the Nymphalidae.




 제가 이번에 설명할 논문은 "Single master regulatory gene coordinates the evolution and development of butterfly color and iridescence" 입니다. 이 논문은 로버트 리드 박사님과 그의 연구팀이 작성했습니다.

  과학자들은 나비들의 날개 색과 무늬들을 결정하는 유전자들은 많으며, 그들이 만들어내는 형질들이 섞여져서 색과 무늬들이 나온다고 생각했었습니다. 논문 자체에서도 여러 유전자들이 날개 무늬와 관련 있음을 알려줍니다. 하지만 연구 결과, 몇몇개의 유전자들만이 실제로 나비 날개의 색 및 무늬 차이의 효과가 크게 나타나게 함을 알게되었습니다. optix (옵틱스) 유전자가 그들 중 하나입니다. 그러나, 옵틱스의 실질적인 기능은 불확실했습니다. 이 논문은 옵틱스 유전자가 멜라닌 억제 및 오모크롬 촉진으로 네발나비과 나비들의 날개 비늘 구조, 날개 색 및 무지개빛 (방향적 구조색)의 결정을 책임짐을 보여줍니다. 팀은 크리스퍼 - 카스9 기술을 이용해 4 종의 네발나비과 나비들의 옵틱스 유전자를 삭제했습니다.


The nymphalids are -


이 글에 나오는 네발나비과 나비들은 - 

  • H. erato
  • A. vanillae
  • V. cardui
  • J. coenia
Next, I'll explain some difficult terms.

다음으로, 어려운 단어들을 해설하겠습니다.
  • Structural coloration: Production of color by microscopically structured surfaces that interfere with light.
  • 구조색: 빛을 간섭하는 마이크로 구조들에 의해 생기는 색.
  • Iridescence: A phenomenon where surfaces change color depending on the angle of view.
  • 방향적 구조색 (흔히 무지개빛이라고 한다): 보는 방향과 각도에 따라 표면의 색이 달라지는 현상.
  • Ommochrome: Biological pigments that occur in the eyes of crustaceans and insects. They have a wide range of colors. In this case, the ommochrome is responsible for the red, orange and brown colors of the butterflies' wings.
  • 오모크롬: 갑각류와 곤충들의 눈에 존재하는 색소로, 넓은 범위의 색을 가지고 있다. 이 경우에는, 나비들의 날개의 빨간색, 주황색, 갈색 색들을 만드는 역할을 한다.
  • Melanin: Biological pigments that are responsible for the dark color (In this case, usually gray and black) of organisms.
  • 멜라닌: 생물체들의 어두운 색 (이 경우에는, 회색과 검은색을 담당한다)을 만드는 색소다.
  • Mosaic: An organism that has two or more populations of cells with different genotypes because of error during mitosis. Mosaics originates from one zygote, which makes them differ from Chimeras (Chimeras originate from multiple zygotes).
  • 모자이크: 수정란이 세포 분열 중 오류로 인해 서로 다른 유전자형을 가진 둘 이상의 세포 집단을 가지게 되는 생물체로, 하나의 수정란에서 시작했기 때문에 둘 이상의 수정란이 혼합된 키메라들과 구분된다.
  • Dorsal: The upper surface of the wing/body of an organism.
  • Dorsal: 생물체의 등쪽, 날개 위쪽을 뜻하는 단어다.
  • VentralThe bottom surface of the wing/body of an organism.
  • Ventral: 생물체의 배쪽, 날개 아래쪽을 뜻하는 단어다.
  • Basal area: The area near the base of the wings.
  • Basal area (기저 구역): 날개가 몸 (머리, 가슴, 배 통틀음)과 가장 가까운 곳 주위 구역.
  • Discal area: The area next to the basal area. You can think of it being the 2/3 part of each wing.
  • Discal area: Basal area 옆 구역으로, 각 날개를 3등분 했을 때, 중간에 해당하는 구역이다. 

                                                                    


Let's talk about H. erato first.


H. erato에 대해 먼저 말하겠습니다.



  You can see that the red ommochrome has changed into black melanin.

  

  빨간색 오모크롬이 검은색 멜라닌으로 바뀌었음을 볼 수 있습니다.



  B shows the obvious color difference between the wild type (Abbreviated as "WT") and the mutant wing patterns. C, C’, C” shows the difference in wing scale morphology (Shape) and color. Each one is the zoomed in version of the previous one. C” shows that the WT wing scale has a pointed structure, while the mutant wing scale has the normal, relatively roundish structure. As mentioned in the article itself, the results showed that the optix gene acts as an "or" switch between ommochrome and melanin.


  B는 야생형 ("WT"로 줄여 표기)과 돌연변이 날개 무늬들의 분명한 색 차이를 보여줍니다. C, C’, C”은 그들의 날개 비늘 모양과 색의 차이를 점점 확대해 보여줍니다. 여기서 C”은 WT 날개 비늘은 뾰족한 모양이며, 돌연변이 비늘은 상대적으로 둥글둥글한, 일반적인 모양을 가짐을 보여줍니다. 논문 자체에서도 말했듯이, 결과들이 옵틱스 유전자가 오모크롬과 멜라닌 사이에 작용하는 "또는" 스위치 역할을 함을 보여주었습니다.



                                                                    



Second is A. vanillae.

두번째는 A. vanillae 입니다.



  Both dorsal and ventral wings show that the orange and brown ommochrome wing scales have turned into black and silver melanin wing scales. The green arrows on the ventral mutant wing show the WT wing scales that turned into silver patches. This was called the "presumptive clone boundaries". I assume it was called that way because the wing scales outside the "clone boundaries" have its while spots unchanged and has minor change in overall wing color (Dark gold-ish forewing top part and hindwing → Light brownish & black forewing top part and black hindwing).


  등쪽과 배쪽 날개에 있는 주황색, 갈색 오모크롬 날개 비늘들이 검은색과 은색 멜라닌을 가진 날개 비늘로 바뀜을 보여줍니다. 돌연변이 나비의 배쪽 날개에 있는 초록색 화살표는 은색으로 변한 야생형 날개 비늘들을 보여줍니다. 이들은 "추정적 클론 경계"라 불렸습니다. 아마 돌연변이의 "클론 경계" 바깥의 날개 비늘들의 하얀색 점들은 바뀌지 않았으며, 전체적인 날개의 색은 변화가 거의 일어나지 않았기 때문으로 저는 생각합니다 (앞날개의 위쪽 어두운 금색 부분과 하얀색 점들을 제외한 뒷날개 전체 → 앞날개의 위쪽은 연한 갈색 & 검은색으로 변하고 뒷날개는 하얀색 점들을 제외하고 검은색이 되었음).



  E shows the zoomed in detail of the knockout clone boundary and the switch of colors between red (WT) and black (Mutant) in the ventral forewing. F was a very interesting picture to see. Some of you reading this might've noticed the three black spots relatively darker than the overall black color of the wing by deletion of the optix gene of the mutant A. vanillae in the first picture that shows A. vanillae. Your eyes weren't wrong. The spots are darker than the usual "black" of the mutant A. vanillae wing. F shows the black spots with white arrows. Also, the black spots can be clearly seen on the ventral wing of the WT A. vanillae. The phenotypic trait of the WT A. vanillae wing didn't change in the case of the mutant A. vanillae wing.

  Here's one more interesting fact: The parts of the A. vanillae wing that didn't change after optix gene knockout are all from the ventral wing. I assume these unchanged patterns are controlled by factors other than the optix gene. G and G’ shows the WT and mutant wing scale morphology from pointed to normal and the color change between WT and the mutant wing scale.


  E는 배쪽 앞날개에 있는 클론 경계를 확대한 모습을 보여주며, 이에서 빨간색 (야생형)에서 검은색 (돌연변이)로 색이 바뀜을 알 수 있습니다. F는 매우 흥미로운 사진이이었습니다. 아마 이 글을 보시는 독자 여러분 중 몇분께서는 A. vanillae를 처음으로 보여주는 사진에서 optix 유전자의 삭제에 의한 돌연변이 A. vanillae의 검은색 배쪽 날개보다 더 어두운 3개의 검은색 점들을 보신 분들도 계실겁니다. 잘못 본 것이 아닙니다. 그 점들은 진짜로 돌연변이 A. vanillae의 검은색 배쪽 날개보다 더 어둡습니다. F에서는 그 점들을 하얀색 화살표로 표시했습니다. 또한, 그 검은 점들은 야생형 A. vanillae의 배쪽 날개에서 눈에 띄게 보입니다. 야생형 A. vanillae 날개의 표현형 형질이 돌연변이의 경우에서도 바뀌지 않은 것입니다.

  흥미로운 점을 또 한가지 더 쓰겠습니다: A. vanillae의 optix 유전자가 삭제된 뒤에도 바뀌지 않은 부분들은 모두 배쪽 날개에 있다는 겁니다. 저는 이러한 변화가 없는 부분들은 optix 유전자 이외의 다른 인자들에 의해 조종되기 때문으로 예상합니다. G와 G’은 날개 비늘의 모양이 뾰족한 모양 (야생형)에서 둥근 일반적인 모양 (돌연변이)로 변화된 모습과 야생형과 돌연변이 날개 비늘의 색 변화를 보여줍니다.



                                                                    


Third is V. cardui.


세번째는 V. cardui 입니다.



  There is a pretty noticeable difference in the mutant ventral wing compared to the previous two butterflies. We can see that the orange part of the WT ventral wing lost its color, meaning the loss of ommochrome has happened as usual. But unlike previous butterflies, hypermelanization (Formation of dark colors in an organism by increased concentration of melanin) was very low. This produced a mutant ventral wing with widespread white colors shown in the picture above.


  돌연변이 나비의 배쪽 날개는 이전 나비들과 다른 특이점을 보여줍니다. 야생형 나비의 배쪽 날개의 주황색 부분들은 색을 잃었으므로, 늘 그랬듯이 오모크롬이 감소/사라졌음을 보여줍니다. 하지만 이전 나비들과 다르게 과멜라닌화 현상 (멜라닌 색소의 농도가 높아짐으로서 생물체한테서 어두운 색이 보이는 현상) 은 매우 적게 나타났습니다. 이는 위 사진과 같이 하얀색이 주로 퍼져있는 돌연변이 배쪽 날개를 만들게 되었습니다.



  B shows melanization of the red/orange parts of the mutant butterflies. C and D shows the phenomenon of loss of ommochrome and low hypermelanization in the ventral forewing and hindwing, respectively.


  B는 돌연변이 나비들의 빨간색/주황색 부분들이 멜라닌화되는 모습을 보여줍니다. C와 D는 배쪽 날개의 앞, 뒷날개의 오모크롬의 감소와 과멜라닌화 현상이 나타나지 않는 모습을 보여줍니다.


  The low hypermelanization didn't happen to all V. cardui butterflies. The team was able to recover late stage pupal V. cardui wings that died before emergence. As shown in the picture above, the mutant late stage pupal ventral V. cardui wings showed hypermelanization in locations where the red ommochromes used to be. The team speculated that this variable strength of ventral wing pattern melanization reflected a dosage effect (The alteration of a phenotype by an increased dosage/amount of the product of a gene), where the phenotypes with stronger melanization representing biallelicoptix deletion clones. However, there's no direct evidence to support the speculation. So the reason for the causation of this phenomenon is yet unknown.


  하지만, 모든 V. cardui 나비들한테서 과멜라닌화 현상이 나타나지 않은 것은 아닙니다. 연구팀은 번데기 말기 시기에 죽은 V. cardui 의 날개를 얻는데에 성공했으며, 위 사진에서 보이듯이, 돌연변이 번데기 말기 시기 V. cardui 는 빨간색 오모크롬이 있던 자리에 과멜라닌화 현상이 나타났습니다. 연구팀은 이러한 배쪽 날개 무늬의 멜라닌화 정도의 차이가 유전자량 효과 (어떤 특정 유전자의 생성물의 양이 증가함으로서 표현형이 변화하는 효과) 를 나타내며, 멜라닌화 현상이 더 강하게 나타난 생물체들은 두 대립 형질 모두 돌연변이가 생겼다고 추측했습니다. 그러나, 이 추측을 뒷받침할만한 증거는 없기 때문에 이 현상이 나타나는 이유는 아직 미확인 상태입니다.



  F shows the conversion of scale morphology from pointed (WT) to normal (Mutant).


  F는 날개 비늘 모양이 뾰족한 모양 (야생형)에서 둥근, 일반적인 모양 (돌연변이)으로 변하는 모습을 보여줍니다.



                                                                    


Fourth is J. coenia.


네번째는 J. coenia 입니다.





  The mutant wings of J. coenia has something unique compared to the previous three. It is structural iridescence. The wings on the very right of picture A are additional phenotypes of mutant J. coenia butterflies. As described in both pictures above, you can see strong, bluish iridescence on the dorsal discal spot and eyespot ring. The ventral wing has less iridescence compared to the dorsal wing. Instead, the ventral wing usually showed melanization of the orange patterns. Heavily "already melanized" WT patterns like the black tips of the forewing didn't go under iridescence.


  돌연변이 J. coenia의 날개들은 이전 나비들과는 다른 특징을 보여줍니다. 눈에 띄는 푸른 방향적 구조색입니다. 사진 A의 맨 오른쪽에 있는 날개들은 돌연변이 J. coenia 나비들의 날개들의 추가적인 표현형을 보여줍니다. 위 두 사진에서 보이듯이, 돌연변이 나비의 등쪽 날개의 2/3 부분과 "눈알" 모양 주변 고리의 구조색이 변형되어 푸른색을 띰을 볼 수 있습니다. 배쪽 날개는 등쪽 날개에 비해 구조색이 덜 변형되었습니다. 그 대신에 배쪽 날개는 주황색 무늬들의 멜라닌화가 주로 일어났습니다. 앞날개의 검은 부분들과 같이 "이미 강하게 멜라닌화된" 야생형 나비의 날개 무늬들은 구조색이 푸른색으로 바뀌지 않았습니다.



  C shows the mutant ventral J. coenia wing shown in the white dotted box in A. As you can see, the orange patterns in the WT ventral wing has undergone melanization.


  C는 A에서 돌연변이 J. coenia의 배쪽 날개에 하얀색 점선으로 이루어진 공간에 보여진 부분을 확대해 보여주었습니다. 야생형 나비의 배쪽 날개에 있던 주황색 무늬들이 멜라닌화되었음을 볼 수 있습니다.



  D shows detail of the iridescence on the ventral hindwing. The "bluish" parts are the parts that has iridescence.


  D는 배쪽 뒷날개에 일어난 구조색의 변형을 확대해 보여줍니다. "푸른" 부분들이 다 구조색이 변형된 부분들입니다.



  E shows the variations of the patterns in dorsal forewings. The left one shows iridescence while the right one shows melanization.


  E는 등쪽 앞날개에 있는 무늬들의 종류를 보여줍니다. 왼쪽은 구조색이 바뀌어 푸른색을 띄고 오른쪽은 무늬들이 멜라닌화되었음을 보여줍니다.



  F shows the difference in morphology of the wing scales between WT and the mutant. The results in J. coenia showed that the optix gene acts as a repressor of structural iridescence in this species.


  F는 야생형과 돌연변이 나비들의 날개 비늘 모양의 차이를 보여줍니다. J. coenia에서 나온 결과들은 optix 유전자가 이 종의 나비들의 날개 비늘의 구조색 변형을 억제하는 기능을 가짐을 보여주었습니다.


                                                                    


  The results from the four nymphalids showed that the optix gene acts as an "or" switch between ommochrome and melanin. But as seen in V. cardui, it may act as an "and" switch in some cases. The optix gene also has an influence to the structure of the wing scales by making them from pointed to normal, round scales.


  4 종의 네발나비과 나비들에서 나온 결과들은 optix 유전자가 오모크롬과 멜라닌 사이에 "또는" 스위치의 역할을 가짐을 보여주었습니다. 하지만 V. cardui 에서 봤듯이, 어떤 경우에는 "그리고" 스위치의 역할을 가지기도 합니다. 또한, optix 유전자는 날개 비늘의 모양을 뾰족한 모양에서 둥근 모양으로 바꿔 날개 비늘 모양/구조에 영향을 줍니다.


                                                                    


  If you look at the bottom wing set of picture A in the case of J. coenia, you'll see that I crossed out the "dorsal" part and fixed it to "ventral". When I was reading the article, I saw picture A having two wing sets both being dorsal wings. I had a bit of suspicion since the previous ones showed both dorsal and ventral wings together. Further reading made me conclude that the second set of wings are ventral. I sent an email to prof. Robert D. Reed and asked him for clarifications.


  J. coenia 의 경우의 사진 A를 보면 제가 "dorsal" (등쪽) 이라 쓴 부분을 "ventral" (배쪽) 이라 고쳐썼음을 볼 수 있을 겁니다. 제가 이 논문을 읽고 있었을 때, 사진 A가 등쪽 날개 두 그룹을 가지고 있음을 보여주었습니다. 이전 나비들은 등쪽과 배쪽 날개 모두 보여주었기 때문에 저는 이에 대해 의심이 들었습니다. 계속 읽어본 결과, 저는 두번째/아래 날개 그룹은 배쪽 날개임을 확신했습니다. 그 뒤, Robert D. Reed 박사님께 이메일을 보내 이에 대해 여쭈어봤습니다.


Here is the email:



 

  A few days later, he replied to me:


  


 

  I was very happy that he read and replied to my email. The reply email confirmed the error and therefore letting me fix the picture's annotation.


I wish this article helped students understand and learn biology better.


<11/17, 2017>


Haein Kim (김해인)



Credits:

Additional sites that helped me understand scientific terms used in the article:


 <동영상이 포함 된 내용을  올립니다.>



The research paper I will explain in this article is 'CRISPR/Cas9 - mediated mutagenesis of the dihydroflavonol - 4 - reductase - B (DFR - B) locus in the Japanese morning glory Ipomoea (Pharbitis) nil'. I will first clarify some words that are difficult to students.


제가 이 글에서 설명할 논문은 'CRISPR/Cas9 - mediated mutagenesis of the dihydroflavonol - 4 - reductase - B (DFR - B) locus in the Japanese morning glory Ipomoea (Pharbitis) nil' 입니다. 먼저, 학생들이 이해하기 어려운 단어들을 설명하겠습니다.

  • Mediate: Bring about a result. (Ex: Physical effects)
  • Mediate: 어떠한 결과를 드러내다.
  • Locus: A fixed position on a chromosome.
  • Locus: 염색체 내에 있는 한 장소 혹은 구간
  • Gene: A locus of DNA.
  • Gene: 유전자. 유전의 기본단위이며, 한 구간 (Locus)에 있는 DNA이라 생각하면 된다.
  • Allele: A variation of a gene.
  • Allele: 대립 형질. 게놈/유전체 내 유전자가 조절하는 하나의 형질에 대한 여러가지 종류.
  • Mutagenesis: A process where genetic information of an organism is changed.
  • Mutagenesis: 돌연변이 생성. 생물의 유전 정보가 바뀌는 과정.

                                                                         



Professor Michiyuki Ono and his team used the CRISPR - Cas9 technology to change the flower petal and stem color of the Japanese morning glory Ipomoea nil (I.nil). The gene dihydroflavonol - 4 - reductase - B (DFR - B) encodes/makes an anthocyanin (A pigment that shows the color purple and blue) biosynthesis enzyme. Anthocyanin is responsible for the purple color of I.nil. Null mutations (Mutations that results in elimination of the target gene's function) in I.nil's DFR - B gene (It will be called InDFR - B from now on. The 'In' is short for I.nil. In other words, InDFR - B is the DFR-B gene of I.nil. The same applies to DFR - A and DFR - C) resulted in anthocyanin - less stems, leaves and flowers. Therefore, that gene was selected as the target gene. The 20bp (Base pair) sequence in the 4th exon (Parts of a gene that encodes a part of the final mature mRNA and therefore contributing in transcription & translation) of InDFR - B was selected as the sgRNA sequence of the CRISPR - Cas9 system.


미치유키 오노 박사님과 그의 연구팀은 크리스퍼 - 카스9 기술을 이용해 일본 나팔꽃 Ipomoea nil (I.nil) 의 꽃잎과 줄기 색을 변형시켰습니다. 유전자 디하이드로플라보놀 - 4 - 리덕테이스 - B (DFR - B)는 안토시아닌 (보라색과 파란색을 띠는 색소) 을 만드는 효소를 만드는 유전자입니다. I.nil의 경우, 안토시아닌은 식물의 보라색/자주색 꽃잎, 줄기 등의 색을 책임집니다. I.nil의 DFR - B 유전자에 대해 그 유전자가 제 기능을 하지 못하게 하는 돌연변이는 안토시아닌이 없는 꽃, 줄기, 잎 등을 보여 그 유전자가 편집 대상으로 선정되었습니다. 그리고 InDFR - B의 넷째 엑손의 (단백질 합성 과정에서 mRNA가 만들어질 때, 단백질 합성에 관여하는 정보가 있어 '성숙한' mRNA에 그 정보가 실리는 DNA의 염기 서열 중 일부분) 20bp (Base pair, 염기쌍시퀀스가 크리스퍼 - 카스9 시스템의 싱글 가이드 RNA (sgRNA) 시퀀스로 선정되었습니다.



There was a problem: I.nil had other DFR species. The DFR in I.nil were InDFR - AInDFR - Band InDFR - C. The two orthologous (Two or more species of sequences that are descended from the same ancestor sequence) sequences of InDFR - B had similar sgRNA sequence patterns. InDFR - A had a 19/20 match and InDFR - C had an 18/20 match with InDFR - B's 20bp sgRNA sequence. This could cause off - target effects which gives us unwanted results and errors. Fortunately, InDFR - B was the only gene that had the TGG PAM sequence which is necessary for the CRISPR - Cas9 system to function. Therefore, chances of possible off - target effects from InDFR - A and InDFR - C were null. Nine more sites were considered as potential off - target sites, but were later revealed to not having any mutations in them.


문제가 한가지 있었습니다. I.nil은 다른 DFR 종류가 있다는 겁니다. I.nil의 DFR 종류들은 InDFR - AInDFR - BInDFR - C입니다. InDFR - B의 20bp sgRNA 시퀀스가 있는 위치에 있는 InDFR - A와 InDFR - C의 염기서열들은 각각 19/20, 18/20 으로 InDFR - B와 일치합니다. 이는 크리스퍼 - 카스9 시스템이 우리가 원치 않는 위치에서 작용을 하는 오프 - 타켓 현상이 일어나게 해 실험 결과에 악영향을 줄 가능성이 있습니다. 다행스럽게도 크리스퍼 - 카스9 시스템이 작동을 하기 위해 필요한 TGG 팸 시퀀스는 InDFR - B에만 있어 InDFR - A와 InDFR - C에서 나타날 수 있는 오프 - 타겟 현상의 출현 가능성은 없는 것으로 결론되었습니다. 그 외 추가적으로 발견된 9개의 구간들이 잠재적인 오프 - 타겟 구간으로 선정되었으나 분석 결과, 그 구간들에서 발견된 돌연변이는 없는 것으로 밝혀졌습니다.


                                                                                               





In the image above, the nucleotides that doesn't match with InDFR - B of InDFR - A  and InDFR - C are shown in black backgrounds. The 20bp sgRNA sequence is shown in blue and the PAM sequence is shown in red. The recognition site sequence / restriction site (The sequence that restriction enzymes recognize and starts functioning) of the restriction enzyme (An enzyme that cleaves DNA into fragments) SpeI is underlined with green and the expected cleavage site of the CRISPR - Cas9 system is indicated with the green triangle

As you can see, the expected cleavage site of the CRISPR - Cas9 system and the recognition site of SpeI overlaps with each other. This means that if proper cleavage in that site by CRISPR - Cas9 and NHEJ (Non Homologous End Joining. It's a method that cells use to repair double strand breaks in DNA) happens, the restriction site would be broken by CRISPR - Cas9. Resulting in no cleavage or 'digestion' of that sequence by SpeI in the restriction digest procedure.

I think this is a very clever way to figure out if the cleavage by CRISPR - Cas9 has occurred as expected. With this method, we can figure out if CRISPR - Cas9 did its job correctly just by analyzing weather or not our target sequence has undergone digestion by SpeI.

Analysis was proceeded via Agarose gel electrophoresis (A method for separating and analyzing DNA, RNA and protiens by their size and charge. Agarose gel is used in this method). Shifting in colors (Example: Purple flower petals turning white) were not observed during plant growth.


위 이미지에서 보면 InDFR - B의 염기서열과 일치하지 않는 InDFR - A와 InDFR - C의 뉴클레오타이드는 검은색 배경으로 칠해져 있으며, InDFR - B의 20bp sgRNA시퀀스는 파란색, 팸 시퀀스는 빨간색으로 나타났습니다. 제한 효소인 SpeI의 인식 부위 / 제한 부위 시퀀스는 초록색 밑줄, 크리스퍼 - 카스9 시스템의 예상 절단 부위는초록색 역삼각형 아래의 구간으로 표시되었습니다.

보시다시피, 크리스퍼 - 카스9 시스템의 예상 절단 부위는 제한 효소 SpeI의 제한 부위와 겹칩니다. 이는 크리스퍼 - 카스9 시스템이 제대로 제 역할을 하고 NHEJ (Non Homologous End Joining 의 약자. 한국어는 비상동말단연결이며, 이는 DNA의 이중가닥 절단을 수선하는 방법 중 하나이다가 발생하면 제한 부위는 크리스퍼 - 카스9 시스템에 의해 절단남을 의미합니다. 그로 인해 제한 효소 SpeI가 그 제한 부위에서 제 기능을 하지 못하게 합니다.

저는 이 방법이 크리스퍼 - 카스9이 예상대로 절단 부위를 제대로 절단했는지 안했는지 구분하는 데에 쓸 수 있는 매우 유용한 방법이라고 생각합니다. 제한 효소 SpeI에 의해 우리의 타겟 시퀀스 (싱글 가이드 RNA 시퀀스) 가 절단되었는지 안되었는지의 여부로 크리스퍼 - 카스9 시스템이 작동을 잘했는지 안했는지의 여부를 알 수 있기 때문입니다. 분석은 아가로스 겔 전기영동법 (DNARNA, 단백질을 크기와 전하량에 따라 분리시키고 분석하는 방법 중 하나로, 아가로스 겔을 이용한다) 으로 시행했으며, 각 개체의 성장기간 동안 꽃잎과 줄기의 색 변화 (예: 자라면서 원래 보라색이었던 꽃잎이 어느날 갑자기 흰색을 띠는 등의 현상. 실험을 통해 만들어진 원래 하얀색이었던 꽃의 꽃잎이 계속 하얀색인 것은 포함되지 않는다) 는 관찰되지 않은 것으로 나타났습니다.


The image above shows us two individuals of I.nil. The left one is a non - transgenic (Having no insertions or deletions) I.nil and the right one is a biallelically mutated (Both alleles having mutations) I.nil. The left one has a purple / violet stem just like the wild type I.nil. However, the right one has a green stem showing phenotypic differences between the two plant individuals.


위 이미지는 두 개의 I.nil 개체를 보여줍니다. 왼쪽은 크리스퍼 - 카스9에 의해 절단 부위가 절단되었으나, 그 뒤 이식 혹은 삭제된 염기가 없는 I.nil이며, 오른쪽은 두 대립 형질 다 돌연변이가 생긴 I.nil입니다. 왼쪽은 자연 상태의 I.nil처럼 보라색 / 자주색 줄기를 가지고 있지만, 오른쪽은 초록색 줄기를 가지고 있어 두 개체 사이에서 발생한 표현형의 차이를 보여줍니다.







NT is short for Non - Transgenic. The maximum of the insertion of nucleotides by NHEJ is 1bp, and the deletion has a range of 2 ~ 98bp. #9-1, #39-1 are biallelic mutants and #8-1, #36-2 are monoallelic (One allele has the mutation) mutants.


NT는 Non - Transgenic의 약자로, 크리스퍼 - 카스9에 의해 절단 부위가 절단되었으나, 그 뒤 이식 혹은 삭제된 염기가 없는 상태를 뜻합니다. NHEJ에 의해 이식된 염기의 최대값은 1bp이며, 삭제된 염기는 2~98bp까지 있는 것으로 나타났습니다. 위 이미지의 #9-1, #39-1은 두 대립 형질 다 돌연변이가 생겼으며, #8-1, #36-2는 한 대립 형질은 돌연변이이고 다른 것은 정상인 것으로 나타났습니다.


                                                                                               




In the image above, you can see three half - ovals that has the letters L1, L2, L3 on them. This is called a meristem. A meristem is a tissue in plants containing undifferentiated cells. They are found in places where plant growth can take place. Each domain where the letters take place are the layers of the meristem. The L1 layer controls most of the flower petal color. The L2 layer controls the stem color and also slightly contributes to the petal color. The L3 layer forms tissues of roots and others. Since the L2 layer slightly contributes to the petal color, #36-2 has a slight shade of purple for the petal color despite having mutations in the L1 layer.

Professor Ono's team concluded the monoallelic mutants #8-1 and #36-2 to be a L2 periclinal chimera and a L1 periclinal chimera respectively. Periclinal means having one component/tissue completely surrounded by the other component/tissue. In genetics, a chimera (Or chimaera) is an organism having two or more sets of DNA and therefore having multiple genotypic traits in one organism. The conclusion makes sense because the two monoallelic mutants are chimeras where the L3 layer is completely surrounded by the L2 layer which is completely surrounded by the L1 layer. 


위 이미지에서 L1L2L3가 적혀있는 반 - 타원이 보이실텐데, 이는 분열조직이라 합니다. 분열조직은 아직 분화되지 않은 세포들로 이루어져 있는 조직으로, 세포 분열 및 분화를 하여 식물을 성장하게 합니다. 글자들이 적혀있는 각 구간들은 분열조직의 입니다. L1층은 꽃잎의 색의 대부분을 조절하고, L2층은 줄기 색을 조절하며, 꽃잎의 색에 조금 관여합니다. L3층은 뿌리 등의 기관들의 조직을 형성합니다. L2층이 꽃잎 색깔에 관여를 하기 때문에 #36-2는 꽃잎의 색이 L1층에 흰색을 띠게 하는 돌연변이를 가진 세포들만 있는데도 불구하고 L2층이 보라색을 띠게 하는 정상 세포들을 가지고 있어 연한 보라색을 띠게 된 것입니다.

오노 박사님의 연구팀은 #8-1과 #36-2를 L2 둘러쌈 키메라L1 둘러쌈 키메라임을 결론지었습니다. 키메라는 여러가지 유전형을 가진 하나의 생물을 뜻하는 단어입니다. #8-1과 #36-2는 L3층이 L2층에 의해 완전히 둘러싸여 있고 또 L2층이 L1층에 의해 완전히 둘러싸여 있는 구조를 가진 키메라라서 이 결론이 맞다는 것을 알 수 있습니다.


The periclinal mutants #8-1, #36-2 bore sectorial flowers that were part white only once in their lifetime. The team thought of two potential reasons for the causation of this phenomenon. The first reason is somaclonal mutations/variations. Somaclonal variation is the variation seen in plants that are produced by plant tissue culture (Techniques used to maintain or grow plant cells and tissues).

The second reason is "Invasion of cells from an adjacent layer". I approached this sentence's meaning as "Cells from layers close to each other in one individual organism like the L1 and L2 layer somehow found a way to 'invade' the other layer and differentiated in that layer. Resulting in white sectorial parts in the flower petals". The white sectors were analyzed and were confirmed to be biallelic. It is interesting how this phenomenon happens only once during the organism's lifetime.


둘러쌈 돌연변이인 #8-1, #36-2는 생애에 딱 한번만 꽃잎에 부채꼴 모양으로 하얀색 부분이 나타나는 것이 관찰되었습니다. 연구팀은 이 현상에 대한 잠재적 이유를 두 가지 냈습니다. 첫 번째 이유는 체세포영양계 변이 때문입니다. 체세포영양계 변이는 식물조직 배양에 의해 나타나는 변이입니다.

두 번째 이유는 "인접한 세포들의 침입" 때문입니다. 저는 이 문장의 의미를 "L1, L2층처럼 하나의 생물 내 인접한 층들 사이 세포들이 상대 층으로 들어가 그 층에서 분화를 해 흰색 부채꼴 모양의 꽃잎이 보이게 했다" 로 이해했습니다. 그 흰색 부분들을 분석한 결과, 두 대립 형질 다 돌연변이가 있는 것으로 드러났습니다. 이 현상은 개체의 생애에 딱 한번만 일어난다는 점이 매우 흥미로웠습니다. 


                                                                       

My explanation ends here. I hope what I wrote helped other students to learn and understand science better.

제 설명은 여기까지입니다. 제가 쓴 내용이 다른 학생들이 과학을 배우고 이해하는데에 도움이 되었으면 합니다.

                                                              

                                                             < 10/17, 2017 >

                                                                Haein Kim

                                                                   김해인


Research paper (해당 논문)https://www.nature.com/articles/s41598-017-10715-1

Images from the Research paper.

First Ipomoea nil image from wikipedia(나팔꽃 이미지): https://en.wikipedia.org/wiki/Ipomoea_nil


Sites that helped me to understand the research paper better:

도움이 된 사이트들:

  • About Chimeras: https://www.britannica.com/topic/chimera-genetics 
  • About Gel electrophoresis: https://en.wikipedia.org/wiki/Gel_electrophoresis
  • About NHEJ: https://en.wikipedia.org/wiki/Non-homologous_end_joining
  • About Exons: https://en.wikipedia.org/wiki/Exon
  • About Somaclonal variations: https://en.wikipedia.org/wiki/Somaclonal_variation
  • About Plant tissue culture: https://en.wikipedia.org/wiki/Plant_tissue_culture
  • About Null mutations: http://medical-dictionary.thefreedictionary.com/null+mutation
  • About Meristems: https://en.wikipedia.org/wiki/Meristem



The article I'll be explaining is "Single master regulatory gene coordinates the evolution and development of butterfly color and iridescence". This article was shared by prof. Robert D. Reed and his team.

  Scientists used to think that multiple genes were responsible for the wing color and patterns of butterflies by interacting with each other. The article itself states that dozens of genes were implicated in wing pattern development as well. But researches showed that only a few genes owned the role of actual, effective causation of the difference in wing color and patterns. The optix gene was one of them. However, the actual developmental function of optix was unclear. This article shows that the optix gene is responsible for the determination of chromatic coloration by repression of melanins and promotion of ommochromesiridescence and wing scale structure in nymphalid butterflies (Also known as "four - footed butterflies")The team used the CRISPR - cas9 technology to drive targeted deletion of the optix gene in four species of the Nymphalidae.


  제가 이번에 설명할 논문은 "Single master regulatory gene coordinates the evolution and development of butterfly color and iridescence" 입니다. 이 논문은 로버트 리드 박사님과 그의 연구팀이 작성했습니다.

  과학자들은 나비들의 날개 색과 무늬들을 결정하는 유전자들은 많으며, 그들이 만들어내는 형질들이 섞여져서 색과 무늬들이 나온다고 생각했었습니다. 논문 자체에서도 여러 유전자들이 날개 무늬와 관련 있음을 알려줍니다. 하지만 연구 결과, 몇몇개의 유전자들만이 실제로 나비 날개의 색 및 무늬 차이의 효과가 크게 나타나게 함을 알게되었습니다. optix (옵틱스) 유전자가 그들 중 하나입니다. 그러나, 옵틱스의 실질적인 기능은 불확실했습니다. 이 논문은 옵틱스 유전자가 멜라닌 억제 및 오모크롬 촉진으로 네발나비과 나비들의 날개 비늘 구조, 날개 색 및 무지개빛 (방향적 구조색)의 결정을 책임짐을 보여줍니다. 팀은 크리스퍼 - 카스9 기술을 이용해 4 종의 네발나비과 나비들의 옵틱스 유전자를 삭제했습니다.


The nymphalids are -


이 글에 나오는 네발나비과 나비들은 - 

  • H. erato
  • A. vanillae
  • V. cardui
  • J. coenia
Next, I'll explain some difficult terms.

다음으로, 어려운 단어들을 해설하겠습니다.
  • Structural coloration: Production of color by microscopically structured surfaces that interfere with light.
  • 구조색: 빛을 간섭하는 마이크로 구조들에 의해 생기는 색.
  • Iridescence: A phenomenon where surfaces change color depending on the angle of view.
  • 방향적 구조색 (흔히 무지개빛이라고 한다): 보는 방향과 각도에 따라 표면의 색이 달라지는 현상.
  • Ommochrome: Biological pigments that occur in the eyes of crustaceans and insects. They have a wide range of colors. In this case, the ommochrome is responsible for the red, orange and brown colors of the butterflies' wings.
  • 오모크롬: 갑각류와 곤충들의 눈에 존재하는 색소로, 넓은 범위의 색을 가지고 있다. 이 경우에는, 나비들의 날개의 빨간색, 주황색, 갈색 색들을 만드는 역할을 한다.
  • Melanin: Biological pigments that are responsible for the dark color (In this case, usually gray and black) of organisms.
  • 멜라닌: 생물체들의 어두운 색 (이 경우에는, 회색과 검은색을 담당한다)을 만드는 색소다.
  • Mosaic: An organism that has two or more populations of cells with different genotypes because of error during mitosis. Mosaics originates from one zygote, which makes them differ from Chimeras (Chimeras originate from multiple zygotes).
  • 모자이크: 수정란이 세포 분열 중 오류로 인해 서로 다른 유전자형을 가진 둘 이상의 세포 집단을 가지게 되는 생물체로, 하나의 수정란에서 시작했기 때문에 둘 이상의 수정란이 혼합된 키메라들과 구분된다.
  • Dorsal: The upper surface of the wing/body of an organism.
  • Dorsal: 생물체의 등쪽, 날개 위쪽을 뜻하는 단어다.
  • VentralThe bottom surface of the wing/body of an organism.
  • Ventral: 생물체의 배쪽, 날개 아래쪽을 뜻하는 단어다.
  • Basal area: The area near the base of the wings.
  • Basal area (기저 구역): 날개가 몸 (머리, 가슴, 배 통틀음)과 가장 가까운 곳 주위 구역.
  • Discal area: The area next to the basal area. You can think of it being the 2/3 part of each wing.
  • Discal area: Basal area 옆 구역으로, 각 날개를 3등분 했을 때, 중간에 해당하는 구역이다. 

                                                                    


Let's talk about H. erato first.


H. erato에 대해 먼저 말하겠습니다.



  You can see that the red ommochrome has changed into black melanin.

  

  빨간색 오모크롬이 검은색 멜라닌으로 바뀌었음을 볼 수 있습니다.



  B shows the obvious color difference between the wild type (Abbreviated as "WT") and the mutant wing patterns. C, C’, C” shows the difference in wing scale morphology (Shape) and color. Each one is the zoomed in version of the previous one. C” shows that the WT wing scale has a pointed structure, while the mutant wing scale has the normal, relatively roundish structure. As mentioned in the article itself, the results showed that the optix gene acts as an "or" switch between ommochrome and melanin.


  B는 야생형 ("WT"로 줄여 표기)과 돌연변이 날개 무늬들의 분명한 색 차이를 보여줍니다. C, C’, C”은 그들의 날개 비늘 모양과 색의 차이를 점점 확대해 보여줍니다. 여기서 C”은 WT 날개 비늘은 뾰족한 모양이며, 돌연변이 비늘은 상대적으로 둥글둥글한, 일반적인 모양을 가짐을 보여줍니다. 논문 자체에서도 말했듯이, 결과들이 옵틱스 유전자가 오모크롬과 멜라닌 사이에 작용하는 "또는" 스위치 역할을 함을 보여주었습니다.


                                                                    


Second is A. vanillae.


두번째는 A. vanillae 입니다.



  Both dorsal and ventral wings show that the orange and brown ommochrome wing scales have turned into black and silver melanin wing scales. The green arrows on the ventral mutant wing show the WT wing scales that turned into silver patches. This was called the "presumptive clone boundaries". I assume it was called that way because the wing scales outside the "clone boundaries" have its while spots unchanged and has minor change in overall wing color (Dark gold-ish forewing top part and hindwing → Light brownish & black forewing top part and black hindwing).


  등쪽과 배쪽 날개에 있는 주황색, 갈색 오모크롬 날개 비늘들이 검은색과 은색 멜라닌을 가진 날개 비늘로 바뀜을 보여줍니다. 돌연변이 나비의 배쪽 날개에 있는 초록색 화살표는 은색으로 변한 야생형 날개 비늘들을 보여줍니다. 이들은 "추정적 클론 경계"라 불렸습니다. 아마 돌연변이의 "클론 경계" 바깥의 날개 비늘들의 하얀색 점들은 바뀌지 않았으며, 전체적인 날개의 색은 변화가 거의 일어나지 않았기 때문으로 저는 생각합니다 (앞날개의 위쪽 어두운 금색 부분과 하얀색 점들을 제외한 뒷날개 전체 → 앞날개의 위쪽은 연한 갈색 & 검은색으로 변하고 뒷날개는 하얀색 점들을 제외하고 검은색이 되었음).



  E shows the zoomed in detail of the knockout clone boundary and the switch of colors between red (WT) and black (Mutant) in the ventral forewing. F was a very interesting picture to see. Some of you reading this might've noticed the three black spots relatively darker than the overall black color of the wing by deletion of the optix gene of the mutant A. vanillae in the first picture that shows A. vanillae. Your eyes weren't wrong. The spots are darker than the usual "black" of the mutant A. vanillae wing. F shows the black spots with white arrows. Also, the black spots can be clearly seen on the ventral wing of the WT A. vanillae. The phenotypic trait of the WT A. vanillae wing didn't change in the case of the mutant A. vanillae wing.

  Here's one more interesting fact: The parts of the A. vanillae wing that didn't change after optix gene knockout are all from the ventral wing. I assume these unchanged patterns are controlled by factors other than the optix gene. G and G’ shows the WT and mutant wing scale morphology from pointed to normal and the color change between WT and the mutant wing scale.


  E는 배쪽 앞날개에 있는 클론 경계를 확대한 모습을 보여주며, 이에서 빨간색 (야생형)에서 검은색 (돌연변이)로 색이 바뀜을 알 수 있습니다. F는 매우 흥미로운 사진이이었습니다. 아마 이 글을 보시는 독자 여러분 중 몇분께서는 A. vanillae를 처음으로 보여주는 사진에서 optix 유전자의 삭제에 의한 돌연변이 A. vanillae의 검은색 배쪽 날개보다 더 어두운 3개의 검은색 점들을 보신 분들도 계실겁니다. 잘못 본 것이 아닙니다. 그 점들은 진짜로 돌연변이 A. vanillae의 검은색 배쪽 날개보다 더 어둡습니다. F에서는 그 점들을 하얀색 화살표로 표시했습니다. 또한, 그 검은 점들은 야생형 A. vanillae의 배쪽 날개에서 눈에 띄게 보입니다. 야생형 A. vanillae 날개의 표현형 형질이 돌연변이의 경우에서도 바뀌지 않은 것입니다.

  흥미로운 점을 또 한가지 더 쓰겠습니다: A. vanillae의 optix 유전자가 삭제된 뒤에도 바뀌지 않은 부분들은 모두 배쪽 날개에 있다는 겁니다. 저는 이러한 변화가 없는 부분들은 optix 유전자 이외의 다른 인자들에 의해 조종되기 때문으로 예상합니다. G와 G’은 날개 비늘의 모양이 뾰족한 모양 (야생형)에서 둥근 일반적인 모양 (돌연변이)로 변화된 모습과 야생형과 돌연변이 날개 비늘의 색 변화를 보여줍니다.


                                                                    


Third is V. cardui.


세번째는 V. cardui 입니다.



  There is a pretty noticeable difference in the mutant ventral wing compared to the previous two butterflies. We can see that the orange part of the WT ventral wing lost its color, meaning the loss of ommochrome has happened as usual. But unlike previous butterflies, hypermelanization (Formation of dark colors in an organism by increased concentration of melanin) was very low. This produced a mutant ventral wing with widespread white colors shown in the picture above.


  돌연변이 나비의 배쪽 날개는 이전 나비들과 다른 특이점을 보여줍니다. 야생형 나비의 배쪽 날개의 주황색 부분들은 색을 잃었으므로, 늘 그랬듯이 오모크롬이 감소/사라졌음을 보여줍니다. 하지만 이전 나비들과 다르게 과멜라닌화 현상 (멜라닌 색소의 농도가 높아짐으로서 생물체한테서 어두운 색이 보이는 현상) 은 매우 적게 나타났습니다. 이는 위 사진과 같이 하얀색이 주로 퍼져있는 돌연변이 배쪽 날개를 만들게 되었습니다.



  B shows melanization of the red/orange parts of the mutant butterflies. C and D shows the phenomenon of loss of ommochrome and low hypermelanization in the ventral forewing and hindwing, respectively.


  B는 돌연변이 나비들의 빨간색/주황색 부분들이 멜라닌화되는 모습을 보여줍니다. C와 D는 배쪽 날개의 앞, 뒷날개의 오모크롬의 감소와 과멜라닌화 현상이 나타나지 않는 모습을 보여줍니다.


  The low hypermelanization didn't happen to all V. cardui butterflies. The team was able to recover late stage pupal V. cardui wings that died before emergence. As shown in the picture above, the mutant late stage pupal ventral V. cardui wings showed hypermelanization in locations where the red ommochromes used to be. The team speculated that this variable strength of ventral wing pattern melanization reflected a dosage effect (The alteration of a phenotype by an increased dosage/amount of the product of a gene), where the phenotypes with stronger melanization representing biallelic optix deletion clones. However, there's no direct evidence to support the speculation. So the reason for the causation of this phenomenon is yet unknown.


  하지만, 모든 V. cardui 나비들한테서 과멜라닌화 현상이 나타나지 않은 것은 아닙니다. 연구팀은 번데기 말기 시기에 죽은 V. cardui 의 날개를 얻는데에 성공했으며, 위 사진에서 보이듯이, 돌연변이 번데기 말기 시기 V. cardui 는 빨간색 오모크롬이 있던 자리에 과멜라닌화 현상이 나타났습니다. 연구팀은 이러한 배쪽 날개 무늬의 멜라닌화 정도의 차이가 유전자량 효과 (어떤 특정 유전자의 생성물의 양이 증가함으로서 표현형이 변화하는 효과) 를 나타내며, 멜라닌화 현상이 더 강하게 나타난 생물체들은 두 대립 형질 모두 돌연변이가 생겼다고 추측했습니다. 그러나, 이 추측을 뒷받침할만한 증거는 없기 때문에 이 현상이 나타나는 이유는 아직 미확인 상태입니다.



  F shows the conversion of scale morphology from pointed (WT) to normal (Mutant).


  F는 날개 비늘 모양이 뾰족한 모양 (야생형)에서 둥근, 일반적인 모양 (돌연변이)으로 변하는 모습을 보여줍니다.

                                                                    


Fourth is J. coenia.


네번째는 J. coenia 입니다.





  The mutant wings of J. coenia has something unique compared to the previous three. It is structural iridescence. The wings on the very right of picture A are additional phenotypes of mutant J. coenia butterflies. As described in both pictures above, you can see strong, bluish iridescence on the dorsal discal spot and eyespot ring. The ventral wing has less iridescence compared to the dorsal wing. Instead, the ventral wing usually showed melanization of the orange patterns. Heavily "already melanized" WT patterns like the black tips of the forewing didn't go under iridescence.


  돌연변이 J. coenia의 날개들은 이전 나비들과는 다른 특징을 보여줍니다. 눈에 띄는 푸른 방향적 구조색입니다. 사진 A의 맨 오른쪽에 있는 날개들은 돌연변이 J. coenia 나비들의 날개들의 추가적인 표현형을 보여줍니다. 위 두 사진에서 보이듯이, 돌연변이 나비의 등쪽 날개의 2/3 부분과 "눈알" 모양 주변 고리의 구조색이 변형되어 푸른색을 띰을 볼 수 있습니다. 배쪽 날개는 등쪽 날개에 비해 구조색이 덜 변형되었습니다. 그 대신에 배쪽 날개는 주황색 무늬들의 멜라닌화가 주로 일어났습니다. 앞날개의 검은 부분들과 같이 "이미 강하게 멜라닌화된" 야생형 나비의 날개 무늬들은 구조색이 푸른색으로 바뀌지 않았습니다.



  C shows the mutant ventral J. coenia wing shown in the white dotted box in A. As you can see, the orange patterns in the WT ventral wing has undergone melanization.


  C는 A에서 돌연변이 J. coenia의 배쪽 날개에 하얀색 점선으로 이루어진 공간에 보여진 부분을 확대해 보여주었습니다. 야생형 나비의 배쪽 날개에 있던 주황색 무늬들이 멜라닌화되었음을 볼 수 있습니다.



  D shows detail of the iridescence on the ventral hindwing. The "bluish" parts are the parts that has iridescence.


  D는 배쪽 뒷날개에 일어난 구조색의 변형을 확대해 보여줍니다. "푸른" 부분들이 다 구조색이 변형된 부분들입니다.



  E shows the variations of the patterns in dorsal forewings. The left one shows iridescence while the right one shows melanization.


  E는 등쪽 앞날개에 있는 무늬들의 종류를 보여줍니다. 왼쪽은 구조색이 바뀌어 푸른색을 띄고 오른쪽은 무늬들이 멜라닌화되었음을 보여줍니다.



  F shows the difference in morphology of the wing scales between WT and the mutant. The results in J. coenia showed that the optix gene acts as a repressor of structural iridescence in this species.


  F는 야생형과 돌연변이 나비들의 날개 비늘 모양의 차이를 보여줍니다. J. coenia에서 나온 결과들은 optix 유전자가 이 종의 나비들의 날개 비늘의 구조색 변형을 억제하는 기능을 가짐을 보여주었습니다.


                                                                    


  The results from the four nymphalids showed that the optix gene acts as an "or" switch between ommochrome and melanin. But as seen in V. cardui, it may act as an "and" switch in some cases. The optix gene also has an influence to the structure of the wing scales by making them from pointed to normal, round scales.


  4 종의 네발나비과 나비들에서 나온 결과들은 optix 유전자가 오모크롬과 멜라닌 사이에 "또는" 스위치의 역할을 가짐을 보여주었습니다. 하지만 V. cardui 에서 봤듯이, 어떤 경우에는 "그리고" 스위치의 역할을 가지기도 합니다. 또한, optix 유전자는 날개 비늘의 모양을 뾰족한 모양에서 둥근 모양으로 바꿔 날개 비늘 모양/구조에 영향을 줍니다.


                                                                    


  If you look at the bottom wing set of picture A in the case of J. coenia, you'll see that I crossed out the "dorsal" part and fixed it to "ventral". When I was reading the article, I saw picture A having two wing sets both being dorsal wings. I had a bit of suspicion since the previous ones showed both dorsal and ventral wings together. Further reading made me conclude that the second set of wings are ventral. I sent an email to prof. Robert D. Reed and asked him for clarifications.


  J. coenia 의 경우의 사진 A를 보면 제가 "dorsal" (등쪽) 이라 쓴 부분을 "ventral" (배쪽) 이라 고쳐썼음을 볼 수 있을 겁니다. 제가 이 논문을 읽고 있었을 때, 사진 A가 등쪽 날개 두 그룹을 가지고 있음을 보여주었습니다. 이전 나비들은 등쪽과 배쪽 날개 모두 보여주었기 때문에 저는 이에 대해 의심이 들었습니다. 계속 읽어본 결과, 저는 두번째/아래 날개 그룹은 배쪽 날개임을 확신했습니다. 그 뒤, Robert D. Reed 박사님께 이메일을 보내 이에 대해 여쭈어봤습니다.


Here is the email:



 

  A few days later, he replied to me:


  


 

  I was very happy that he read and replied to my email. The reply email confirmed the error and therefore letting me fix the picture's annotation.


I wish this article helped students understand and learn biology better.


<11/17, 2017>


Haein Kim (김해인)



Credits:

Additional sites that helped me understand scientific terms used in the article:



The research paper I will explain in this article is 'CRISPR/Cas9 - mediated mutagenesis of the dihydroflavonol - 4 - reductase - B (DFR - B) locus in the Japanese morning glory Ipomoea (Pharbitis) nil'. I will first clarify some words that are difficult to students.


제가 이 글에서 설명할 논문은 'CRISPR/Cas9 - mediated mutagenesis of the dihydroflavonol - 4 - reductase - B (DFR - B) locus in the Japanese morning glory Ipomoea (Pharbitis) nil' 입니다. 먼저, 학생들이 이해하기 어려운 단어들을 설명하겠습니다.

  • Mediate: Bring about a result. (Ex: Physical effects)
  • Mediate: 어떠한 결과를 드러내다.
  • Locus: A fixed position on a chromosome.
  • Locus: 염색체 내에 있는 한 장소 혹은 구간
  • Gene: A locus of DNA.
  • Gene: 유전자. 유전의 기본단위이며, 한 구간 (Locus)에 있는 DNA이라 생각하면 된다.
  • Allele: A variation of a gene.
  • Allele: 대립 형질. 게놈/유전체 내 유전자가 조절하는 하나의 형질에 대한 여러가지 종류.
  • Mutagenesis: A process where genetic information of an organism is changed.
  • Mutagenesis: 돌연변이 생성. 생물의 유전 정보가 바뀌는 과정.

                                                                         



Professor Michiyuki Ono and his team used the CRISPR - Cas9 technology to change the flower petal and stem color of the Japanese morning glory Ipomoea nil (I.nil). The gene dihydroflavonol - 4 - reductase - B (DFR - B) encodes/makes an anthocyanin (A pigment that shows the color purple and blue) biosynthesis enzyme. Anthocyanin is responsible for the purple color of I.nil. Null mutations (Mutations that results in elimination of the target gene's function) in I.nil's DFR - B gene (It will be called InDFR - B from now on. The 'In' is short for I.nil. In other words, InDFR - B is the DFR-B gene of I.nil. The same applies to DFR - A and DFR - C) resulted in anthocyanin - less stems, leaves and flowers. Therefore, that gene was selected as the target gene. The 20bp (Base pair) sequence in the 4th exon (Parts of a gene that encodes a part of the final mature mRNA and therefore contributing in transcription & translation) of InDFR - B was selected as the sgRNA sequence of the CRISPR - Cas9 system.


미치유키 오노 박사님과 그의 연구팀은 크리스퍼 - 카스9 기술을 이용해 일본 나팔꽃 Ipomoea nil (I.nil) 의 꽃잎과 줄기 색을 변형시켰습니다. 유전자 디하이드로플라보놀 - 4 - 리덕테이스 - B (DFR - B)는 안토시아닌 (보라색과 파란색을 띠는 색소) 을 만드는 효소를 만드는 유전자입니다. I.nil의 경우, 안토시아닌은 식물의 보라색/자주색 꽃잎, 줄기 등의 색을 책임집니다. I.nil의 DFR - B 유전자에 대해 그 유전자가 제 기능을 하지 못하게 하는 돌연변이는 안토시아닌이 없는 꽃, 줄기, 잎 등을 보여 그 유전자가 편집 대상으로 선정되었습니다. 그리고 InDFR - B의 넷째 엑손의 (단백질 합성 과정에서 mRNA가 만들어질 때, 단백질 합성에 관여하는 정보가 있어 '성숙한' mRNA에 그 정보가 실리는 DNA의 염기 서열 중 일부분) 20bp (Base pair, 염기쌍시퀀스가 크리스퍼 - 카스9 시스템의 싱글 가이드 RNA (sgRNA) 시퀀스로 선정되었습니다.



There was a problem: I.nil had other DFR species. The DFR in I.nil were InDFR - AInDFR - B and InDFR - C. The two orthologous (Two or more species of sequences that are descended from the same ancestor sequence) sequences of InDFR - B had similar sgRNA sequence patterns. InDFR - A had a 19/20 match and InDFR - C had an 18/20 match with InDFR - B's 20bp sgRNA sequence. This could cause off - target effects which gives us unwanted results and errors. Fortunately, InDFR - B was the only gene that had the TGG PAM sequence which is necessary for the CRISPR - Cas9 system to function. Therefore, chances of possible off - target effects from InDFR - A and InDFR - C were null. Nine more sites were considered as potential off - target sites, but were later revealed to not having any mutations in them.


문제가 한가지 있었습니다. I.nil은 다른 DFR 종류가 있다는 겁니다. I.nil의 DFR 종류들은 InDFR - AInDFR - BInDFR - C입니다. InDFR - B의 20bp sgRNA 시퀀스가 있는 위치에 있는 InDFR - A와 InDFR - C의 염기서열들은 각각 19/20, 18/20 으로 InDFR - B와 일치합니다. 이는 크리스퍼 - 카스9 시스템이 우리가 원치 않는 위치에서 작용을 하는 오프 - 타켓 현상이 일어나게 해 실험 결과에 악영향을 줄 가능성이 있습니다. 다행스럽게도 크리스퍼 - 카스9 시스템이 작동을 하기 위해 필요한 TGG 팸 시퀀스는 InDFR - B에만 있어 InDFR - A와 InDFR - C에서 나타날 수 있는 오프 - 타겟 현상의 출현 가능성은 없는 것으로 결론되었습니다. 그 외 추가적으로 발견된 9개의 구간들이 잠재적인 오프 - 타겟 구간으로 선정되었으나 분석 결과, 그 구간들에서 발견된 돌연변이는 없는 것으로 밝혀졌습니다.


                                                                                               





In the image above, the nucleotides that doesn't match with InDFR - B of InDFR - A  and InDFR - C are shown in black backgrounds. The 20bp sgRNA sequence is shown in blue and the PAM sequence is shown in red. The recognition site sequence / restriction site (The sequence that restriction enzymes recognize and starts functioning) of the restriction enzyme (An enzyme that cleaves DNA into fragments) SpeI is underlined with green and the expected cleavage site of the CRISPR - Cas9 system is indicated with the green triangle. 

As you can see, the expected cleavage site of the CRISPR - Cas9 system and the recognition site of SpeI overlaps with each other. This means that if proper cleavage in that site by CRISPR - Cas9 and NHEJ (Non Homologous End Joining. It's a method that that cells use to repair double strand breaks in DNA) happens, the restriction site would be broken by CRISPR - Cas9. Resulting in no cleavage or 'digestion' of that sequence by SpeI in the restriction digest procedure.

I think this is a very clever way to figure out if the cleavage by CRISPR - Cas9 has occurred as expected. With this method, we can figure out if CRISPR - Cas9 did its job correctly just by analyzing weather or not our target sequence has undergone digestion by SpeI.

Analysis was proceeded via Agarose gel electrophoresis (A method for separating and analyzing DNA, RNA and protiens by their size and charge. Agarose gel is used in this method). Shifting in colors (Example: Purple flower petals turning white) were not observed during plant growth.


위 이미지에서 보면 InDFR - B의 염기서열과 일치하지 않는 InDFR - A와 InDFR - C의 뉴클레오타이드는 검은색 배경으로 칠해져 있으며, InDFR - B의 20bp sgRNA시퀀스는 파란색, 팸 시퀀스는 빨간색으로 나타났습니다. 제한 효소인 SpeI의 인식 부위 / 제한 부위 시퀀스는 초록색 밑줄, 크리스퍼 - 카스9 시스템의 예상 절단 부위는초록색 역삼각형 아래의 구간으로 표시되었습니다.

보시다시피, 크리스퍼 - 카스9 시스템의 예상 절단 부위는 제한 효소 SpeI의 제한 부위와 겹칩니다. 이는 크리스퍼 - 카스9 시스템이 제대로 제 역할을 하고 NHEJ (Non Homologous End Joining 의 약자. 한국어는 비상동말단연결이며, 이는 DNA의 이중가닥 절단을 수선하는 방법 중 하나이다가 발생하면 제한 부위는 크리스퍼 - 카스9 시스템에 의해 절단남을 의미합니다. 그로 인해 제한 효소 SpeI가 그 제한 부위에서 제 기능을 하지 못하게 합니다.

저는 이 방법이 크리스퍼 - 카스9이 예상대로 절단 부위를 제대로 절단했는지 안했는지 구분하는 데에 쓸 수 있는 매우 유용한 방법이라고 생각합니다. 제한 효소 SpeI에 의해 우리의 타겟 시퀀스 (싱글 가이드 RNA 시퀀스) 가 절단되었는지 안되었는지의 여부로 크리스퍼 - 카스9 시스템이 작동을 잘했는지 안했는지의 여부를 알 수 있기 때문입니다. 분석은 아가로스 겔 전기영동법 (DNARNA, 단백질을 크기와 전하량에 따라 분리시키고 분석하는 방법 중 하나로, 아가로스 겔을 이용한다) 으로 시행했으며, 각 개체의 성장기간 동안 꽃잎과 줄기의 색 변화 (예: 자라면서 원래 보라색이었던 꽃잎이 어느날 갑자기 흰색을 띠는 등의 현상. 실험을 통해 만들어진 원래 하얀색이었던 꽃의 꽃잎이 계속 하얀색인 것은 포함되지 않는다) 는 관찰되지 않은 것으로 나타났습니다.


The image above shows us two individuals of I.nil. The left one is a non - transgenic (Having no insertions or deletions) I.nil and the right one is a biallelically mutated (Both alleles having mutations) I.nil. The left one has a purple / violet stem just like the wild type I.nil. However, the right one has a green stem showing phenotypic differences between the two plant individuals.


위 이미지는 두 개의 I.nil 개체를 보여줍니다. 왼쪽은 크리스퍼 - 카스9에 의해 절단 부위가 절단되었으나, 그 뒤 이식 혹은 삭제된 염기가 없는 I.nil이며, 오른쪽은 두 대립 형질 다 돌연변이가 생긴 I.nil입니다. 왼쪽은 자연 상태의 I.nil처럼 보라색 / 자주색 줄기를 가지고 있지만, 오른쪽은 초록색 줄기를 가지고 있어 두 개체 사이에서 발생한 표현형의 차이를 보여줍니다.







NT is short for Non - Transgenic. The maximum of the insertion of nucleotides by NHEJ is 1bp, and the deletion has a range of 2 ~ 98bp. #9-1, #39-1 are biallelic mutants and #8-1, #36-2 are monoallelic (One allele has the mutation) mutants.


NT는 Non - Transgenic의 약자로, 크리스퍼 - 카스9에 의해 절단 부위가 절단되었으나, 그 뒤 이식 혹은 삭제된 염기가 없는 상태를 뜻합니다. NHEJ에 의해 이식된 염기의 최대값은 1bp이며, 삭제된 염기는 2~98bp까지 있는 것으로 나타났습니다. 위 이미지의 #9-1, #39-1은 두 대립 형질 다 돌연변이가 생겼으며, #8-1, #36-2는 한 대립 형질은 돌연변이이고 다른 것은 정상인 것으로 나타났습니다.


                                                                                               




In the image above, you can see three half - ovals that has the letters L1, L2, L3 on them. This is called a meristem. A meristem is a tissue in plants containing undifferentiated cells. They are found in places where plant growth can take place. Each domain where the letters take place are the layers of the meristem. The L1 layer controls most of the flower petal color. The L2 layer controls the stem color and also slightly contributes to the petal color. The L3 layer forms tissues of roots and others. Since the L2 layer slightly contributes to the petal color, #36-2 has a slight shade of purple for the petal color despite having mutations in the L1 layer.

Professor Ono's team concluded the monoallelic mutants #8-1 and #36-2 to be a L2 periclinal chimera and a L1 periclinal chimera respectively. Periclinal means having one component/tissue completely surrounded by the other component/tissue. In genetics, a chimera (Or chimaera) is an organism having two or more sets of DNA and therefore having multiple genotypic traits in one organism. The conclusion makes sense because the two monoallelic mutants are chimeras where the L3 layer is completely surrounded by the L2 layer which is completely surrounded by the L1 layer. 


위 이미지에서 L1L2L3가 적혀있는 반 - 타원이 보이실텐데, 이는 분열조직이라 합니다. 분열조직은 아직 분화되지 않은 세포들로 이루어져 있는 조직으로, 세포 분열 및 분화를 하여 식물을 성장하게 합니다. 글자들이 적혀있는 각 구간들은 분열조직의 입니다. L1층은 꽃잎의 색의 대부분을 조절하고, L2층은 줄기 색을 조절하며, 꽃잎의 색에 조금 관여합니다. L3층은 뿌리 등의 기관들의 조직을 형성합니다. L2층이 꽃잎 색깔에 관여를 하기 때문에 #36-2는 꽃잎의 색이 L1층에 흰색을 띠게 하는 돌연변이를 가진 세포들만 있는데도 불구하고 L2층이 보라색을 띠게 하는 정상 세포들을 가지고 있어 연한 보라색을 띠게 된 것입니다.

오노 박사님의 연구팀은 #8-1과 #36-2를 L2 둘러쌈 키메라L1 둘러쌈 키메라임을 결론지었습니다. 키메라는 여러가지 유전형을 가진 하나의 생물을 뜻하는 단어입니다. #8-1과 #36-2는 L3층이 L2층에 의해 완전히 둘러싸여 있고 또 L2층이 L1층에 의해 완전히 둘러싸여 있는 구조를 가진 키메라라서 이 결론이 맞다는 것을 알 수 있습니다.


The periclinal mutants #8-1, #36-2 bore sectorial flowers that were part white only once in their lifetime. The team thought of two potential reasons for the causation of this phenomenon. The first reason is somaclonal mutations/variations. Somaclonal variation is the variation seen in plants that are produced by plant tissue culture (Techniques used to maintain or grow plant cells and tissues).

The second reason is "Invasion of cells from an adjacent layer". I approached this sentence's meaning as "Cells from layers close to each other in one individual organism like the L1 and L2 layer somehow found a way to 'invade' the other layer and differentiated in that layer. Resulting in white sectorial parts in the flower petals". The white sectors were analyzed and were confirmed to be biallelic. It is interesting how this phenomenon happens only once during the organism's lifetime.


둘러쌈 돌연변이인 #8-1, #36-2는 생애에 딱 한번만 꽃잎에 부채꼴 모양으로 하얀색 부분이 나타나는 것이 관찰되었습니다. 연구팀은 이 현상에 대한 잠재적 이유를 두 가지 냈습니다. 첫 번째 이유는 체세포영양계 변이 때문입니다. 체세포영양계 변이는 식물조직 배양에 의해 나타나는 변이입니다.

두 번째 이유는 "인접한 세포들의 침입" 때문입니다. 저는 이 문장의 의미를 "L1, L2층처럼 하나의 생물 내 인접한 층들 사이 세포들이 상대 층으로 들어가 그 층에서 분화를 해 흰색 부채꼴 모양의 꽃잎이 보이게 했다" 로 이해했습니다. 그 흰색 부분들을 분석한 결과, 두 대립 형질 다 돌연변이가 있는 것으로 드러났습니다. 이 현상은 개체의 생애에 딱 한번만 일어난다는 점이 매우 흥미로웠습니다. 


                                                                       

My explanation ends here. I hope what I wrote helped other students to learn and understand science better.

제 설명은 여기까지입니다. 제가 쓴 내용이 다른 학생들이 과학을 배우고 이해하는데에 도움이 되었으면 합니다.

                                                              

                                                             < 10/17, 2017 >

                                                                Haein Kim

                                                                   김해인


Research paper (해당 논문)https://www.nature.com/articles/s41598-017-10715-1

Images from the Research paper.

First Ipomoea nil image from wikipedia(나팔꽃 이미지): https://en.wikipedia.org/wiki/Ipomoea_nil


Sites that helped me to understand the research paper better:

도움이 된 사이트들:

  • About Chimeras: https://www.britannica.com/topic/chimera-genetics 
  • About Gel electrophoresis: https://en.wikipedia.org/wiki/Gel_electrophoresis
  • About NHEJ: https://en.wikipedia.org/wiki/Non-homologous_end_joining
  • About Exons: https://en.wikipedia.org/wiki/Exon
  • About Somaclonal variations: https://en.wikipedia.org/wiki/Somaclonal_variation
  • About Plant tissue culture: https://en.wikipedia.org/wiki/Plant_tissue_culture
  • About Null mutations: http://medical-dictionary.thefreedictionary.com/null+mutation
  • About Meristems: https://en.wikipedia.org/wiki/Meristem


Last time, we learned about enzymes and their functions.

Now, we are here to talk about how to use some mathematics to understand more about their interactions! Shall we start?

                                          


Enzyme kinetics is the study of chemical reactions catalyzed by enzymes.

It concentrates a lot on the speed - Or I should say "Rate".

By doing this, we can study how we can alter the rate of reaction or it's functionality.

In enzyme kinetics, we have some terms to learn about before we do anything else. The first one is this: E + S ↔ ES → E + P

Where E responds to the enzyme, S as the substrate, ES as the ES complex, and P as the product. The enzyme - substrate complex can rewind back to an enzyme and its substrate. The Enzyme and it's product can do the same too.

However, because products are usually very stable, they hardly rewind back.

Assuming that you know kinetics, let's call the forward reaction of E + S → ES "Reaction 1". And "Reaction 2" for the forward reaction of ES → E + P.

Then, the rate of our reactions will be like this:

Rate1 = K1[E][S] and Rate2 = K2[ES]

Where Rate1 is the rate of reaction 1 and vice versa.

Same for K1 and K2. Got it? Ok.

Next is something called "Vmax". And... What is it?

Say that we have an enzyme that can catalyze 10 reactions per second.

If we had four of them, what would be the maximum rate?

40 reactions per second, right? That's what we call "Vmax". 

The max rate that the enzyme can catalyze per second. That's Vmax.

By the way, Vmax will always occur in every situation. Keep that in mind please.

                                          


Next is a diagram where you plot the reaction rate(V) and the substrate concentration. 


We start with no reaction rate and no substrate.

Then as we add more substrate, our rate goes up, and it becomes more flatter in the end, approaching a certain value. The point where it flattens out is when the enzyme is saturated with substrate.

Which means it's reached it's Vmax.

Now, if you look at the y - axis, you can see 1/2 of the Vmax. 

It's corresponding X axis is also shown as well too.

This X axis - Or substrate concentration is called Km.

Km, is also called the Michaelis Constant. We will talk about why it's called like that later. So this Km, is the substrate concentration when the rate is 1/2 Vmax. When people talk about this, some say that Km is 1/2 Vmax.

That is Wrong! I mean like, Really WRONG!

Km is a Substrate Concentration. Remember this and you'll be fine! 

                                          



Next is the Michaelis Menten Equation!

E + S ↔ ES → E + P

Since we have reaction 1 as the forward for E + S ↔ ES, let's get ourselves a reverse reaction and call it "Reaction -1". We can make a "Reaction -2" for the same reason, but as I said - "Products hardly reverse back". Let's sign that out.

Now, we are going to have something called the Steady - State Assumption.

At steady state, [ES] is constant. Meaning that the rate of formation of [ES] is equal to the rate of dissociation of [ES].

K1[E][S] = K-1[ES] + K2[ES]

Let's group the right side with [ES].

K1[E][S] = (K-1+ K2)[ES]

Now let's move the rate constants on the right and the others on the left!

We will get [E][S]/[ES] = (K-1 + K2)/K1

Turns out, is that this Km is actually (K-1 + K2)/K1!

Then we will get Km =[E][S]/[ES]

Let's move the [ES] from the right to the left side.

[ES]Km = [E][S]

Now let's define a new term called [E]T or "total enzyme" as [E]T = [E] + [ES]

Then we can substitute the [E] from the right to [E]T - [ES]. 

That gives us [ES]Km = ([E]T - [ES])[S].

If we distribute the right side, it's: [ES]Km = [E]T[S] - [ES][S].

By moving the [ES][S] to the left side and grouping the [ES] together, we get

[ES](Km + [S]) = [E]T[S].

If we leave [ES] on the left side only, then we get [ES] = [E]T[S]/(Km + [S]).

Our rate is equal to the rate of formation of product, Or K2[ES].

So Vo = K2[ES]. At high [S], The enzyme will be saturated with substrate. So the Vo = Vmax. And since all the substrate are bonded with the enzyme, there are no free enzymes in the solution. Meaning that [E]T = [ES] as well.

So it also means that Vmax = K2[E]T as well.

Let's multiply K2 on both sides of [ES] = [E]T[S]/(Km + [S]).

Then we get K2[ES] = K2[E]T[S]/(Km + [S]).

By using our terms above, we get Vo = Vmax[S]/(Km + [S]).

This, is the Michaelis Menten Equation!

                                          

In our usual graph, which is in a hyperbolic curve, is not that quite accurate.

For example, if you got the Vmax wrong because of where the curve was headed to, then you would get the Km wrong and everything!

This is where the Lineweaver - Burk plot comes in.

It's derivation is quite simple. 

First, get our Michaelis Menten equation Vo = Vmax[S]/(Km + [S]).

Second, take the inverse of it, making 1/Vo = (Km + [S])/Vmax[S].

Third, rewrite the equation to 1/Vo = Km/Vmax[S] + [S]/Vmax[S].

Simplifying the equation gives us 1/Vo = Km/Vmax * 1/[S] + 1/Vmax.

And that... Is the same way of saying y = mx + b!

Which gives us a straight line on the graph shown below.

With this, we can get a more accurate value of Km and Vmax!

The X - Intercept is -1/Km, and the Y - Intercept is 1/Vmax.

And there are the other values shown in the picture above.

                                         



In enzyme kinetics, we also learn about Inhibitors as well.

Inhibitors are molecules that decreases the enzyme's activity.

The first type of inhibitor is called a Competitive Inhibitor.

Competitive inhibitors - Like the name says, "Competes" with the substrate for the enzyme. And whoever comes first gets the enzyme.

Normally it's for the active site, but it can also be the Allosteric site as well too.

If we write this out, it will be: E + I ↔ EIE + S ↔ ES ↔ E + P.

At steady state, our Km will still be [E][S]/[ES]. 

However, now we will have a new term called KI. Which is [E][I]/[EI].

Now let's remove some terms that aren't necessary.

Let's divide Km by KI. Then we get Km/KI = [S][EI]/[ES][I].

Now it looks like we have nothing else to simplify... Or is it?

Let's look at our toolbox of equations! 

Since[EI] is also a part of [E]T, [E]T = [E] + [ES] + [EI].

If we substitute [EI], then Km/KI = ([E]T - [E] - [ES]) [S]/[ES][I].

Simplifying it gives us Km/KI = [E]T[S]/[ES][I] - [E][S]/[ES][I] - [ES][S]/[ES][I]

Km = [E][S]/[ES], so [E]/[ES] = Km/[S].

And Vmax = K2[E]T, Vo = K2[ES]. Therefore, Vmax/Vo = [E]T/[ES].

Applying it gives us Km/KI = Vmax/Vo *[S]/[I] - Km/[S] * [S]/[I] - [S]/[I].

Grouping the [S]/[I] and moving it to the left side equals...

Km/KI * [I]/[S] = (Vmax/Vo - Km/[S] - 1).

After this, let's put the Vmax/Vo in one side and everything else in the other.

Vmax/Vo = Km/KI *[I]/[S] + Km/[S] + 1.

By grouping, we get Vmax/Vo = Km(1/KI * [I]/[S] + 1/[S]) + 1.

And then Vmax/Vo = Km([I]/KI + 1) 1/[S] + 1.

Dividing both sides by Vmax is: 1/Vo = Km([I]/KI + 1)/Vmax * 1/[S] + 1/Vmax. 

Lastly, we substitute [I]/KI + 1 into a constant called α.

So it becomes 1/Vo = Km*α/Vmax *1/[S] + 1/Vmax.

This is the Lineweaver burk equation of a competitive inhibitor.

If we graph this, we get the fallowing picture below:

The 1/Vmax stays the same. So there is no difference to the Y - Intercept.

For the X - Intercept however, is different.

If we solve what the X intercept is for the Competitive inhibitor, we get -1/Kmα.

And since α is larger then 1, our X intercept is going to be larger then the original X intercept.

In short, Vmax does not change. However, the apparent Km increases.

How does that happen? The Substrate and the Inhibitor are competing for the enzyme. So if we have a LOT of substrate then the inhibitor, we will be able to reach the original Vmax by beating the inhibitor on the way to the enzyme.

So there is no change to the apparent Vmax.

In order to reach the same Vmax, more substrate is needed to reach the same 1/2Vmax. So the enzyme has a low substrate affinity. Therefore the apparent Km increases.

The reason why I used the word "Apparent" is because the Vmax and the Km itself doesn't change. The inhibitors are making it look like the Vmax and Km changed when it's actually decreasing it's activity.

Vmax and Km are Constants. And Constants stay Constant.

                                          



Next are Uncompetitive Inhibitors.

These guys don't "Compete" with the substrate.

Instead, they bind to the Allosteric site when the Enzyme is associated with the substrate. Making an ESI complex. 


Writing it out will show this:

E + S ↔ ES ↔ E + P, ES + I ↔ ESI.

I won't do the LBP derivation all over again. Instead, I'll show it to you directly -

1/Vo = Km/Vmax *1/[S] + α`/Vmax.

Whereas α` = 1 + [I]/K`I. What is this α` and K`I?

K`I is the KI(inhibitor constant) when the inhibitor is uncompetitive,

And α` is α with [I]/K`I in it instead of [I]/KI.

They're just here to distribute which inhibitor they are in.

If we graph it, it will look like this:

It's Y intercept is α`/Vmax. And since α` is also larger then 1, The Y intercept is higher then the original one.

For the X intercept, we get -α`/Km. That makes the X intercept more lower then the original one.

So it means that the apparent Vmax and Km both decreases in an uncompetitive inhibitor. How does that happen?

The uncompetitive inhibitor binds to an ES complex only. So no matter how many substrate you add, the inhibitor is going to stick anyways. So the apparent Vmax decreases.

And as the ESI complexes gets formed, ES is decreasing over time.

So the system tries to balance the equilibrium by accepting more substrate.

This leads to a higher affinity - and thus, low Km.

                                          

The third are mixed inhibitors.

These guys are basically a "mix" of the two inhibitors above.

So if we write them out, it's like this:

E + S ↔ ES  E + P, E + I ↔ EI, ES + I  ESI.

It's Vmax will be decreased due to the effects of the ESI.

However for the Km, it can increase, decrease, or stay the same as before. Because it's a mixture of the competitive inhibitor's behavior(increase) and the uncompetitive inhibitor's behavior(decrease).

It's lineweaver burk equation is 1/Vo = Km*α/Vmax * 1/[S] + α`/Vmax.

In a graph, here's what it looks like:

The X intercept is -α`/Km*α. Which can be smaller or bigger then the original X intercept. It depends which is bigger - α or α`.

If α is bigger, then the X intercept is bigger then the original.

If α` is bigger though, then the X intercept is smaller then the original.

In this situation, α is bigger then α`.

For the Y intercept however, is always α`/Vmax. So the Y intercept is always bigger then the original.

                                          

The last are Noncompetitive inhibitors.

They are a type of mixed inhibitors. However we call it a noncompetitive inhibitor when α = α`!

It's graph looks like this:


The X intercept is the same as the original one since α = α`.

The Y intercept, of course is α`/Vmax. Which is bigger then the original.

                                          

And that is it! The end of Enzyme Kinetics!

I might post some more stuff about things like cooperativity.

Until then, stay tooned!

                                  2016. 5/29

                                  <해인이가>

Image credits section:

Khan academy MCAT videos.

Wikipedia: Main page, Enzyme kinetics section.

Wikibooks: Main page.

EDX: Principles of Biochemistry.

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So far, you know that a lot of chemical reactions are happening in your body.

But all of these reactions are slow. 

Even a simple reaction will take a lot of time! 

Because we all know that reactions occurs when two molecules "crash" in the correct order and all those other right conditions.

What makes all these reactions go fast in a second? 

The answer is right here in this article everybody. 

They are Enzymes. Our body's well - known catalyst!

                                          

Now, an enzyme is a type of catalyst. A catalyst is a substance which makes reactions go faster. In order for enzymes to do it's job, it needs something to react with. That is the Substrate.


The Active site is the part that enzymes bind with the substrates. 

And the others are all called Allosteric sites. We will talk about them later.

Here's a good thing to know - Most enzymes are proteins. So they are made out of amino acids. However, some proteins are made out of RNA molecules which are called Ribozymes. So not every enzymes are not proteins.

                                          

Say, can it be possible to mark how enzymes work in a graph?

The answer is Yes. We can show that with the Reaction coordinate diagram!


This is the Reaction Coordinate Diagram. And it shows how much energy we need/use to complete a reaction.

The X coordinate shows us the process of the reaction and the Y coordinate shows the amount of energy that we need for the reaction to occur.

As you know A is the reactant and B is the product.

For the reaction to be spontaneous, A has a higher energy level then B does.

It might be nice if the reactions just go to A to B straight, 

but that's not the case in most reactions. 

Instead, they need to overcome an "Energy barrier" to be a different molecule. The highest point of the energy barrier is called the Transition state. (‡

The transition state is where the most amount of energy is used and when the molecule is the most unstable.

The Free energy of activation, or activation energy (ΔG‡ or EA) is the energy barrier from A --> B. And it's shown as the difference between A and the transition state. 

Lastly, the Standard free energy change (ΔG˚ or ERXN) is the net change in energy levels between A and B. And it's the energy that is released once the reaction is over. And Enzymes lower the reaction's activation energy to speed up the reaction.

                                           

There are two ways to represent how enzymes bind and react to substrates.

One is called the Lock/Key model and the other one is the induced fit model.



The Lock and Key model was introduced by Emil Fischer. It shows that the enzyme and the substrate binds together perfectly like puzzle pieces. 


The Induced fit model was shown by Daniel Koshland. 

And It is a modification of the Lock and key model. 

The induced fit model shows that the substrate and enzyme does not quite fit perfectly. So the amino acid (side chains) of the active site changes it's position depending on the substrate while not giving any harm to the enzyme itself. After the reaction is done, the active site returns into it's normal position.

The induced fit model is used more often because it shows better how the enzyme and substrate binds.

                                          

Sometimes, an Inhibitor can bind to the enzyme and change the active site.

If that happens, the original substrate cannot bind to the enzyme.

So the substrate competes with the Inhibitor for their own purposes.

We call this Competitive Inhibition.

In competitive inhibition, just like the name says, the substrate and the inhibitor "race" each other to go to the active site.

And whoever goes to the active site first gets the enzyme. 

So if the substrate got in first, the reaction will occur and the inhibitor cannot bind to the enzyme and vice versa.

But the inhibitor doesn't always go to the active site. 

It can go to the allosteric site as well too!

Note: The Allosteric site includes every place of the enzyme except for the active site.

When in a competitive inhibition but the inhibitor goes to the allosteric site, we call this the Allosteric competitive inhibition.

In allosteric competitive inhibition, the same thing happens just like the competitive inhibition, but the inhibitor goes to the allosteric site instead of the active site.

However, For Non competitive inhibition, the substrate and the inhibitor doesn't necessarily compete each other.

Instead, the substrate can bind the enzyme until the inhibitor sticks to the enzyme's allosteric site. (Active site if there's no substrate)

In other words, Both of them can bind, but if that's the case, nothing happens.

                                         



Sometimes, enzymes are named for their functions

An example is DNA Polymerase, which is named because it is a polymer that works on DNA. The suffix "~ase" is tagged along with enzyme names.

If you're familiar with Amino acids, you will know that we can classify amino acids like aromatic, aliphatic groups.

Just like that, we can also classify enzymes as well!

Number 1: Transferase.

Transferases transfer specific functional groups from one molecule to another.

                           A + BX --> AX + B

Where A and B are the molecules and X is a functional group.

A good example of a transferase is an enzyme called Peptidyl transferase.

What this enzyme does is to link amino acids that is derived from the TRNA in translation to a growing polypeptide chain. And it also makes sure that they don't jumble up as well.

For short, Ala - Gly - His + TRNA - Ser --> Ala - Gly - His - Ser + TRNA

Notice how the Serine residue moved from one molecule to another.

Number 2: Ligase.

Ligases Joins two molecules together into one.

                                A + B --> AB

Speaking of Ligases, doesn't it gives you the feeling of our old friend?

That's right! It's DNA Ligase!

As we all know, DNA Ligase Fills up the Okazaki Fragments of the lagging strand at DNA replication. In other words, DNA Ligase joins the two strands of DNA to a single strand.

Number 3: Oxidoreductase.

Long name, but if we look closely, the name has the word "Oxido" and "Reduct". By that, we can figure out that this enzyme uses Oxidation and Reduction!

And that's true. They do use OR to catalyze reactions.

As you may know, Oxidation is when a molecule loses electrons and Reduction is when a molecule gains electrons.

                          A + B: --> A: + B 

NOTE: ":" is a pair of electrons.

An example of a oxidoreductase is something called Lactate Dehydrogenase.

Which it was a pain to learn about because it was in cellular respiration.

Lactate Dehydrogenase (LDH) helps the oxidation/reduction when Pyruvate and NADH reacts to make Lactic acid and NAD+ in Lactic acid fermentation.

              Pyruvate + NADH --> Lactic acid + NAD+

Click to Learn about Cellular respiration/Lactic acid fermentation.

Click to Learn about LDH in Wikipedia.

Number 4: Isomerase.

Isomerases helps to convert a molecule to one of it's Isomers.

                                 A --> B

An enzyme called Phosphoglucose Isomerase catalyzes the reaction of glucose 6 phosphate to one of it's isomer, fructose 6 phosphate.

                    Glucose 6 - P --> Fructose 6 - P 

Number 5: Hydrolase.

Hydrolases Uses water(H2O) to cleave a molecule into two other molecules.

                             A + H20 --> B + C

You might know what hydrolysis is if you learned about Amino acids.

Hydrolysis is the reaction where we use H2O to cleave peptide bonds of a polypeptide chain. And in hydrolysis, there is an enzyme called Serine Hydrolase. (Or Serine Protease) 

This enzyme uses the nucleophilic amino acid Serine at the active site to catalyze hydrolysis.

                       Pro - Cys + H2O --> Pro + Cys

Number 6: Lyase.

Lyases are a bit special. These guys don't use hydrolysis nor oxidation to catalyze a reaction. Instead, what they do is to break bonds of a single molecule and then form a double bond or a ring structure to make it stable.

This type of reaction needs one molecule for the forward reaction, but two for the opposite reaction. 

                                 A --> B + C

For an example of lyase, there's an enzyme called Argininosuccinate lyase.

And it catalyzes the formation of the amino acid Arginine and the dicarboxylic acid Fumarate from the molecule Argininosuccinate.

There, ASL breaks N - C bond and makes a new double bond between the "Broken" molecules to make it stable.

                  Argininosuccinate --> Arginine + Fumarate

And.... That's the 6 types of enzymes shown today!

                                          

Sometimes, enzymes cannot do it's work all by itself alone. 

They need some "Helpers" to do their job properly.

These "Helpers" are non protein molecules. We call them Cofactors.

We can go further into Cofactors and we can see that there are three different types of Cofactors that work differently.   

     


The first are Co - enzymes. They are Organic(C based) cofactors that are loosely bound to enzymes. Co - enzymes carries molecules or electrons.

We can see that in the lactic acid fermentation process, lactate dehydrogenase uses NADH as a coenzyme.

Another type of a coenzyme is Coenzyme A. (CoA) Which carries acyl groups instead of electrons.

     

The next are Prosthetic groups, and these can be organic or inorganic. 

Unlike Coenzymes, which are loosely bound, Prosthetic groups are either tightly bound or even covalently bonded to the enzyme.  

Do you know about Hemoglobin? It's a tetramer protein that carries oxygen around our body. In hemoglobin, there is a molecule called heme.

Heme is a prosthetic group that consists a Fe2+ (Ferrous) ion in the center of an organic heterocyclic ring called a porphyrin. Which is made by four pyrrolic groups connected by Methine bridges. (=CH-)

The last ones are Metal Ions. And they are inorganic

These Metal ions are Iron, Magnesium, copper, etc.

One more thing to say - Have you ever heard about Apoenzymes and Holoenzymes? Apoenzymes are enzymes that needs a Cofactor but doesn't have one. Apoenzymes are inactive - So it doesn't do anything unless the cofactor is bound to the enzyme.

Holoenzymes on the other hand, are Apoenzymes that has a Cofactor on it's side. To make it short, Apoenzyme + Cofactor = Holoenzyme.

                                          

When environmental factors, like PH or temperature changes too much that the protein can't stand, they undergo denaturation. And we've already seen how proteins undergo denaturation in amino acids.

Enzymes work the same way as well too.


For example, in DNA replication, DNA polymerase has to "Deal" with the negatively charged phosphate groups on DNA.

To stabilize the charges, DNA polymerase binds to the cofactor Mg2+ ion with one of it's aspartate residues. Which is deprotonated at normal PH conditions.

At normal PH, the negatively charged Aspartate residue will attract the positively charged Mg2+ ion through electrostatic interactions.

By using that Mg2+ ion, the DNA polymerase can stabilize the charge and can do it's job properly.

However if the PH is changed to low PH, the aspartate residue will be protonated and become into aspartic acid.

If that happens, the aspartic acid can't hold to the Mg2+ ion, and it can't function it's job. Note that H+ is not a cofactor. And a lot of H+ makes it a very acidic environment. Which is bad in this situation.

The other one is Temperature.

You may know what happens to proteins when they are in a very high temperature. They vibrate violently, breaking all the bonds that consist secondary, tertiary and quaternary structures. 

Just like that, enzymes denature in temperature changes as well too.

                                          

That's it for now in about enzymes and it's functions.

In the next time, things will be a bit easy to understand If you learn some basic stuff about Kinetics! The next is all about Enzyme Kinetics you know!

Click this to Learn Kinetics!

                                          

                                  ~Credits Section~

All Images by Khan academy, Wikipedia, and Wikibooks.

I do not own any of these sites.

If you want to see where the images came from or learn more, go to 

Wikipediahttps://en.wikipedia.org/wiki/Main_Page

Khan academy - MCAThttps://www.khanacademy.org/test-prep/mcat

Khan academy - Biologyhttps://www.khanacademy.org/science/biology

Wikibookshttps://en.wikibooks.org/wiki/Main_Page

                                    2016. 2/15

                              By Jane Kim <해인이가>









Welcome back everyone! Long time no see!

You know, I had to come in this late and write the article.... Because I wanted to tell you about the many things I've learned! 

Amino acid's left/right hand, a fact that has PH within it, and the one you have been longing to wait for.... How does AA's turn into well - shaped proteins?

We will be learning those soon and see some others as well!

Alright, Let's go! Into this wonderfully small world!

                                        

The first thing I want you to know is something called "Fischer Projection".

Fischer Projection is a 2D model for a 3D structure.


In Fischer projection, there are two types of AA's - The L and D types!

As you can see in the picture above, the fischer projection shows amino acids in four spots. The Amino group(Amine, or NH2), the carboxylic acid group(COOH), the side chain(R), and the bonded Hydrogen atom.

The L and D amino acids are like our left and right hands. 

Try to overlap your hands together. Do they overlap them perfectly? 

The answer is NO, and just like our hands, L/D AA's also cannot overlap each other perfectly. So we can say that L/D AA's are Non - superimposable to each other. 

The difference in structure is that the NH2 is on the left side for the L - AA while the NH2 is on the right side for the D - AA.

This makes the L amino acid to act as part of the system of proteins, enzymes, and lots of others that has to do with living things.

But for D amino acids, however, are not yet well - known. So it's hard to tell.

(If I get some more info, I'll tell you about it. Ok?) 

                                                    

Now, here is the time I can tell you about Iso - electric point and zwitterions!


If we take a glycine molecule and show it, you will normally have NH2 and COOH. However, there can be a slight change to the structure. That is, when the COOH "knocks off" or donate the H+ to the NH2, making it into COO- and NH3+. This new glycine molecule has both positive and negative charge, while having a net charge of 0.(1 - 1) 

Like form B, we call this a Zwitterion!

Now, let me do a quick review of the difference between PH, PKa, and Ka.

Ka is the strength of an acid, or the "Degree of acid dissociation". 

PKa is the negative log of Ka, so the stronger the acid, the larger the Ka, the smaller the PKa, and vice versa.

PH is the measure(amount) of H+ in the solution. It is described as the negative log of [H+] ---> (The concentration of H+).

If you go back to the picture, you can see that the zwitterion can change it's 'form' depending on the value of PKa.

For form A, it has a PKa of 2.34, which means it's in a acidic solution.

This also tells us that the concentration of H+ has increased as well too.

Because a lot of H+ is in the area, the COO- can form a bond with the H+, making it back to COOH. And the net charge result is 1. (0 + 1)

On the other hand, for form C, it has a PKa of 9.6. 

This means it's in a basic - solution, and has a high PH which means that the concentration of OH- has increased.

Because of the OH- in the solution, the NH3+ can give it's H+ and become into NH2 again. So it's net charge is -1. (0 - 1)

Here's the fun part, you can now know at what PH can this glycine molecule become into a zwitterion! 

How do we do it? By using something called a Iso - electric point!

The Iso - electric point, also named PI for short, and this is the PH point where the AA has a net charge of 0.

How do we find PI? By calculating the total of our PKa's!

PI = (PKa1 + PKa2) / 2

With this equation, let's find the PI of glycine!

The equation is like this: (2.34 + 9.6) / 2 = 11.94 / 2 = 5.97

So we can see that 5.97 is our PI! Yay!

Oh, and one more thing. Different side chains(Acidic/basic) can change the PKa's to be different. And that's why all of the AA's PI's are different.

                                       

As we all know, we have 21 Amino acids in nature. 

Then, how do we sort these individual AAs in groups?

There's a lot of ways to classify Amino acids, but in this case, i'm gonna use it as this way:     

Amino acids --> Hydrophobic / Hydrophilic

Hydrophobic: Aliphatic group, Aromatic group, cyclic group.

Hydrophilic: Amide, S or Se containing group(neutral group), acidic, and basic group.  

Hydrophobic AAs are non - polar and does not like to form a bond with water while Hydrophilic AAs are polar and likes to bond and hang out with water.

We can split hydrophobic AAs to Aliphatic and aromatic groups. 

Both Aliphatic and aromatic groups has side chains are made out of Carbon and Hydrogen. the difference is - Aromatic groups has a cyclic hexagon - like structure(ex: benzene) while aliphatic groups doesn't.

Aliphatic groups includes Amino acids Glycine, Alanine, Valine, Methionine, Leucine, Isoleucine. Even though Methionine does have a sulfur atom, it is bonded to two Carbon atoms instead of an R - S - H bond. Making it nonreactive and non polar.

Aromatic groups includes AAs Phenylalanine, Tryptophan, Histidine ,and Tyrosine(Non polar because of it's benzene).

The Cyclic group includes one Amino acid - Proline. (Because of its odd, cyclic structure)

For the Neutral - or Hydroxyl, Alcohol, Amide, Sulfur, Selenium containing group, is in the hydrophilic group. Neutral groups includes Serine, Threonine, Asparagine, Glutamine, Cysteine(Cystine), and Tyrosine(slightly polar because of it's OH). 

Acidic groups have carboxylic acid(COOH) on their side chain, which is very acidic(strong hydrogen donor). Acidic members are Aspartic acid and Glutamic acid. In their deprotonated state, they become Aspartate and Glutamate.

Basic groups contain Amine groups as their side chain, which is basic.

Basic groups are amino acids Histidine(It's side chain is basic and aromatic), Lysine, Arginine.

These are the ways of classifying AAs in my way.

                                       



Say, after listening to all this amino acid stuff, doesn't it make you wonder?

About how does Amino acids turn and become proteins.

As it turns out, they go into a four step process for their well done protein structure - Primary, Secondary, Tertiary, and Quaternary!

Primary structure is about everything we learned - Amino acids, linked in peptide bonds. The very basics of proteins lives in here.

Secondary structure is where these primary structures folds themselves into Alpha helix(α helix) and Beta sheets.(β sheet)


An Alpha helix gets folded in a way that the C = O bond of the "nth" amino acid can form a Hydrogen bond with the N - H of the "n + 4 th" amino acid.

Which makes the Hydrogen bonds, drawn in dotted lines in the structure above.


For Beta sheets, however, they form a Hydrogen bond with the amino acid that's on the different strand (or "team") rather then the amino acid that's on the same strand.

Each of the chain of AAs in beta sheets are called Beta strands.

There are two kinds of Beta sheets - Parallel and Antiparallel.

Parallel Beta sheets has beta strands that starts and ends on the same side, while Antiparallel Beta sheets has beta strands that starts at a different side.

Just like on the Beta sheet picture above.

Now, what happens if a beta sheet is on 3D?


The Beta strands will be shown as arrows in the direction Nterm -> Cterm

and each Beta sheets will be shown as arrows more or equal to two.

The picture above is one of the beta sheets in 3D. Kinda looks like a hairpin, right? Well, just like that, the structure's name is called a Beta hairpin!

This Beta hairpin occurs when two antiparallel beta strands are together.

Do you see the paperclip - like loop in the 3D picture? That's called a Beta turn. It's made out of 2 ~ 5 amino acids and mostly consists Proline or Glycine.

These loops are important to keep the beta strands to be holded together and even to link with alpha helices. If they fold more, then what forms is called the Tertiary structure.

In Tertiary structure, there are two kinds - Globular and Fibrous. We are going to learn about Globular Tertiary structure. (Because I did not learn deeply to Fibrous Tertiary structures yet.)


Starting from Tertiary structure, you can call the whole thing a protein from now on. Globular proteins fold themselves as a ball shape in order to let it's hydrophobic stuff to the inside and the hydrophilic stuff on the outside.

This is called Hydrophobic interactions.

As well as the Hydrogen bond from the secondary structure, there are other bonds in tertiary structure to keep it stable.

This include Ionic bonds, and Disulfide bridges.(which is pretty hard to make, since they only happen in oxidizing environments) 

Quaternary structure are multiples of these tertiary proteins that are together.

Quaternary structures use the bonds that stabilizes tertiary structures.

An Example of a Globular quaternary structure is Hemoglobin. Which is a four - protein quaternary structure.

In quaternary structures, each protein that makes the quaternary structure is called a Subunit. When two of them are together, we call it a dimer.

For three of them, it's a trimer. Four is tetramer. And more then that are all called multimers. Each with their own name.

                                           

Normal proteins are formed with the steps above. But sometimes there can be an error in the step that causes the protein to be "Denatured".

Denaturation occurs if the protein gets enough "stress" to make it's structures to be broken and lowered down to the previous structure.

One thing to note though, Denaturation does not effect the Primary structure.

Now here are the Top 3 ways to Denature a protein:

Number 1: Heat.

Heat makes a lot of (Kinetic) energy that is strong enough to make the molecules vibrate rapidly. This breaks the Hydrogen bond of the secondary, tertiary, quaternary structure.

Number 2: Acid/Bases.

If an acid makes the solution to go below the PI, the concentration of H+ will increase. Making the negatively charged anion to be neutral.(Same with bases) Breaking the ionic bond of the tertiary, quaternary structure.

Number 3: Alcohols.

Alcohols like ethanol(CH3CH2OH), has this OH(Hydroxyl group) that can break the old hydrogen bond and form a new one. Making the secondary, tertiary, and quaternary structure to be denatured.

                                            

This is it, The end of this article!

The next one will be all about Enzymes.

Happy new year everyone, and happy studying!

NOTE: All pictures from Khan academy and Wikipedia. 

I do not own this picture nor site.

If you want more information, go to this link: 

https://www.khanacademy.org/test-prep/mcat/biomolecules

https://en.wikipedia.org/wiki/Main_Page

                                    2016 1/05

                                    By Jane Kim

                                    <해인이가>

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Alright! I'm in a spectacular feeling about this biology article!

This time, we are going to learn about the building blocks of proteins,

Amino Acids!

21 of these little Amino acids all have different things that it can do - 

And we, are going study the basics of this adventure!

                                        

First of all, I would like you to know where amino acids are even a part of the body. Let's give you an example of a protein called "Hemoglobin".

Hemoglobin is a protein that's inside the red blood cells, and it's job is to pick up oxygens and bring it to the tissues(cells) all over the body.

The tissues uses the oxygen to generate ATP, which makes energy in order to live. And Amino acids are the ones that makes this important hemoglobin.

Here is the structure of a amino acid:

Amino acids are made by containing amine(NH2) and a carboxylic acid(-COOH) and a R and H.

Can you see the letter R? That's the side chain of the amino acid.

It's difference is what makes the amino acids all different.

Also, if you learned chemistry, you would know that there is a Carbon in the center part of the others. 

The carbon that is bonded with the NH2, R, and another carbon (alongside with a H atom that is not seen) is called the (chiral) alpha carbon.

"Chiral" means a carbon atom bonded to four unique other elements.

And the other carbon that's connected to the O, OH, and the alpha carbon is the carbonyl carbon.

Now, in order for these individual amino acids to become into a protein, The AAs should be linked together. A bond of two amino acids are called a "Peptide bond".


As you can see in the picture above, the first step is to "pluck" off the H from the N and combine it with OH-.

With the formation of a water molecule, connect the NH+ and the carbonyl C. Then a bond will show between the carbonyl carbon and the nitrogen. That's a peptide bond. And the molecule that's been formed is called a Dipeptide.

When these AAs join together, a polypeptide chain is formed.

For simplicity, We only write the N, alpha C and carbonyl C.

And so the polypeptide chain is like this: N-C-C-N-C-C-N-C-C

The side that the N is on is called the N terminal.

And the side that the carbonyl C is the C terminal.

We can imagine that the N-C-C to act as a "group" and call it a residue.

                                        

Just like AAs can make peptide bonds, you can also break down the peptide bond as well. in order to do that, we need to use a way called Hydrolysis.

There are two ways of Hydrolysis:

One is called "Acid Hydrolysis". And just like the name, we use strong acids with heat to break the peptide bonds.

The other one is called "Proteolysis". And for that, we use a special enzyme called protease. Unlike acid hydrolysis, proteolysis can let you pick which peptide bond to break.

For example, a protease called Trypsin cuts the C term of arginine and lysine. So if we have a polypeptide chain like this:

***- ^^^- arginine - @@@- !!! - lysine - &&& And input trypsin inside,

we would have it like this: ***- ^^^- arginine   @@@- !!! - lysine  &&&

Using proteolysis has a big advantage for experiments because you can cut wherever you want for the research. Pretty cool, right?

                                           

Now, we will talk about our most special Amino acids of them all - The AA show!

It looks like our contestants are Histidine, Proline, Glycine, and Cysteine.

What could possibly make them special?  

Well, let's look at our first stop, Histidine!



The thing that makes Histidine special is that it's pKa is approximately 6.5 which is close to physiological PH.

Now you might say "Umm, ok, the pka and ph is close.. So what?"

According to the Henderson - Hasselbalch equation, the pKa = PH + log [HA] / [A-].

If we make pKa > PH, or making the solution more acidic, the log [HA] / [A-] > 0. Which means that [HA] > [A-]. Meaning that a stronger acid will cause the formation of [HA], the protonated form.

If we make pKa < PH, AKA making the solution more basic, the log [HA] / [A-] < 0. So [HA] < [A-], meaning a stronger base will cause the formation of [A-], the deprotonated form.

Look here for more: 

http://www.mhhe.com/physsci/chemistry/carey5e/Ch27/ch27-1-3.html

For Histidine, since it's pKa is close to physiological PH, it can switch from protonated to deprotonated from and vice versa with just small changes of PH. Making it to a H+ donor/acceptor without wasting energy and Enable things like transporting protons to one molecule to another.  

                                             

Next is Proline and Glycine, the alpha(α) helix breakers!



Let's look at Proline first.

Unlike all the other Amino acids, you can see that proline has its R group to have a cyclic structure that has a bond with the amine(NH2).

So we say that Proline has a secondary alpha amino group.

Because of it's odd side chain, proline is very rigid then others. So it doesn't have a free rotation alongside the other atoms.

On the other hand, Glycine is the opposite of proline's structure.

Glycine's side chain is DA most simplest side chain in the HISTORY of side chains! It only has a H for it's side chain!

Because of it's simple side chain, it's very flexible then others.

But there is a catch for that simple side chain as well.

Remember the part where I told you about the "Chiral" alpha carbon?

Like I said, "Chiral" means four different elements bonded into a C.

But for Glycine, since it's side chain is a H atom, we have a double H for the alpha carbon. So we cannot say that the carbon is chiral for glycine.

                                        

Now, about this time now, I think you should know what an alpha helix is.

An alpha helix is a coiled up polypeptide chain. It kinda looks like this:


Even though Proline and Glycine has a role of forming alpha helixes, they introduces a Beta turn to the alpha helix.

A beta turn is a bend(or turn) in the primary structure.

For Proline, because of it's odd secondary alpha amino group, it makes a bend when making peptide bonds. Glycine also makes bends because of it's flexibility,

                                          

Our last Amino acid is Cysteine!


And so, what cysteine has is this little trick on it's sleeve to form a disulfide bridge when each of them are close to each other as a polypeptide chain.


It would be good to learn redox reactions before this.

https://www.khanacademy.org/science/chemistry/oxidation-reduction/redox-oxidation-reduction/v/introduction-to-oxidation-and-reduction

Now, you can see that Cysteine has a thiol group( C - SH) as it's side chain.

The Cysteine molecules are in a reduced environment. So if we put this in a oxidizing environment, the side chains will lose an H and the S will form a bond.

We call that bond a Disulfide bridge.

And I should really say - Cysteine has name issues!

Some are asking is "Cysteine" or "Cystine" the correct way to spell the name.

And the thing is, both of them are right.

The name "Cysteine" is used for it's usual form, where the disulfide bridge is not formed yet.

The name "Cystine" is used when it has a disulfide bridge, in an oxidized state.

So, you can think about it like this:

1, Cysteine is used for it's usual form, and inside it's reduced - intracellular space.

2, Cystine is used for it's bonded form, and inside it's oxidized - extracellular space.

                                                                   

And with that, the Amino acids show will end.

Amino acids are pretty fun, right?

I asked a question to my dad about why Amino acids are left out of the central dogma. (Because I was curious)

And the answer was simple - Since Amino acids makes proteins, we can say that they are the smallest protein unit. So it's not "left out".

Well, that's it for today's article. And I hope to see you again in the next one about Amino acids - DA protein makers! Swag! 

                                      2015, 12/30   

                                   Jane Kim (김해인)

                                            



 

                                                                                                                                                           Report-05 2015-11-2

This time, we are here to talk about to see if the 3:1 ratio still works when there are two different characters which have a phenotype when shown.
And I'll tell you one thing, you'll see something familiar when you know it.
Since i'm assuming that you've seen the last writing and had a lot of thoughts to it, I only have one thing to say - there are two characters, they are the texture and color of the seed. The round gene is R, wrinkled is r, yellow is Y, and green is y.
NOTE: R and Y is dominant and r and y is recessive.
If we are all set, let's go for it!

Let's say that we have two pure(original) pea plants that is round and yellow, and are wrinkled and green.
Since we said that the pea plants are "pure", that means that there are no heterozygotes in the genotypes. Am I right?
Now, here is what we are going to have something called "Dihybrid cross".
This word means that we are going to have a hybridization within the alleles that has an complete dominant alleles and complete recessive alleles.
Now with all that information, I'm sure you all might notice what the whole story goes - one pea pod is going to have a gene of RRYY and a another pea pod is rryy.


Right now, I know you now got frustrated with this whole "two genes in one" kinda stuff, but this is just a fancy way of saying "I have a RR gene and a YY gene inside my individual chromosomes in my cell" and vice versa.
For a picture(If you want,) it's like this -
RRYY { (RR) (YY) } rryy { (rr) (yy) }
The {} is a cell, and () is a gene locus(~location).

Now, let's see the first picture of Mendel's second law. When you have a RRYY and a rryy, the first thing to do is to know what kind of gene possibilities you can have when you calculate the genotypes of the F1 generation with the punnet square.
Since we know that neither the characters 'texture' and 'color' cannot be apart from the pea plant's gene, the RRYY is only going to have a RY gene passed out to the offspring while the rryy is going to pass down the ry gene only.
This means by definition that we are not going to have any biological errors, the F1 generation is going to have a genotype of RrYy and a phenotype of round and yellow pea plants, because R and Y are dominant to the r and y.
Then, what about the F2 generation? What possible phenotypes could they produce to us?

We do know that the pea plants are going to self-fertilize, which means that the F1 generation pea plants are going to fertilize with each other. This means a RrYy gened pea plant and the exact gened RrYy pea plant is going to fertilize for a F2 generation.
Let's look at the punnet square at the bottom of the picture.
This time since we are dealing with a genotype of RrYy, we are going to have 4 outcomes - RY, Ry, rY, ry.
if we punnet square them, we would get nine genotypes of the F2 generations - RRYY, RRYy, RRyy, RrYY, RrYy, Rryy, rrYY, rrYy, rryy.
But unlike the genotypes, we would get four phenotypes of the F2 generations because the genotypes that has at least one to two dominant allele in a character would show the dominant stuff and covers the recessive ones - Round and yellow, wrinkled and yellow, round and green, wrinkled and green.
*NOTE*: When you look at the punnet square, you would see that there are 16 genotypes, not 9, this is just because there are a bit more of the same genes(example: there are two pea plants that has the genotype RRYy).

Say, can I give you a question?
What do you think what would happen if the R gene will ALWAYS stick to the Y gene and vice versa? What would happen if this pea plant has self fertilized?
This is sudden, but a noting word. What I meant about the R gene always sticking with the Y gene is that the locus(location) of the R gene and the locus of the Y gene is on the same chromosome.
What's wrong with that kind of style is that when the cells undergo meiosis, the "R and Y" pair and the "r and y" pair will never come apart - therefore it never makes a random combination of the genes, not making any of the offspring a heterozygote. That kind of genotyped pea plant will always have either a RY or a ry to give to the offspring when self fertilized, and we cannot say this as Mendel's law of independent assortment AKA "Mendel's second law".
Go to this site for the same information (Sorry that it's in korean, english users!) http://study.zum.com/book/18094


So now that we have proved that we cannot have the texture charactered gene and the color charactrized gene in the same chromosome(or "sticking together forever like best pals."), I would give you a easy question for you -
Do you think the answer will be the same if we had more Ry genes to be put in the punnet square? the answer is NO! we cannot have more or lesser genes then the average ones we already have.
Therefore, we can say that the ratio of the RY, Ry, rY, ry gene possibilities 'mix' is 1 : 1 : 1 : 1
Look at the second picture, if you can see the purple circles with down arrows and a number 1 on top of them, that is the stuff I was talking about the whole time.
This will help you too.https://www.ndsu.edu/pu…/~mcclean/plsc431/mendel/mendel3.htm

Now, look at the third picture that has a lot more red text then any of the other pictures. Because we will see some obvious but reasonable stuffs.
Let's remember the thoughts, The parent pea plants are RRYY and rryy, which can only hand down RY and ry to the F1 generation.
The F1 generation is RrYy, so it can give out either the RY, Ry, rY, ry genes to it's second generation(F2). Look at the purple circle with the RY, Ry, ry, ry genes, do you see something familiar?
That's right! The genes RY and ry is handed down from the parent genes RRYY and rryy. So we can call that the RY and ry gene are parental genes.
Then, what about the Ry and rY genes?
Those genes are from the F1 generation which is a union of the two parental genes RRYY and rryy.


Therefore, we could say that the Ry and rY genes are recombinant genes.
Then, what is the ratio between the parental and recombinant genes?
Let's think about the information we currently have for this question.
We do know that the ratio between RY, Ry, rY, ry is 1 : 1 : 1 : 1,
And we know that RY and ry are parental genes, and Ry and rY are recombinant genes.
This information we currently have can tell us that the ratio between the parental genes : recombinant genes is going to be the same as the ratio between RY and ry : Ry and rY.
This gives us a answer of 2 : 2 because all of the ratios of the genotypes for the punnet square is 1.
if we simplify this, it will be 1 : 1, or 50% over 50%. That is the ratio.

We did the ratio of the purple circles, now let's do the ratio of the punnet square.
we can see that 9 of the 16 possibilities of the pea plants are round and yellow, and 3/16 is wrinkled and yellow, another 3/16 is round and green, and the last 1/16 is wrinkled and green.
in Mendel's first law, or Mendel's law of dominance and inheritance shows us a ratio of 3 : 1 whereas Mendel's second law (Mendel's law of independent assortment) gives us a ratio of 9 : 3 : 3 : 1.
You might think that 3 : 1 and 9 : 3 : 3 : 1 has nothing in connection, but look at the punnet square. Because i'll give you two questions.
What is the ratio between the round and wrinkled pea plants? It's 3 : 1.
What is the ratio between the yellow and green pea plants? It's also a 3 : 1 ratio.
So, the 3 : 1 ratio still works, but only the difference it has is that since we have two traits(characters), we have to kinda "Double" the 3 : 1 ratio into a 9 : 3 : 3 : 1 ratio.
Check out for wikipedia, the free encyclopedia about Mendel's laws for more information!
https://en.wikipedia.org/wiki/Mendelian_inheritance


Well, this is all for today's dihybrid cross of two pea plants that each has two character traits - texture and color.
I'm just still 12, so advanced biology users! Please tell me if there is anything wrong with my article, thank you!


By Jane Kim
<<해인이가>>



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DNA, DNA, DNA......... It's all about the DNA!

Remember the time when I showed you how the famous DNA duo Crick and Watson made the first DNA structure? Sure you do!

I have something to ask, is DNA the genetic material?

All of you who learned biology would say "Of course, who in the world would not know that?" That's right, it's now too obvious that DNA is the genetic material.

However, in 1869 after DNA was first showed to biologists, some of them still thought that the proteins was the guy who was carrying all the information because DNA is a lot more simpler(only has four base pairs and some phosphates) then proteins do.

Luckily, some great scientists showed evidence and experiments that DNA was our little genetic material guy - And I'll tell you this: Crick and Watson did not solve this problem.

Wanna know who did this and what experiments it was? Let's start!


Now, somebody. A guy named Griffith had an experiment with mice and pneumonia bacterias and did it quite well.

First, look here: https://en.wikipedia.org/wiki/Bacterial_capsule

This will show you some structures of bacteria cells. 

The pneumonia cells also have characters in their DNA, and the one we are going to use is something called the rough and smooth cells.

The bacterias has a polysaccharide coat which acts like a protection gear protecting the bacteria from enemies. For example, a mammal immune system.

Bacterias with the smooth side progeny(or gene) has polysaccharide coats while the rough side progeny(gene) doesn't have them.

What Griffith did in his experiment is to have four examples of pneumonia with different gene types or progenies in each one. Then he injected those to a mouse.

The first one was the smooth type pneumonia which was highly virulent because they have almost no problem entering the body with the help of their polysaccharide coat. As a result, the mouse died.

The second one is a rough type pneumonia which was not that virulent because unlike the smooth pneumonia, the rough one didn't have the polysaccharide coat with them. The mouse lived in result.

The third one was the smooth type, only the difference now is that Griffith had heat - killed this bacteria. The mouse has lived again, and it's nonvirulent.

The last one is a mixture of a rough type pneumonia with a heat - killed smooth pneumonia, a mixture of two nonvirulent bacterias. But guess what? The mouse has died from this mixture!

Weird, we would expect the mouse to live in this one because both of the mixture is nonvirulent.... So Griffith examined the blood from the dead mice from the fourth experiment, And a shocking news came to him.

There was suddenly a live smooth pneumonia bacteria living!

But what we had injected to the mouse was the rough bacteria with the heat killed smooth bacteria, so if we say mutations are not possible in this experiment, this means that there was a chance that the mixture could have changed to become into a smooth bacteria. 

The heat killed smooth guy has already lost it's polysaccharide coat, and it's dead. So there's no way it would have rebirthed and lived.

But the rough type bacteria could have picked up something from the heat - killed bacteria and turned into a smooth bacteria! That could work!

Griffith has discovered that the rough pneumonia has been transformed into a smooth bacteria by a "transformation principle". These days, we can now know that the transformation principle was actually a happening by the rough type pneumonia who has taken the DNA of the heat killed pneumonia when it lived from the heat, making it from a rough type into a smooth type.







And then, for further more of this, Avery, McLeod and McCarty has experimented with the heat killed pneumonia to be with some special enzymes and mixed it with the rough type pneumonia to inject it to a mouse. 

In this one, we are going to have two enzymes. A enzyme that eats proteins and an enzyme that eats DNA. 

The experiment has told us this: When the heat killed bacteria was with the protein eating enzyme and did the steps above, the mouse died. So the transformation still occurred.  

However, when it was with the DNA eating enzyme and did the same steps, the mouse had lived! The transformation has not occurred!

These data has proved well enough for the "DNA is the genetic material" argues, but there is just one more experiment I want to tell you. 

Something called a Hershey - Chase experiment has filled up some few gaps of the Avery - McLeod - McCarty experiment, making it a stronger proof that DNA is the genetic material.

First things first, do you know about phages?

Phages, short for "Bacteriophage" are tiny viruses that hunts for bacterias. Once they found their target, phages enters their DNA inside the bacteria, and the DNA will form into new phages, which will go outside from the bacteria, making the bacteria unable to do it's original job.

Phages has a protein body(or phage coat) on the outside, but in the head part, they have their DNA inside the protein phage coats. That's a basic intuition of phages.

The second thing to know is that proteins has sulphur, but not phosphorus inside them while DNA has phosphorus, but not sulphur inside them.

Hershey and Chase wanted to label the phages whether to know if the genetic material is the protein or the DNA, so they kept two 'sides' of phages.

One side was to keep and grow phages in radioactive 35 - sulphur and the other side was to do the same thing in radioactive 32 - phosphorus.

Within this, Hershey and Chase could label the protein side has the sulphur, and the DNA side as the phosphorus.

Hershey and Chase kept and grew the phages from the S side and the P side, then they let the phages attack a bacteria for just enough time for the phage DNA to enter the bacteria. Then they blended it to separate the phage coat from the bacteria and centrifuged it.

The results had a supernatent and a precipitate, the supernatent was the phage coat and the precipitate was the bacteria pellet. Look here for more:

https://en.wikipedia.org/wiki/Precipitation_(chemistry)

I'll give you some interesting news: When the bacteria and the phage was blended, the bacteria had to give out it's DNA. So the bacteria pellet must have either of the 'labels' which is either sulfur or phosphorus. 

So that means depending on what came out of the bacteria pellet will determine the result of the experiment, and will also proof to the protein side or the DNA side for the question "What is the genetic material?".

Guess what? The radioactive 35 - Sulphur came out of the supernatent while the radioactive 32 - Phosphorus came out of the bacteria pellet!

It means that DNA is indeed, the genetic material! 


The experiment of Griffith, the Avery - McLeod - McCarty experiment, and the Hershey - Chase experiment all tells us that DNA is the genetic material that we need to have in order to live. And that's how we now "obviously" know that DNA is our most important guy in our cell, the genetic material.

It's pretty fascinating, the structure of four nitrogeneous pairs and phosphorus has some sort of mechanism to code the start of life... The why of this still hasn't been solved yet, but i'm sure somebody(who can be me or you) could figure it out for a another dimension!

That's all for now in this article, and happy biologying!

                               By Jane Kim,   <<해인이가>>


     




 






I'm in a desert, and I can choose either a compass or a map.

What would you choose if you were me?

I'd choose a map. And not just a regular map, i'm choosing a gene map!

And what I would like to show you this time is to know how to make this gene map when you have 3 character traits, along learning what is a gene order, and something called interference and COC.(And of course, our tester is a fruit fly)

Challenge accepted, let's do this!


First, I would like to show you a few stuff before we do anything.

We will use not 1, not 2, but 3 characters for this example.

And these characters are called Vermilion, crossveinless, and cut genes.

Vermilion: Fruit flies who has this gene has a brighter red eye color while the wild types has a dark red eye. And it's a recessive mutant.

Crossveinless: Most fruit flies that we know has a horizontal line and a bridge like vertical line on it's wings. However, crossveinless fruit flies does not have the "Vertical" lines on it's wings, and it's also recessive.

Cut: Normal fruit flies has round and smooth wings, but flies who has this character has it's wings that looks like it has been kinda shattered like glass or snipped by scissors. Again, it's a recessive mutant.

Now, look at the pictures with black backgrounds. There will be 3 of them.

If I write that in a simple form, it's just like this:


V, CV+, CT+ = 580       V+, CV, CT = 592      V, CV, CT+ = 45               V+, CV+, CT = 40        V, CV, CT = 89        V+, CV+, CT+ = 94 

V, CV+, CT = 3           V+, CV, CT+ = 5      Total progeny: 1448

NOTE: The genes that has + with them is the wild type, and the ones without the + is recessive mutant genotype.








Say, do you remember when we first learned about the term Centimorgans?

We normally write it as cM or m.u(map units). 




Well, this time might be the perfect time in order to use them in real life!

Sometimes you'd might want to know exactly how far is the distance between a gene loci from another, and want to know what gene is in where(To be specific, it's "what order"). A gene map can give you that much information about the genes in that way. It kinda looks like this:

                A                    34cM(mu)         B  12cM(mu)  C

   ---------|-----------------------------|-----------|--------

                |--------------------46cM(mu)-------------|

To make this, we have to know the distance between the three genes in order to make a gene map. 

To do that, we need to know something called a "gene order".

Then you might say "Ok, isn't the gene order V, CV, CT? All of the genotypes you showed me just looks like that!"

Well..... That could look like that at first, but you should know that there is a catch from the double mutant genotyped chromosomes.

Double crossovers moves their middle genes to it's sister chromatid, so we have to know the actual gene order. And we have to use that gene order once we have found it out.




How to find the gene order is easier then you would think. 


First, you need to find the parental genotypes and the double crossovered(DC) genotypes. 


Second, compare them and find the odd one.

Third, put the odd one in the middle, then put the rest of the genes where it was. 


And fourth, With the newly found gene order, find the distance between each gene loci. Sounds kinda hard, right? Follow the steps and you'll see.


Step 1: In order to find the parental genotype and the DC genotype, we just need to see what is the top 1 and 2 highest and lowest number of fruit fly progenies. The highest ones are the parental genotypes, and the lowest are the DC genotypes. In this example, we can now know that the genotypes V, CV+, CT+ and V+, CV, CT are the parental genes while the genotypes V, CV+, CT and V+, CV, CT+ are the DC recombinant genes.

Step 2: We have to compare the parental and DC genotypes together. 

That means we have to compare V, CV+, CT+  and V, CV+, CT together and also compare V+, CV, CT and V+, CV, CT+ together as well. 

We can directly see that the CT gene is the odd one out, therefore we put it in the middle and putting the other genes back in it's original location.

So we get V, CT, CV as our gene order, which means that the CV gene and CT genes swaps places.

Now, since we've figured out the gene order, the only thing left to solve is the distance between the gene locis! The only thing is, how?

Well first, let's arrange the stuff that we have just earned. 

If we put that in a gene map, it would look something like this:

                V                   ?cM(mu)           CT  ?cM(mu) CV

   ---------|-----------------------------|-----------|--------

                |--------------------?cM(mu)--------------|







Remember in the 20th line where I have wrote all the possible progenies?

As you all have learned that the homologous chromosome pairs has to undergo "Genetic Recombination" for other progenies to appear in their phenotypes, they can do the same thing! A easy way to figure out the map units is to figure out where the crossing over happens, what kind of genotypes can be formed, and make that into a percent! 

According to the gene map above and the total genotype progenies, we can see that the distance between V and CT or CT and CV can undergo genetic recombination.  

                V                   ?cM(mu)          CT+  ?cM(mu) CV+

   ---------|-----------------------------|-----------|--------

                                        X                         X

               V+                   ?cM(mu)          CT  ?cM(mu)  CV

   ---------|-----------------------------|-----------|--------  Hmmmm...... What expected genotypes can occur when crossover from the left happens? Right! We can get V, CT, CV and V+, CT+, CV+

Then what about the recombination on the right? 

We would get a result of V, CT+, CV and V+, CT, CV+

And what if both happened? A result of V, CT, CV+ and V+, CT+, CV would occur, and the other two left are the parental genes.

To figure out the map units, we need to add up all the genotype's progeny numbers, divide that to the total progeny number(Which is 1448), and multiply it by 100. 

So for the V ~ CT range, we just need to know what are the progeny numbers from each genotypes that can occur from that range's genetic recombination.

But there is a catch - I just noticed this. You were probably just thinking to add the progeny numbers of the genotypes V, CT, CV and V+, CT+, CV+(89, 94) and solve the question, right?

Well, that is wrong! Think about the genotypes that has undergone a double crossover, that kind of genotype does both of the recombination V ~ CT and CT ~ CV. So you should put their progeny numbers as well.








Then we can know that the distance between V ~ CT is: (89 + 94 + 3 + 5) / 1448 = ~0.132 

0.132 * 100 = 13.2%!

CT ~ CV = (45 + 40 + 3 + 5) / 1448 = 0.064

0.064 * 100 = 6.4%

Then what is the distance between V and CV? 

We just have to add the two distances together!

13.2 + 6.4 = 19.6

Which means our gene map looks like.......!

                V                   13.2cM(mu)       CT  6.4cM(mu)   CV

   ---------|-----------------------------|--------------|-----

                |--------------------19.6cM(mu)--------------|

Yeah! It's all done! The gene map is finished!

But, even with all this excitement, we still have one more mission left.

I would like you to learn a bit about probability for this one, assuming that you have, here it is.

Let's just say that we don't know the progeny number of the double recombinant genotypes. If that's so, then we would figure out the number with this "algorithm". Like below. 

0.132 * 0.064 = 0.0084 

0.0084 * 1448 = 12

(We did this based by the progenies of each recombination)

We got 12 as our answer, right? 

Then count the total number of progenies from this chart:

V, CV+, CT+ = 580       V+, CV, CT = 592      V, CV, CT+ = 45               V+, CV+, CT = 40        V, CV, CT = 89        V+, CV+, CT+ = 94 

V, CV+, CT = 3           V+, CV, CT+ = 5      Total progeny: 1448

Since V, CV+, CT  and V+, CV, CT+ is the double recombinant genotypes, we would get our answer as 3 + 5 = 8

Now what just happened here? We thought the total number of progenies was 12, when we actually got 8!


All of this is because of Interference.

When meiosis happens and genetic recombination occurs, if a gene loci(Or "spot) has done genetic recombination, it decreases the probability that the gene loci next to it will also undergo recombination.

The interference tells us how strong a crossover from one DNA gene loci place interferes with the others.

To figure out the interference, we need to figure out something called COC, and it's short for "Coefficient of coincidence".

How to find the interference is to follow this formula: 1 - COC

COC = Observed # of DC progenies / Expected # of DC progenies.

Luckily, we have already figured out the COC value, which is 8 / 12 = 0.66

Then the Interference value is 1 - 0.66 = ~0.33

And that's why the observed & expected progenies are different.








So a long talk! But we finally figured out what is the gene map for this fruit fly and about interference!

Oh, and guess what? Here are the links to some of the helpful sites that will help you clear any more confusions!

https://en.wikipedia.org/wiki/Coefficient_of_coincidence

https://www.youtube.com/watch?v=J-aTSriYnak

https://www.ndsu.edu/pubweb/~mcclean/plsc431/linkage/linkage3.htm

I will be writing a another article when I learn some other interesting stuff!   

Thanks for reading this!         

                                By Jane Kim    <<해인이가>>












update: 2015-10-12

Replication!

The source of how we grow, and how we get healed, and how we are even can be twins!

Yes, just like the examples I showed you, even tiny tiny cells in your body replicates and even the DNA inside of it!

Technically, if a DNA is replicating, it means something good might happen!
And this, my friends. Is the stuff we will learn about in today.
The "Advanced source" of DNA replication. Let's get ourselves an egg roll, shall we?

The first thing you know about DNA is the writing I showed you in the last one, whereas the DNA has a part going in the 5` to 3` direction, and the 3` to 5` direction.
DNA has it's deoxyribose backbone and it's four nitrogenous pairs - Adenine, Thymine, Cytosine, Guanine. Because of that, you can easily tell which base pairs are going to be with which base pairs.
There are 8 basic enzymes in order to start you DNA's replication, and those are - Helicase, SSB's, Topoisomerases, RNA primase, DNA polymerase, Sliding DNA clamps, RNAse H, DNA ligase.
And below are the jobs that these enzymes do.


Helicase: This enzyme uses ATP hydrolysis energies to break apart the DNA strands. Their limit is the replication fork, which tells the enzymes where to replicate and where to leave it be.

SSB("Single Strand Binding" proteins): it's job is to make sure that the opened up strands of DNA's doesn't get twisted back to it's normal shape.

Topoisomerases: It makes sure that the other DNA's doesn't get unwounded or get to coiled with each other(Supercoils). This enzyme also uses ATP for it's job, like the Helicase.

RNA primase: Before the other enzymes could start putting in some DNA, it has to have a starting line first, and that 'starting line' is the RNA parts by the RNA primase. Without it, the other enzymes can't do their work of putting the DNA parts.

DNA polymerase and the Sliding DNA clamp duo: Those two are excellent combos. DNA polymerase puts the DNA parts after the 'starting line RNA' that RNA primase has left, and the Sliding DNA clamp makes sure that the DNA polymerase doesn't get side - tracked.

RNAse H: This guy over here removes the RNA parts left by RNA primase.

The second DNA polymerase: Like the first DNA polymerase, this enzyme places DNA blocks in the parts where the RNAse H has removed the RNA parts. It makes Okazaki Fragments in the process, which is a gap between the DNA parts from the RNA places and the original DNA parts.

DNA ligase: This little enzyme fills out the gaps of the Okazaki Fragments, and finishes the replication!

For beginners, teachers shows these enzymes as if they are coming and going on their own, for the sake of simplicity. However, they don't actually just come and go out of nowhere! They are all doing their work based on the structure of the main Holoenzyme!
Look at the 3rd picture which looks like an ant's head with some sort of green stuff stuck on it's antenna, did you got that picture?
Now, I will show you what is the Holoenzyme and what's it's structure.


DNA clamp holder: That's the place where the holoenzyme holds the DNA clamp and uses it and reuses it over and over again.

γ complex(clamp holder): When replication starts, it makes the DNA clamps and holds it in the clamp holder.

τ protein: This protein bundles to hold the DNA polymerases.

That's about it, that's the holoenzyme, and the replication picture is the picture below the holoenzyme picture.
Oh, and one more thing! remember this: 5` to 3` direction is where the enzymes go, so the DNA's have a leading strand and the lagging strand.
The leading strand has no problems for replicating, because it's strand is in the 5` to 3` direction. But the lagging strand is a 3` to 5` direction, so you have to do stuff all over and over again, making a lot of Okazaki Fragments.

Now, with all of this replication stuff done, do you have any questions?
If not, then you could go for a dessert break. And if you have one, you are a wise student! Ok, what's the question? I'll give you one!
Have you ever thought what was the actual stuff that the DNA polymerase places down after the RNA parts? Sure, it's DNA, but what does the actual backbone looks like, and the actual structure, and it's molecule! What are those kinds of stuff, anyways?
And here is the answer - Deoxyribonucleotide triphosphate
This thing - When you look at it, if kinda looks familiar. That's right! it kinda looks like the structure of a DNA nucleotide!
The thing is, instead of having one phosphate group connected with the 5` end, this DTP(Deoxyribonucleotide triphosphate) has 3 phosphate groups!
And I'll tell you why.

DTP is like the DNA nucleotide with 3 phosphate groups instead of one.
Those 3 phosphate groups have a name - The α group phosphate, β group phosphate, and the γ group phosphate.
Whenever replication occurs, the DNA polymerase places this molecule one by one. But here is the thing, whenever the DTP lands on a another DTP, guess what happens to the triphosphate groups?
The β group phosphate, and the γ group phosphate separates from the molecule!
Which gives us a result of the normal DNA nucleotide and 2 individual phosphate molecules!

Well, that's all about replication that I have learned today.
Already on my 3rd page of this activity! What would the 4th one be?
Well, better go and find the flour of videos soon!
See you at next lesson!

By Jane Kim
<<해인이가>>

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by jane kim(김해인) 2015.10.11에 작성: Vector개념을 배우고 이를 코드로 작성해 보았습니다..

코드 실행: https://www.khanacademy.org/computer-programming/pvector-instance/6479887750266880


/**

Try to catch the ball! It's harder when you move, but easy when you stay..... :)


Line 37: The reason why I used constrain() is because sometimes(or mostly) when you move your mouse in further directions, the ball can get off the canvas. Making the user a bit fustrated.

So, I just used constrain() to command the code where the ball can go without further ado.(it can only move inside the canvas!)

*/


background(97, 237, 255);


//I'm puting this in a function just in case somebody tweaks this code and gets an error. (Nobody wants that to happen!)

var ball = function(x, y) {

    translate(x, y);

    noStroke();

    fill(255, 0, 0);

    ellipse(0, 0, 30, 30);

    fill(255, 255, 255);

    arc(0, 0, 30, 30, 0, 180);

    fill(0, 0, 0);

    ellipse(0, 0, 5, 5);

};


mouseMoved = function() {

    background(97, 237, 255);

    

    //Creating two new PVectors, mouse and center.

    /* The mouse is a vector showing us where the mouse is, and the center is just the center of the canvas.(In vector scale!) */

    var mouse = new PVector(mouseX, mouseY);

    var center = new PVector(width/2, height/2);

    

    /* The mouse.sub(center) is in order to let the ball to fallow the mouse. Without it, it will go away twice from the position the mouse is currently on.

NOTE: mouse.mult(3) is to make distances from the mouse and the ball 3 times further.*/

    mouse.sub(center);

    mouse.mult(3);

    

    //Since translate() is not with either pushMatrix() or popMatrix(), we have to reset the code by using resetMatrix().

    resetMatrix();

    translate(width/2, height/2);

    ball(constrain(mouse.x, -185, 185), constrain(mouse.y, -185, 185));

};


/**

Note: This is just a practice program for my vector skills, not for khantober!

*/

by jane kim

Another great day with Biology!           update 2015-10-03

This time was also about the structure of DNA, but in an 'advanced' way.
I learned some more stuff about this
Even though some chemical stuff and some hard concepts were pretty difficult and was hard to know, I did it with the help of the internet.
(Note: I'd recommend you to learn chemistry at least in a periodic table level. For maximum, I think learning until pi and sigma bonds is a good choice.)
All right! Let's get this started!

The grey ball is carbon, blue is nitrogen, orange is phosphate, and red is oxygen, got it?

When you look inside a single DNA and take a small chunk and zoom it, you can see the carbon and phosphate backbone in a phosphodiester bond. 

Also, you can see the nitrogenous bases bonded with the 1st carbon in a glycosidic bond with the nitrogen in the base.
Well, now it's time to go in further! Look at the second picture with models being a bit smaller then the 3rd one, do you see it?
From what I have understood, because the angle of the glycosidic bonds are both different (Honestly, I'm still working on that part. If anyone knows something right or wrong, please tell me!) they make "Grooves" or maybe "Distances" from one backbone to another.
And also, for ones who don't understand, the 'grooves' are distances from the left and right top backbone and the distances from the left and right bottom backbone.

There are 2 grooves, a shorter one and a longer one. We call the shorter side a Minor groove, and we call the longer side the Major groove.
The Minor groove is 1.429 nm, and the Major groove is 1.974nm, just for a tip.
So far, so good. This is the easy part.


Now is the time for the tricky part! GO to the 3rd picture, with the same model, but in a more zoomed - in state.
Ignore the CH3 and the extra H for a bit, but look at the molecules with the letters HA and HD. They mean Hydrogen Acceptors and Hydrogen Donors.       (The grey ball is carbon, blue is nitrogen, and red is oxygen. Again, you didn't forget about it, right?)

Here is the problem - Look at the periodic table, people. You will see what's the problem.
Did you get it? Yes it is, Both N and O are supposed to accept H from somebody else, but a nitrogen is a donor! How silly is that? 
So I had to come up with some Crazy theories that was up in my mind and other stuff..... and I eventually searched Google for the answer. And it was really helpful!


http://atlasgeneticsoncology.org/Educ/DNAEngID30001ES.html


If you look at the donoring Nitrogen, you can see it only has a single bond with the whole other carbonic stuff and is holding 3 H's (I think it has 3 H's because of the C) and the fallowing oxygen is lacking a H, so the N and O is sharing the H together - therefore making the N a donor, and the O as a acceptor! 


Now, let's think for a bit. Let's think that we are some sort of protons in the cell in order to do stuff like transcription, translation and replication.
If we encounter a DNA for the job, there are always 2 choices. 

One is a side that has less information, and hard to tell if it's a A or T base pair or a C or G base pair, the other one is a side that has lots of information and is easy to know what kind of base pair the DNA is holding.

Which one do you pick? You'd mostly pick the second one, for sure.
And that's how the ACTUAL protons and mechanisms picks the sides for their own jobs!
The cellular stuffs and jobs are mostly(but not always) being done in the major groove because of that same reason.


Well, there we have it! We have this beautiful, beautiful structure of DNA learned!

Even though this is all a chemical bond of Carbon, Nitrogen, Oxygen and Hydrogen - The thing that is mystical is that the fact that those chemicals joins themselves as life, and thoughts, and an organism....

And maybe, that's the beauty of Biology! Well, at least I think so! ;D
My learning never stops here, more interesting facts and news are awaiting!
I'll make a another one when i'm a crackin' again!

By Jane Kim             <<해인이가>>


<<Would you say, professor. About my question, please look at the picture below.>>                                                       

You can see that this is a picture of a adenine and a thymine group together.   Around the chemical stuff like O, N and H there are colored circles. You can compare it to four different analysis. Red is a Hydrogen acceptor, blue is a Hydrogen donor, sky green is Hydrogen, and yellow is CH3.  Ignore the CH3 for a bit and look at all the Oxygens and Nitrogens. You will see a NH that is blue, which represents that it's a Hydrogen donor. And, that's strange. Because in the periodic table, all of them(N and O) are supposed to be 'HA' s instead of 'HD's. 

When I looked at the picture, I made an assumption. In order to prove that N is a hydrogen donor, it has to have some sort of bond to something else. And that was the O which has a H bond with the N.  The first thing to do is to know how many H does the N and O (Which is a HD and a HA) has.  At first glance, I thought that the N has 2 H and the O has 1 hydrogens sticking with it. (Because the O has the only H because of the bond with the N)  But then, it just didn't made sense. If the N has 2 hydrogens and the O has 1, that means both of them lacks a hydrogen. Therefore making them both hydrogen acceptors. And as we all know, that's not right. So, what I did is to make this kind of thought.        

Number 1: The N has to have 3 hydrogens because it can only share it's H if it's shell has the perfect amount of H(which is 3 of them).  

Number 2: Since the O is the molecule that accepts the Hydrogen, it has to have only 1 Hydrogen molecule with it.

Number 3: Since the H that the O has is because of the hydrogen bond the N has made, we have to find the H that the O has before the bond. In other words, we will disqualify the H from the bond about saying "I'm a member of the Oxygen!".

You know, I have this problem professor. I realized that the O and the N are both stuck with C molecules. It's not seen in the picture, but it's just not seen for simplicity.  So, I thought that the only possible way for the N and O molecule to get their missing H is to actually have a Covalent bond with the Carbon molecules!        

 In that way, the nitrogen and the oxygen molecules can get their last hydrogen, making the Nitrogen have 3 hydrogen molecules, and the oxygen having 2 hydrogen molecules.(One is from the covalent bond with the C, the other one is from the bond with the N).

Professor, what do you think? Is this the correct answer? Or is there some sort of other way for the O and N to have their hydrogens?     I'm willing to know, and I'd be grateful if you'd kindly answer my question. 



'생명과학 > Biology' 카테고리의 다른 글

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Let's say we want to make that program that generates a world of monkeys. Your program could generate a thousand Monkey objects, each with a height value between 200 and 300 (as this is a world of monkeys that have heights between 200 and 300 pixels).

var randomHeight = random(200, 300);

Does this accurately depict the heights of real-world beings? Think of a crowded sidewalk in New York City. Pick any person off the street and it may appear that their height is random. Nevertheless, it’s not the kind of random that random() produces. People’s heights are not uniformly distributed; there are a great deal more people of average height than there are very tall or very short ones. To simulate nature, we may want it to be more likely that our monkeys are of average height (250 pixels), yet still allow them to be, on occasion, very short or very tall.

       표준편차의 이해: http://bcho.tistory.com/972

     

A distribution of values that cluster around an average (referred to as the “mean”) is known as a “normal” distribution. It is also called the Gaussian distribution (named for mathematician Carl Friedrich Gauss) or, if you are French, the Laplacian distribution (named for Pierre-Simon Laplace). Both mathematicians were working concurrently in the early nineteenth century on defining such a distribution.

When you graph the distribution, you get something that looks like the following, informally known as a bell curve:

Graph of a standard bell curveA standard bell curve

The curve is generated by a mathematical function that defines the probability of any given value occurring as a function of the mean (often written as μ, the Greek letter mu) and standard deviation (σ, the Greek letter sigma).

The mean is pretty easy to understand. In the case of our height values between 200 and 300, you probably have an intuitive sense of the mean (i.e. average) as 250. However, what if I were to say that the standard deviation is 3 or 15? What does this mean for the numbers? Looking at graphs can give us a hint. The graph above shows us the distribution with a very low standard deviation, where the majority of the values cluster closely around the mean. The graph below shows us a higher standard deviation, where the values are more evenly spread out from the average:

Graph of a bell curve with a higher standard deviationA bell curve with a higher standard deviation

Not familiar with the concept of "standard deviation"? Don't worry! You can studyVariance and standard deviation separately on Khan Academy before continuing.

The numbers work out as follows: Given a population, 68% of the members of that population will have values in the range of one standard deviation from the mean, 98% within two standard deviations, and 99.7% within three standard deviations. Given a standard deviation of 5 pixels, only 0.3% of the monkey heights will be less than 235 pixels (three standard deviations below the mean of 250) or greater than 265 pixels (three standard deviations above the mean of 250).


Calculating Mean and Standard Deviation

Consider a class of ten students who receive the following scores (out of 100) on a test:

85, 82, 88, 86, 85, 93, 98, 40, 73, 83

The mean is the average: 81.3

The standard deviation is calculated as the square root of the average of the squares of deviations around the mean. In other words, take the difference from the mean for each person and square it (variance). Calculate the average of all these values and take the square root as the standard deviation.

ScoreDifference from MeanVariance
8585 - 81.3 = 3.7(3.7)^2 = 13.69
8282 - 81.3 = 0.7(0.7)^2 = 0.49
8888 - 81.3 = 6.7(6.7)^2 = 44.89
etc.... ... 
 Average Variance:228.81

The standard deviation is the square root of the average variance: 15.13

Want to understand standard deviation better? You can study Variance and standard deviation in more depth here on Khan Academy.


Luckily for us, to use a normal distribution of random numbers in a program here, we don't have to do any of these calculations ourselves. Instead, we can make use of the Randomobject provided by ProcessingJS.

To use Random, we must first instantiate a new Random object, passing in 1 as the parameter. We call that variable "generator" because what we've created can be basically thought of as a random number generator.

var generator = new Random(1);

If we want to produce a random number with a normal (or Gaussian) distribution each time we run through draw(), it’s as easy as calling the function nextGaussian().

var num = generator.nextGaussian();
println(num);

So, now, what are we supposed to do with this value? What if we wanted to use it, for example, to assign the x-position of a shape we draw on screen?

The nextGaussian() function returns a normal distribution of random numbers with the following parameters: a mean of zero and a standard deviation of one. Let’s say we want a mean of 200 (the center horizontal pixel in a window of width 400) and a standard deviation of 60 pixels. We can adjust the value to our parameters by multiplying it by the standard deviation and adding the mean.

var standardDeviation = 60;
var mean = 200;
var x = standardDeviation * num + mean;

Now, we can create our program that draws semi-transparent circles according to a normal distribution. The darkest spot will be near the center, where most of the values cluster, but every so often circles are drawn farther to the right or left of the center.


// Adapted from Dan Shiffman, natureofcode.com


var generator = new Random(1);


<코드>

var draw = function() {

    // The nextGaussian() function returns a normal distribution of random numbers with the following parameters: a mean of zero and a standard deviation of one

    var num = generator.nextGaussian();

    var standardDeviation = 60;

    var mean = 200;

    

    // Multiply by the standard deviation and add the mean.

    var x = standardDeviation * num + mean;

    

    noStroke();

    fill(214, 159, 214, 10);

    ellipse(x, 200, 16, 16);

};


    <출력>



<코드>

var generator = new Random(1);

var standardDeviation = 2;

var mean = 0;


var Walker = function() {

    this.x = width/2;

    this.y = height/2;

};


Walker.prototype.display = function() {

    strokeWeight(3);

    stroke(0, 0, 0);

    point(this.x, this.y);

};

// Randomly move up, down, left, right, or stay in one place

Walker.prototype.walk = function() {

    var xStepSize = standardDeviation * generator.nextGaussian() + mean;

    var yStepSize = standardDeviation * generator.nextGaussian() + mean;

    this.x += xStepSize;

    this.y += yStepSize;

};


var w = new Walker();


var draw = function() {

    w.walk();

    w.display();

};

<출력>



      

2# 잘못된 코드

var generator = new Random(1);

var standardDeviation = 2;

var mean = 0;

var num = generator.nextGaussian();


var Walker = function() {

    this.x = width/2;

    this.y = height/2;

};


Walker.prototype.display = function() {

    strokeWeight(3);

    stroke(0, 0, 0);

    point(this.x, this.y);

};


// Randomly move up, down, left, right, or stay in one place

Walker.prototype.walk = function() {


    var xStepSize = standardDeviation*num + mean;

    var yStepSize = standardDeviation*num + mean;

  

    this.x += xStepSize;

    this.y += yStepSize;

};


var w = new Walker();


var draw = function() {

    w.walk();

    w.display();

};

<출력>


3#절충형

var generator = new Random(1);

var standardDeviation = 2;

var mean = 0;



var Walker = function() {

    this.x = width/2;

    this.y = height/2;

};


Walker.prototype.display = function() {

    strokeWeight(3);

    stroke(0, 0, 0);

    point(this.x, this.y);

};


// Randomly move up, down, left, right, or stay in one place

Walker.prototype.walk = function() {

    var num = generator.nextGaussian();

    var xStepSize = standardDeviation * num + mean;

    var yStepSize = standardDeviation * generator.nextGaussian() + mean;

  

    this.x += xStepSize;

    this.y += yStepSize;

};


var w = new Walker();


var draw = function() {

    w.walk();

    w.display();

};


<출력>


<^^>





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