Question

Answer 1

a) Xét gen B:

\(N=\dfrac{2L}{3,4}=1680\left(nu\right)\)

\(\left\{{}\begin{matrix}A+G=50\%N\\A=15\%N\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}A=T=15\%N=252\left(nu\right)\\G=X=35\%N=588\left(nu\right)\end{matrix}\right.\)

Xét gen b:

\(N=\dfrac{2L}{3,4}=1280\left(nu\right)\)

\(A=T=G=X=320\left(nu\right)\)

 

Answer 2

b) Kì giữa I, NST kép : BBbb

\(\left\{{}\begin{matrix}A=T=2\left(252+320\right)=1144\left(nu\right)\\G=X=2\left(588+320\right)=1816\left(nu\right)\end{matrix}\right.\)

Kì cuối I, 2 loại BB và bb

- Tế bào BB:

\(\left\{{}\begin{matrix}A=T=2.252=504\left(nu\right)\\G=X=588.2=1176\left(nu\right)\end{matrix}\right.\)

- Tế bào bb

\(A=T=G=X=320.2=640\left(nu\right)\)