\(N_{gen1}=N_{gen2}=\dfrac{2L}{3,4}=2400\left(nu\right)\)
\(a,\)Xét gen 1.
\(5\%N=120\left(nu\right)\)
Theo bài ta có hệ: \(\left\{{}\begin{matrix}A-G=120\\2A+2G=2400\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}A=T=660\left(nu\right)\\G=X=540\left(nu\right)\end{matrix}\right.\)
Xét gen 2.
\(A_{gen1}-A_{gen2}=180\rightarrow\) \(A_{gen2}=660-180=480\left(nu\right)\)
\(\Rightarrow G=X=\dfrac{N}{2}-480=720\left(nu\right)\)
\(b,\) \(H_{gen1}=2.660+3.540=2940\left(lk\right)\)
\(H_{gen2}=2.480+3.720=3120\left(lk\right)\)
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