\(a,N=\dfrac{2L}{3,4}=1200\left(nu\right)\)
\(b,\)Theo bài ta có hệ: \(\left\{{}\begin{matrix}A=2G\\2A+2G=1200\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}A=400\left(nu\right)\\G=200\left(nu\right)\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{400}{1200}.100\%\simeq33,3\%\) \(\rightarrow G=50\%-33,3\%=16,7\%\)
\(c,\) \(TH1:A_1=C_1=100\left(nu\right)\)
\(\Rightarrow\left\{{}\begin{matrix}A_2=T_1=400-100=300\left(nu\right)\\C_2=G_1=200-100=100\left(nu\right)\end{matrix}\right.\)
\(TH2:A_2=X_2=100\left(nu\right)\)
Tương tự: \(\Rightarrow\left\{{}\begin{matrix}A_1=T_2=300\left(nu\right)\\C_1=G_2=100\left(nu\right)\end{matrix}\right.\)
\(d,\) \(H=2A+3G=1400\left(lk\right)\)
\(LKHT=2.\left(N-1\right)=2398\left(lk\right)\)
\(C=\dfrac{N}{20}=60\left(ck\right)\)
\(e,M=N.300=360000(dvC)\)
a) Tổng số nu : \(\dfrac{2040.2}{3,4}=1200\left(nu\right)\)
b) Có : \(\left\{{}\begin{matrix}A=2G\\A+G=\dfrac{1200}{2}=600\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}A=T=400\left(nu\right)\\G=C=200\left(nu\right)\end{matrix}\right.\)
% từng loại nu : \(\left\{{}\begin{matrix}A\%=\%T=\dfrac{400}{1200}.100\approx34\%\\\%G=\%C=50\%-34\%=16\%\end{matrix}\right.\)
c) Cho mạch đó là mạch 1, ta có A1 = C1 = 100 nu
Theo NTBS :
A1 = T2 = 100 nu
T1 = A2 = A - A1 = 300 nu
G1 = C2 = G - G2 = 100 nu
C1 = G2 = 100 nu
d) Số lk H : \(H=2A+3G=1400\left(lk\right)\)
Số lk hóa trị : \(H_{HT}=N-2=1200-2=1198\left(lk\right)\)
Số chu kì xoắn : \(C=\dfrac{N}{20}=60\left(cki\right)\)
e) Khối lượng gen : \(M=300N=360000\left(đvC\right)\)
