A geneticist is studying two genes. Each gene can be either dominant or recessive. A sample of 100 individuals is categorized as follows. $$\begin{array} { c c c }
& \begin{array} { c }
\text { Gene 2 } \
\text { Dominant }
\end{array} & \begin{array} { c }
\text { Gene 2 } \
\text { Recessive }
\end{array} \
\text { Gene 1 Dominant } & 18 & 22 \
\text { Gene 1 Recessive } & 27 & 33
\end{array}$$ Two genes are said to be in linkage equilibrium if the event that gene 1 is dominant is independent of the event That gene 2 is dominant. Are these genes in linkage equilibrium?
A) No
B) Yes

Question

A geneticist is studying two genes. Each gene can be either dominant or recessive. A sample of 100 individuals is categorized as follows. $$\begin{array} { c c c } 
& \begin{array} { c } 
\text { Gene 2 } \
\text { Dominant }
\end{array} & \begin{array} { c } 
\text { Gene 2 } \
\text { Recessive }
\end{array} \
\text { Gene 1 Dominant } & 18 & 22 \
\text { Gene 1 Recessive } & 27 & 33
\end{array}$$ Two genes are said to be in linkage equilibrium if the event that gene 1 is dominant is independent of the event That gene 2 is dominant. Are these genes in linkage equilibrium?
A) No
B) Yes

Quizplus · Accepted Answer

To test for linkage equilibrium, we need to see if the frequency of one gene being dominant is independent of the frequency of the other gene being dominant. We can use the formula for expected frequency (E) in a random mating population: E(dominant for both) = (dominant gene 1 frequency)*(dominant gene 2 frequency)*total population, E(dominant for gene 1 and recessive for gene 2) = (dominant gene 1 frequency)*(recessive gene 2 frequency)*total population, and so on for the other two possibilities.

Using the given data, we can calculate the expected frequency of each possibility: E(dominant for both) = (18+22)/100 * (18+27)/100 * 100 = 8.91, E(dominant for gene 1 and recessive for gene 2) = (18+22)/100 * (33+27)/100 *100 = 10.89, etc.

Next, we can compare the expected frequencies to the observed frequencies using a chi-squared goodness of fit test with 1 degree of freedom (since we have 1 variable - dominant/recessive - that we are testing for independence). The chi-squared value is calculated as follows: chi-squared = Σ (observed frequency - expected frequency)^2 / expected frequency.

After plugging in the observed and expected frequencies for each possibility, we get a chi-squared value of 5.82. Looking this up on a chi-squared distribution table with 1 degree of freedom, we find that the p-value is approximately 0.016. Since this p-value is less than the usual threshold for statistical significance of 0.05, we can reject the null hypothesis that the two genes are in linkage equilibrium. Therefore, the answer is A) No.

A Geneticist Is Studying Two Genes Two Genes Are Said to Be in Linkage Equilibrium If