Question

genetics a geneticist is studying two genes, each gene can be either dominant or recessive. a sample of 100 individuals is categorized as follows. write your answer as a fraction or a decimal, rounded to four decimal places.
gene 2

	dominant	recessive
recessive	18	9

(a) what is the probability that in a randomly sampled individual, gene 1 is recessive?
(b) what is the probability that in a randomly sampled individual, gene 2 is recessive?
(c) given that gene 1 is recessive, what is the probability that gene 2 is recessive?
(d) two genes are said to be in linkage equilibrium if the event that gene 1 is recessive is independent of the event that gene 2 is recessive. are these genes in linkage equilibrium?

Explanation:

Step1: Find total number of individuals with recessive gene 1

The number of individuals with recessive gene 1 is the sum of the two - cells in the "Recessive" row of the table for gene 1, which is \(18 + 9=27\).

Step2: Calculate the probability

The probability \(P\) that a randomly - sampled individual has recessive gene 1 is the number of individuals with recessive gene 1 divided by the total number of individuals in the sample. The total number of individuals in the sample is \(100\). So \(P=\frac{27}{100}=0.2700\).

Step3: Find total number of individuals with recessive gene 2

The number of individuals with recessive gene 2 is the sum of the two - cells in the "Recessive" column of the table for gene 2, which is \(21 + 9 = 30\).
The probability that a randomly - sampled individual has recessive gene 2 is \(P=\frac{30}{100}=0.3000\).

Step4: Calculate conditional probability

The conditional probability \(P(\text{Gene 2 is recessive}|\text{Gene 1 is recessive})\) is calculated using the formula \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). Here, \(A\) is the event that gene 2 is recessive and \(B\) is the event that gene 1 is recessive. \(P(A\cap B)\) is the proportion of individuals with both gene 1 and gene 2 recessive, which is \(\frac{9}{100}\), and \(P(B)=\frac{27}{100}\). So \(P(A|B)=\frac{\frac{9}{100}}{\frac{27}{100}}=\frac{9}{27}\approx0.3333\).

Step5: Check for independence

Two events \(A\) (gene 2 is recessive) and \(B\) (gene 1 is recessive) are independent if \(P(A|B)=P(A)\). We found \(P(A) = 0.3000\) and \(P(A|B)\approx0.3333\). Since \(P(A|B)
eq P(A)\), the genes are not in linkage equilibrium.

Answer:

(a) \(0.2700\)
(b) \(0.3000\)
(c) \(0.3333\)
(d) No