Genetics: A geneticist is studying two genes. Each gene can be either dominant or recessive. A sample of 100individuals is categorized as follows. Write your answer as a fraction or a decimal, rounded to four decimal places.

Gene 2
Dominant Recessive
Gene 1 Dominant 57 24
Recessive 15 4

(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 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 is recessive is independent of the event that gene 1 is recessive. Are these genes in linkage equilibrium?

1 answer

To answer the questions, let’s first summarize the data by the total number of individuals in each category:

  • Gene 1 Dominant, Gene 2 Dominant (DD): 57
  • Gene 1 Dominant, Gene 2 Recessive (DR): 24
  • Gene 1 Recessive, Gene 2 Dominant (RD): 15
  • Gene 1 Recessive, Gene 2 Recessive (RR): 4

Total individuals: 57 + 24 + 15 + 4 = 100

(a) What is the probability that in a randomly sampled individual, gene 1 is recessive?

The number of individuals where gene 1 is recessive consists of the individuals categorized as "Recessive (RD) and Recessive (RR)":

\[ \text{Number of individuals with gene 1 recessive} = 15 + 4 = 19 \]

The probability \( P(\text{Gene 1 is recessive}) \) is: \[ P(\text{Gene 1 is recessive}) = \frac{\text{Number of individuals with gene 1 recessive}}{\text{Total number of individuals}} = \frac{19}{100} = 0.19 \]

(b) What is the probability that in a randomly sampled individual, gene 2 is recessive?

The number of individuals where gene 2 is recessive consists of the individuals categorized as "Recessive (DR) and Recessive (RR)":

\[ \text{Number of individuals with gene 2 recessive} = 24 + 4 = 28 \]

The probability \( P(\text{Gene 2 is recessive}) \) is: \[ P(\text{Gene 2 is recessive}) = \frac{\text{Number of individuals with gene 2 recessive}}{\text{Total number of individuals}} = \frac{28}{100} = 0.28 \]

(c) Given that gene 1 is recessive, what is the probability that gene 2 is recessive?

To find this conditional probability \( P(\text{Gene 2 is recessive} | \text{Gene 1 is recessive}) \), we use:

\[ P(\text{Gene 2 is recessive} | \text{Gene 1 is recessive}) = \frac{P(\text{Gene 1 is recessive and Gene 2 is recessive})}{P(\text{Gene 1 is recessive})} \]

From part (a), we have \( P(\text{Gene 1 is recessive}) = 0.19 \).

The number of individuals with both gene 1 and gene 2 recessive is 4 (from the RR category).

Thus, the probability \( P(\text{Gene 1 is recessive and Gene 2 is recessive}) = \frac{4}{100} = 0.04 \).

Now substituting these values:

\[ P(\text{Gene 2 is recessive} | \text{Gene 1 is recessive}) = \frac{0.04}{0.19} \approx 0.2105 \]

(d) Are these genes in linkage equilibrium?

To check if gene 1 recessive is independent of gene 2 recessive, we need to see if the following holds:

\[ P(\text{Gene 2 is recessive} | \text{Gene 1 is recessive}) = P(\text{Gene 2 is recessive}) \]

From above, we have:

  • \( P(\text{Gene 2 is recessive} | \text{Gene 1 is recessive}) \approx 0.2105 \)
  • \( P(\text{Gene 2 is recessive}) = 0.28 \)

Since \( 0.2105 \neq 0.28 \), we conclude that the events are not independent.

Thus, the genes are not in linkage equilibrium.

Summary:

(a) Probability that gene 1 is recessive: 0.19
(b) Probability that gene 2 is recessive: 0.28
(c) Probability that gene 2 is recessive given gene 1 is recessive: 0.2105
(d) The genes are not in linkage equilibrium.