Question

Two theoretical genetic strains of a virus \(\left(a^{-} b^{-} c^{-} \text {and } a^{+} b^{+} c^{+}\right)\) were used to simultaneously infect a culture of host bacteria. Of 10,000 plaques scored, the following genotypes were observed. Determine the genetic map of these three genes on the viral chromosome. Decide whether interference was positive or negative.

Short answer

Answer: The genetic map for the three genes is ordered as a - 13.1 cM - c - 0.6 cM - b. The interference is negative, as the observed number of double recombinants is greater than the expected number, indicating that one recombination event increases the probability of a second recombination event happening nearby.

Step-by-step solution

Organize the given data

First, we need to tabulate the given data and express it in terms of recombination frequencies. Let's represent the wild type as (+) and the mutant type as (-). Here are the observed genotypes and their occurrences out of 10,000 plaques scored: - a+b+c+: 3,140 - a-b-c-: 3,080 - a+b-c-: 1,740 - a-b+c+: 1,660 - a-b-c+: 675 - a+b-c+: 635 - a+b+c-: 25 - a-b+c-: 35

Determine the recombination frequencies

Now, we will calculate the recombination frequency for each possible pair of genes. The recombination frequency is the proportion of recombinant offspring as a result of recombination events between the two genes. - Recombination frequency between a and b: Since a+b-c- and a-b+c- genotypes are recombinants for a and b, we calculate the ratio in the following way: \(\frac{1,740 + 1,660}{10,000} = 0.340\) - Recombination frequency between a and c: Similarly, using the genotypes a-b-c+ and a+b-c+, we calculate: \(\frac{675 + 635}{10,000} = 0.131\) - Recombination frequency between b and c: Using the genotypes a+b+c- and a-b+c-, we calculate: \(\frac{25 + 35}{10,000} = 0.006\)

Create the genetic map

To create the genetic map, we multiply each recombination frequency by 100 to convert it to a map distance in centiMorgans (cM). - Map distance between a and b: \(0.340 \times 100 = 34 \text{ cM}\) - Map distance between a and c: \(0.131 \times 100 = 13.1 \text{ cM}\) - Map distance between b and c: \(0.006 \times 100 = 0.6 \text{ cM}\) Since the distance between a and c is smaller than the distance between a and b, we can order the genes as follows: a - c - b. So, the genetic map is: a - 13.1 cM - c - 0.6 cM - b

Determine interference

To determine if interference is positive or negative, we need to look at the actual number of double recombinants observed versus the expected number of double recombinants, assuming no interference. The expected number of double recombinants is calculated by multiplying the total number of progeny (10,000) by the product of the recombination frequencies between genes a and c, and genes b and c. Expected double recombinants \( = 10,000 \times 0.131 \times 0.006 = 7.86 \approx 8\) The actual observed number of double recombinants is 25+35=60. Since the observed number of double recombinants (60) is greater than the expected number (8), the interference is negative. This means that the occurrence of one recombination event increases the probability of a second recombination event happening nearby.