Question

In a cross between a strain of large guinea pigs and a strain of small guinea pigs, the \(F_{1}\) are phenotypically uniform, with an average size about intermediate between that of the two parental strains. Among $1014 \mathrm{F}_{2}$ individuals, 3 are about the same size as the small parental strain and 5 are about the same size as the large parental strain. How many gene pairs are involved in the inheritance of size in these strains of guinea pigs?

Short answer

Approximately 2 gene pairs are involved in the inheritance of size in these strains of guinea pigs.

Step-by-step solution

Identify the total number of phenotypes in the F2 generation

From the given data, we know that there are \(\text{3 small}\), \(\text{5 large}\), and \((1014-3-5) = 1006\) intermediate individuals in the \(F_2\) generation.

Calculate the fraction of extreme phenotypes in the F2 generation

Since there are \(3\) small and \(5\) large-sized individuals, the fraction of extreme phenotypes in the \(F_2\) generation is: $$ \frac{3+5}{1014} = \frac{8}{1014} \approx 0.0079 $$

Determine the number of gene pairs

To find the number of gene pairs involved in the variation of size in these guinea pigs, we can use the following formula: $$ \frac{1}{4^{n}} = \text{fraction of extreme phenotypes} $$ where \(n = \) the number of gene pairs. In this case, the fraction of extreme phenotypes is \(0.0079\). We can calculate the number of gene pairs as follows: $$ \frac{1}{4^{n}} = 0.0079 $$ Taking the logarithm of both sides: $$ \log_{4}\left(\frac{1}{4^{n}}\right) = \log_{4}(0.0079) \\ -n = \log_{4}(0.0079) \\ n = -\log_{4}(0.0079) \\ n\approx 2 $$

Answer the question

Based on our calculations, there are approximately \(2\) gene pairs involved in the inheritance of size in these strains of guinea pigs.