Question

36) Wild type mice (+/+) were crossed to brown/brown homozygote recessive mutant mice. When 36) 100 million offspring were examined, 100 individuals with the brown phenotype were counted. Indicate the mutation rate for the brown gene. A) 0.001 mutations per gene per gamete per generation B) 5x10-0 mutations per gene per gamete per generation C) 1x10-0 mutations per gene per gamete per generation D) 1x10-8 mutations per gene per gamete per generation E) There is not enough information to do this calculation

Answer 1

Total number of offspring, N = 100 million = 100 x 10⁶ = 10⁸ individuals ( total number of genes = 2 x 10⁸ )

No. of individuals with the brown phenotype, p = 100 individuals ( number of brown genes = 2 x 100 = 200 genes)

Mutation rate for the brown gene ( $\mu$ )= no. of individuals with the brown phenotype/total number of individuals = no. of brown genes/total number of genes per gamete per generation = p/N = 200/2 x 10⁸ = 100/10⁸ = 1 x 10^-6 mutations per gene per gamete per generation.

hence correct answer is C