Once again, first determine the genotypes of the parents. We know that the female is heterozygous for the rate of head bobbing but homozygous recessive for the color of her throat patch. So, what must her genotype be?





The male is also heterozygous for head bobbing but is homozygous dominant for throat patch color. What, then, is the male's genotype?





Now that you have the genotypes of the parents, perform a punnett square to determine what the F1 offspring would be (remember to place the gametes of one parent across the top and the gametes of the other parent down the left side. It does not matter which parent's gametes are placed across the top and which are placed down the side.):





Your punnett square should look similar to the one below. If it does not, review the material on gamete formation and how to set up the punnett square and try again.

Fr Fr fr fr
FR FFRr FFRr FfRr FfRr
FR FFRr FFRr FfRr FfRr
fR FfRr FfRr ffRr ffRr
fR FfRr FfRr ffRr ffRr


Now simply count up the number of offspring of each genotype and phenotype produced.




Return to the question page and select the correct answers.