Genetic Assignment Name Factor V is a component of the blood that is required for normal blood clotting. A mutation in the gene encoding this protein leads to increased risk of forming blood clots leading to deep vein thrombosis and potentially pulmonary embolisms and/or heart attack or stroke. In working the following problem use V to indicate the normal allele and Vm to indicate the mutant allele. This is a case of incomplete dominance where each copy of the mutant allele that an individual carries increases the risk of inappropriate blood clots: VV-normal risk, VV-higher risk, VmVm-highest risk. A receptor protein located on the surface of the epithelial cells lining the circulatory system is required to effectively remove circulating cholesterol from blood. High levels of circulating cholesterol increase the risk of blood clots leading to heart attack or stroke. A mutant form of the receptor exists that is nonfunctional; it is present in the cell membrane but does not bind cholesterol and remove it from the blood. In working the following problem use R to indicate the normal receptor and Rm to indicate the mutant allele. One way to measure the phenotypic risk of circulatory problems as it relates to these two genes is to count the number of mutant alleles a person carries. The lowest risk is represented by Zero (no mutant alleles) and the highest risk is represented by a Four (2 mutant alleles in the genotype for each gene). Consider a mating between two individuals heterozygous for both genes. 1. What are the genotypes of the male and female partner in this mating? (2 Points) 2. If the two genes are unlinked, what would be the genotypes of all possible gametes? (2 Points) Male Gametes Female 5. What is the probability (x/16) that the couple would produce a child with a phenotype risk number of-2. (be careful, there are multiple ways to arrive at a risk number of 2) (Bonus: 0.5 Points) 6. Could both parents of the man in this cross have been homozygous for one of the non-mutant alleles? Give one possible combination of genotypes that could have existed in the parents of these individuals (show a genotype for each parent-a cross). (Bonus: 0.5 Points)