Genetic linkage mapping for a large number of families identifies 4% recombination between the genes for rh blood type and elliptocytosis. at the rh locus, alleles r and r control rh+ and rh- blood types. allele e producing elliptocytosis is dominant to the wild-type recessive allele e.

tom and terri each have elliptocytosis, and each is rh+.

tom's mother has elliptocytosis and is rh- while his father is healthy and has rh+.

terri's father is rh+ and has elliptocytosis; terri's mother is rh- and is healthy.

genetic linkage mapping for a large number of fami

what is the probability that the first child of tom and terri will be rh- and have elliptocytosis?

enter your answer to four decimal places (example 0.2365).

0.2404

Explanation:

The genes R/r and E/e are linked and there is 4% recombination between them.

The possible genotypes and phenotypes are:

Tom and Terri each have elliptocytosis (they are E_), and each is Rh+ (they are R_).

Tom's mother has elliptocytosis (E_) and is Rh- (rr), so she has the genotype Er/_r. His father is healthy (ee) and has Rh+ (R_), so he has the genotype eR/e_. Tom must have inherited his E allele from his mother and his R allele  from his father, so he has the genotype eR/Er.

Terri's father is Rh+ (R_) and has elliptocytosis (E_), while Terri's mother is Rh- (rr) and is healthy (ee) with the genotype er/er. Terry could only receive the chromosome er from her mother, and because she is heterozygous for both genes the dominant alleles were both received from her father. Terri's genotype is ER/er.

The frequency of recombination is 4%, so 4% of the produced gametes will be recombinant. There are two possible recombinant gametes, so each will appear 2% of the times (a frequency of 0.02).

Tom will produce the following gametes:

Terri will produce the following gametes:

A child Rh- with elliptocytosis has the genotype rrE_. This can happen from the independent combination of the following gametes from Tom and Terri respectively:

And the total probability of having a rrE_ child will be 0.2304 + 0.0096 + 0.0004 = 0.2404

lets get this bread mom

b is dominant over the allele for the white fur.