Considering that Hemophilia B is an X-linked recessive disease, than for it to be observed in female offspring, the mother must be homozygous recessive because females have XX chromosomes and if the father transfers only the dominant gene then the female offspring will always be heterozygous dominant.

In the case of a father who carries the dominant gene on his X choromosome and a mother who is heterozygous for the disease meaning she has both the dominant and the recessive gene, if we calculate the percentages for this situation, we see that for the first generation, one female offspring is heterozygous dominant, one female offspring if homozygous, one male offspring carries the dominant gene on his X chromosome and the other male offspring carries the recessive gene on his X choromosome which is the one that carries the disease.

I hope this answer helps.

6 0

3 years ago

In mice, black color (B) is dominant to white (b). At a different locus, a dominant allele (A) produces a band of yellow just be

artcher [175]

Answer:

Explanation:

Heterozygous mice at both loci would be BbAa genotype.

B (black) is dominant over b (white) but the presence of A results in agouti.

Crossing two heterozygous:

BbAa x BbAa

Progeny

9 B_A_ - Agouti

3 bbA_ - agouti

3 B_aa - Black

1 bbaa - white

Hence, 3 out of 16 progeny will be black in color.

See the attached image for the Punnet's square analysis of the cross.

4 0

3 years ago

How do we combine forces?

AleksandrR [38]

Answer:

You first kill/dominate the other and then you dictate whatever is left.....

3 0

2 years ago