Answer:
a: 1/4
b: 1/2
c: 1/4
d: 0
Explanation:
Since the wife had a color-blind father (X^cY), she should be a carrier for the disease. Her genotype would be "X^cX". Fathers do not transmit the X-linked disorders to their sons, so the genotype of the husband is "XY".
a. The probability that the couple's first child will be a normal son: 1/4
b. The probability that the couple's first child will be a normal daughter: 1/2
c. The probability that the couple's first child will be a color-blind son: 1/4
d. The probability that the couple's first child will be a color-blind daughter: Zero
Answer:
28 units
Explanation:
This disorder follows quantitative inheritance. It is controlled by three genes which do not show the usual dominant-recessive relationship . The six alleles individually contribute to the effect which add up to produce the cumulative phenotype. Dominant allele contributes 6 units of risk whereas recessive allele contributes 2 units of risk.
Individual with genotype AABbCc has four dominant alleles (AABC) and two recessive alleles (bc). So their total risk units =
(6*4) + (2*2) = 24 + 4
= 28 units
Answer:
13
Explanation:
it get recycled again when it becomes old