Question

The ability of the human body to break down the red color in beets is controlled by an autosomal dominant allele. The inability is recessive, detected by red coloration of the urine (we will call this phenotype 'secretor'). If a nonsecretor woman with a secretor father has children with a nonsecretor man who in a previous relationship had a secretor daughter, what is the probability that their first child will be a secretor girl?A)1/2B)1/4C)1/8D)3/8E)1/3

Answer 1

Answer:

c. 1/8

Explanation:

Let's assume that the recessive allele "s" gives the "secretor" phenotype in the homozygous state while the dominant allele "S" is responsible for the "nonsecretor" phenotype.

According to the given information, the nonsecretor woman has a “secretor” father (ss). This makes the woman heterozygous dominant (Ss) for the trait. Likewise, the man has a “secretor” daughter (ss) from the previous relationship. Therefore, the man is also heterozygous dominant (Ss) for the trait.

A cross between two heterozygous dominant parents (Ss x Ss) would produce progeny in following ratio= 3/4 Nonsecretor (1/4 SS and 1/2 Ss): 1/4 Secretor (1/4 ss)

During each reproductive cycle, there are 1/2 chances for a couple to have a girl child (XX x XY= 1/2 XX: 1/2 XY). Therefore, the probability that their first child will be a “secretor girl”= 1/4 x 1/2= 1/8 .