A man and a woman both of whom are deaf are each homozygous recessive in one of

Question

A man and a woman both of whom are deaf are each homozygous recessive in one of three different “hearing” genes, A, B and C and heterozygous in the other two genes (see the genotypes below). Recessive homozygosity for any one of these three genes causes deafness. In addition recessive homozygosity for any two or three genes will result in embryonic lethality and spontaneous abortion. Given the genotypes of the parents below, what is the probability that a live-born child will be deaf. (Hint: Think sample space. . . .not all of the genotypes will be live born, only a portion. A branch diagram again would be real useful for sorting out the different genotypes and their proportions.)

Mother: Aa, Bb, cc

Father: Aa, bb, Cc

Explanation / Answer

when Aa Bb cc will be crossed with AabbCc

probability of dominant and recessive phenotype for each is

Aa with Aa dominant probability is 0.75 and recessive is 0.25

Bb and bb in this case 0.50 dominant and 0.50 recessive

Cc and cc 0.50 dominant and 0.50 recessive

so probabiltiy of dominant one are:-

0.75*0.50*0.50=0.1875 IS THE PROBABILITY THAT THEY WILL SHPOW DOMINANT GENOTYPE and 0.8125 for recessive. so these many child will be deaf

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