A very rare dominant allele that causes the little finger to be crooked has a pe

Question

A very rare dominant allele that causes the little finger to be crooked has a penetrance value of 80%. If a homozygous unaffected person has children with a heterozygote carrying this mutant allele, what is the probability that an offspring will have little fingers that are crooked? Give the phenotypes of the parents as well as the genotypic and phenotypic ratios of the offspring for the following crosses: I_A I x I^B I^B I^A I^B x I^B I If a mother and her child belong to blood group O, then the father could not be of what blood group ?

Explanation / Answer

Answer: Probability of an offspring having crooked finger =40%

Crooked finger allele=C

Normal finger allele=c

cc (Homozygous unaffected) x Cc (heterozygous individual)

F1: Cc(crooked), Cc (crooked), cc(normal), cc(normal)

50% are crooked finger individuals and 50% are normal

Penetrance= 80%

Therefore= 0.5 x 0.8= 0.4

40% of the offsprings will have crooked finger.

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