Question

In humans, an allele that causes its possessor to have six fingers per hand is dominant to an allele that causes five fingers. Use F for the dominant allele and f for the recessive allele and answer the following questions. What is the genotype of someone who is heterozygous at this locus? What is the phenotype of someone who is heterozygous at this locus? What is the genotype of someone who is homozygous recessive at this locus? What is the phenotype of someone who is homozygous recessive at this locus? If a man and a woman, both homozygous recessive at this locus, have a child, what is the probablility that this child will have six fingers per hand? If a man who is heterozygous at this locus and a woman with five fingers per hand have a child, what is the probablility that the child will have six fingers per hand?

Answer 1

According to the question,

Dominant allele causing the possessor to have six fingers per hand = F

Recessive allele causing the possessor to have five fingers per hand = f

Now,

Genotype of someone who is heterozygous at this locus:

Being heterozygous means having two different alleles at each of the two locus for a gene on homologous chromosomes. Therefore an individual who is heterozygous will have the dominant allele on one locus and the recessive allele on the other.

Therefore their genotype will be = Ff

Phenotype of someone who is heterozygous at this locus:

Phenotype is the observable trait produced as the result of the underlying genotype. The genotype of a heterozygous individual at this locus is Ff. Since, F is the dominant allele and is responsible for having six fingers per hand, it will be the one that is expressed.

Therefore, the individual will have six fingers per hand, and that is their phenotype.

Genotype of someone who is homozygous recessive at this locus:

A homozygous individual will have the same allele in the locus of both homologous chromosome. For a homozygous recessive individual, the allele present in both locus will be the recessive allele. As the recessive allele is denoted by f, the required genotype is ff.

Thus, genotype of a homozygous recessive individual = ff.

Phenotype of someone who is homozygous recessive at this locus:

The genotype of homozygous recessive individual is ff. Since only recessive allele is present, it will be the one expressed and the individual will have the recessive phenotype of five fingers per hand.

Thus, the required phenotype is five fingers per hand.

Probability that the child will have six fingers per hand for homozygous recessive parents:

If both the man and woman is homozygous recessive (ff), then the child would not inherit the dominant F allele responsible for six fingers per hand from either parents. Therefore there is no chance that the child will have six fingers per hand.

Since,

Genotype of the father = ff

Genotype of the mother = ff

All possible offspring phenotype and genotype can be illustrated by the following Punnett Square:

Sperm  → Ovum ↓	f	f
f	ff (five fingers per hand)	ff (five fingers per hand)
f	ff (five fingers per hand)	ff (five fingers per hand)

Thus, as evident, the probability of the child having six fingers per hand = 0/4 = 0. So there is no chance that the child will have six fingers per hand.

Probability that the child will have six fingers per hand for heterozygous father and five fingers per hand mother:

In this case,

Genotype of the father = Ff

Genotype of the mother = ff

[Since, the woman has five fingers per hand and it is a recessive trait, therefore the woman must have recessive allele in both the homologous chromosome].

Now,

All possible offspring phenotype and genotype can be illustrated by the following Punnett Square:

Sperm  → Ovum ↓	F	f
f	Ff (six fingers per hand)	ff (five fingers per hand)
f	Ff (six fingers per hand)	ff (five fingers per hand)

Therefore, the probability that the child will have six fingers per hand = 2/4 = 1/2 = 0.5.

Thus, there is a 1 in 2 or a 50% chance that the child will have six fingers per hand.