Answer:
Possible genotypes and their frequencies
Frequency of babies with blues eye having genotype "bb" is [tex]0.5[/tex]
Frequency of babies with brown eye having genotype "Bb" is [tex]0.41[/tex]
Frequency of babies with brown eye having genotype "BB" is [tex]0.084[/tex]
Explanation:
In a normal human being, brown eye color is dominant over blue eye color.
Thus B is dominant over b
Now,
Babies with blue eye colour will have genotype "bb" because it is a recessive trait thus both the alleles must be the recessive allele.
Therefore,
[tex]q^{2} = \frac{10}{20} \\=0.5[/tex]
Thus
[tex]q=\sqrt{0.5} \\= 0.707\\=0.71[/tex]
While the genotype of brown babies can be either homozygous dominant or heterzygous dominant
Now as per Hardy Weinberg's equation -
[tex]p+q=1[/tex]
so [tex]p=1-0.71[/tex]
[tex]p = 0.29[/tex]
Thus frequency of babies with genotype "BB" is equal to [tex]p^{2}[/tex]
[tex]= 0.084[/tex]
Now frequency of babies with genotype "Bb" is
[tex]1-p^{2} -q^{2} =2pq[/tex]
[tex]2pq = 0.41[/tex]
Thus,
Frequency of babies with blues eye having genotype "bb" is [tex]0.5[/tex]
Frequency of babies with brown eye having genotype "Bb" is [tex]0.41[/tex]
Frequency of babies with brown eye having genotype "BB" is [tex]0.084[/tex]
