Question

During ur last shift, 20 babies were born. 10 had blue eyes, and 10 had brown eyes

B- Brown b- Blue
Given the phenotypes, what are the possible genotypes for these babies

Answer 1

Each person then has 4 total alleles that determine their eye color. The B allele (brown) is always dominant. The blue eye trait is always recessive.

Answer 2

Possible genotypes and their frequencies

Frequency of babies with blues eye having genotype "bb" is
`0.5`

Frequency of babies with brown eye having genotype "Bb" is
`0.41`

Frequency of babies with brown eye having genotype "BB" is
`0.084`

Step-by-step 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,

`q^(2) = (10)/(20) \\=0.5`

Thus

`q=√(0.5) \\= 0.707\\=0.71`

While the genotype of brown babies can be either homozygous dominant or heterzygous dominant

Now as per Hardy Weinberg's equation -

`p+q=1`

so
`p=1-0.71`

`p = 0.29`

Thus frequency of babies with genotype "BB" is equal to
`p^(2)`

`= 0.084`

Now frequency of babies with genotype "Bb" is

`1-p^(2) -q^(2) =2pq`

`2pq = 0.41`

Thus,

Frequency of babies with blues eye having genotype "bb" is
`0.5`

Frequency of babies with brown eye having genotype "Bb" is
`0.41`

Frequency of babies with brown eye having genotype "BB" is
`0.084`