Question

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

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

Answer:

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 $0.084$

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,

$q^{2} = \frac{10}{20} \\=0.5$

Thus

$q=\sqrt{0.5} \\= 0.707\\=0.71$

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 $0.084$