Sickle cell disease (SCD) is caused by a recessive version of the hemoglobin gene. In the US, SCD occurs in about 0.2% of the newborn babies. In some African countries, 4% of the newborn babies have sickle cell. Out of a random sample of 10,000 newborn babies in the US, how many would you expect to be homozygous for the normal, dominant hemoglobin genotype assuming Hardy Weinberg equilibrium?

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Arclight Arclight · Answer 1 · 2019-09-12T06:48:08+02:00

Answer:

Number of babies with dominant hemoglobin genotype will be [tex]9120[/tex]

Explanation:

Given -

[tex]0.2[/tex]% of babies have SCD which means they have recessive version of hemoglobin

Let the recessive version of hemoglobin be represented by "h" and normal version of hemoglobin be represented by "H"

Babies having SCD will have genotype "hh" i.e homozygous recessive.

As per Hardy Weinberg's equation -

[tex]q^2= 0.002[/tex]

Thus, frequency of allele for recessive version of hemoglobin (h) is

[tex]\sqrt{q^2} \\= \sqrt{0.002} \\= 0.045\\[/tex]

Frequency of allele for normal version of hemoglobin (H) is

[tex]1-0.045\\= 0.955[/tex]

Thus, Frequency of population with normal version of hemoglobin i.e HH is equal to

[tex]p^{2}[/tex]

[tex]= 0.955^2\\= 0.912[/tex]

hence, number of babies with dominant hemoglobin genotype will be

[tex]0.912 * 10000[/tex]

[tex]= 9120[/tex]