In a population, if the proportion of individuals who have sickle-cell disease is 0. 50, and the population is assumed to be at Hardy-Weinberg equilibrium, what is the expected frequency of the HbS allele

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Question:

In a population, if the proportion of individuals who have sickle-cell disease is 0. 50, and the population is assumed to be at Hardy-Weinberg equilibrium, what is the expected frequency of the HbS allele

Answer:

According to Hardy-Weinberg equilibrium, the expected frequency of the HbS allele is 0.71 as the proportion of individuals who have sickle-cell disease is 0. 50.

What is Hardy-Weinberg equilibrium?

“The Hardy-Weinberg equilibrium is a principle stating that the genetic variation in a population will remain constant from one generation to the next in the absence of disturbing factors.”

What are the equations to solve Hardy-Weinberg equilibrium?

There are two equations necessary to solve a Hardy-Weinberg Equilibrium question:

  •  is the frequency of the dominant allele.  
  • is the frequency of the recessive allele.  
  • ² is the frequency of individuals with the homozygous dominant genotype.  
  • 2 is the frequency of individuals with the heterozygous genotype.
  • ² is the frequency of individuals with the homozygous recessive genotype.
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What is an allele?

“An allele is one of two or more versions of DNA sequence (a single base or a segment of bases) at a given genomic location.”

What is sickle-cell disease?

Sickle-cell disease is an inherited disease in which the red blood cells have an abnormal crescent shape, block small blood vessels, and do not last as long as normal red blood cells.”

Given, the proportion of individuals who have sickle-cell disease is 0. 50.

∴ The homozygous recessive individuals (aa) are represented by the   term in the Hardy-Weinberg equilibrium equation.

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