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Gaucher disease type 1 (GD1 ) is a recessive genetic disease that affects 1 in 900 individuals in a particular population. GD1 is caused by a mutation in the enzyme glucocerebrosidase. Assuming the population is in Hardy-Weinberg equilibrium, calculate the frequency of the wild-type (nonmutant) allele for the enzyme glucocerebrosidase.

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hihirasohail

Answer:

Using the Hardy-Weinberg equilibrium equation, the frequency for the wild-type allele (referred to as p, see explanation below) comes out to be [tex]\frac{2}{3}[/tex]

Explanation:

An individual will have to be homozygous recessive for the mutation in the enzyme glucocerebrosidase to have the Gaucher disease type 1 (GD1). This percentage will be [tex]q^{2}[/tex]

[tex]q^{2}[/tex] = [tex]\frac{1}{900}[/tex] × 100 = [tex]\frac{1}{9}[/tex]

[tex]q[/tex] = [tex]\sqrt{\frac{1}{9} }[/tex] = [tex]\frac{1}{3}[/tex]

We know that [tex]\\ p + q = 1[/tex] because the sum of both allele frequencies has to be 100%

So [tex]\\ 1 - \frac{1}{3} = \frac{2}{3}[/tex]

Hence, [tex]\\ p = \frac{2}{3}[/tex]

Hope that answers the question, have a great day!

Gaucher disease is a disorder that gets genetically inherited and is caused due to the deficit of enzyme glucocerebrosidase. This deficiency results in fat accumulation within the spaces of the spleen, bones, liver etc.

The frequency of the wild type allele for the enzyme is  [tex]\dfrac{2}{3}[/tex].

The frequency can be estimated by:

According to the Hardy Weinberg equation:

[tex]p^{2} + 2pq + q^{2} = 1[/tex]

Where,

  • Purebred dominant frequency = [tex]p^{2}[/tex]
  • Frequency of heterozygous = 2pq
  • Purebred recessive frequency = [tex]q^{2}[/tex]

The person suffering from the GD 1 will have a mutation in their recessive purebred allele and will be given by [tex]q^{2}[/tex].

Step 1: As there is a chance of one person getting diseased from the population of 900:

[tex]\begin{aligned}q^{2} &= \dfrac{1}{900} \times 100\\\\q^{2}&= \dfrac{1}{9}\\\\q &= \dfrac{1}{3}\end{aligned}[/tex]

We know that the total of both allele frequencies should be 100% then,

[tex]p+q = 1[/tex]

Step 2: Putting the value of q from above:

[tex]\begin{aligned} p+q &= 1\\&= 1 - \dfrac{1}{3} \\&= \dfrac{2}{3}\end{aligned}[/tex]

Therefore,

[tex]p = \dfrac{2}{3}[/tex]

Thus, the frequency of the wild-type is [tex]\dfrac{2}{3}[/tex].

Learn more about Gaucher disease type 1 (GD 1) here:

https://brainly.com/question/6374221

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