# USMLE Step 1 – Genetics and the Hardy Weinberg equation Tyler York The Hardy Weinberg equation plays an important role in population genetics. It can be used to calculate the frequency of each of the three genotypes from knowledge of the frequency of the individual alleles, and vice versa. This video also gives examples on how to apply the equation in autosomal recessive and X-linked recessive disorders.

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## Full USMLE Step 1 – Genetics and the Hardy Weinberg equation video transcript:

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Hello everyone this is Sujata. In this video we will discuss genetics Hardy Weinberg equation. What is the Hardy Weinberg law or equation? This equation states that P square +2 pq plus Q square equals one where P&Q are two alleles of a gene, P being the normal or disease free allele.
00:00:27
Whereas Q being the abnormal or disease causing allele P square or homozygous for the normal allele gives the disease free population state whereas P + Q equals to 1. Hardy Weinberg's equation or law in autosomal recessive diseases.
00:00:50
In autosomal recessive diseases, we need 2 diseased allele or two diseased chromosomes to get the disease, whereas we need only one diseased allele to get a carrier state. Therefore P square which is the homozygous normal state will be a totally disease Free State. Q square which is homozygous for the diseased allele will give you the frequency of the diseased population in an autosomal recessive disease.
00:01:20
Two PQ on the other hand, will give you the frequency of carrier population of an autosomal recessive disease. Let's look at an application of Hardy Weinberg equation in autosomal recessive disease. Tay Sachs is an autosomal recessive disorder. It has a frequency of 1 in 400 in the Ashkenazi Jewish population.
00:01:46
How will you calculate the carrier frequency of Tay Sachs disease? Since the frequency of Tay Sachs is 1 in 400 and we know it is an autosomal recessive disorder, that means one in 400 stands for the value of Q square or homozygous, diseased or homozygous abnormal chromosome. So from that we can calculate the value of Q which will be one in 20.
00:02:16
To calculate the carrier frequency of an autosomal recessive disorder, we need 2P Q. We already have the value for Q and the value for P can be found from the formula. P + Q = 1, which will be approximately 1. So the carrier frequency will come out to 2 * 1 * 1 in 20, which is 1 in 10.
00:02:46
X linked recessive disorders. Males carry only 1X chromosome while females carry 2X chromosomes. So if a male carries the diseased X chromosome then he will suffer from the X linked recessive disorder on the other hand since females carry 2X chromosomes.
00:03:03
One of them will be a normal and the other is likely to be abnormal. So females are likely to be carriers and not suffer from X linked recessive disorders. Therefore, in X linked recessive disorders the incidence in males is equal to Q. The incidence in females is equal to Q square because they need 2 diseased genes to get the disease.
00:03:31
And on the other hand, the frequency of carriers will be equal to two PQ. Let us look at application of Hardy Weinberg equation in X linked recessive disorder. Color blindness is an X linked recessive disease with the frequency of 1 in 400 males. Calculate the carrier frequency of color blindness.
00:04:00
The carrier frequency of an X linked recessive disorder is given by two PQ. The incidence of Indiana males is Q, so one in 400 will be equal to Q2. PQ will be 2 * 1 * 1 in 400 equals to one in 200. So the carrier frequency of color blindness will be one in 200.
00:04:31
Hope this video helped you understand the application of Harley Weinberg equation. Thanks for watching.
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