{"id":1463,"date":"2020-08-07T00:27:59","date_gmt":"2020-08-07T00:27:59","guid":{"rendered":"https:\/\/meddists.com\/learn\/pre-clinical\/medical-genetics\/population-genetics\/hardy-weinberg-law-practice-questions\/"},"modified":"2021-02-04T10:27:19","modified_gmt":"2021-02-04T09:27:19","slug":"hardy-weinberg-law-practice-questions","status":"publish","type":"page","link":"https:\/\/meddists.com\/learn\/pre-clinical\/medical-genetics\/population-genetics\/hardy-weinberg-law-practice-questions\/","title":{"rendered":"Hardy-Weinberg law practice questions"},"content":{"rendered":"\n

Question 1<\/h2>\n<\/span>\n\n\n

The frequency of the ‘D’ allele is 0.34 in a population of 1378 people. Assuming that it is an ideal population,<\/p>\n\n\n\n

a. How many homozygous recessive people are in the population?<\/strong><\/p>\n\n\n\n

As discussed previously, we take the dominant allele frequency as p and recessive as q. So;<\/p>\n\n\n\n

p=0.34, and since p+q=1<\/strong>, q=1-p=1-0.34=0.66<\/p>\n\n\n\n

If they ask for homozygous recessive people, they are basically asking for genotypic frequency(a genotype makes up a person, not an allele). So they are asking for the frequency of ‘dd’ which is q2<\/sup><\/p>\n\n\n\n

q2<\/sup>=0.662<\/sup>=0.4356. Therefore, the number of homozygous recessive people is 0.4356\u00d71378=600 people<\/p>\n\n\n\n

b. How many heterozygous people are in the population?<\/strong><\/p>\n\n\n\n

In this case, they are asking for the genotypic frequency of ‘Dd’ which is 2pq=2\u00d70.34\u00d70.66=0.4488<\/p>\n\n\n\n

Therefore, the number of heterozygous people is 0.4488\u00d71378=618 people<\/p>\n\n\n\n

Just a reminder that the allelic and genotypic frequencies must never be greater than one. They are relative frequencies and probabilities and as you remember from the Biostatistics course, probability can never be greater than one.<\/em><\/p>\n\n\n<\/span>\n

Question 2<\/h2>\n<\/span>\n\n\n

In a Hardy-Weinberg population, 800 people express the dominant phenotype for a certain trait while the remaining 300 expresses the recessive phenotype.<\/p>\n\n\n\n

What are the allelic frequencies?<\/strong><\/p>\n\n\n\n

If 800 people express the dominant phenotype, this is the sum of the homozygous dominant and heterozygous people i.e p2<\/sup>+2pq=800\/1100=0.7272 <\/strong>and q2<\/sup>= 300\/1100=0.2727<\/strong>(keep in mind that these are relative frequencies which means we compare it to the total population. This is why we divide by 1100)<\/em><\/p>\n\n\n\n

If q2<\/sup>=0-2727, then q=\u221a0.2727=0.522. In this case, p=1-q=1-0.522= 0.478<\/p>\n\n\n\n

Therefore, the allelic frequencies are p=0.478 and q=0.522<\/p>\n\n\n<\/span>\n

Question 3<\/h2>\n<\/span>\n\n\n

In a population, 1 out of 7 people has sickle cell.<\/p>\n\n\n\n

What is the probability of a couple to have a child with sickle cell if one of the parents is a heterozygote and the other one is healthy?<\/strong><\/p>\n\n\n\n

Sickle cell is an autosomal recessive disease i.e only homozygous recessive people will have the disease. Therefore q2<\/sup>=1\/7=0.1428 and q= \u221a0.1428= 0.3778<\/p>\n\n\n\n

p+q=1<\/strong> so p=1-q=1-0.3778=0.6222<\/p>\n\n\n\n

The only way for a heterozygote to have a homozygous recessive child is by mating with another heterozygote<\/strong>. Which means that the other ‘healthy’ parent has to be a carrier. What is the probability of this healthy person to be a carrier? <\/strong>Remember that healthy people involve the homozygotes dominant and the heterozygotes(carriers). So;<\/p>\n\n\n\n

All healthy people= p2<\/sup>+2pq= (0.622)2<\/sup>+2(0.6222)(0.3778) =0.8570<\/p>\n\n\n\n

Probability of a healthy person to be a carrier is 2pq\/(p2<\/sup>+2pq) <\/strong>=0.4701\/0.8570= 0.5485<\/p>\n\n\n\n

Let us make a cross between the heterozygotes<\/p>\n\n\n\n

Parental Phenotypes:                Carrier\u00d7Carrier<\/p>\n\n\n\n

Parental Genotypes:                   AS\u00d7AS<\/p>\n\n\n\n

Parental gametes:                      \u24b6\u24c8\u00d7\u24b6\u24c8<\/p>\n\n\n\n

 <\/td>\u24b6<\/td>\u24c8<\/td><\/tr>
\u24b6<\/td>AA<\/td>AS<\/td><\/tr>
\u24c8<\/td>AS<\/td>SS<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

F1 phenotype: 1 healthy(non carrier-AA), 2 healthy(carrier-AS) and one sickler(SS)<\/p>\n\n\n\n

Probability of this couple to have a child who is a sickler is:<\/p>\n\n\n\n

 the probability of parent A to be a carrier\u00d7probability of parent B to be a carrier\u00d7probability of the child to be sick<\/strong><\/p>\n\n\n\n

We multiply all three because as stated previously, both parents must be carriers for the child to inherit the trait.<\/p>\n\n\n\n

  • We multiply because the events are independent events(refer to the Biostatistics course)<\/em><\/li>
  • For the first parent, the probability is 1(we were told this in the question so we are sure of it)<\/em><\/li>
  • The probability for the second parent is 0.5485(as calculated above)<\/em><\/li>
  • The probability of the child is 1\/4<\/li><\/ul>\n\n\n\n

    1\u00d70.5485\u00d7(1\/4)= 0.13705<\/p>\n\n\n<\/span>\n

    Question 4<\/h2>\n<\/span>\n\n\n

    In a population, 1 out of 7 people has sickle cell.<\/p>\n\n\n\n

    What is the probability of a couple to have a child with sickle cell if the mother has a sister with sickle cell? The grandparents and parents of the child are healthy.<\/strong><\/p>\n\n\n\n

    This is similar to the previous question except with a twist; one of the aunties have sickle cell. And as we already know that the only way for a heterozygote to have a homozygous recessive child is by mating with another heterozygote, it means that the maternal grandparents are heterozygotes( carriers). Lets us quickly do the cross below;<\/p>\n\n\n\n

    Parental Phenotypes:                Carrier\u00d7Carrier<\/p>\n\n\n\n

    Parental Genotypes:                   AS\u00d7AS<\/p>\n\n\n\n

    Parental gametes:                      \u24b6\u24c8\u00d7\u24b6\u24c8<\/p>\n\n\n\n

     <\/td>\u24b6<\/td>\u24c8<\/td><\/tr>
    \u24b6<\/td>AA<\/td>AS<\/td><\/tr>
    \u24c8<\/td>AS<\/td>SS<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    F1 phenotype: 1 healthy(non carrier-AA), 2 healthy(carrier-AS) and one sickler(SS)<\/p>\n\n\n\n

    Probability of the mother to be a heterozygote is 2\/3(out of all three healthy children, two are heterozygotes)<\/p>\n\n\n\n

    Probability of the father to be  heterozygote is 2pq\/(p2<\/sup>+2pq) <\/strong>=0.4701\/0.8570= 0.5485(refer to previous question)<\/em><\/p>\n\n\n\n

    Probability of the child to be a sickler = 2\/3 \u00d7 0.5485 \u00d7 1\/4 =0.0914 (refer to the previous question)<\/em><\/p>\n\n\n<\/span>\n

    Question 5<\/h2>\n<\/span>\n\n\n

    In a population of 200 people in Hardy-Weinberg equilibrium, the allelic frequency of the recessive allele for haemophilia is 0.19.<\/p>\n\n\n\n

    How many hemophilic men and women are present in the population?<\/strong><\/p>\n\n\n\n

    Haemophilia is an X-linked recessive disease. This means that only homozygous recessive women will have the disease.<\/p>\n\n\n\n

    As males are hemizygous for this trait (possess only one copy of the gene), only recessive males will express this trait(bear in mind that this trait is present only on the X chromosome)<\/p>\n\n\n\n

    Therefore, the fraction of affected women=homozygous recessive women=q2<\/sup>=0.192<\/sup>=0.0361 and number of affected women=200\u00d70.0361=7 women<\/p>\n\n\n\n

    The fraction of affected men=recessive men=q=0.19 and number of affected men=0.19\u00d7200=38 men (we do not do q2 <\/sup>because these men possess only one copy of this allele as they have only one X chromosome but the women have two alleles on both X chromosomes)<\/em><\/p>\n\n\n<\/span>\n

    Question 6<\/h2>\n<\/span>\n\n\n

    In a population of 1000 people, 500 of them have blood group O, 120 have AB, 234 have A and the rest have B.<\/p>\n\n\n\n

    What are the allelic frequencies?<\/strong><\/p>\n\n\n\n

    The Hardy-Weinberg law was created for non-polymorphic genes but the ABO genes are polymorphic(details on Polymorphism are discussed in the Polymorphism course)<\/em> so we have to slightly modify the law, but the principle stays the same. Remember that the IA<\/sup> and IB<\/sup> gene is dominant while i is recessive. As usual, we will use ‘p’ for the dominant allele and ‘q’ for the recessive. We will use ‘r’ for the second dominant allele. In this case,<\/p>\n\n\n\n

    • p is IA<\/sup><\/li>
    • r is IB<\/sup><\/li>
    • q is i<\/li><\/ul>\n\n\n\n

      Therefore, p+q+r=1<\/strong><\/p>\n\n\n\n

      For the genotypic frequencies, let us recall that<\/p>\n\n\n\n

      Blood group A can be IA<\/sup> IA <\/sup>(p2<\/sup>)<\/strong> or IA<\/sup>i (2pq)<\/strong><\/p>\n\n\n\n

      Blood group B can be IB<\/sup> IB <\/sup>(r2<\/sup>)<\/strong> or IB<\/sup>i (2rq)<\/strong><\/p>\n\n\n\n

      Blood group AB is IA<\/sup> IB <\/sup>(2pr)<\/strong><\/p>\n\n\n\n

      Blood group O is ii (r2<\/sup>)<\/strong><\/p>\n\n\n\n

      Therefore, (p2<\/sup>)+(2pq)+(r2<\/sup>)+(2rq)+(2pr)+(r2<\/sup>)=1<\/strong><\/p>\n\n\n\n

      In this case, r2 <\/sup>= 500\/1000 =0.5  and    r=\u221a0.5=0.7071<\/p>\n\n\n\n

      2pr=120\/1000=0.12  if r= 0.7071, p=0.0848<\/p>\n\n\n\n

      If p+q+r=1, <\/strong>q=1-(0.0848+0.7071)=0.2081<\/p>\n\n\n\n

      To conclude, the allelic frequencies are;<\/strong> p=0.0848, q=0.2081 and r=0.7071<\/p>\n\n\n\n

      Genotypic frequencies are p2<\/sup>, 2pq, r2<\/sup>, 2rq, 2pr <\/strong>and r2<\/sup><\/strong> . To find them, just substitute the numbers above. <\/p>\n\n\n<\/span>\n

      Question 7 <\/h2>\n<\/span>\n\n\n

      A study found that 1 in every 13 African-Americans is born with Sickle cell disease. <\/p>\n\n\n\n

      Assuming Hard-Weinberg equilibrium, which of the following statements is true?<\/strong><\/p>\n\n\n\n

      a. 40% of them do not have sickle cell disease<\/p>\n\n\n\n

      b. The allelic frequency for the dominant allele is 0.723<\/p>\n\n\n\n

      c. 40% of them are carriers for the trait<\/p>\n\n\n\n

      d. The allelic frequency for the dominant allele is 0.923<\/p>\n\n\n\n

      Sickle cell disease(SCD) is an autosomal recessive disease. This means that q2=1\/13=0.077 and q=0.277<\/p>\n\n\n\n

      Since p+q=1, then p=1-q=1-0.277=0.723
      So the allelic frequency of the dominant allele is 0.723<\/p>\n\n\n\n

      p2=0.523
      2pq=0.4<\/p>\n\n\n\n

      So 40% of them are carriers for the trait
      40%+52.3%=92.3% of them do not have SCD<\/p>\n\n\n\n

      The answers for this question are options b and c<\/strong><\/p>\n\n\n<\/span>\n

      Question 8<\/h2>\n<\/span>\n\n\n

      In a population, 5% of the men have color blindness. <\/p>\n\n\n\n

      Which of the following statements are true?<\/strong><\/p>\n\n\n\n

      a. 9.5% of the women are carriers for the trait <\/p>\n\n\n\n

      b. 95% of the women are carriers for the trait <\/p>\n\n\n\n

      c. 10% of the women do not have colour blindness<\/p>\n\n\n\n

      d. 0.25% of the women have color blindness<\/p>\n\n\n\n

      Colour blindness is an X-linked recessive disease. This means that if 5% of the men have it, then q=0.05 as men have only one X chromosome<\/p>\n\n\n\n

      If p+q=1, then p=1-q=1-0.05=0.95 so 95% of men are healthy
      p2=0.009025=0.9025%
      2pq= 0.0095=9.5% of women are carriers
      q2=0.00025=0.25% of women have colour blindness<\/p>\n\n\n\n

      The answers for this question are options a and d<\/strong><\/p>\n\n\n<\/span>\n

      Question 9<\/h2>\n<\/span>\n\n\n

      In a Hardy-Weinberg population of 5000 people, 34% of them have blood group O and the frequency of the B allele is 0.2. <\/p>\n\n\n\n

      Which of the following is true?<\/strong><\/p>\n\n\n\n

      a. The frequency of the recessive allele is 0.583<\/p>\n\n\n\n

      b. 31% of people have blood group A<\/p>\n\n\n\n

      c. 4.7% of people have blood group A<\/p>\n\n\n\n

      d. The frequency of the recessive allele is 0.217<\/p>\n\n\n\n

      Genotypic frequency of O=0.34
      if p=IA, r=IB and q=i, then q2=0.34
      q=0.583 so the frequency of the recessive allele is 0.583<\/p>\n\n\n\n

      r=0.2 and p+q+r=1
      p=1-r-q=0.217<\/p>\n\n\n\n

      People with blood group A= IAIA + IAi = p2 + 2pq= 0.311= 31.1%<\/p>\n\n\n\n

      The answers for this question are options a and b<\/strong><\/p>\n\n\n\n

      <\/p>\n<\/span>

      Question 1<\/h2>

      Question 2<\/h2>

      Question 3<\/h2>

      Question 4<\/h2>

      Question 5<\/h2>

      Question 6<\/h2>

      Question 7 <\/h2>

      Question 8<\/h2>

      Question 9<\/h2><\/div>","protected":false},"excerpt":{"rendered":"

      Question 1 The frequency of the ‘D’ allele is 0.34 in a population of 1378 people. Assuming that it is an ideal population, a. How many homozygous recessive people are in the population? As discussed previously, we take the dominant allele frequency as p and recessive as q. So; p=0.34, and since p+q=1, q=1-p=1-0.34=0.66 If […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1449,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-1463","page","type-page","status-publish","hentry"],"yoast_head":"\nHardy-Weinberg law practice questions – Meddists<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/meddists.com\/learn\/pre-clinical\/medical-genetics\/population-genetics\/hardy-weinberg-law-practice-questions\/\" \/>\n<meta name=\"twitter:label1\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data1\" content=\"7 minutes\" \/>\n<script type=\"application\/ld+json\" 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