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Problem 1

Problem 1. Consider the populations that have the genotypes shown in the following table: A. Which of the populations are in Hardy-Weinberg equilibrium B. What are the p and q in each population?

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Problem 1

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  1. Problem 1 • Consider the populations that have the genotypes shown in the following table: • A. Which of the populations are in Hardy-Weinberg equilibrium • B. What are the p and q in each population? • In population 5, it is discovered that the A to a mutation rate is 5x10-6 and that the reverse mutation is negligible. What must be the fitness of the a/a pehnotype? (q2=u/s where u is the mutation rate and s is the coefficient of selection)

  2. Answer Problem 1 A. Which of the populations are in Hardy-Weinberg equilibrium • B. What are the p and q in each population? • For each the p and q must first be calculated • #1 p=1+1/2(0)=1 q=1-1=0 • #2 p=0+1/2(1)=.5 q=1-.5=.5 • #3 q=1+1/2(0)=1 p=1-1=0 • #4 p=0.5+1/2(0.25)=0.625 q=1-0.625=0.375 • #5 p=0.986049+1/2(0.013902)=0.993 q=1-0.993=0.007 • From the p and q simply calculate A/A= p2, A/a=2pq, a/a=q2 • #1 A/A= p2=1x1=1 so yes • #2 A/a=2pq=2x.5x.5=0.5 so no • #3 a/a=q2=1x1=1 so yes • #4 A/A=p2=0.625x0.625=0.391 so no • #5 A/A=p2=0.993x0.993=0.986 A/a=2pq=2x0.993x0.007=0.014 a/a=q2=0.007x0.007=0.000049 so yes

  3. Problem 1 • In population 5, it is discovered that the A to a mutation rate is 5x10-6 and that the reverse mutation is negligible. What must be the fitness of the a/a phenotype? (q2=u/s where u is the mutation rate and s is the coefficient of selection) • we know that the fitness is related to the coefficient of selection by the equation f=1-s. So s=1-f. Therefore 0.000049=5x10-6/s s=0.102 and f=0.898 • So the fitness of the a/a phenotype is 0.898

  4. I II III Problem 2 • A woman (II2 in the pedigree) wishes to know the probability that she is a carrier of Duchenne muscular dystrophy. • a. What is the probability if she has another affected male child • b. What is the probability if she has another unaffected male child • c. What is the probability if she has another unaffected female child • d. If she has another unaffected male child what is the probability she will have another affected child

  5. I II III Answer Problem 2 • A woman (II2 in the pedigree) wishes to know the probability that she is a carrier of Duchenne muscular dystrophy. This problem is best solved using Bayes’ Theorem It is easiest to make a table Ancestral information Considers children So the probability she is a carrier is 1/9

  6. Answer Problem 2 a. What is the probability if she has another affected male child If she has an affected child then she is a carrier =1 I II III

  7. I II III Answer Problem 2 • b. What is the probability if she has another unaffected male child This problem is best solved using Bayes’ Theorem It is easiest to make a table If she has another unaffected male child then the conditional probability will change So the probability she is a carrier is 1/17

  8. Answer Problem 2 • c. What is the probability if she has another unaffected female child This problem is best solved using Bayes’ Theorem It is easiest to make a table I If she has a female child no additional information is gained because the disease is X-linked recessive II III So the probability she is a carrier is 1/9

  9. Answer Problem 2 • d. If she has another unaffected male child what is the probability she will have another affected child This problem is best solved using Bayes’ Theorem It is easiest to make a table I If she has another unaffected male child then the conditional probability will change II III So the probability she is a carrier is 1/17 if she had four unaffected boys. To pass on the recessive allele is ½ and the chance the child is male is 1/2. So the chance she has an affected child is 1/17X1/2X1/2=1/68

  10. Problem 3 • What is the probability that the a healthy member of the general population is a carrier of cystic fibrosis if she tests negative on the common mutation screening analysis (DF508 mutation). It is known that the incidence of cystic fibrosis in the general population is 1 in 1600 and the common mutation test detects 75 per cent of all cycstic fibrosis alleles.

  11. Answer Problem 3 • First you should make a table • The prior probability. • Is a carrier • We know that the incidence of cystic fibrosis in this case is 1/1600. Because cystic fibrosis is an autosomal recessive disease a/a=q2=1/1600 so q=.025 and p=0.975 (p+q=1). The possibility of being heterozygous is 2pq=2x.025x0.975=0.049 • Is not a carrier • 1-0.049=0.951 • Conditional probability • Is a carrier • fequals chance she does not have the mutation detected by the test which 25% • Is not a carrier • equal 1 • Joint probability • just priorxconditional • Is a carrier • 0.25x0.049=0.012 • is not carrier • 1x0.951 • Posterior • Is not carrier • 0.951/(0.012+0.951)=0.987 • Is a carrier • 0.012/(0.012+0.951)=0.012

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