18. An autosomal recessive disease has an incidence of 1/10,000. What is the app

Question

18. An autosomal recessive disease has an incidence of 1/10,000. What is the approximate frequency of heterozygote carriers for this disease? Assume Hardy-Weinberg conditions apply A. 1/50,or 2% B. 1,100, or 1% C. 1,1000, or 0.1% D. 1110, or 10% E. 1/4,or 25% 19. If a population is at Hardy-Weinberg equilibrium... (i) can you calculate the genotype frequencies if you know the allele frequencies? (i) can you calculate the allele frequencies if you know the genotype frequencies? A. () -yes; () NO B. (i) yes )-yes C. (i) = NO; (ii) = yes D. (i) NO; () NO E. only for x-linked genes 20. If a population is NOT at Hardy-Weinberg (and you don't know why it is not at HW equil.) (i) can you calculate the genotype frequencies if you know the allele frequencies? (ii) can you calculate the allele frequencies if you know the genotype frequencies? A. ()- yes: () NO B. (i) yes; (i) yes C. (i)NO; (i) yes D. () NO; (i) NGO E. () yes; () only if there's no inbreeding

Explanation / Answer

18. Population is at Hardy-Weinberg equilibrium so here we use Hardy- Weinberg equation i.e. p²+2pq+q²=1

where, p²=Frequency of homozygous dominant genotype

2pq= Frequency of heterozygous genotype

q²= Frequency of homozygous recessive genotype

So here, q² is given i.e. 1/10000, q= ?1/10000 =1/100=0.01

At Hardey-Weinberg equilibrium, p+q=1, where p is frequency of dominant allele and q is frequency of recessive allele.

So   p=1-q

   p=1-0.01

1-0.01=0.99

p=0.99,so p²=0.98

p²+2pq+q²=1 ,so 2pq = 1-p²-q²,                            q²=1/10000=0.0001

2pq = 1-0.98-0.0001

1-0.98-0.0001=0.0199 approximately 0.02 means 2% or 1/50.

Here frequency of heterozygous carriers fo autosomal recessive disease is 2%.

19.

(i) Yes we can calculate genotype frequency if we know the allele frequency if population is in Hardey-Weinberg equilibrium,

Hardey -Weinberg equation ,p²+2pq+q²=1

p+q=1

if we know p we can calculate p²(p² is square of p) in same way we can calculate q.

(ii) Yes we can calculate allele frequency if we know genotype frequency.

If we know the p² i.e genotype ferequency of homozygous dominant ,we can calculate allele frequency of dominant allele i.e p (p is square root of p²) same way we can calculate recessive allele frequency.

20.

(i) No we can not calculate the genotype frequency if we know allele frequency if population is not at Hardey-Weinberg equilibrium because it do not follow Hardey Weinberg equation i.e p²+2pq+q²=1

(ii) Yes we can calculate allele frequency if we know genotype frequency,

frequency of dominant allele = 2*number of homozygous dominant genotype+number of heterozygous genotype/2*total number of individual

Frequency of recessive allele= 2* number of homozygous recessive genotype+number of heterozygous genotype/2*total number of individual

   

    

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