will rate help greatly appreciated 15. An x-linked recessive disease, caused by

Question

will rate help greatly appreciated

15. An x-linked recessive disease, caused by a single gene, affects in 1/25 men (1 in 25 0.04). What is the frequency of affected females (assume Hardy-Weinberg can be applied). A. 0.0016 B. 0.04 C. 0.2 D. 0.32 E. 0.64 16. Which of the following would NOT be expected to cause allele frequencies to vary over time A. Migration B. Random mating C. Mutation D. Genetic Drift E. Natural selection 17. A certain population is at Hardy-Weinberg equilibrium. In this population, there is an autosomal recessive condition. Without knowing the allele frequencies, what can you conclude about the genotype frequencies? A. The frequency of those affected by the condition is less than or equal to the frequency of carriers B. The frequency of those affected by the condition is greater than the frequency of carriers C. The frequency of heterozygotes is greater than the frequency of homozygotes D. The frequency of those homozygous for the dominant allele is less than the frequency of those homozygous for the recessive allele E. You can't conclude any of the above without knowing allele frequencies 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. I/100, or 1% C. 1 /1000, or 0.1% D. 1 /10, 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? (ii) can you calculate the allele frequencies if you know the genotype frequencies? A. (i) yes; (ii) NO B. (i) yes; (ii)-yes C. (i) NO; (ii) yes

Explanation / Answer

Q15-

As we know that in x-linked recessive diseases one partilcular sex is affected and exhibits criss cross pattern.

Here in this question it is already given that frequency of males is 0.04%

Since males have only one x chromosome (xy) so the frequency will remain 0.04 for males.

In case of females,they have two x chromosomes(xx),so the frequency of affected females will be

0.04*0.04= 0.0016.

Q16-

mutation,natural selection, migration and genetic drift all are factors which destablize a population away from equilibrium.

According to Hardy-weinberg law, when mating is at random in large populations with no disruptive circumstances, both type of frequencies i.e genotype and allele frecquencies will remain constant as they are in equilibrium.

Q17-

in case of autosomal recessive condition we cannot calculate genotype without knowing the allelic frequencies.The reserch is going on this area,may be someday we find a solution to this problem.

Q18-

According to hardy weinberg law

p+q=1

q = 1/10000 = 0.00001,

hence p =1-q = 1-0.00001 =0.9999

now p2 +q2 +2pq =1

=2*.9999*0.00001 =1.9

=2% approx.

Q19-

yes you can calculate genotype frequencies from allelic frequencies and vice versa by rearranging the formulla,needs simple math.

so this is all buddy,plz provide your valuable feedback and will assist you further if needed.

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