BIOCHEMISTRY NBME (Genetics) Exam Questions and Answers 100% Pass.
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Course
Biochem
Institution
Biochem
BIOCHEMISTRY NBME (Genetics) Exam
Questions and Answers 100% Pass.
Hardy-Weinberg Principle - answerAllele frequencies remain constant, a population is in
genetic equilibrium. Used to predict allele variations in a population.
p2 = AA
pq = Aa
q2 = aa
Hardy-Weinberg Conditions - answerNo gene...
BIOCHEMISTRY NBME (Genetics) Exam
Questions and Answers 100% Pass.
Hardy-Weinberg Principle - answer✔Allele frequencies remain constant, a population is in
genetic equilibrium. Used to predict allele variations in a population.
p2 = AA
pq = Aa
q2 = aa
p + q = 1 - answer✔A = 91
aa = 9
What is the allele frequency of each?
p2 = 0.49
2pq = 0.42
p2 = 0.3
A population incidence of 1/2500. What is the allele frequency of the heterozygous carriers?
Hint: Cystic Fibrosis is a recessive trait - answer✔q2 = 1/50
2pq = 2(1)(1/50) = 1/25
0.025% of 1/40,000 of the population are homozygous for Hyperprolinemia. If a man mates,
what is the chance the child will have the mutation for this disease? - answer✔p2 = 1
(homozygous dominant)
2pq = 2(1)(1/200) = 1/100 (heterozygous carriers)
Child chance:
p = 1/2 chance from dad
pq = 1/100 heterozygous chance
q = 1/2 chance from mom
1/2 x 1/100 x 1/2 = 1/400 = 0.25%
Normal mother with Cystic Fibrosis, what are the chances of having a normal baby? What are
the chances of having carrier children?
Hint: X-linked recessive disease - answer✔CC = 25% (normal dominant)
cc = 25% (sick recessive)
Cc = 50% (sick heterozygous)
Carriers = <75% (Cc and cc) because mom is said to be normal
What is the allele frequency from 0.01% of autosomal recessive disease? - answer✔=>
population frequency is the same as allele frequency, because affected males and homozygous
females with the disease die in utero or in vitro.
p2 = 0.0001 = 0.01 or 1%
Achondroplasia is an autosomal dominant disease, if 2% of the population has this condition,
what is the allele frequency?
What are the chances that the first child will have Achondroplasia? - answer✔=> 2pq = 2(1/200)
= 1/100 = 0.01 or 1%
=> The chance of the child:
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