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#1




Genetics for you!
1. The incidence of cutaneous albinism among a population of Europeans is 1/2500. Which of the following is the predicted incidence of carriers of a cutaneous albinism mutation in this population?
A. 1/25 B. 1/50 C. 2/2500 D. 1/2500 E. (1/2500)2 2. Hitchhiker's thumb (distal hyperextensibility of the thumb) is an autosomal recessive trait. If in a given population, about 1 in 100 individuals have hitchhiker's thumb, approximately how many people are carriers for the gene for hitchhiker's thumb? A. 1 in 100 B. 9 in 100 C. 10 in 100 D. 18 in 100 E. 26 in 100
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#5




Guys keep it up u all doing quite well
More answers!
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#6




1 a 1/25
2 d 18/100 
#7




1) 1/25 for sure
2) not sure. Please explain thank you. 
#8




Quote:
2 D explanation q2 = 1/100 so q =1/10 Carrier = 2qp =2*1/10 * 1 = 1/5 which equals 20% amongest answers D 18/100 is most approximate to 20 % answer E equals 25%
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#9




Ok guys answers are
1.A 2.D 1 is obvious So explanation for 2 is HardyWeinberg equilibrium (HWE) states that under certain conditions, if the population is large and randomly mating, the genotypic frequencies of the population will remain stable from generation to generation. All of the conditions that must be met for a population to be in HWE focus on keeping the allele frequencies constant, such as no mutation, no selection, etc. HWE can be expressed by the following equations. In a large, randomly mating population, there is a gene with two alleles, A and a. The A allele is completely dominant to the a allele. Let: The frequency of A allele = p The frequency of a allele = q Then: The allele (gene) frequencies at that locus can be expressed as: p + q = 1 The genotypic frequencies at that locus can be expressed as: p2 + 2pq + q2 =1 Where: p2= frequency of individuals with genotype AA, 2pq= frequency of individuals with genotype Aa, and q2= frequency of individuals with genotype aa. The equations for allele frequencies and for genotypic frequencies are related through the frequency of individuals with the homozygous recessive genotype q2. The square root of q2 yields q, which can be used to solve for p. Thus, if the frequency of an autosomal recessive disease is known in a population, it can be used to find the allele frequencies in the current generation, which can then be used to solve for other variables. In this problem, the frequency of hitchhiker's thumb is equal to 1/100. This number is equal to q2. The frequency of the allele for hitchhiker's thumb, q, can be found by taking the square root of q2. In this problem, the frequency is q = 0.1. The frequency of the normal allele, p, is equal to 0.9, since 1  q = p. Finally, the frequency of those who are heterozygous for hitchhiker's thumb is expressed by 2pq and equals 2(0.9)(0.1), or 0.18, or 18 in 100.
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#10




Thank you keep it up

#11




I kind of hate hardy weinberg cant understand it

#12




I kinda get confused when to consider p=1
Is it when the incidence is v low we consider it 1?? 
#13




Quote:
But you are absolutely right that it only can be used when prevalence is low . It said that if prevalence is greater than 1/100 or q is greater than 1/10 u should use the complete haerdy Weinberg equation
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