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2. In a similar population of 1000 bison (but one with different allele frequencies than the one in question #1), which you AREALDX KNOW is in HW equilibrium, there are 160 bb bison. What are the allele frequencies of B and b?

2. In a similar population of 1000 bison (but one with different allele frequencies than the one in question #1), which you AREALDX KNOW is in HW equilibrium, there are 160 bb bison. What are the allele frequencies of B and b?

Human Anatomy & Physiology (11th Edition)
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
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Please answer these question. The first page is question 1 and the second one is question 2. 

2. In a similar population of 1000 bison (but one with different allele frequencies than the one in question #1), which you AREALRX KNOW is in HW equilibrium, there are 160 bb bison. What are the allele frequencies of B and b? Ansuer this...
Transcribed Image Text:2. In a similar population of 1000 bison (but one with different allele frequencies than the one in question #1), which you AREALRX KNOW is in HW equilibrium, there are 160 bb bison. What are the allele frequencies of B and b? Ansuer this...
1. In a population of 1000 bison, there are two alleles at the B locus. It acts incompletely dominantly, so that you are able to figure out each animal's genotype simply by observing its phenotype. How convenient. You find 665 BB, 225 Bb, and 110 bb bison. Total population= 1000 So genotypic frequency of BB= 665/1000 =0.665 Genotypic frequency of BB= 225/1000 =0.225 Genotypic frequency of bb= 110/1000 =0.11 a) What are the allele frequencies of B and b? (round off the number) The gene or allele frequency of B would be = 0.665+ (0.225/2) =0.78 The gene or allele frequency of b would be = 0.11 + (0.225/2) =0.22 Using the allele frequencies, what numbers (not just fractions, but numbers of actual bison in this population of 1000 and yes you can round off) would you expect to be BB, Bb, and bb? b) we have got the allele frequency of B & b. So the expected number of genotype would be =(0.78+0.22)² =(0.78)²+ 2×0.78×0.22 + (0.22)? =0.61+0.34+0.05| BB would be =0.61 ×1000= 610 Bb would be =0.34 ×1000= 340 bb would be =0.05 ×1000= 50 с) Do you think this bison population is in HW equilibrium? The observed and expected frequency of genotypes are not exactly same. So this population is not in Hardy Weinberg equilibrium. d) No matter what your answer to c, if they weren't in HW equilibrium, name 5 possible reasons why. Ansuer this..
Transcribed Image Text:1. In a population of 1000 bison, there are two alleles at the B locus. It acts incompletely dominantly, so that you are able to figure out each animal's genotype simply by observing its phenotype. How convenient. You find 665 BB, 225 Bb, and 110 bb bison. Total population= 1000 So genotypic frequency of BB= 665/1000 =0.665 Genotypic frequency of BB= 225/1000 =0.225 Genotypic frequency of bb= 110/1000 =0.11 a) What are the allele frequencies of B and b? (round off the number) The gene or allele frequency of B would be = 0.665+ (0.225/2) =0.78 The gene or allele frequency of b would be = 0.11 + (0.225/2) =0.22 Using the allele frequencies, what numbers (not just fractions, but numbers of actual bison in this population of 1000 and yes you can round off) would you expect to be BB, Bb, and bb? b) we have got the allele frequency of B & b. So the expected number of genotype would be =(0.78+0.22)² =(0.78)²+ 2×0.78×0.22 + (0.22)? =0.61+0.34+0.05| BB would be =0.61 ×1000= 610 Bb would be =0.34 ×1000= 340 bb would be =0.05 ×1000= 50 с) Do you think this bison population is in HW equilibrium? The observed and expected frequency of genotypes are not exactly same. So this population is not in Hardy Weinberg equilibrium. d) No matter what your answer to c, if they weren't in HW equilibrium, name 5 possible reasons why. Ansuer this..
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