Please answer (d) 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. Howconvenient.You find 665 BB, 225 Bb, and 110 bb bison.Total population= 1000So genotypic frequency of BB= 665/1000 =0.665Genotypic frequency of BB= 225/1000 =0.225Genotypic frequency of bb= 110/1000 =0.11a)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) =0.78The gene or allele frequency of b would be = 0.11 + (0.225/2) =0.22Using the allele frequencies, what numbers (not just fractions, but numbers of actual bison inthis 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= 610Bb would be =0.34 ×1000= 340bb would be =0.05 ×1000= 50с)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 inHardy Weinberg equilibrium.d)No matter what your answer to c, if they weren't in HW equilibrium, name 5 possible reasonswhy.Ansuer this..

The Hardy-Weinberg equilibrium is a principle stating that the genetic variation in a population…

opulation of 1000 bison, there are two alleles at the B locus. It acts incompletely dominantly, ou are able to figure out each animal's genotype simply by observing its phenotype. How ent. 1 665 BB, 225 Bb, and 110 bb bison. -pulation= 1000 typic frequency of BB= 665/1000 =0.665 pic frequency of BB= 225/1000 =0.225 pic frequency of bb= 110/1000 =0.11 hat are the allele frequencies of B and b? (round off the number) e or allele frequency of B would be = 0.665+ (0.225/2) =0.78 e or allele frequency of b would be = 0.11 + (0.225/2) =0.22

opulation of 1000 bison, there are two alleles at the B locus. It acts incompletely dominantly, ou are able to figure out each animal's genotype simply by observing its phenotype. How ent. 1 665 BB, 225 Bb, and 110 bb bison. -pulation= 1000 typic frequency of BB= 665/1000 =0.665 pic frequency of BB= 225/1000 =0.225 pic frequency of bb= 110/1000 =0.11 hat are the allele frequencies of B and b? (round off the number) e or allele frequency of B would be = 0.665+ (0.225/2) =0.78 e or allele frequency of b would be = 0.11 + (0.225/2) =0.22

Biology (MindTap Course List)

11th Edition

ISBN:9781337392938

Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Publisher:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Chapter19: Evolutionary Change In Populations

Section: Chapter Questions

Problem 3TYU: The MN blood group is of interest to population geneticists because (a) people with genotype MN...

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