Testing the goodness of fit between the data and the Hardy Weinberg equilibrium model generated expectations.
4.1 In a species of bird, feather color is controlled by genes at a single locus, with the red feather allele dominant to the yellow feather allele. A population has 22 red and 14 yellow birds, with 9 of the red birds having a homozygous dominant genotype. Is this population in equilibrium?
Calculate p and q from the number of individuals of each genotype:
p = ____
q = ____
Calculate the expected frequency of each genotype if the population is in equilibrium:
_____ = Frequency of homozygous dominant individuals
_____ = Frequency of heterozygous individuals
_____ = Frequency of homozygous recessive individuals
Calculate the expected number of individuals of each genotype in a population of 36 birds if the gene is in equilibrium:
_____ = Number of homozygous dominant individuals
_____ = Number of heterozygous individuals
_____ = Number of homozygous recessive individuals
Test how well your data fits the expected values from the equilibrium model:
_____ = Chi-square test statistic
_____ = P value
_____ (y/n) in equilibrium?
4.2 In a species of mouse, tail length is controlled by genes at a single locus, with the long tail allele dominant to the short tail allele. A population has 49 long tail and 25 short tail mice, with 22 of the long tail mice having a homozygous dominant genotype. Is this population in equilibrium?
Calculate p and q from the number of individuals of each genotype:
p = _____
q = _____
Calculate the expected frequency of each genotype if the population is in equilibrium:
_____ = Frequency of homozygous dominant individuals
_____ = Frequency of heterozygous individuals
_____ = Frequency of homozygous recessive individuals
Calculate the expected number of individuals of each genotype in a population of 74 mice if the gene is in equilibrium:
_____ = Number of homozygous dominant individuals
_____ = Number of heterozygous individuals
_____ = Number of homozygous recessive individuals
Test how well your data fits the expected values from the equilibrium model:
_____ Chi-square test statistic
______ P value
_____ (y/n) in equilibrium?
4.3 In humans, the hitchhiker’s thumb trait is controlled
by genes at a single locus, with the non-hitchhiker’s thumb allele
dominant to the hitchhiker’s thumb allele. A population has 46
people that do not have the hitchhiker’s thumb and 21 that do. Of
the humans without a hitchhiker’s thumb, 19 have a homozygous
dominant genotype. Is this population in equilibrium?
Calculate p and q from the number of individuals of each genotype:
p = _____
q = _____
Calculate the expected frequency of each genotype if the population is in equilibrium:
_____ = Frequency of homozygous dominant individuals
_____ = Frequency of heterozygous individuals
_____ = Frequency of homozygous recessive individuals
Calculate the expected number of individuals of each genotype in a population of 67 humans if the gene is in equilibrium:
_____ = Number of homozygous dominant individuals
_____ = Number of heterozygous individuals
_____ = Number of homozygous recessive individuals
Test how well your data fits the expected values from the equilibrium model:
_____ Chi-square test statistic
______ P value
_____ (y/n) in equilibrium?
4.4 In a certain species of prickly pear, having straight
or curved spines is a trait controlled by genes at a single locus,
with the straight spine allele dominant to the curved spine allele.
A population of prickly pears has 37 individuals with straight
spines and 42 individuals with curved spines. Of the prickly pears
with straight spines, 12 have a homozygous dominant genotype. Is
this population in equilibrium?
Calculate p and q from the number of individuals of each genotype:
p = _____
q = _____
Calculate the expected frequency of each genotype if the population is in equilibrium:
_____ = Frequency of homozygous dominant individuals
_____ = Frequency of heterozygous individuals
_____ = Frequency of homozygous recessive individuals
Calculate the expected number of individuals of each genotype in a population of 79 prickly pear cacti if the gene is in equilibrium:
_____ = Number of homozygous dominant individuals
_____ = Number of heterozygous individuals
_____ = Number of homozygous recessive individuals
Test how well your data fits the expected values from the equilibrium model:
_____ Chi-square test statistic
______ P value
_____ (y/n) in equilibrium?
4.1) No of homozygous dominant individuals = 9
Total no of individuals = 36
Hence p2 = PP = 9/36 = 0.25
Therefore, p = 0.5
q2 = 14/36 = 0.38
q = 0.62
Hence p + q is not equal to 1 in this case hence the population is not in hardy weinberg equilibrium.
frequency of genotype AA = PP = 9/36 = 0.25
frequency of genotype aa = q2 = 14/36 = 0.38
frequency of genotype Aa= 13/36= 0.36
If the gene is in equilibrium,
No of AA = 9, No of aa = 14, No of Aa = 13
p2 + 2pq + q2 = 1
chi square test static = (O-E)2 / E = 0.00038
This is not in hardy weinberg equilibrium
4.2) No of homozygous dominant = 22
total no = 74
PP = 22/74 = 0.29
p = 0.54
q = 0.57
Not in hardy weinberg equlibrium
Frequency of AA = 22/74 = 0.29 , Aa= 27/74= 0.36 , aa = 25/74 = 0.33
No of AA = 22 , Aa= 27 and aa= 25
Chi square = 0.00129
4.3) PP = 19/ 67 = 0.28 =p = 0.52
pp= 21/67= 0.31 =q= 0.55
p= 0.52, q = 0.55, hence not in hardy weinberg equilibrium since p=+ q is not equal to 1.
Frequency of AA= 0.28, Aa = 27/67= 0.40 , aa= 0.31
No of AA= 19, Aa= 27, aa = 21
chi square= 0.00034
4.4) PP= 12/ 79 = 0.15 = p= 0.38
pp= 42/79= 0.53= q= 0.72
p= 0.38, q= 0.72
hence not in hardy weinberg equilibrium
Frequency of AA= 0.15, Aa= 25/79= 0.31 , aa= 0.53
No of AA= 12, Aa= 25, aa= 42
chi square= 0.0003
This is not in hardy weinberg equilibrium.
Testing the goodness of fit between the data and the Hardy Weinberg equilibrium model generated expectations....
please click on the photo to see all of it The basic equations of Hardy-Weinberg Equilibrium p² + 2pq + q2 = 1 p+q=1 p= frequency of the dominant allele in the population 9 = frequency of the recessive allele in the population př= percentage of homozygous dominant individuals q* = percentage of homozygous recessive individuals 2pq - percentage of heterozygous individuals 1. You have sampled a population in which you know that the percentage of the homozygous recessive genotype...
Calculating expected genotypic frequencies and individuals in a population from allele frequencies: 3.1 In a population of peas, the frequency of the dominant allele for a purple flower is 0.77 and the frequency of the recessive allele for a white flower is 0.23. What would the genotypic frequencies be if the population is in equilibrium? _____ = Frequency of homozygous dominant individuals _____ = Frequency of heterozygous individuals _____ = Frequency of homozygous recessive individuals How many individuals would you...
Hardy-Weinberg Practice Problems: You need to list equations used and provide steps of problem solving. Providing answer itself is not enough for full grade. 1. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Calculate the frequency of the heterozygous genotype, homozygous dominant genotype and homozygous recessive genotype. 2. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 49%....
2.3 Problem 3 The Hardy-Weinberg equation is useful for predicting the percent of a hu- man population that may be heterozygous carriers of recessive alleles for certain genetic diseases. Phenylketonuria (PKU) is a human metabolic dis- order that results in mental retardation if it is untreated in infancy. In the United States, one out of approximately 10.000 babies is born with the disor- der. Approximately what percent of the population are heterozygous carriers of the recessive PKU allele? If you...
Equations: p + q=1 p2 + 2(p)(q) + q2 1) In a population with Hardy-Weinberg equilibrium, the frequency of a dominant allele for a gene is 85% (0.85) while the frequency of the recessive allele for the same gene is 15% (0.15). 0.0005 0.013 0.675 0.723 0.255 0.50 0.023 2) proportion of individuals will be homozygous recessive? 0.0005 0.013 0.675 0.723 0.255 0.50 0.023 3) proportion of individuals will be homozygous dominant? 0.0005 0.723 0.013 0.675 0.255 0.50 0.023 4)...
In a population that meets Hardy- Weinberg equilibrium assumptions, 81% of the individuals are homozygous for the recessive allele. What percentage of the individuals would be expected to be heterozygous for this locus in the next generation?
reting Data: Hardy-Weinberg Equation 2 of 10 you use the Hardy Weinberg equation to answer questions about a hypotheticalcat population Part A A hypothetical population of 500 cats has two wees, Tandt for a gene that codes for tail length (Tis completely dominantot) The table below presents the phenotype of cats with each possible genotype, as well as the number of individuals in the population with each genotype. Assume that this population is in Hardy-Weinberg equilibrium Recall that the Hardy...
Determining if allele frequencies are changing from one generation to the next (microevolution) from the number of individuals of each genotype present: The following steps are used to determine if allele frequencies are changing: Calculate Allele frequency from the number of individuals of each genotype Calculate expected genotypic frequencies and individuals in a population from allele frequencies: Test the goodness of fit between the data and the Hardy Weinberg equilibrium model generated expectations. The following problems are the calculations used...
.1. The Hardy-Weinberg principle and its equations predict that frequencies of alleles and genotypes remain constant from generation to generation in populations that are not evolving. What five conditions does this prediction assume to be true about such a population? a._______ b._______ c._______ d._______ e._______ 2. Before beginning the activity, answer the following general Hardy-Weinberg problems for practice (assume that the population is at Hardy-Weinberg equilibrium).a. If the frequency of a recessive allele is 0.3, what is the frequency of the dominant...
The occurrence of the NN blood group genotype in the US population is 1 in 400, consider NN as the homozygous recessive genotype in this population. You sample 1,000 individuals from a large population for the MN blood group, which can easily be measured since co-dominance is involved (i.e., you can detect the heterozygotes). They are typed accordingly: BLOOD TYPE GENOTYPE NUMBER OF INDIVIDUALS RESULTING FREQUENCY M MM 490 0.49 MN MN 420 0.42 N NN 90 0.09 Using the...