3. Frequency Calculation In a population of 80 rabbits, 20 are homozygous dominant, 40 are heterozygotes, and 20 are albino homozygous recessive. a. Provide the frequencies of each genotype and the frequencies of each allele in the gene pool. Use proper notation (for example, f(AA)). Answer: b. Why do the frequencies add up to 1? Answer:

3. Frequency Calculation In a population of 80 rabbits, 20 are homozygous dominant, 40 are heterozygotes, and 20 are albino homozygous recessive. a. Provide the frequencies of each genotype and the frequencies of each allele in the gene pool. Use proper notation (for example, f(AA)). Answer: b. Why do the frequencies add up to 1? Answer:

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter19: Population Genetics And Human Evolution

Section: Chapter Questions

Problem 11QP: Using the HardyWeinberg Law in Human Genetics In a given population, the frequencies of the four...

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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. a) What are the allele frequencies of B and b? (round off the number) b) 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? c) Do you think this bison population is in HW equilibrium? d) No matter what your answer to c, if they weren't in HW equilibrium, name 5 possible reasons why.
1.) Consider flower color in a hypothetical population of 7,000 flowers, red flowers (4,000), blueflowers (2,000) are dominant over pink flowers. If there are 1,000 pink flowers, find thefollowing:a. Frequency of dominant (red flower) alleleb. Frequency of recessive (pink flower) allelec. Frequency of each genotype (homozygous dominant, homozygous recessive,heterozygous)d. Make a Punnett square and use the rule of multiplication.
b. a. in a population. Hardy-Weinberg p+q=1 (Rr), and 16% white (rr). This population is in a Hardy-Weinberg equilibrium. 42. In a population of cattle, the following color distribution was noted: 36% red (RR), 48% roan baith p²+2pq+q² =1 Determine the frequencies of the R alleles and the r alleles? assumptions are met? What will be the distribution of genotypes in the next generation if the Hardy-Weinberg a. 36% red; 48% roan; 16% white b. 30% red; 50% roan; 20% white c. 30% red; 60% roan; 10% white d. 40% red; 40% roan; 20% white 10 u ti- sor
1. The ability to taste the compound PTC is controlled by a dominant allele T, while individuals homozygous for the recessive allele (t) cannot taste PTC. In a population consisting of 500 individuals, 347 are tasters and 153 are non-PTC tasters. Calculate the frequency of the T and t alleles in this population, and frequency of the genotypes.(Please train yourself to use the Hardy-Weinberg equation.)To present your answers, follow the format in the picture below.
26. Snow geese (Chen caerulescens) come in two color types, white “snows” and “blues” with dark bodies. A single gene controls coloration, where the dark (“blue”) allele (D) is dominant. A population of 30,012 geese includes 9236 dark individuals. Genetic testing reveals that 7636 of the 9236 dark individuals are heterozygous (Dd). What is the actual frequency of allele D? 0.819 0.181 0.435 0.347
1.: Male side-blotched lizards display throat color polymorphisms; some males have orange throats, some have yellow throat stripes, and some have blue throats. The following table gives genotypes observed in a California population in 1993. Genotype Individuals 2 10 oy ob 14 yy 32 yb bb 42 Please calculate: a) The allele frequencies. b) The observed heterozygosity. Note that you have more than one type of heterozygote at this locus. c) The expected heterozygosity.
4. Attached earlobes are dominant over free hanging earlobes. Complete the Punnett Square for the following individuals: Mom-Bb and Dad=bb Provide the genotype and phenotype ratios using the Punnett square given below. Genotypes 5. When two alleles are observed independently in a phenotype is called multiple alleles. Human blood group ABO is a good example of multiple alleles. The three alleles are: IA, IB, i. Mother's blood type is AB (IAI³) and father's blood type is B (I³i). Provide the genotype and phenotype ratios using the Punnett square given below. Genotypes Phenotypes Genotypes Phenotypes 6. When the allele of a trait is found on X or Y chromosome, it becomes sex-linked. The results differ from males to females. One such example is of red green colorblindness. The mother is a carrier (X© Xº ), and the father is colorblind (X© Y). Provide the genotype and phenotype ratios using the Punnett square given below. Phenotypes
p? + 2pq + q? = 1 (to determine the genotype frequencies) and p+q =1 (to determine the allele frequencies) p = frequency of the dominant allele in the population q = frequency of the recessive allele in the population p = percentage of homozygous dominant individuals q?: = percentage of homozygous recessive individuals 2pq = percentage of heterozygous individuals Homework Problems For these problems you will need to use a calculator and can round for the percentages. the nearest who number Example problem: You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following: A. The frequency of the "aa" genotype. Answer: 36% (given) B. The frequency of the "a" allele. Answer: The frequency of aa is 36%, which means that q? = 0.36. If q? = 0.36, then q = 0.6, because 0.62 is equal to 0.36. Since q equals the frequency of the a allele, then the frequency is 60%. C. The frequency of the "A" allele.…
1. For a single locus with two alleles, A₁ and A₂: (a) Draw a graph (using graph paper) showing both the frequency of A₁ A2 heterozy- gotes and A₂ A₂ homozygotes, at Hardy-Weinberg frequencies, as functions of p (the frequency of A₁). Note that both p and the genotype frequencies should have values between 0 and 1. (b) Find the value of p above which A₁ A2 genotypes are more common than A₂42 genotypes. You can solve this algebraically, or estimate it from your graphs. 2. Consider three loci, A, B, and C, each with two alleles, with the frequencies of A₁, B₁, and C₁ all being We look at a population and find that there are four distinct haplotypes, shown here, each with a frequency of: A₁ B1 TT A1 C₁ B1 A₂ B₂ AT C₂ A₂ B₂ C₁ C₂ Of the three pairs of loci (AB, AC, and BC) which pair(s) are in Gametic Equilibrium (D = 0) and which are in Gametic Disequilibrium (D ‡0)? [Hint: Consider each pair separately, ignoring the other locus. For example: for the BC pair, consider the four…
1. Widow's peak allele (W) is dominant to straight hair line (allele w). In a population of 1000, there are 760 persons with widow's peak. a. What is the frequency of genotype WW? b. What is the frequency of the heterozygotes?
. A farmer have 2000 cows. A total of 1500 of the cows are browh coated and the rest are white coated. Allele B for the brown coat is dominant, while allele b for the white coat is recessive: Assume that the population of the cows is at genetic equilibrium and bred randomly. Calculate the frequency for the dominant and recessive alleles of the population. i. Calculate the genotype frequencies in the F1 generation if the cows are left to breed randomly. ii. If the farmer adds another 1000 homozygous dominant brown cows into the populations, calculate the new dominant allele frequency. ii. iv. State three conditions for the Hardy Weinberg equilibrium to be achieved.
I 2. In frost moths, two alleles of one gene determine the character difference of spotted versus striped wings and two alleles of a separate, independent gene determine the character difference of orange wing background versus white wing background. The results for four matings of moth phenotypes are shown below. [Yes - it's possible for the same phenotypes to have different genotypes]. Mating Parental Phenotypes 1 2 3 4 spotted/white x striped/orange spotted/white x striped/orange spotted/orange x striped/ white spotted/orange x striped/orange Spotted/ orange 122 812 422 623 Number of progeny Spotted/ Striped / white orange 131 128 0 430 0 637 a. Determine which alleles are dominant and assign a letter to each. Striped / white 133 0 0 0 b. What are the genotypes of each parent in each cross? If more than one genotype is possible, list the alternative possibilities.