33. In humans, the presence of chin and cheek dimples is dominant to the absence of dimples, and the ability to taste the compound PTC is dominant to the inability to taste the compound. Both traits are autosomal, and they are unlinked. The frequencies of alleles for dimples are D=0.62 and d=0.38. For tasting, the allele frequencies are T= 0.76 and t= 0.24 Determine the frequency of genotypes for each gene and the frequency of each phenotype. What are the expected frequencies of the four possible phenotype combinations: dimpled tasters; undimpled tasters, dimpled non-tasters, and undimpled non-tasters?

33. In humans, the presence of chin and cheek dimples is dominant to the absence of dimples, and the ability to taste the compound PTC is dominant to the inability to taste the compound. Both traits are autosomal, and they are unlinked. The frequencies of alleles for dimples are D=0.62 and d=0.38. For tasting, the allele frequencies are T= 0.76 and t= 0.24 Determine the frequency of genotypes for each gene and the frequency of each phenotype. What are the expected frequencies of the four possible phenotype combinations: dimpled tasters; undimpled tasters, dimpled non-tasters, and undimpled non-tasters?

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 9QP: Using the HardyWeinberg Law in Human Genetics Suppose you are monitoring the allelic and genotypic...

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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…
3. In domestic chicken, feather colour and shape are two traits which are controlled by autosomal alleles that are co-dominant. Table 2 shows the alleles and phenotype of both traits. The heterozygous state for feather colour is grey while the heterozygous state for feather shape is mildly frizzled. TABLE 2 Trait Effect Produces black feathers in homozygotes Produces white feathers in homozygotes Produces straight feathers in homozygotes Produces frizzled feathers in homozygotes Allele Feather colour CW Feather shape SF a) A farmer bred a black frizzled rooster with a grey mildly frizzled hen. Draw a genetic diagram to show the expected results of this cross. Include in your answer the phenotypic ratios. b) Is it necessary to do a test cross to determine the genotypes of the offspring? Explain your answer. c) The farmer would like all his chickens grey and mildly frizzled. What would you advise him to do? Explain your answer.
3. In domestic chicken, feather colour and shape are two traits which are controlled by autosomal alleles that are co-dominant. Table 2 shows the alleles and phenotype of both traits. The heterozygous state for feather colour is grey while the heterozygous state for feather shape is mildly frizzled. TABLE 2 Effect Produces black feathers in homozygotes Produces white feathers in homozygotes Produces straight feathers in homozygotes Produces frizzled feathers in homozygotes Trait Allele Feather colour CW Feather shape SF A farmer bred a black frizzled rooster with a grey mildly frizzled hen. Draw a genetic diagram to show the expected results of this cross. Include in your answer the phenotypic ratios. а) b) Is it necessary to do a test cross to determine the genotypes of the offspring? Explain your answer. The farmer would like all his chickens grey and mildly frizzled. What would you advise him to do? Explain your answer. c)
In humans, there is a gene that influences the ability to smell the thioesters present in urine after asparagus is eaten. The ability to smell these compounds is an autosomal dominant trait. A is dominant and confers the ability to smell these thioesters a is recessive and does NOT confer the ability to smell these thioesters p = freq (A) = frequency of allele A in the population = number of A alleles/ total number of alleles %3D q = freq (a) = frequency of allele a in the population where p + q = 1 and, therefore, p? + 2pq + q? = 1 and %3D %D p? = freq (AA) = frequency of genotype AA = number of AA individuals/ total number of individuals %3D 2pq = freq (Aa) = frequency of genotype Aa q? = freq (aa) = frequency of genotype aa Phenotype Number of People Able to smell thioesters 378 Unable to smell thioesters 1323 Total 1701 1. Calculate p and q for this population. 2. Calculate p?, 2pq, and q? for this population.
In domestic chicken, feather colour and shape are two traits which are controlled by autosomal alleles that are co-dominant. Table 2 shows the alleles and phenotype of both traits. The heterozygous state for feather colour is grey while the heterozygous state for feather shape is mildly frizzled. TABLE 2 Trait Feather colour Allele CB CW SS SF Effect Produces black feathers in homozygotes Produces white feathers in homozygotes Produces straight feathers in homozygotes Produces frizzled feathers in homozygotes Feather shape A farmer bred a black frizzled rooster with a grey mildly frizzled hen. Draw a genetic diagram to show the expected results of this cross. Include in your answer the phenotypic ratios. Is it necessary to do a test cross to determine the genotypes of the offspring? Explain your answer. The farmer would like all his chickens grey and mildly frizzled. What would you advise him to do? Explain your answer.
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
5A. In shorthorn cattle, the polled condition is dominant over horned. Also, the heterozygous condition (RW) of red coat (R) and white coat (W) is roan, an example of codominance. (The roan coat contains both red and white fur because neither allele is dominant over the other). If a homozygous polled, white male is bred to a horned, red female, what will be the appearance of the F1 generation? Horned: r: Roan Coat: a. Possible genotypes for Polled condition Coat:_ ossible geno c. Male genotype: d. Female genotype: Male possible gametes Female possible gametes e. f. Genotypic Ratio: g. Phenotypic Ratio 5B, Breed two of the F1 generation to determine the possibilities in the F2 generation. Horned: a. Possible genotypes for Polled condition b. Possible genotypes for Red coat: White Coat:_ Roan Coat: c. Male genotype: d. Female genotype Male possible gametes Female possible gametes: f. Genotypic Ratio: g. Phenotypic Ratio
5A. In shorthorn cattle, the polled condition is dominant over horned. Also, the heterozygous condition (RW) of red coat (R) and white coat (W) is roan, an example of codominance. (The roan coat contains both red and white fur because neither allele is dominant over the other). If a homozygous polled, white male is bred to a horned, red female, what will be the appearance of the F1 generation? a. Possible genotypes for Polled condition: Horned: b. Possible genotypes for Red coat:White Coat: Roan Coat: c. Male genotype Male possible gametes: Female genotypeFemale possible gametes: e. f. Genotypic Ratio: g. Phenotypic Ratio:
B. Labrador retrievers have 3 varieties of fur colour: yellow, chocolate or black. Two genes are involved, each with a dominant (Y and B) and a recessive allele (y and b). The allele Y codes for the ability to produce hair pigment: Yy and YY dogs have dark pigmented fur but all yx, individuals are lighter coloured and yellow. The B allele codes for the ability to make hair with the darkest colouration: BB and Bb individuals have black fur, bb dogs have a lighter shade which is chocolate. In the blank Punnet Square below, calculate and state the ratio of phenotypes that occur with a cross between two individuals with the genotype XyBb
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
7) Skin color is a complex trait and while just a few genes account for the major difference between light and dark skin color, dozens of other genes participate. Assume in this hypothetical scenario there are only two genes for skin color, A and B. Each gene exhibits incomplete dominance. AABB (homozygous dominant for both genes) results in the darkest skin color and aabb (homozygous recessive for both genes) results in the lightest skin color. Tomas is heterozygous for both genes; his partner, Helga, is heterozygous for the A gene and homozygous recessive for the B gene. Construct a Punnett Square showing the cross of this couple. What are the expected genotypes and phenotypes of the children this couple could have?
10. About 70% of Americans get a bitter taste form the substance called phenylthiocarbamide (PTC). It is tasteless to the rest. The taster allele is dominant to non-taster. Also, normal skin pigmentation is dominant to albino. A normally pigmented woman who is taste-blind for PTC has an albino-taster father. She marries an albino man who is a taster, though the man's mother is a non-taster. Show the expected offspring of this couple. Genotypes: Phenotypes: Genotypic Ratio: Phenotypic Ratio: