Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
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Chapter 3, Problem 12P
One of your fellow students tells you that there is no way to know that the spotted and dotted patterns on the lentils in Fig.3.4a are due to codominant alleles (Cs and CD) of the single gene C. He claims that spotting could be controlled by gene S, with a completely dominant allele S that directs spotting and recessive allele s that directs no spots. Likewise, he claims that dotting could be controlled by a separate gene D, with a completely dominant allele D that directs doting and recessive allele d that directs no dots. Is he correct, or does the information in Fig. 3.4a argue against this idea? Explain.
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Let's continue with these two genes that control the color and texture of tomatoes.
Allele "R" generates a red coloration, and is dominant. Allele "r" generates a green coloration and is recessive.
Allele "S" generates a smooth texture, and is dominant. Allele "s" generates a rough texture and is recessive.
You now decide to cross your dihybrid RrSs but this one with a tomato with Rrss.
RrSs x Rrss
What is the probability of getting Red tomatoes with Rough texture in this new cross?
Assume complete dominance and 100% penetrance and expressivity.
9/16
3/8
1/4
1/16
3/16
1/8
x 1/2
Let's continue with these two genes that control the color and texture of tomatoes.
Allele "R" generates a red coloration, and is dominant. Allele "r" generates a green coloration and is recessive.
Allele "S" generates a smooth texture, and is dominant. Allele "s" generates a rough texture and is recessive.
You now decide to cross your dihybrid RrSs but this one with a tomato with Rrss.
RrSs x Rrss
What is the probability of getting Red tomatoes with Rough texture in this new cross?
Assume complete dominance and 100% penetrance and expressivity.
O O O O
1/8
3/16
9/16
3/8
1/16
1/4
1/2
In 1927, the ophthalmologist George Waaler tested9049 schoolboys in Oslo, Norway, for red-green colorblindness and found 8324 of them to be normal and725 to be color blind. He also tested 9072 schoolgirlsand found 9032 that had normal color vision while40 were color blind.a. Assuming that the same sex-linked recessive allelec causes all forms of red-green color blindness,calculate the allele frequencies of c and C (the allelefor normal vision) from the data for the schoolboys.(Hint: Refer to your answer to Problem 12a.)b. Does Waaler’s sample demonstrate Hardy-Weinbergequilibrium for alleles of this gene? Explain youranswer by describing observations that are eitherconsistent or inconsistent with this hypothesis.On closer analysis of these schoolchildren, Waalerfound that there was actually more than one c allelecausing color blindness in his sample: one kind forthe prot type (cp) and one for the deuter type (cd).(Protanopia and deuteranopia are slightly differentforms of red-green…
Chapter 3 Solutions
Genetics: From Genes to Genomes
Ch. 3 - For each of the terms in the left column, choose...Ch. 3 - In four-oclocks, the allele for red flowers is...Ch. 3 - The Aa heterozygous snapdragons in Fig. 3.3 are...Ch. 3 - Recall from Chapter 2 Fig. 2.20 that Mendels R...Ch. 3 - In the fruit fly Drosophila melanogaster, very...Ch. 3 - A cross between two plants that both have yellow...Ch. 3 - In radishes, color and shape are each controlled...Ch. 3 - A wild legume with white flowers and long pods is...Ch. 3 - Assuming no involvement of the Bombay phenotype in...Ch. 3 - Several genes in humans in addition to the ABO...
Ch. 3 - Alleles of the gene that determines seed coat...Ch. 3 - One of your fellow students tells you that there...Ch. 3 - In a population of rabbits, you find three...Ch. 3 - In clover plants, the pattern on the leaves is...Ch. 3 - Fruit flies with one allele for curly wings Cy and...Ch. 3 - In certain plant species such as tomatoes and...Ch. 3 - In a species of tropical fish, a colorful orange...Ch. 3 - People heterozygous for normal and nonfunctional...Ch. 3 - Using old Fugate family Bibles and the Perry...Ch. 3 - A rooster with a particular comb morphology called...Ch. 3 - A black mare was crossed to a chestnut stallion...Ch. 3 - Filled-in symbols in the pedigree that follows...Ch. 3 - You perform a cross between two true-breeding...Ch. 3 - a. How would you describe inheritance of flower...Ch. 3 - Suppose the intermediate called Colorless...Ch. 3 - Explain the difference between epistasis and...Ch. 3 - The dominant allele H reduces the number of body...Ch. 3 - Secretors genotypes SS and Ss secrete their A and...Ch. 3 - Normally, wild violets have yellow petals with...Ch. 3 - A woman who is blood type B has a child whose...Ch. 3 - The following table shows the responses of blood...Ch. 3 - Three different pure-breeding strains of corn that...Ch. 3 - In mice, the AY allele of the agouti gene is a...Ch. 3 - A student whose hobby was fishing pulled a very...Ch. 3 - Suppose that blue flower color in a plant species...Ch. 3 - This problem examines possible biochemical...Ch. 3 - Considering your answers to Problem 36, does the...Ch. 3 - You picked up two mice one female and one male...Ch. 3 - Figure 3.21 and Fig. 3.28b both show traits that...Ch. 3 - Three genes in fruit flies affect a particular...Ch. 3 - The garden flower Salpiglossis sinuata painted...Ch. 3 - In foxgloves, three different petal phenotypes...Ch. 3 - In a culture of fruit flies, matings between any...Ch. 3 - Prob. 44PCh. 3 - A couple wants to know the probability that their...Ch. 3 - This problem illustrates why classical geneticists...Ch. 3 - Prob. 47PCh. 3 - Familial hypercholesterolemia FH is an inherited...Ch. 3 - You have come into contact with two unrelated...Ch. 3 - Polycystic kidney disease is a dominant trait that...Ch. 3 - Identical monozygotic twins have similar, but not...Ch. 3 - Using each of the seven coat color genes discussed...
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- A mutant allele in persons with familial hypercholesterolemia (FH) causes death due to a lack of liver receptors for LDL. Susceptible persons have half the normal number of receptors, while other individuals have the normal number of receptors. In a phenotypically normal couple, the man had a female 1 cousin (on his father’s side) who died from FH; the woman had a maternal uncle with FH. a. What is the probability that neither of the couple might be susceptible. b. What is the probability that one of them might be susceptible, but the other is not. c. What is the probability that they will have an FH child if a test discloses that both of them are susceptible.arrow_forwardIn chickens, a dominant hereditary abnormality (e.g., creepers) causes death when the genotype is homozygous (CC). The recessive condition (cc) at this locus produces a normal phenotype. Another gene locus with co dominant alleles is known to govern feather color such that the genotype FF = black, ff = splashed white, and Ff = blue. Also, a completely dominant gene W produces white skin. Yellow skin is produced by the homozygous genotype ww. QUESTIONS: In CcFfWw x ccffWw, what proportion of the offspring is: b.2 normal, white feathered, white skinned b.3 yellow skinnedarrow_forwardIn aliens, the genes A, B, C, D and E are dominant to their respective alleles a, b, c, d and e: A = one eye a = two eyes B = pointed ears b = round ears C = red skin c = green skin D = two antennae d = three antennae E = square head e = triangular head An alien with genotype AA bb Cc dd Ee marries an alien with genotype Aa Bb Cc Dd Ee. If the genes segregate independently, what is the probability that they will have a child with one eye, pointed ears, red skin, three antennae and a triangular head? A. 9/512 B. 9/128 C. 27/1024 D. 3/64arrow_forward
- In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: Rose crossed with pea produces 20 walnut offspring. O A. Rr pp x Rr Pp B. Rr pp x rr PP C. Rr pp x rr Pp D. RR pp x rr PP O E. RR pp x rr Pp Reset Selection Mark for Douiouuarrow_forwardAs seen in the photo, Labradors come in three colors-- black, brown and yellow. What is the genetic basis for these different coat colors? One gene produces melanin, a pigment which is deposited in the dog's fur and makes the color dark. With this gene, allele B (black) is dominant to allele b. Only in the case of a recessive homozygote (bb) will the dog's phenotype be brown. The regulatory gene is separate from the melanin gene but it acts as a switch, either turning the melanin gene on or turning it off. Allele E is dominant and allows for the melanin to be deposited in the dog's fur ("on" switch), but if the switch gene is a recessive homozygote, the melanin is blocked ("off" switch) and a yellow dog is the result! 1. Two other Labradors mate and produce puppies. Their genotypes are Bbee and BbEe. What color are each parent and what are the phenotypic rations of their offspring in the F1 Generation? Show your work with a Punnett square.arrow_forwardAs seen in the photo, Labradors come in three colors-- black, brown and yellow. What is the genetic basis for these different coat colors? One gene produces melanin, a pigment which is deposited in the dog's fur and makes the color dark. With this gene, allele B (black) is dominant to allele b. Only in the case of a recessive homozygote (bb) will the dog's phenotype be brown. The regulatory gene is separate from the melanin gene but it acts as a switch, either turning the melanin gene on or turning it off. Allele E is dominant and allows for the melanin to be deposited in the dog's fur ("on" switch), but if the switch gene is a recessive homozygote, the melanin is blocked ("off" switch) and a yellow dog is the result! Review the information on Labrador retrievers above. What are the phenotypic ratios of the F1 generation offspring of two dihybrids? Make sure to match the numbers with coat colors (e.g, which number in the ration goes with which color). Use a Punnett square to…arrow_forward
- A mutant allele in persons with familial hypercholesterolemia (FH) causes death due to a lack of liver receptors for LDL. Susceptible persons have half the normal number of receptors, while other individuals have the normal number of receptors. In a phenotypically normal couple, the man had a female 1 cousin (on his father’s side) who died from FH; the woman had a maternal uncle with FH. a. Calculate the probability that neither of the couple might be susceptible. b. Calculate the probability that one of them might be susceptible, but the other is not. c. Calculate the probability that they will have an FH child if a test discloses that both of them are susceptible.arrow_forwardIn the yawncat (a rare hypothetical animal), the dominant allele R causes solid tail color, and the recessive allele r results in white spots on a colored background. The black coat color allele B is dominant to the brown allele b, but the B and b alleles can only be expressed if the animal has an mm genotype at a third gene locus. Animals that are M_ are yellow regardless of which allele from the B locus is present. A mating between a solid yellow-tailed male yawncat and a solid brown-tailed female yawncat produces 16 offspring with the following tail phenotypes: six solid yellow, two spotted yellow, three solid black, one spotted black, three solid brown, and one spotted brown. What is the probability of the next offspring from these same two parents having a spotted brown tail? O A. 1/16 B. 9/16 С. 3/16 D. 1/2 O E. 1/4arrow_forwardIn peaches fuzzy skin (F) is completely dominant to smooth, nectarine skin (f). The gland at the base of the leaves is controlled by a separate, unlinked locus. The gland in heterozygotes is round, while homozygotes with the Go allele have oval glands, and homozygotes with the the GN allele have no glands. A homozygous fuzzy, no-gland peach variety is bred to a smooth, oval-gland variety. The F1 offspring are interbred to produce the F2 generation. What fraction of the F2 offspring will be smooth with round glands? A) 3/16 B) 7/16 C) 4/16 D) 6/16 E) 9/16 F) 2/16 G) 12/16 H) 1/16arrow_forward
- In chickens, comb shape is determined by genes at two loci (Gene 1 has alleles R, r and Gene 2 has alleles P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_P). The genotype R_pp gives a rose comb phenotype; rrP_gives a pea comb phenotype; and rrpp gives a single comb phenotype. Based on this information give specific genotypes for comb shape of the parents in the crosses listed below: A) Walnut crossed with single produces 1 walnut, 1 rose, 1 pea and 1 single offspring. B) Rose crossed with pea produces 20 walnut offspring. C) Pea crossed with single produces 1 single offspring.arrow_forward(1.7 In 'summer squash', white colour in fruits is controlled by a dominant allele, W, and yellow colour by a recessive allele, w. At a different locus, a dominant allele, S, produces disc-shaped fruits while the recessive allele, s, produces round-shaped fruits. If disc-shaped, white homozygous fruits (WWSS) are crossed to round- shaped, yellow homozygous fruits (wwss), all F₁ progenies are disc- shaped and white (WwSs). If F₁ progenies are crossed randomly, what is the expected ratio of the phenotypes in the F2 generation?arrow_forwardIn poultry, two independently assorting genes together determine the structure of the comb. Thus, when R is the only dominant allele present, the fowl possesses a rose comb; when P is only dominant allele present, a ‘pea’ comb is produced; when both dominants are present, a ‘walnut’ comb is produced; when only the recessive alleles are present, a ‘single’ comb is obtained. (a) What are the possible genotypes of all the phenotypes obtainable? Determine the genotype of the parent birds in the following crosses: (b) Rose x walnut produced 3/8 walnut, 3/8 rose, 1/8 rose, 1/8 pea and 1/8 single offspring. (c) Walnut x single produced ¼ walnut and ¼ pea, ¼ rose and ¼ single offspring (d) Rose x pea produced 6 walnut and 5 rose offspring.arrow_forward
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