Concepts of Genetics (11th Edition)
11th Edition
ISBN: 9780321948915
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Textbook Question
Chapter 4, Problem 41ESP
Students taking a genetics exam were expected to answer the following question by converting data to a “meaningful ratio” and then solving the problem. The instructor assumed that the final ratio would reflect two gene pairs, and most correct answers did. Here is the exam question:
“Flowers may be white, orange, or brown. When plants with white flowers are crossed with plants with brown flowers, all the F1 flowers are white. For F2 flowers, the following data were obtained:
Convert the F2 data to a meaningful ratio that allows you to explain the inheritance of color. Determine the number of genes involved and the genotypes that yield each
- (a) Solve the problem for two gene pairs. What is the final F2 ratio?
- (b) A number of students failed to reduce the ratio for two gene pairs as described above and solved the problem using three gene pairs. When examined carefully, their solution was deemed a valid response by the instructor. Solve the problem using three gene pairs.
- (c) We now have a dilemma. The data are consistent with two alternative mechanisms of inheritance. Propose an experiment that executes crosses involving the original parents that would distinguish between the two solutions proposed by the students. Explain how this experiment would resolve the dilemma.
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. A corn geneticist has three pure lines of genotypes a/a ; B/B ; C/C, A/A ; b/b ; C/C, and A/A ; B/B ; c/c. All the phenotypes determined by a, b, and c will increase the market value of the corn; so, naturally, he wants to combine them all in one pure line of genotype a/a ; b/b ; c/c. a. Outline an effective crossing program that can be used to obtain the a/a ; b/b ; c/c pure line. b. At each stage, state exactly which phenotypes will be selected and give their expected frequencies. c. Is there more than one way to obtain the desired genotype? Which is the best way?Assume independent assortment of the three gene pairs. (Note: Corn will self or cross-pollinate easily.
Mendel crossed two Pea plants for plant height and flower color Tall plant (T) is dominant to Short Plant (t). Purple Flower (P) is dominant to white flower (p). Using the following information perform the dihybrid cross using punnett squares that will predict all possible genotypes of the offspring and list the number and description of the phenotypes of the offspring. A. One plant homozygous dominant for plant height and flower color crossed with another plant homozygous recessive for plant height and heterozygous for flower color.
Two true-breeding varieties of maize, one 11 cm high and the other 47 cm high were crossed and the resultant F1 hybrids were then crossed to generate the F2 . In the F2 there were a total of 13,923 plants with a continuous variation in heights between the two extremes and with only 3 plants as large as 47 cm high and 5 plants of 11 cm high.SO In a cross between a 29cm plant and a 20cm plant what would be the genotypes giving the smallest number of different phenotypes? specify the phenotypes observed.
Chapter 4 Solutions
Concepts of Genetics (11th Edition)
Ch. 4 - In the guinea pig, one locus involved in the...Ch. 4 - In some plants a red flower pigment, cyanidin, is...Ch. 4 - Below are three pedigrees. For each trait,...Ch. 4 - Prob. 1CSCh. 4 - Prob. 2CSCh. 4 - Prob. 3CSCh. 4 - HOW DO WE KNOW? In this chapter, we focused on...Ch. 4 - CONCEPT QUESTION Review the Chapter Concepts list...Ch. 4 - In shorthorn cattle, coat color may be red, white,...Ch. 4 - In foxes, two alleles of a single gene, P and p,...
Ch. 4 - In mice, a short-tailed mutant was discovered....Ch. 4 - List all possible genotypes for the A, B, AB, and...Ch. 4 - With regard to the ABO blood types in humans,...Ch. 4 - In a disputed parentage case, the child is blood...Ch. 4 - The A and B antigens in humans may be found in...Ch. 4 - In chickens, a condition referred to as creeper...Ch. 4 - In rabbits, a series of multiple alleles controls...Ch. 4 - Three gene pairs located on separate autosomes...Ch. 4 - As in Problem 12, flower color may be red, white,...Ch. 4 - Horses can be cremello (a light cream color),...Ch. 4 - With reference to the eye color phenotypes...Ch. 4 - Pigment in mouse fur is only produced when the C...Ch. 4 - In rats, the following genotypes of two...Ch. 4 - Given the inheritance pattern of coat color in...Ch. 4 - In a species of the cat family, eye color can be...Ch. 4 - In a plant, a tall variety was crossed with a...Ch. 4 - In a unique species of plants, flowers may be...Ch. 4 - Five human matings (15), identified by both...Ch. 4 - A husband and wife have normal vision, although...Ch. 4 - In humans, the ABO blood type is under the control...Ch. 4 - In Drosophila, an X-linked recessive mutation,...Ch. 4 - Another recessive mutation in Drosophila, ebony...Ch. 4 - In Drosophila, the X-linked recessive mutation...Ch. 4 - While vermilion is X-linked in Drosophila and...Ch. 4 - In a cross in Drosophila involving the X-linked...Ch. 4 - Consider the three pedigrees below, all involving...Ch. 4 - In goats, the development of the beard is due to a...Ch. 4 - Predict the F1 and F2 results of crossing a male...Ch. 4 - Two mothers give birth to sons at the same time at...Ch. 4 - Discuss the topic of phenotypic expression and the...Ch. 4 - Prob. 35PDQCh. 4 - Labrador retrievers may be black, brown...Ch. 4 - A true-breeding purple-leafed plant isolated from...Ch. 4 - In Dexter and Kerry cattle, animals may be polled...Ch. 4 - A geneticist from an alien planet that prohibits...Ch. 4 - The following pedigree is characteristic of an...Ch. 4 - Students taking a genetics exam were expected to...Ch. 4 - In four oclock plants, many flower colors are...Ch. 4 - Proto-oncogenes stimulate cells to progress...Ch. 4 - Below is a partial pedigree of hemophilia in the...
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- Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are dominant to green (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares would you need to complete a Punnett square analysis of this cross?arrow_forwardPedigree analysis is a fundamental tool for investigating whether or not a trait is following a Mendelian pattern of inheritance. It can also be used to help identify individuals within a family who may be at risk for the trait. Adam and Sarah, a young couple of Eastern European Jewish ancestry, went to a genetic counselor because they were planning a family and wanted to know what their chances were for having a child with a genetic condition. The genetic counselor took a detailed family history from both of them and discovered several traits in their respective families. Sarahs maternal family history is suggestive of an autosomal dominant pattern of cancer predisposition to breast and ovarian cancer because of the young ages at which her mother and grandmother were diagnosed with their cancers. If a mutant allele that predisposed to breast and ovarian cancer was inherited in Sarahs family, she, her sister, and any of her own future children could be at risk for inheriting this mutation. The counselor told her that genetic testing is available that may help determine if this mutant allele is present in her family members. Adams paternal family history has a very strong pattern of early onset heart disease. An autosomal dominant condition known as familial hypercholesterolemia may be responsible for the large number of deaths from heart disease. As with hereditary breast and ovarian cancer, genetic testing is available to see if Adam carries the mutant allele. Testing will give the couple more information about the chances that their children could inherit this mutation. Adam had a first cousin who died from Tay-Sachs disease (TSD), a fatal autosomal recessive condition most commonly found in people of Eastern European Jewish descent. Because TSD is a recessively inherited disorder, both of his cousins parents must have been heterozygous carriers of the mutant allele. If that is the case, Adams father could be a carrier as well. If Adams father carries the mutant TSD allele, it is possible that Adam inherited this mutation. Because Sarah is also of Eastern European Jewish ancestry, she could also be a carrier of the gene, even though no one in her family has been affected with TSD. If Adam and Sarah are both carriers, each of their children would have a 25% chance of being afflicted with TSD. A simple blood test performed on both Sarah and Adam could determine whether they are carriers of this mutation. Would you decide to have a child if the test results said that you carry the mutation for breast and ovarian cancer? The heart disease mutation? The TSD mutation? The heart disease and the mutant alleles?arrow_forwardPedigree analysis is a fundamental tool for investigating whether or not a trait is following a Mendelian pattern of inheritance. It can also be used to help identify individuals within a family who may be at risk for the trait. Adam and Sarah, a young couple of Eastern European Jewish ancestry, went to a genetic counselor because they were planning a family and wanted to know what their chances were for having a child with a genetic condition. The genetic counselor took a detailed family history from both of them and discovered several traits in their respective families. Sarahs maternal family history is suggestive of an autosomal dominant pattern of cancer predisposition to breast and ovarian cancer because of the young ages at which her mother and grandmother were diagnosed with their cancers. If a mutant allele that predisposed to breast and ovarian cancer was inherited in Sarahs family, she, her sister, and any of her own future children could be at risk for inheriting this mutation. The counselor told her that genetic testing is available that may help determine if this mutant allele is present in her family members. Adams paternal family history has a very strong pattern of early onset heart disease. An autosomal dominant condition known as familial hypercholesterolemia may be responsible for the large number of deaths from heart disease. As with hereditary breast and ovarian cancer, genetic testing is available to see if Adam carries the mutant allele. Testing will give the couple more information about the chances that their children could inherit this mutation. Adam had a first cousin who died from Tay-Sachs disease (TSD), a fatal autosomal recessive condition most commonly found in people of Eastern European Jewish descent. Because TSD is a recessively inherited disorder, both of his cousins parents must have been heterozygous carriers of the mutant allele. If that is the case, Adams father could be a carrier as well. If Adams father carries the mutant TSD allele, it is possible that Adam inherited this mutation. Because Sarah is also of Eastern European Jewish ancestry, she could also be a carrier of the gene, even though no one in her family has been affected with TSD. If Adam and Sarah are both carriers, each of their children would have a 25% chance of being afflicted with TSD. A simple blood test performed on both Sarah and Adam could determine whether they are carriers of this mutation. Would you want to know the results of the cancer, heart disease, and TSD tests if you were Sarah and Adam? Is it their responsibility as potential parents to gather this type of information before they decide to have a child?arrow_forward
- Pedigree analysis is a fundamental tool for investigating whether or not a trait is following a Mendelian pattern of inheritance. It can also be used to help identify individuals within a family who may be at risk for the trait. Adam and Sarah, a young couple of Eastern European Jewish ancestry, went to a genetic counselor because they were planning a family and wanted to know what their chances were for having a child with a genetic condition. The genetic counselor took a detailed family history from both of them and discovered several traits in their respective families. Sarahs maternal family history is suggestive of an autosomal dominant pattern of cancer predisposition to breast and ovarian cancer because of the young ages at which her mother and grandmother were diagnosed with their cancers. If a mutant allele that predisposed to breast and ovarian cancer was inherited in Sarahs family, she, her sister, and any of her own future children could be at risk for inheriting this mutation. The counselor told her that genetic testing is available that may help determine if this mutant allele is present in her family members. Adams paternal family history has a very strong pattern of early onset heart disease. An autosomal dominant condition known as familial hypercholesterolemia may be responsible for the large number of deaths from heart disease. As with hereditary breast and ovarian cancer, genetic testing is available to see if Adam carries the mutant allele. Testing will give the couple more information about the chances that their children could inherit this mutation. Adam had a first cousin who died from Tay-Sachs disease (TSD), a fatal autosomal recessive condition most commonly found in people of Eastern European Jewish descent. Because TSD is a recessively inherited disorder, both of his cousins parents must have been heterozygous carriers of the mutant allele. If that is the case, Adams father could be a carrier as well. If Adams father carries the mutant TSD allele, it is possible that Adam inherited this mutation. Because Sarah is also of Eastern European Jewish ancestry, she could also be a carrier of the gene, even though no one in her family has been affected with TSD. If Adam and Sarah are both carriers, each of their children would have a 25% chance of being afflicted with TSD. A simple blood test performed on both Sarah and Adam could determine whether they are carriers of this mutation. If Sarah carries the mutant cancer allele and Adam carries the mutant heart disease allele, what is the chance that they would have a child who is free of both diseases? Are these good odds?arrow_forwardHemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?arrow_forwardThe text outlines some of the problems Frederick William I encountered in his attempt to breed tall Potsdam Guards. a. Why were the results he obtained so different from those obtained by Mendel with short and tall pea plants? b. Why were most of the children shorter than their tall parents?arrow_forward
- An organism of the genotype AaBbCc was testcrossed to a triply recessive organism (aabbcc). The genotypes of the progeny are presented in the following table. 20 AaBbCc 20 AaBbcc 20 aabbCc 20 aabbcc 5 AabbCc 5 Aabbcc 5 aaBbCc 5 aaBbcc (a) If these three genes were all assorting independently, how many genotypic and phenotypic classes would result in the offspring, and in what proportion, assuming simple dominance and recessiveness in each gene pair? (b) Answer part (a) again, assuming the three genes are so tightly linked on a single chromosome that no crossover gametes were recovered in the sample of offspring. (c) What can you conclude from the actual data about the location of the three genes in relation to one another?arrow_forwardA cross was made between a plant that has blue flowers and purple seeds and a plant with white flowers and green seeds. The F1 generation was then allowed to self-fertilize. The following data were obtained. F1 Generation: All offspring have blue flowers with purple seeds. F2 Generation: 208 blue flowers, purple seeds 13 blue flowers, green seeds 19 White flowers, purple seeds 60 white flowers, green seeds Total: 300 The scientist needs to test if the two genes segregate independently and has formulated the null-hypothesis that there is no significant difference between the observed and expected data. What are the degrees of freedom for this experiment?arrow_forwardTwo true-breeding varieties of maize, one 11 cm high and the other 47 cm high were crossed and the resultant F1 hybrids were then crossed to generate the F2. In the F2 there were a total of 13,923 plants with a continuous variation in heights between the two extremes and with only 3 plants as large as 47 cm high and 5 plants of 11 cm high. What would be the size and genotype of the F1 from a cross between a true-breeding 11 cm plant and a true-breeding 47 cm plant?arrow_forward
- Two true-breeding varieties of maize, one 11 cm high and the other 47 cm high were crossed and the resultant F1 hybrids were then crossed to generate the F2 . In the F2 there were a total of 13,923 plants with a continuous variation in heights between the two extremes and with only 3 plants as large as 47 cm high and 5 plants of 11 cm high. What would be the size and genotype of the F1 from a cross between a true-breeding 11 cm plant and a true-breeding 47 cm plant? Please answer the following problem & EXPLAIN your answer showing ALL WORKING.arrow_forwardTwo true-breeding varieties of maize, one 11 cm high and the other 47 cm high were crossed and the resultant F1 hybrids were then crossed to generate the F2 . In the F2 there were a total of 13,923 plants with a continuous variation in heights between the two extremes and with only 3 plants as large as 47 cm high and 5 plants of 11 cm high.So What would be the size and genotype of the F1 from a cross between a true-breeding 11 cm plant and a true-breeding 47 cm plant?arrow_forwardPart A: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a single plant will be AB? Enter your answer as a decimal to three places (for example: 0.120). Part B: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a single plant will be Ab? Part C: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a single plant will be aB? Part D: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a signal plant will be ab?arrow_forward
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