Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Videos
Textbook Question
Chapter 4, Problem 33P
Dr. O. Sophila, a close friend of Dr. Ara B. Dopsis, reviews the
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
When many families were tested for the ability to tastethe chemical phenylthiocarbamide, the matings weregrouped into three types and the progeny were totaled,with the results shown below:ChildrenNumber NonParents of families Tasters tastersTaster × taster 425 929 130Taster × nontaster 289 483 278Nontaster × nontaster 86 5 218With the assumption that PTC tasting is dominant (P)and nontasting is recessive (p), how can the progenyratios in each of the three types of mating be accountedfor?
A genetic engineer is going to cross two watermelon plants to produce seeds for a spring planting. He is breeding for size, and wants to have as many watermelons with the phenotype for long shape as possible. In watermelons, the allele for short shape (R) is dominant to the allele for long shape (r). Would crossing a watermelon homozygous recessive for the trait with a watermelon heterozygous for the trait give the most long watermelons possible? Explain your answer using Punnett Squares.
This is a modified question from the textbook, Campbell Biology in Focus (3rd edition), question #5 on page 256.
A man with normal vision marries a woman with normal vision whose father was red-green color-blind (a recessive, sex-linked condition). Assuming that the woman's mother had normal vision and did not carry the recessive gene, what is the probability that the married couple will have a color-blind daughter? That their first son will be color-blind? That their second son will be color-blind? Include Punnett square for each question in your answer. You may find the below diagram helpful (the orange square means the individual has the phenotype of color-blindness, while the beige means the individual is a carrier of the mutated color-blind gene where XN and Xn represent dominant and recessive allele, respectively).
Chapter 4 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 4 - 1. Define and distinguish incomplete penetrance...Ch. 4 -
2. Define and distinguish epistasis and...Ch. 4 - When working on barley plants, two researchers...Ch. 4 - Fifteen bacterial colonies growing on a complete...Ch. 4 - 5. In a type of parakeet known as a “budgie,”...Ch. 4 - 6. The and blood groups are given below for four...Ch. 4 - The wild-type color of horned beetles is black,...Ch. 4 - 8. Two genes interact to produce various...Ch. 4 - Prob. 9PCh. 4 - 10. In rats, gene produces black coat color if the...
Ch. 4 - 11. In the rats identified in Problem, a third...Ch. 4 - Using the information provided in Problems 10 and...Ch. 4 - 13. Total cholesterol in blood is reported as the...Ch. 4 - 14. Flower color in snapdragons results from the...Ch. 4 - 5. A plant line with reduced fertility comes to...Ch. 4 - Prob. 16PCh. 4 - The coat color in mink is controlled by two...Ch. 4 - Prob. 18PCh. 4 - 19. Feather color in parakeets is produced by the...Ch. 4 - Brachydactyly type D is a human autosomal dominant...Ch. 4 - 21. A male and a female mouse are each from...Ch. 4 - Xerodermapigmentosum (XP) is an autosomal...Ch. 4 - 23. Three strains of green-seeded lentil plants...Ch. 4 - Blue flower color is produced in a species of...Ch. 4 - 25. The following crosses are performed between...Ch. 4 - Two pure-breeding strains of summer squash...Ch. 4 - Marfan syndrome is an autosomal dominant disorder...Ch. 4 - 28. Yeast are single-celled eukaryotic organisms...Ch. 4 - Prob. 29PCh. 4 - Dr. Ara B. Dopsis and Dr. C. Ellie Gans are...Ch. 4 - Human ABO blood type is determined by three...Ch. 4 - In rabbits, albinism is an autosomal recessive...Ch. 4 - Dr. O. Sophila, a close friend of Dr. Ara B....Ch. 4 - In a breed of domestic cattle, horns can appear on...Ch. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - 37. Epistatic gene interaction results in a...Ch. 4 - 38. Draw a pedigree containing two parents and...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- A recessive maternal effect mutant in zebrafish, called ichabod, results in embryos lacking heads that are non-viable. You have been instructed to identify females that are homozygous for the ichabod mutant allele. At your disposal are a tank of wild-type fish (males and females), a tank of male and female parental fish that are all heterozygous for the ichabod mutant allele (ichabod/+), and a tank of F1 fish derived from a cross between a heterozygous male and heterozygous female (ichabod/+). Which of the following would be a way to identify females that are homozygous mutant, i.e. ichabod/ichabod? Select all answers that would work.arrow_forwardE. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings, 853 virescentwhite seedlings, and 260 yellow seedlings (E. W. Lindstrom. 1921. Genetics 6:91–110). a. Give the genotypes for the green, virescent-white, and yellow progeny. b. Explain how color is determined in these seedlings. c. Is there epistasis among the genes that determine color in the corn seedlings? If so, which gene is epistatic and which is hypostatic?arrow_forwardHemophilia and colorblindness are both recessive conditions caused by mutations in 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 the normal alleles and one chromosome with two mutant alleles), and the daughter’s sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has colorblindness only, and one has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness? Show your analysis or calculation, not just the final number.arrow_forward
- Hemophilia and colorblindness are both recessive conditions caused by mutations in 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 the normal alleles and one chromosome with two mutant alleles), and the daughter’s sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has colorblindness only, and one has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness? Show your analysisarrow_forwardIn a certain type of pea plant, the gene for yellow color (Y) is dominant to the gene for green color (y), and the gene for round shape (R) is dominant to the gene for wrinkled shape (r). If a homozygous dominant plant is mated with a homozygous recessive plant (P-gen), what proportion in the F2 generation will be homozygous recessive? a) 1/4 b) 1/16 c) 3/4 d) 3/8arrow_forwardA plant geneticist is examining the mode of inheritance of flower color in two closely related species of exotic plants. The first species may have two pure-breeding lines—one produces a distinct red flower; and the other produces flowers with no color at all, or very pale yellow flowers. However, she cannot be sure. A cross of these varieties produces all pink-flowered progeny. The second species exhibits similar pure-breeding varieties; that is, one variety produces red flowers; and the other produces an albino or very pale yellow flower. A cross of these two varieties, however, produces orange-flowered progeny exclusively. Analyze the mode of inheritance of flower color in these two plant species.arrow_forward
- In corn, the gene for kernel color has two alleles: purple (P) or red (p). There is also a gene where the dominant allele (I) prevents kernel color, whereas the recessive allele (i) allows color to develop. If you cross plants heterozygous for both traits, what will be the expected phenotypic ratio of the offspring? What is the probability that the genotype of an offspring will be PPii? A child with Type O blood is born to a mother with Type A blood. What is the genotype of the child? What is the genotype of the mother? What are the possible genotypes of the father?arrow_forwardIn garden peas, long stems are dominant to short stems, and yellow seeds are dominant to green seeds. 100 long/yellow pea plants, all of which had one short/green parent, are interbred (bred to each other). 1600 progeny result. Please answer the following questions about these progenies. A. Assuming that these two genes are unlinked, about how many long/green pea plants would you expect to find among the offspring? B. What ratio of yellow to green seed color would you expect among the offspring? C. What would you expect the overall phenotypic ratio among the 1600 offspring to be (taking into consideration both traits)?arrow_forwardParts a-c of this question address the crossing of parents with the following genotypes: A/a; B/b; C/c; D/d; E/e; F/f x a/a; b/b; c/c; d/d; e/e; f/f a. What is the probability that an F1 progeny that is homozygous recessive at every locus will be produced? For values greater than 0, please answer as a fraction. b. What is the probability that an F1 progeny that is homozygous dominant at every locus will be produced? For values greater than 0, please answer as a fraction. c. Would one of the parents in this cross be considered a test cross parent (tester)? Yes, the parent on the left, with genotype A/a; B/b; C/c; D/d; E/e; F/f would be the test cross parent (tester). Yes, the parent on the right, with genotype a/a; b/b; c/c; d/d; e/e; f/f would be the test cross parent (tester). Yes, both parents would be considered a test cross parent (tester) No, neither parent would be considered a test cross parent (tester) None of the above and I have explained my…arrow_forward
- In cultivated flowers called “stocks,” the recessive genotype of one locus (aa) prevents the development of pigment in the flower thus producing a white color. In the presence of the dominant allele A, alleles at another locus may be expressed as follows: B_ = red. bb = cream. a) When cream stocks of the genotype Aabb are crossed to red stocks of the genotype AaBb, what genotypic and phenotypic proportions are expected in the progeny? b) When dihybrid red stocks are crossed together, what phenotypic ratio is expected among the progeny? c) If red stocks crossed to white stocks produced progeny with red, cream and white flowers, what are the genotypes of the parents? Show the results of the cross based on your answer.arrow_forwardIn classical Mendelian genetics, how can one check the genotype of a parent (A) expressing the characters of a dominant allele? Select one: a. By performing a back cross with a recessive homozygote parent (B). If the A parent is homozygote for the dominant allele, then all the individuals from the F1 will display the dominant character. If the parent A was, instead, a heterozygote, then 50% of the F1 progeny will express the recessive character (homozygote recessive) and 50% the dominant one (heterozygotes). b. It is impossible to check such genotype without using specific molecular assays. c. By performing a back cross with a dominant homozygote parent (B). If the A parent is homozygote for the dominant allele, then all the individuals from the F1 will display the dominant character.arrow_forwardWhite fruit color in squash is due to a dominant allele. Yellow fruit occurs in plants which are homozygous for the recessive allele. If pollen from the anthers of a heterozygous white-fruited plant is placed on the pistil of the yellow-fruited plant; show, using ratios, the genotypes and phenotypes you would expect the seeds from this cross to produce. a. Expected genotypic ratio: Expected phenotypic ratio:arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning
Human Biology (MindTap Course List)BiologyISBN:9781305112100Author:Cecie Starr, Beverly McMillanPublisher:Cengage Learning
Concepts of BiologyBiologyISBN:9781938168116Author:Samantha Fowler, Rebecca Roush, James WisePublisher:OpenStax College
Human Heredity: Principles and Issues (MindTap Co...
Biology
ISBN:9781305251052
Author:Michael Cummings
Publisher:Cengage Learning
Human Biology (MindTap Course List)
Biology
ISBN:9781305112100
Author:Cecie Starr, Beverly McMillan
Publisher:Cengage Learning
Concepts of Biology
Biology
ISBN:9781938168116
Author:Samantha Fowler, Rebecca Roush, James Wise
Publisher:OpenStax College
How do Plants Handle Stress?; Author: Alex Dainis;https://www.youtube.com/watch?v=TYsnveEHqec;License: Standard Youtube License