Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 4, Problem 33P
Dr. O. Sophila, a close friend of Dr. Ara B. Dopsis, reviews the
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Part 1 - With use of Punnett diagrams discuss how the results of experiments carried out by Austrian monk Gregor Mendel on pea plants brought about the Law of Dominance, the Law of Segregation, and the Law of Independent assortment.
Part 2 - In humans, the gene that produces the disease Sickle Cell Anaemia is recessive to the gene for healthy haemoglobin production.
a) How can two phenotypically healthy parents have a child who suffers from Sickle cell Anaemia? In your answer use suitable letters to show the genotypes of the parents and construct a Punnett diagram to show all the possibilities of the children's genotypes.
b) Based on the outcome of your Punnett diagram, state and explain the probability of a child being a sufferer of Sickle Cell Anaemia and the probability of a child not suffering from the disease.
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E. 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?
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...
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- Hemophilia 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_forwardWhat would justify the following ratio appearing after phenotyping the outcome of a crossing trial: 8.9: 2.9: 3.2:1? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a b C d Obviously this represents independent assortment based on crossing dihybrid heterozygotes. Obviously this represents gene linkage based on test crossing a dihybrid heterozygote. Obviously this represents the results of a trihybrid test cross. Obviously this represents independent assortment based on crossing monohybrid heterozygotes.arrow_forward
- The following pedigree shows the inheritance of deafmutism.a. Provide an explanation for the inheritance of thisrare condition in the two families in generations I andII, showing the genotypes of as many persons aspossible; use symbols of your own choosing.b. Provide an explanation for the production of onlynormal persons in generation III, making sure that yourexplanation is compatible with the answer to part aarrow_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_forwardArabidopsis thaliana is a diploid plant model organism with 2n 10. Please select the number options to match the following number of copies of each gene an Arabidopsis thaliana 1 leaf cell has number of chromosomes an Arabidopsis thaliana leaf cell contains 1. two 2. five number of chromosomes an Arabidopsis thaliana gamete cell contains 3. ten pairs of homologous chromosomes an Arabidopsis thaliana 2 leaf cell containsarrow_forward
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