Introduction To Genetic Analysis
12th Edition
ISBN: 9781319114787
Author: Anthony J.F. Griffiths, John Doebley, Catherine Peichel, David A. Wassarman
Publisher: W. H. Freeman
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 2, Problem 18P
(1)
Summary Introduction
To determine: The generation-I parent is heterozygous in the pedigree.
Introduction: An autosomal recessive trait that is the individuals with a shaded in the symbol are homozygous recessive (aa) and show the recessive trait. A half-shaded symbol represents heterozygous individuals (Aa), and an unshaded symbol represents individuals who are homozygous dominant (AA).
(2)
Summary Introduction
To determine: The mendelian ratio in generation IV.
Introduction:
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
Mendel crossed pea plants that produced round seeds with those that produced wrinkled seeds. From a total of 7324 F2 seeds, 5474 were round and 1850 were wrinkled. Using the symbols W and w for genes, (a) symbolize the original P cross; (b) the gametes; and (c) F1 progeny, (d) Represent cross between two F1 plants (or one selfed); (e) symbolize the gametes; and (f) summarize the expected F2 results under the headings: phenotypes, genotypes, genotypic frequency, and phenotypic ratio.
Gray seed color in peas is dominant to white. Assume that Mendel conducted a series of experiments where plants with gray seeds were crossed with each other and the following progeny were produced: 320 gray and 80 white. (a) What is the most probable genotype of each parent? (b) Based on your answer in (a) above, what genotypic and phenotypic ratios are expected in the progeny?
In the pedigree shown in Figure Q19–14, the first born in each of three generations is the only person affected by a dominant genetically inherited disease, D. your friend concludes that the first child born has a greater chance of inheriting the mutant D allele than do later children.A. According to Mendel’s laws, is this conclusion plausible?B. What is the probability of obtaining this result by chance?
Chapter 2 Solutions
Introduction To Genetic Analysis
Ch. 2 - Prob. 1PCh. 2 - Prob. 2PCh. 2 - Prob. 3PCh. 2 - Prob. 4PCh. 2 - Prob. 5PCh. 2 - Prob. 6PCh. 2 - Prob. 7PCh. 2 - Prob. 8PCh. 2 - Prob. 9PCh. 2 - Prob. 10P
Ch. 2 - Prob. 11PCh. 2 - Prob. 12PCh. 2 - Prob. 13PCh. 2 - Prob. 14PCh. 2 - Prob. 15PCh. 2 - Prob. 17PCh. 2 - Prob. 18PCh. 2 - Prob. 19PCh. 2 - Prob. 20PCh. 2 - Prob. 21PCh. 2 - Prob. 22PCh. 2 - Prob. 23PCh. 2 - Prob. 24PCh. 2 - Prob. 25PCh. 2 - Prob. 26PCh. 2 - Prob. 27PCh. 2 - Prob. 28PCh. 2 - Prob. 29PCh. 2 - Prob. 30PCh. 2 - Prob. 31PCh. 2 - Prob. 32PCh. 2 - Prob. 33PCh. 2 - Prob. 34PCh. 2 - Prob. 35PCh. 2 - Prob. 36PCh. 2 - Prob. 37PCh. 2 - Prob. 38PCh. 2 - Prob. 39PCh. 2 - Prob. 40PCh. 2 - Prob. 41PCh. 2 - Prob. 42PCh. 2 - Prob. 43PCh. 2 - Prob. 44PCh. 2 - Prob. 45PCh. 2 - Prob. 46PCh. 2 - Prob. 47PCh. 2 - Prob. 48PCh. 2 - Prob. 49PCh. 2 - Prob. 50PCh. 2 - Prob. 51PCh. 2 - Prob. 52PCh. 2 - Prob. 53PCh. 2 - Prob. 54PCh. 2 - Prob. 55PCh. 2 - Prob. 56PCh. 2 - Prob. 56.1PCh. 2 - Prob. 56.2PCh. 2 - Prob. 56.3PCh. 2 - Prob. 56.4PCh. 2 - Prob. 56.5PCh. 2 - Prob. 56.6PCh. 2 - Prob. 56.7PCh. 2 - Prob. 56.8PCh. 2 - Prob. 56.9PCh. 2 - Prob. 56.10PCh. 2 - Prob. 56.11PCh. 2 - Prob. 56.12PCh. 2 - Prob. 56.13PCh. 2 - Prob. 56.14PCh. 2 - Prob. 56.15PCh. 2 - Prob. 57PCh. 2 - Prob. 58PCh. 2 - Prob. 59PCh. 2 - Prob. 60PCh. 2 - Prob. 61PCh. 2 - Prob. 62PCh. 2 - Prob. 63PCh. 2 - Prob. 64PCh. 2 - Prob. 65PCh. 2 - Prob. 67PCh. 2 - Prob. 68PCh. 2 - Prob. 69PCh. 2 - Prob. 70PCh. 2 - Prob. 71PCh. 2 - Prob. 72PCh. 2 - Prob. 73PCh. 2 - Prob. 74PCh. 2 - Prob. 75PCh. 2 - Prob. 76PCh. 2 - Prob. 77PCh. 2 - Prob. 78PCh. 2 - Prob. 79PCh. 2 - Prob. 80PCh. 2 - Prob. 81PCh. 2 - Prob. 82PCh. 2 - Prob. 83PCh. 2 - Prob. 84PCh. 2 - Prob. 85PCh. 2 - Prob. 86PCh. 2 - Prob. 87PCh. 2 - Prob. 88PCh. 2 - Prob. 89PCh. 2 - Prob. 90PCh. 2 - Prob. 91PCh. 2 - Prob. 1GSCh. 2 - Prob. 2GSCh. 2 - Prob. 3GS
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
- For the following cross, show the P generation Genotypes and the Phenotypic ratio that would be seen in the F1 and F2. Remember, to produce the F2 generation you want to cross Heterozygotes from the F1. d) Genes 1 and 2 exhibit Epistasis (9:6:1) and Gene 3 is an Autosomal Dominant. In the P generation, the Male is Homozygous Recessive for the Genes showing Epistasis. Use E1, E2 and E3 to represent the Phenotypes shown by Epistasis. Report your results in the following format: P = aabb x AABB, F1 = 100%AaBb (Phenotype), and %3! F2 = 9/16 A_B_ (Phenotype), 3/16 aaB (Phenotype), 3/16 A_bb (Phenotype), 1/16 aabb (Phenotype)arrow_forwardThe following family pedigree shows individuals afflicted with a metabolic disease. Suppose the male (b) and the female (c) have 2 girls and 2 boys, indicate directly on the pedigree the symbols for these children that would be most consistent with the mode of inheritance observations in the pedigree? II II a b d e IVarrow_forwardThis is a typical pedigree for a family that carries neurofibromatosis. Is individual I-2 most likely homozygous or heterozygous? Why? Explain.arrow_forward
- What pattern of Mendelian inheritance is represented in this pedigree and explain how you made this determination.(please refer to image attached)arrow_forward*Cystic fibrosis is a rare autosomal recessive condition. phenotypically normal man whose father had cystic fibrosis marries a phenotypically normal woman from outside the family.² a) Draw the pedigree as far as described. b) If the frequency of heterozygotes in the general population is 1/50, what is the probability that the couple's first child will have cystic fibrosis? c) If the first child does have cystic fibrosis, what is the probability that the second child will be normal?arrow_forwardQ6. Guinea pigs, which were homozygous for long, black hair were crossed with ones which were homozygous for short, white hair. All the F1 offspring had short, black hair. (a) Using suitable symbols, draw a genetic diagram showing parents genotypes, to explain this result in F1 offspring. (b) Complete the Punnett square to show the results of interbreeding the F1 offspring. Gametesarrow_forward
- A type of tomato (Solanum lycopersicum) produces fruit in three possible colors: red, orange, and green. You cross a true-breeding red-fruited plant with a true-breeding green-fruited plant, and all the F1 offspring are red. You intercross the red F1s, and the resulting F2 generation consists of 61 red-, 13 orange-, and 6 green-fruited plants. a) Assign the phenotypes to a modified 9:3:3:1 ratio (do not just calculate the actual ratio of the phenotypes). Show your work.arrow_forwardIn Figure 4-6, why does the diagram not show meiosesin which two crossovers occur between the same twochromatids (such as the two inner ones)?arrow_forwardExplain why it is possible for the proband in the following pedigree to have children of blood types A, B, and AB. Considering epistatic genes, what are the possible genotypes of II-2?arrow_forward
- The accompanying pedigree shows a family in which one child (II-1) has an autosomal recessive condition. On the basis of this fact alone, provide the following information. 1) What is the chance that among the three children in generation II who have the dominant phenotype, one of them is AAAA and two of them are AaAa? (Hint: Consider all possible orders of genotypes.) Express your answer to two decimal places.arrow_forwardIn pedigrees, individuals are usually specified by using a Roman numeral for their generation in the chart and an Arabic number for their position (reading left to right) within that generation. If we use the letter c for the allele that causes cystic fibrosis, what are the genotypes of individuals III-3 and III-4 (the third and fourth individuals shown in generation III) in the pedigree that shows this disease?arrow_forwardWe have dealt mainly with only two genes, but the sameprinciples hold for more than two genes. Consider thefollowing cross:A/a ; B/b ; C/c ; D/d ; E/e × a/a ; B/b ; c/c ; D/d ; e/ea. What proportion of progeny will phenotypicallyresemble (1) the first parent, (2) the second parent,(3) either parent, and (4) neither parent?b. What proportion of progeny will be genotypically thesame as (1) the first parent, (2) the second parent,(3) either parent, and (4) neither parent?Assume independent assortment.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Human Biology (MindTap Course List)BiologyISBN:9781305112100Author:Cecie Starr, Beverly McMillanPublisher:Cengage Learning
Human Biology (MindTap Course List)
Biology
ISBN:9781305112100
Author:Cecie Starr, Beverly McMillan
Publisher:Cengage Learning
How to solve genetics probability problems; Author: Shomu's Biology;https://www.youtube.com/watch?v=R0yjfb1ooUs;License: Standard YouTube License, CC-BY
Beyond Mendelian Genetics: Complex Patterns of Inheritance; Author: Professor Dave Explains;https://www.youtube.com/watch?v=-EmvmBuK-B8;License: Standard YouTube License, CC-BY