True-breeding diabetes-resistant male monkeys were bred to true-breeding wildtype female monkeys. All the F1 monkeys are phenotypically wildtype. The F1 monkeys are bred to each other, producing eight F2 babies. Only one of the eight F2 monkeys has the diabetes resistance. Is it possible for this condition to be controlled by a single autosomal gene? Use chi-square analysis to test the hypothesis, show your work, and explain your conclusion.

True-breeding diabetes-resistant male monkeys were bred to true-breeding wildtype female monkeys. All the F1 monkeys are phenotypically wildtype. The F1 monkeys are bred to each other, producing eight F2 babies. Only one of the eight F2 monkeys has the diabetes resistance. Is it possible for this condition to be controlled by a single autosomal gene? Use chi-square analysis to test the hypothesis, show your work, and explain your conclusion.

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Description: True-breeding diabetes-resistant male monkeys were bred to true-breeding wildtype femalemonkeys. All the F1 monkeys are phenotypically wildtype. The F1 monkeys are bred to each other,producing eight F2 babies. Only one of the eight F2 monkeys has th

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter11: Genome Alterations: Mutation And Epigenetics

Section: Chapter Questions

Problem 16QP: Familial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior...

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You believe that the gene you're investigating, eyes absent, is an autosomal recessive mutant allele F7 ge in Drosophila. You set up a cross between a true-breeding mutant male and a true-breeding wildtype female to obtain the F1 generation. You then cross the F1 flies. In the F2 offspring, you observe the following: 237 wild-type males 180 wild-type females 70 mutant males 83 mutant females Based on these findings, is the mutant allele autosomal recessive? How do you know?
Multiple crosses were made between true-breeding lines of black and yellow Labrador retrievers. All the F1 progeny were yellow. When these progeny were intercrossed, they produced an F2 consisting of 121 yellow, 9 black and 30 chocolate. What epistatic ratio and what kind of epistasis is approximated in the F2? Propose a biochemical pathway for coat color in Labrador retrievers based on the type of epistasis. Correlate each genotype with the phenotype that would occur in your pathway. Also show the frequency of each genotype. A-B- A-bb aaB- aabb
Familial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior history of the disease. Consider the following pedigree (the darkly colored symbols represent affected individuals): a. Circle the individual(s) in which the mutation most likely occurred. b. Is the person who is the source of the mutation affected by retinoblastoma? Justify your answer. c. Assuming that the mutant allele is fully penetrant, what is the chance that an affected individual will have an affected child?
Your internship with Dr. Nefario at Gru Industries is going well so far, but a recent mission to obtain a shrink ray requires a special kind of Minion. You identify that the ideal Minion phenotypes for this mission are two eyes, pale yellow, and short. From your previous work, you’ve found that:Two-eyed (D) is completely dominant over one-eyed (d)Yellow (Y) is incompletely dominant with white (y)Tall (T) is completely dominant over short (t) a. The only reproductive individuals you have at the moment (it was surprise mission!) are a true breeding two-eyed, yellow, short Minion and a one heterozygous for each trait. What genotypic ratios would you expect for this crossassuming that each locus is on a different chromosome? What total proportion would you expect to be the desired phenotype for the mission? b. The mission has since been delayed due to lack of financial support, so you’ve got some time. All the individuals from the F1 cross above are now reproductive. Which genotypes would…
IN DROSOPHILA, AN X-LINKED RECESSIVE MUTATION, Xm CAUSES MINIATURE WINGS. LIST THE F₂ PHENOTYPIC RATIOS IF: A MINIATURE-WINGED FEMALE IS CROSSED WITH A NORMAL MALE AND A MINIATURE-WINGED MALE IS ● ● CROSSED WITH A NORMAL FEMALE. WHAT WOULD THE PHENOTYPIC RATIO FROM (A) BE IF THE MINIATURE- WINGED GENE WERE AUTOSOMAL? ASSUME IN ALL CASES THAT THE P1 INDIVIDUALS ARE TRUE-BREEDING.
A male fly that is homozygous for both red eyes (A) and normal wings (B) mates with a female fly that has apricot (a) eyes and pointed wings (b). Both traits are controlled by autosomal alleles. Using the information provided above, answer the following two questions (Question 1 AND 2): What is the expected phenotype for the flies in the F1 generation? What is the expected genotype for the flies in the F1 generation?
Consider two hypothetical recessive autosomal genes a and b, where a heterozygote is testcrossed to a double- homozygous mutant. Predict the phenotypic ratios under the following conditions: (a) a and b are located on separate autosomes. (b) a and b are linked on the same autosome but are so far apart that a crossover always occurs between them. (c) a and b are linked on the same autosome but are so close together that a crossover almost never occurs.
The fruit fly, Drosophila melanogaster, a vestigial female fly is mated to a male that is scarlet eyes and bristled. Phenotypically wild type F1 female progeny were mated to fully homozygous (mutant) males, and the following progeny (of 1000 total) were observed (shown on table) a. Which gene is in the middle? b. What are the map distances between the three genes for this question? c. Calculate and show the work for coefficient of coincidence d. Calculate the interference. What does this mean and what does it tell you?
For this problem, use the information given in problem 15. A young woman with blood type O gave birth to a baby with blood type O. In a court case, she claims that a certain young man is the father of her child. The man has type A blood. Could he be the father? Can it be proven on this evidence alone that he is the father? ^Need answers to that one, including genotype, phenotype gametes and punnett square. This is question 15 as a little background dont need this solved . In humans, there are three alleles for blood type: A, B, and O. The allele for blood type A and the allele for blood type B show incomplete dominance. A person with both alleles has blood type AB. Both A and B dominate type O. A person with alleles for blood types A and O marries someone with alleles for blood types B and O. List the types of blood their offspring could have and the probability for each blood type in the offspring
The insulin-like growth factor 2 (Igf2) gene is imprinted and paternally expressed. Mice that express only a mutant version of the gene (igf2-) are born small (dwarf). If a geneticist crosses heterozygous female and male mice (Igf2 igf2-), what is the expected ratio of normal to dwarf mice among the offspring? Explain your answer.
Consider Dr. Whitt’s comments about the relationship between the wild-type and mutant RB1 alleles and what you now know about Knudson’s two-hit hypothesis. Do these ideas seem to confirm or contradict your earlier prediction (based on the pedigree) about whether the mutant RB1 allele is dominant or recessive to the wild-type allele? Explain
A researcher crosses mice with brown eyes and long tails, and the F1 progeny were recovered in the following numbers and phenotypic classes: F1: 6 apricot, short : 30 brown, long : 15 brown, short : 9 apricot, long You know the genes encoding these traits are autosomal, completely dominant and assort independently. You want to use a chi-square test to analyse these results. a) Making use of the appropriate genetic convention for naming alleles, give the genotype of the male parent in this cross. b) What is your null hypothesis for the chi-square test? c) Give the expected number of individuals in the "brown, long" class. d) You obtain a value of 3.47 for the chi-square test. What conclusion can you make from the results of the chi-square test? P df 0.995 0.975 0.9 0.5 0.1 0.05* 0.025 0.01 0.005 1 0.000 0.000 0.016 0.455 2.706 3.841 5.024 6.635 7.879 2 0.010 0.051 0.211 1.386 4.605 5.991 7.378 9.210 10.597 0.072 0.216 0.584 2.366 6.251 7.815 9.348 11.345 12.838 4 0.207 0.484 1.064 3.357…