A researcher crosses mice with apricot eyes and long tails, and the F1 progeny were recovered in the following numbers and phenotypic classes: F1: 21 apricot, short : 30 brown, long : 19 brown, short : 110 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) What is your null hypothesis for the chi-square test? b) You obtain a value of 11.33 for the chi-square test. What conclusion can you make from the results of the chi-square test? c) Give the expected number of individuals in the "apricot, long" class. d) Making use of the appropriate genetic convention for naming alleles, give the genotype of any of the progeny in the "apricot, long" class. P df 0.995 0.975 0.9 0.5 0.1 0.05* 0.025 0.01 0.005 0.000 0.000 0.016 0.455 2.706 3.841 5.024 6.635 7.879 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 7.779 9.488 11.143 13.277 14.860

A researcher crosses mice with apricot eyes and long tails, and the F1 progeny were recovered in the following numbers and phenotypic classes: F1: 21 apricot, short : 30 brown, long : 19 brown, short : 110 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) What is your null hypothesis for the chi-square test? b) You obtain a value of 11.33 for the chi-square test. What conclusion can you make from the results of the chi-square test? c) Give the expected number of individuals in the "apricot, long" class. d) Making use of the appropriate genetic convention for naming alleles, give the genotype of any of the progeny in the "apricot, long" class. P df 0.995 0.975 0.9 0.5 0.1 0.05* 0.025 0.01 0.005 0.000 0.000 0.016 0.455 2.706 3.841 5.024 6.635 7.879 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 7.779 9.488 11.143 13.277 14.860

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter5: The Inheritance Of Complex Traits

Section: Chapter Questions

Problem 7QP: Sunflowers with flowers 10 cm in diameter are crossed with a plant that has 20-cm flowers. The F1...

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A researcher crosses mice with apricot eyes and long tails, and the F1 progeny were recovered in the following numbers and phenotypic classes:F1: 21 apricot, short : 30 brown, long : 19 brown, short : 110 apricot, longYou 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) What is your null hypothesis for the chi-square test? b) You obtain a value of 11.33 for the chi-square test. What conclusion can you make from the results of the chi-square test? c) Give the expected number of individuals in the “apricot, long” class. d) Making use of the appropriate genetic convention for naming alleles, give the genotype of any of the progeny in the “apricot, long” class.
In rabbits grey hair is dominant to white & black eyes are dominant to red. -Cross 2 parents who are both heterozygous for both traits. 1st Question to answer ---- How many of the offspring could have grey hair & black eyes? 2nd Question to answer ---- How many of the offspring could have grey hair & red eyes? 3rd Question to answer ---- How many of the offspring could have white hair & black eyes? 4th Question to answer ---- How many of the offspring could have white hair & red eyes?
. Consider the imaginary trait, bristles with split-ends, a trait hypothesized to be X-linked dominant. In a cross, the P1 virgin females were homozygous wild type while the males had bristles with split-ends. The F1 84 males were all wild type while the 90 females all had split-ends. In addition, the data for the F2 generation revealed 26 wild type males, 35 wild type females, 29 split-end males and 40 split-end females. Is this what is to be expected? Use chi-square to prove your position.
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, longYou 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?
A genetic cross was made between two true-breeding parental fly strains. One parent strain displays red eyes and curly wings, while the other parent strain displays white eyes and vestigial wings. The resulting F1 generation flies were allowed to self-cross to produce F2 generation. The following data were obtained: F1 generation: all have red eyes and curly wings. F2 generation: 295 were red eyes and curly wings, 95 were red eyes and vestigial wings, 89 were white eyes and curly wings, 31 were white eyes and vestigial wings With this data above – please answer the following questions: Propose a hypothesis that explains the above inheritance pattern, test the goodness of fit between the data and your hypothesis using a chi square test. Use the chi square table from your textbook/online. Use the information obtained from the chi square table to explain what your calculated chi square results mean.
A cross between two pure breeding pea plants in which one parent produced yellow seeds and the other produced green seeds yielded 70 yellow seed F2 offspring and 30 green seed F2 offspring. what is the degrees of freedom, level of significance, and evaluation of X2. Do you accept or reject the hypothesis ?
A genetic cross was made between two types of pea plants. One type produces round and green seeds, while the other type produces wrinkled and yellow seeds. A plant breeder crossed these two types and then allowed the F1 generation to self-fertilize. The following data were obtained: F1 generation: all produce wrinkled and yellow seeds. F2 generation: 157 produce wrinkled and yellow seeds 57 produce wrinkled and green seeds 54 produce round and yellow seeds 20 produce round and green seeds a) Propose a hypothesis that you think is consistent with the observed data. b) Test the goodness of fit between the data and your hypothesis using a chi square test. c) Explain what your calculated chi square results mean.
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Imagine that you set up a three-point mapping cross to determine the order of three X-linked genes. You have recessive mutations available for all three genes: yellow, white, and echinus. A heterozygous F1 Drosophila female is crossed to hemizygous mutant male. The observed numbers and phenotypes of F2 are as follows. 4685 yellow, white, echinus 4759 wild type 80 yellow 70 white, echinus 193 yellow, white 207 Echinus 3yellow, echinus 3white Based on this data which of the above phenotypic classes represent the PARENTAL types? Each answer yields a clue useful in following question. Choose all that apply