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 th 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

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 th 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

Name: A researcher crosses mice with brown eyes and long tails, and the F1
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Description: 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 tr

Concepts of Biology

1st Edition

ISBN:9781938168116

Author:Samantha Fowler, Rebecca Roush, James Wise

Publisher:Samantha Fowler, Rebecca Roush, James Wise

Chapter8: Patterns Of Inheritance

Section: Chapter Questions

Problem 2ACQ: Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are...

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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.
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?
. 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.
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?
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 seperate the hemophilia locus from the locus of the color blindness.
An organism of the genotype AaBbCc was testcrossed to a triplyrecessive organism (aabbcc). The genotypes of the progeny are inthe following table.AaBbCc 20 AaBbcc 20aabbCc 20 aabbcc 20AabbCc 5 Aabbcc 5aaBbCc 5 aaBbcc 5 a.) Assuming simple dominance and recessiveness in each genepair, if these three genes were all assorting independently,how many genotypic and phenotypic classes would result inthe offspring, and in what proportion? b.) Answer part (a) again, assuming the three genes are sotightly linked on a single chromosome that no crossovergametes were recovered in the sample of offspring. c.) What can you conclude from the actual data about thelocation of the three genes in relation to one another?
Most people with the dominant mutant polydactyly allele have extra digits but at least 25% have the normal number of digits. What is the genetic explanation for this observation?
In a diploid plant species, an F1 with the genotype Mm Rr Ss is test crossed to a pure breeding recessive plant with the genotype mm rr ss. The offspring genotypes are as follows: Genotype Number Mm Rr Ss 687 Mm Rr ss 5 Mm rr Ss 68 Mm rr ss 196 mm Rr Ss 185 mm Rr ss 72 mm rr Ss 8 mm rr ss 679 Total 1900 1. What is the gene order of these linked genes?
For this particular family, what is the recombination rate between the D17S74 marker and the breast cancer gene? Lets say that at age 45 the third granddaughter III-3, is diagnosed with breast cancer. Now recalculate the recombination rate between the marker locus D17S74 and the breast cancer gene? Use only the 5 women in your analysis:the two daugthers in generation II and the three granddaughters in generation II
Incomplete dominance and epistasis are both terms thatdefine genetic relationships. What is the most basicdistinction between these terms?
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 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.