student crossed a female worm homozygous for the dpy-17 e164 allele with a wild-type male worm heterozygous for dpy-17 e164. He then scored all of the developing eggs/larva for the Dpy phenotype. He found 88 with the Dpy phenotype and 98 wild type. Do a Chi square test on phenotypic data to determine if your data supports the hypothesis that the mutation segregates as a single-gene mutation. Show your work in a table and include Null hypothesis Χ2 value dof p value

student crossed a female worm homozygous for the dpy-17 e164 allele with a wild-type male worm heterozygous for dpy-17 e164. He then scored all of the developing eggs/larva for the Dpy phenotype. He found 88 with the Dpy phenotype and 98 wild type. Do a Chi square test on phenotypic data to determine if your data supports the hypothesis that the mutation segregates as a single-gene mutation. Show your work in a table and include Null hypothesis Χ2 value dof p value

BIOLOGY:CONCEPTS+APPL.(LOOSELEAF)

10th Edition

ISBN:9781305967359

Author:STARR

Publisher:STARR

Chapter13: Patterns In Inherited Traits

Section: Chapter Questions

Problem 6GP: In sweet pea plant, an allele for purple flowers (P) is dominant when paired with a recessive allele...

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In corn the genes for tassel length (alleles T and t) and rust resistance (alleles R and r) are known to be on separate chromosomes. In the course of a routine mating check experiment, a breeder got the following results upon crossing T/t:R/r with a double recessive strain (t/t;r/r): genotype Number of progeny T/t; R/r 98 t/t; r/r 104 T/t; r/r 3 t/t; R/r 5 In addition to these data she noted that the cobs contain only about 50% of the expected kernels. Note that in corn, each kernel is the result of one mating or can be considered one progeny. Hence 50% of the matings are not producing viable offspring. Draw a diagram showing the arrangement of the alleles on the parental chromosomes. Based on your hypothesis from part b, reassign genotypes to the progeny classes. Enter your new genotypes into the table above as an additional column. If the karyotype was done on the progeny what unusual feature would be observed?
In corn the genes for tassel length (alleles T and t) and rust resistance (alleles R and r) are known to be on separate chromosomes. In the course of a routine mating check experiment, a breeder got the following results upon crossing T/t:R/r with a double recessive strain (t/t;r/r): genotype Number of progeny T/t; R/r 98 t/t; r/r 104 T/t; r/r 3 t/t; R/r 5 In addition to these data she noted that the cobs contain only about 50% of the expected kernels. Note that in corn, each kernel is the result of one mating or can be considered one progeny. Hence 50% of the matings are not producing viable offspring. Based on the location of these genes on separate chromosomes what is the expected result? Suggest a hypothesis of a genetic process that could be used to explain the potential karyotype of one of these parents. What are the actual genotypes of the parents?
Cross Cross A Cross A Cross B Cross B Phenotype F1 generation F2 generation F1 generation F2 generation Male red eyes 132 150 0 99 Female red eyes 135 295 110 101 Male white eyes 0 147 105 93 Female white eyes 0 0 0 95 Using “+” to indicate the wildtype red-eyed allele and “w” to indicate the mutant white-eyed allele, state the genotypes of the following: Wildtype red-eyed and white-eyed parental flies from cross A and cross B. Males and females from the F1 generation flies from cross A and cross B Males and females, F2 generation flies from cross A and cross B.
Homozygous wild-type male mice (AA BB CC) were crossed with triplemutant female mice (aa bb cc), forming an F1 generation with the followinggenotype (Aa Bb Cc). The F1 males were crossed with triple mutantfemales, forming the following F2 phenotypes”“a B c” 3“A b C” 3“a b c” 8“A B c” 5“a b C” 5“A B C” 8“a B C” 6“A b c” 6 44 Determine the sequence of the genes
A cross between a wild-type fruit fly strain and various true-breeding mutant strains (all at a single locus) for bristle mutations resulted in the following phenotypes: Mutant Bristle Phenotype 1 0.20 shorter 2 0.30 longer 3 0.75 shorter 4 bent and damaged, non-functional 5 no bristles Using Muller's Morphs classification, Mutant 1 could be considered what type of "morph?" Question 6 options: Gain-of-function morph Antimorph Neomorph Amorph Hypomorph
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?
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. In 1932, H. J. Muller suggested a genetic test to determine whether a particular mutation whose phenotypiceffects are recessive to wild type is a null (amorphic)allele or is instead a hypomorphic allele of a gene.Muller’s test was to compare the phenotype of homozygotes for the recessive mutant alleles to the phenotype of a heterozygote in which one chromosomecarries the recessive mutation in question and thehomologous chromosome carries a deletion for aregion including the gene.In a study using Muller’s test, investigators examined two recessive, loss-of-function mutant alleles ofrugose named rg41 and rgγ3. The eye morphologiesdisplayed by flies of several genotypes are indicated inthe following table. Df(1)JC70 is a large deletion thatremoves rugose and several genes to either side of it.Genotype Eye surface Cone cells per ommatidiumwild type smooth 4rg41/rg41 mildly rough 2–3rg41/Df(1)JC70 moderately rough 1–2rg γ3/rg γ3 very rough 0–1rg γ3/Df(1)JC70 very rough 0–1a. Which allele…
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.
In sweet pea plant, an allele for purple flowers (P) is dominant when paired with a recessive allele for red flowers (p). An allele for long pollen grains (L) is dominant when paired with a recessive allele for round pollen grain (l). Bateson and Punnett crossed a plant having purple flowers/long pollen grains with one having white/flowers/round pollen grains. All F1 offspring had purple flowers and long pollen grains. Among the F2 generation, the researchers observed the following phenotypes: 296 purple flowers/long pollen grains 19 purple flowers/round pollen grains 27 red flowers/long pollen grains 85 red flowers/round pollen grains What is the best explanation for these results?
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Genes D, E and F are located on the same chromosome. From a three-point testcross mapping experiment, the following phenotypes were observed in the offspring: D E F 365 d E F 3 d e f 367 D e f 4 d E f 110 D E f 25 D e F 105 d e F 21 1. Which is the correct gene sequence? A. EFD B. DEF C. EDF D. FED 2. What is the distance between D and E in mu? 3. What is the distance between F and E in mu? 4. What is the distance between D and F in mu? 5. What is the coefficient of coincidence?