Introduction To Genetic Analysis
12th Edition
ISBN: 9781319114787
Author: Anthony J.F. Griffiths, John Doebley, Catherine Peichel, David A. Wassarman
Publisher: W. H. Freeman
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Chapter 3, Problem 48P
Summary Introduction
To explain: The chi-square or χ2test.
Introduction: The χ2 test plays an important role in experimental situations, where observed results are compared with those predicted by a hypothesis. It is used to decide whether or not an observed experimental deviation is reasonably compatible with a working hypothesis.
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A presumed dihybrid in Drosophila, BbFf is testcrossed with bbff. (B-black body, b-brown body, F-forked bristles, f-unforked bristles) The results for the testcross are below:
Black, forked: 230
Black, unforked: 210
Brown, forked: 240
Brown, unforked: 250
Determine the expected values and ratios for each genotype.
The data set attached summarizes F2 numbers from an F1 cross arising from two, true-breeding Drosophila strains (P generation), which differ with respect to two mutant traits.
Here are the hypothesis:
Leg length - The wild-type and mutant alleles for leg length are incomplete dominant relative to each other.
Justification: The data set includes three phenotypic categories for leg length: wild type (long leg), medium leg, and truncated wings. The presence of three distinct phenotypes suggests an incomplete dominance pattern, where the heterozygous individuals exhibit an intermediate leg length phenotype (medium leg). The absence of purebred short-legged individuals supports the idea that the long leg allele is dominant over the short leg allele. This shows that mode of inheritance is incomplete dominance of the alleles relative to each other. Since the data does not mention any specific differences between males and females, we can assume that the mode of inheritance for the trait is…
Female Drosophila with cinnabar eye (cn) and vestigial wings (vg) were mates to males with roof wings(fr). The F1 were all wild type ñ. When the F1 females were tested and crossed with males homozygous for all three traits the following result were obtained Give the phenotype of the offspring ( this time simply follow the given sequence above and separate the genes with single space):
Chapter 3 Solutions
Introduction To Genetic Analysis
Ch. 3 - Prob. 1PCh. 3 - Prob. 2PCh. 3 - Prob. 3PCh. 3 - Prob. 4PCh. 3 - Prob. 5PCh. 3 - Prob. 6PCh. 3 - Prob. 7PCh. 3 - Prob. 8PCh. 3 - Prob. 9PCh. 3 - Prob. 10P
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- Female Drosophila with cinnabar eye (cn) and vestigal wings (vg) were mated to males with roof wings (fr). The F1 were all wild type. When the F1 females were tested and crossed with male homozygous for all thrre traits, the following result were obtained. For this problem, rf, cn, vg is for roof wing / cinnabar eye / vestugal wing rf + cn + vg all wild type (no definite phenotype described). What is the value of coincidence and interference?arrow_forwardWhat is the genetic distance between the eye colour locus (w) and the bristle locus (sn)? answer in Mu (Map units) and to 2 decimal places? The following table summarises the results of the 2022-2023 Drosophila three-point cross involving the loci white eyes (w), miniature wings (m) and singed bristles (sn). The data is also available in the Excel file 'Drosophila Counts 2022-2023'. We strongly recommend working in Excel during this exercise.Phenotype Count+ + + 584w m sn 324w + + 227+ m sn 150+ m + 134w + sn 196 + + sn 134 w m + 92arrow_forwardConsider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL. A. What would be the physical phenotype of these flies? B. If PC was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PC products? C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker? D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?arrow_forward
- Consider the data in the following table, derived from crossing a female fruit fly that was heterozygous for the sex-linked recessive loci, cut wings (ct), raspberry eyes (ra), and forked (f) bristles, with a wild-type male. Female progeny were all wild-type. Male progeny are listed below. Phenotype Number of males ct ra f 70 + ra f 76 + + + 74 ct + f 356 + ra + 345 ct ra + 3 ct + + 72 + + f 4 Total 1,000 What is the map distance between cut wings and raspberry eyes? 29.2 15.1 30.6 15.5arrow_forwardThe allele b gives Drosophila flies a black body and b+ gives brown, the wild-type phenotype. The allele wx of a separate gene gives waxy wings and wx+ gives non-waxy, the wild-type phenotype. The allele cn of a third gene gives cinnabar eyes and cn+ gives red, the wild-type phenotype. A female heterozygous for these three genes is testcrossed, and 1000 progeny are classified with the following phenotypes. 382 cinnabar 379 black, waxy 69 waxy, cinnabar 67 black 48 waxy 44 black, cinnabar 5 wild type 6 black, waxy, cinnabar Based on this data, what is the correct map of these genes in terms of order and distance?arrow_forwardThe following table summarises the results of the 2022-2023 Drosophila three-point cross involving the loci white eyes (w), miniature wings (m) and singed bristles (sn). The data is also available in the Excel file 'Drosophila Counts 2022-2023'. We strongly recommend working in Excel during this exercise.Phenotype Count+ + + 584w m sn 324w + + 227+ m sn 150+ m + 134w + sn 196 + + sn 134 w m + 92The aim of today's tutorial is to use the above data to establish the genetic map for these three loci. We have been told that all three loci are on the X chromosome; however, as scientists, we shouldn't simply take someone else's word for it - we need to test whether our data does, in fact, indicate linkage.As a first step, we will therefore conduct a chi-square test in order to test our data against the null hypothesis of independent assortment (i.e., "no…arrow_forward
- Female Drosophila with cinnabar eye (cn) and vestigal wings (vg) were mated to males with roof wings (fr). The F1 were all wild type. When the F1 females were tested and crossed with male homozygous for all thrre traits, the following result were obtained. For this problem, rf, cn, vg is for roof wing / cinnabar eye / vestugal wing rf + cn + vg all wild typr(no definite phenotypr described.arrow_forwardFemale Drosophila with cinnabar eye (cn) and vestigal wings (vg) were mated to males with roof wings (fr). The F1 were all wild type. When the F1 females were tested and crossed with male homozygous for all thrre traits, the following result were obtained. For this problem, rf, cn, vg is for roof wing / cinnabar eye / vestugal wing rf + cn + vg all wild typr(no definite phenotypr described. Guve the phenotype and genotype of the male parent.arrow_forwardFemale Drosophila with cinnabar eye (cn) and vestigial wings (vg) were mated to males with roof wings (rf). The F1 were all wild-type. When the F1 females were test crossed with males homozygous for all three traits the following result were obtained. For this problem, rf cn vg is for roof wing / cinnabar eye / vestigial wing rf + cn + vg + all wild (no definite phenotype described) Give the genotype of the offspring (this time simply follow the given sequence above and separate the genes with single space) F2: Phenotype Frequency Genotype cinnabar, vestigial 382 roof 401 cinnabar 3 roof, vestigial 4 roof, cinnabar, vestigial 59 Wild 67 roof, cinnabar 44 vestigial 40 2. Give the phenotype and genotype of the female parent: phenotype: ____ genotype: _______ 3. Give the phenotype and genotype of the male parent: phenotype: _______ genotype: _______ 4. What…arrow_forward
- A complementation analysis was performed using Drosophila which normally have black wings. Six mutants for clear wings were discovered. The mutants were bred in a complementation test to produce the following: Mutant 2 4 # 1 1 1 1 2 1 1 1 1 1 1 4 1 1 Where a "0" (zero) means clear wing offspring and a "1" (one) means wildtype (black wing offspring). How many genes can you identify that affect the production of wing colour in Drosophila? O a. 3 O b.4 О с. 5 O d. 6 O e. none of the above 3. 5arrow_forwardDrosophila females heterozygous for three recessive mutations, a, b, and c , were crossed to males homozygous for all three mutations.The cross yielded the following results: in the image Q. Construct a linkage map showing the correct order of these genes and estimate the distances between them.arrow_forwardFrom a Drosophila testcross, the number of each phenotype obtained was as follows: W+ m f+ 218 W m+ f 236 W+ m+ f 168 W m f+ 178 W+ m f 95 W m+ f+ 101 W+ m+ f+ 3 W m f 1 Total, 1000 Construct a genetic map.arrow_forward
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