Chi-squared analysis for linkage between eyeless and wingless mutations (testing the null hypothesis that the mutations are unlinked) using the data in the previous question gives a chi squared value of 3.598. Use the Chi squared table included in the "equations and tables for population genetics" in the Week 4 module to answer this question: which of the following correctly reports the p- value and appropriate conclusion to draw?
Chi-squared analysis for linkage between eyeless and wingless mutations (testing the null hypothesis that the mutations are unlinked) using the data in the previous question gives a chi squared value of 3.598. Use the Chi squared table included in the "equations and tables for population genetics" in the Week 4 module to answer this question: which of the following correctly reports the p- value and appropriate conclusion to draw?
Human Anatomy & Physiology (11th Edition)
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Genetic Variation
Genetic variation refers to the variation in the genome sequences between individual organisms of a species. Individual differences or population differences can both be referred to as genetic variations. It is primarily caused by mutation, but other factors such as genetic drift and sexual reproduction also play a major role.
Quantitative Genetics
Quantitative genetics is the part of genetics that deals with the continuous trait, where the expression of various genes influences the phenotypes. Thus genes are expressed together to produce a trait with continuous variability. This is unlike the classical traits or qualitative traits, where each trait is controlled by the expression of a single or very few genes to produce a discontinuous variation.
Question
15,16
Transcribed Image Text:Chi-squared analysis for linkage between eyeless and wingless mutations
(testing the null hypothesis that the mutations are unlinked) using the data in the
previous question gives a chi squared value of 3.598. Use the Chi squared table
included in the "equations and tables for population genetics" in the Week 4
module to answer this question: which of the following correctly reports the p-
value and appropriate conclusion to draw?
<0.05 (5%); cannot reject null hypothesis, so conclude that genes are probably unlinked
between 0.3 (30%) and 0.5 (50%); cannot reject null hypothesis, so conclude that genes
are probably unlinked
between 0.3 (30%) and 0.05 (5%); cannot reject null hypothesis, so conclude that genes
are probably unlinked
<0.05 (5%); reject null hypothesis and conclude that genes are linked
between 0.3 (30%) and 0.05 (5%); reject null hypothesis and conclude that genes are
linked
between 0.3 (30%) and 0.5 (50%); reject null hypothesis and conclude that genes are
linked
Transcribed Image Text:A wild-type fruit fly is crossed to a fly that is homozygous for recessive
mutations in two different genes resulting in wingless and eyeless phenotypes.
The F1 generation is 100% phenotypically wild type. F1 flies are mated together
to produce the following F2 generation:
wild type: 367
wingless (normal eyes): 119
eyeless (normal wings): 134
eyeless and wingless: 52
In a chi squared test for linkage between the eyeless and wingless mutations,
how many wingless F2 offspring with normal eyes are expected according to the
null hypothesis?
126
168
134
119
Expert Solution
Step 1: Introduction.
Q 15: answer :- To determine the appropriate conclusion based on the chi-squared value and p-value, we should compare the calculated chi-squared value (3.598) with a critical chi-squared value at a specific significance level (alpha). In this case, alpha is typically set at 0.05 (5%).
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