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(#8) According to genetics theory, red-green colorblindness in humans is a recessive sex- linked characteristic. In this case, the gene is carried on the X chromosome only. We will denote an X chromosome with the colorblindness gene by X, and one without the gene by X. Women have two X chromosomes, and they will be red-green colorblind only if both chromosomes have the gene, designated XX. A woman can have normal vision but still carry the colorblind gene if only one of the chromosomes has the gene, denoted by XX. A man carries an X and a Y chromosome; if the X chromosome carries the colorblind gene (XY), the man is colorblind. According to genetics theory, if a man with normal vision (X,Y) and a woman carrier (XX.) have a child, the probabilities that the child will have red-green colorblindness, will have normal vision and not carry the gene, or will have normal vision and carry the gene are given by the equally likely events in table 1. Mother Xc X. TABLE 1 Red-Green Colorblindness Father Red-green colorblind Normal vision, noncarrier Normal vision, carrier _X.. X. X. Χ.Χ. Y XX XX P(child has normal vision and is not a carrier) = P(X,Y) + P(X, X, ) — — P(child has normal vision and is a carrier) = P(X, X.) = P(child is red-green colorblind) = P(X,Y)= To test this genetics theory, Genetics Labs took a random sample of 200 children whose mothers were carriers of the colorblind gene and whose fathers had normal vision. The results are shown in table 2. Test the hypothesis that the population follows the distribution predicted by the genetics theory (see table 1). Use α = .01. TABLE 2 35 105 60

(#8) According to genetics theory, red-green colorblindness in humans is a recessive sex- linked characteristic. In this case, the gene is carried on the X chromosome only. We will denote an X chromosome with the colorblindness gene by X, and one without the gene by X. Women have two X chromosomes, and they will be red-green colorblind only if both chromosomes have the gene, designated XX. A woman can have normal vision but still carry the colorblind gene if only one of the chromosomes has the gene, denoted by XX. A man carries an X and a Y chromosome; if the X chromosome carries the colorblind gene (XY), the man is colorblind. According to genetics theory, if a man with normal vision (X,Y) and a woman carrier (XX.) have a child, the probabilities that the child will have red-green colorblindness, will have normal vision and not carry the gene, or will have normal vision and carry the gene are given by the equally likely events in table 1. Mother Xc X. TABLE 1 Red-Green Colorblindness Father Red-green colorblind Normal vision, noncarrier Normal vision, carrier _X.. X. X. Χ.Χ. Y XX XX P(child has normal vision and is not a carrier) = P(X,Y) + P(X, X, ) — — P(child has normal vision and is a carrier) = P(X, X.) = P(child is red-green colorblind) = P(X,Y)= To test this genetics theory, Genetics Labs took a random sample of 200 children whose mothers were carriers of the colorblind gene and whose fathers had normal vision. The results are shown in table 2. Test the hypothesis that the population follows the distribution predicted by the genetics theory (see table 1). Use α = .01. TABLE 2 35 105 60

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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(#8) According to genetics theory, red-green colorblindness in humans is a recessive sex- linked characteristic. In this case, the gene is carried on the X chromosome only. We will denote an X chromosome with the colorblindness gene by X and one without the gene by X. Women have two X chromosomes, and they will be red-green colorblind only if both chromosomes have the gene, designated XX. A woman can have normal vision but still carry the colorblind gene if only one of the chromosomes has the gene, denoted by XX. A man carries an X and a Y chromosome; if the X chromosome carries the colorblind gene (XY), the man is colorblind. According to genetics theory, if a man with normal vision (XY) and a woman carrier (XX) have a child, the probabilities that the child will have red-green colorblindness, will have normal vision and not carry the gene, or will have normal vision and carry the gene are given by the equally likely events in table 1. TABLE 1 Red-Green Colorblindness Mother Xc X _X. Father Red-green colorblind Normal vision, noncarrier Normal vision, carrier X.X. Χ.Χ. Y XY XY P(child has normal vision and is not a carrier) = P(X_Y) + P(X_X) = x.)=1/12/2 P(child has normal vision and is a carrier) = P(X__X_) P(child is red-green colorblind) = P(X,Y)= To test this genetics theory, Genetics Labs took a random sample of 200 children whose mothers were carriers of the colorblind gene and whose fathers had normal vision. The results are shown in table 2. Test the hypothesis that the population follows the distribution predicted by the genetics theory (see table 1). Use α = .01. TABLE 2 = 35 105 60
Transcribed Image Text:(#8) According to genetics theory, red-green colorblindness in humans is a recessive sex- linked characteristic. In this case, the gene is carried on the X chromosome only. We will denote an X chromosome with the colorblindness gene by X and one without the gene by X. Women have two X chromosomes, and they will be red-green colorblind only if both chromosomes have the gene, designated XX. A woman can have normal vision but still carry the colorblind gene if only one of the chromosomes has the gene, denoted by XX. A man carries an X and a Y chromosome; if the X chromosome carries the colorblind gene (XY), the man is colorblind. According to genetics theory, if a man with normal vision (XY) and a woman carrier (XX) have a child, the probabilities that the child will have red-green colorblindness, will have normal vision and not carry the gene, or will have normal vision and carry the gene are given by the equally likely events in table 1. TABLE 1 Red-Green Colorblindness Mother Xc X _X. Father Red-green colorblind Normal vision, noncarrier Normal vision, carrier X.X. Χ.Χ. Y XY XY P(child has normal vision and is not a carrier) = P(X_Y) + P(X_X) = x.)=1/12/2 P(child has normal vision and is a carrier) = P(X__X_) P(child is red-green colorblind) = P(X,Y)= To test this genetics theory, Genetics Labs took a random sample of 200 children whose mothers were carriers of the colorblind gene and whose fathers had normal vision. The results are shown in table 2. Test the hypothesis that the population follows the distribution predicted by the genetics theory (see table 1). Use α = .01. TABLE 2 = 35 105 60
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