INTRO TO GEN ANALYSIS W/ACHIEVE ACCESS
12th Edition
ISBN: 9781319423865
Author: Griffiths
Publisher: MAC HIGHER
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Chapter 2, Problem 58P
Summary Introduction
To determine: The probability that a man and a woman who are heterozygous for a recessive allele for albinism will possess the dizygotic children (twins) having the same
Introduction: The congenital disorder of humans, which is known as albinism, is characterized by the absence of pigmentation. The deficiency is caused by the defect in melanin or reduced production of melanin. Thus, the defected individuals will not have pigmented hair, skin, and eyes.
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Chapter 2 Solutions
INTRO TO GEN ANALYSIS W/ACHIEVE ACCESS
Ch. 2 - Prob. 1PCh. 2 - Prob. 2PCh. 2 - Prob. 3PCh. 2 - Prob. 4PCh. 2 - Prob. 5PCh. 2 - Prob. 6PCh. 2 - Prob. 7PCh. 2 - Prob. 8PCh. 2 - Prob. 9PCh. 2 - Prob. 10P
Ch. 2 - Prob. 11PCh. 2 - Prob. 12PCh. 2 - Prob. 13PCh. 2 - Prob. 14PCh. 2 - Prob. 15PCh. 2 - Prob. 17PCh. 2 - Prob. 18PCh. 2 - Prob. 19PCh. 2 - Prob. 20PCh. 2 - Prob. 21PCh. 2 - Prob. 22PCh. 2 - Prob. 23PCh. 2 - Prob. 24PCh. 2 - Prob. 25PCh. 2 - Prob. 26PCh. 2 - Prob. 27PCh. 2 - Prob. 28PCh. 2 - Prob. 29PCh. 2 - Prob. 30PCh. 2 - Prob. 31PCh. 2 - Prob. 32PCh. 2 - Prob. 33PCh. 2 - Prob. 34PCh. 2 - Prob. 35PCh. 2 - Prob. 36PCh. 2 - Prob. 37PCh. 2 - Prob. 38PCh. 2 - Prob. 39PCh. 2 - Prob. 40PCh. 2 - Prob. 41PCh. 2 - Prob. 42PCh. 2 - Prob. 43PCh. 2 - Prob. 44PCh. 2 - Prob. 45PCh. 2 - Prob. 46PCh. 2 - Prob. 47PCh. 2 - Prob. 48PCh. 2 - Prob. 49PCh. 2 - Prob. 50PCh. 2 - Prob. 51PCh. 2 - Prob. 52PCh. 2 - Prob. 53PCh. 2 - Prob. 54PCh. 2 - Prob. 55PCh. 2 - Prob. 56PCh. 2 - Prob. 56.1PCh. 2 - Prob. 56.2PCh. 2 - Prob. 56.3PCh. 2 - Prob. 56.4PCh. 2 - Prob. 56.5PCh. 2 - Prob. 56.6PCh. 2 - Prob. 56.7PCh. 2 - Prob. 56.8PCh. 2 - Prob. 56.9PCh. 2 - Prob. 56.10PCh. 2 - Prob. 56.11PCh. 2 - Prob. 56.12PCh. 2 - Prob. 56.13PCh. 2 - Prob. 56.14PCh. 2 - Prob. 56.15PCh. 2 - Prob. 57PCh. 2 - Prob. 58PCh. 2 - Prob. 59PCh. 2 - Prob. 60PCh. 2 - Prob. 61PCh. 2 - Prob. 62PCh. 2 - Prob. 63PCh. 2 - Prob. 64PCh. 2 - Prob. 65PCh. 2 - Prob. 67PCh. 2 - Prob. 68PCh. 2 - Prob. 69PCh. 2 - Prob. 70PCh. 2 - Prob. 71PCh. 2 - Prob. 72PCh. 2 - Prob. 73PCh. 2 - Prob. 74PCh. 2 - Prob. 75PCh. 2 - Prob. 76PCh. 2 - Prob. 77PCh. 2 - Prob. 78PCh. 2 - Prob. 79PCh. 2 - Prob. 80PCh. 2 - Prob. 81PCh. 2 - Prob. 82PCh. 2 - Prob. 83PCh. 2 - Prob. 84PCh. 2 - Prob. 85PCh. 2 - Prob. 86PCh. 2 - Prob. 87PCh. 2 - Prob. 88PCh. 2 - Prob. 89PCh. 2 - Prob. 90PCh. 2 - Prob. 91PCh. 2 - Prob. 1GSCh. 2 - Prob. 2GSCh. 2 - Prob. 3GS
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- A woman homozygous dominant for albinism marries a man who is homozygous recessive for albinism. What are the possible genotypes and phenotypes percentages for their offspring?arrow_forwardFor a recessive condition, two normal heterozygous individuals have children. What is the likelihood of their children being affected by this condition? What is the likelihood of their children being carriers without the condition? What is the likelihood of their asymptomatic children being carriers? Suppose that an individual with the condition has children with a heterozygous individual, what is the likelihood of their children being carriers?arrow_forwardPls answer the very last question. “ suppose that lll-2 and lll-5 were to have a child. What is the probability that their first child will be have the disorder?”arrow_forward
- A woman homozygous dominant for albinism marries a man who is homozygous recessive for albinism. What are the possible genotypes and phenotypes for their offspring? What are the ratios?arrow_forwardRed-green color blindness is inherited as an X-linked recessive (Xc). If a color-blind man marries a woman who is heterozygous for normal vision, what would be the expected phenotypes of their children with reference to this character? In your answer, specify in your phenotype descriptions the gender of the children. (For example, don’t just say 75% of the children would be colorblind – you would instead say 100 % of the daughters would be colorblind and 50% of the sons would be colorblind. Note that this is not a correct answer; it is just to give you an idea of how to explain the correct phenotypes of the cross.)___arrow_forwardin humans brown eyes (B) are dominant over blue (b).A brown eye man marries a blue-eyed woman and they have three children , two of whom are brown eyed and one of whom is blue eyed.Draw the punnett square that illustrate this marrage.what is the man's genotype ? what are the genotype of the children?arrow_forward
- In addition to the allelic pair determining pattern baldness in man (B,b), consider early baldness to be due to another autosomal allele (E) on a different pair of chromosomes and also dominant in males but recessive in females. The phenotype for ee may be late or nonbaldness depending on sex and the genotype for B, b alleles. Two doubly heterozygous persons marry. What is the phenotype of the male parent? What is the phenotype of the female parent? Give the phenotypic ratio expected among male children of couples such as this one. Show corresponding genotypes for each phenotype mentioned in your phenotypic ratio. Give the phenotypic ratio expected among female children of couples such as this one. Show corresponding genotypes for each phenotype mentioned in your phenotypic ratio.arrow_forwardIn humans, the genes for colorblindness and hemophilia are both located on the X chromosome with no corresponding gene in the Y. These are both recessive alleles. If a man and a woman, both with normal vision, marry and have a colorblind son, draw the Punnet square that illustrates this. If the man dies and the woman remarries a colorblind man, draw a Punnet Square showing the type of children that could be expected from the second marriage. How many/what percentages of each could be expected.arrow_forwardThe most common form of colorblindness is a recessive, sex-linked hereditary con dition caused by a defect on the X chromosome. Females are XX, while males are XY. Individuals inherit one chromosome from each parent, with equal probability; for example, an individual has a 50% chance of inheriting their father's X chromosome, and a 50% chance of inheriting their father's Y chromosome. If a male has an X chromosome with the defect, he is colorblind. However, a female with only one defective X chromo some will not be colorblind. Thus, colorblindness is more common in males than females; 7% of males are colorblind but only 0.5% of females are colorblind. (a) Assume that the X chromosome with the wild-type allele is X+ and the one with the disease allele is X. What is the expected frequency of each possible female genotype: X+X+, X+X¯, and X-X-? What is the expected frequency of each possible male genotype: X+ Y and X-Y? (b) Suppose that two parents are not colorblind. What is the…arrow_forward
- Suppose a woman has four sons, and two are colorblind but have normal blood clotting and two have hemophilia but normal color vision. What is the probable genotype of the woman?arrow_forwardCystic fibrosis is a recessive human condition. A male with Cystic fibrosis and a woman with a dominant phenotype have sevral children, in which one displays Cystic fibrosis. What can you conclude about the genotype of the maternal parent and what is the probability that a child who does not display Cystic fibrosis is heterozygous?arrow_forwardColour-blindness is the result of an X-linked recessive allele, Xc. The allele for normal eyesight is XC. (a) A woman with normal colour vision whose father was colour-blind marries a colour-blind man. Give the genotypes and phenotypes of their children. What ratio of their children can be expected to be colour-blind? (b) A man with normal colour vision whose father was colour-blind marries a woman carrier of the colour-blind allele. What is the likelihood that their children will be colour-blind? Carriers of the trait?arrow_forward
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