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Ch, 15 - The Chromosomal Basis of Inheritance
Chapter 15, Problem 1

A man with hemophilia (a recessive, sex-linked condition) has a daughter without the condition. She marries a man who does not have hemophilia. What is the probability that their daughter will have hemophilia? Their son? If they have four sons, what is the probability that all will be affected?

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Identify the genotypes of the parents: The father has hemophilia, which is a recessive, sex-linked condition, represented as XhY. The mother does not have hemophilia but her father does, so she is a carrier, represented as XHXh. The man she marries does not have hemophilia, so his genotype is XHY.
Determine the possible gametes: The mother can produce XH or Xh eggs, and the father can produce XH or Y sperm.
Set up a Punnett square to determine the genotypes of their children: The possible combinations are XHXH (normal daughter), XHXh (carrier daughter), XHY (normal son), and XhY (son with hemophilia).
Calculate the probability of each child having hemophilia: Daughters cannot have hemophilia as they would need two Xh alleles, and the father can only provide XH or Y. Therefore, the probability of a daughter having hemophilia is 0%. Sons have a 50% chance of having hemophilia, as they will have hemophilia if they inherit the Xh allele from their mother and the Y allele from their father.
Calculate the probability that all four sons will have hemophilia: Since each son has a 50% chance of having hemophilia, the probability that all four sons will have hemophilia is (1/2) * (1/2) * (1/2) * (1/2) = 1/16 or 6.25%.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hemophilia and Sex-Linked Inheritance

Hemophilia is a genetic disorder caused by a deficiency in clotting factors, leading to excessive bleeding. It is inherited in a recessive manner and is linked to the X chromosome. Males, having one X and one Y chromosome, are more likely to express the condition if they inherit the affected X. Females have two X chromosomes, so they can be carriers without showing symptoms if they have one normal X.
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Sex-Linked Inheritance

Punnett Square and Probability

A Punnett square is a tool used in genetics to predict the genotypes of offspring from parental genotypes. By mapping the possible alleles from each parent, one can calculate the probability of inheriting specific traits. In this case, it helps determine the likelihood of their children inheriting hemophilia based on the parents' genotypes.
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Punnett Square Probability

Independent Assortment and Probability of Multiple Offspring

Independent assortment refers to the way genes segregate independently during gamete formation. This principle allows for the calculation of probabilities for multiple offspring. For example, if the probability of a son having hemophilia is known, the probability that all four sons will be affected can be calculated by raising the individual probability to the power of four, reflecting the independence of each son's genetic outcome.
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Related Practice
Textbook Question

Pseudohypertrophic muscular dystrophy is an inherited disorder that causes gradual deterioration of the muscles. It is seen almost exclusively in boys born to apparently unaffected parents and usually results in death in the early teens. Is this disorder caused by a dominant or a recessive allele? Is its inheritance sex-linked or autosomal? How do you know? Explain why this disorder is almost never seen in girls.

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Textbook Question

A planet is inhabited by creatures that reproduce with the same hereditary patterns seen in humans. Three phenotypic characters are height (T=tall, t=dwart), head appendages (A=antennae, a=no antennae), and nose morphology (S=upturned snout, s=downturned snout). Since the creatures are not 'intelligent,' Earth scientists are able to do some controlled breeding experiments using various heterozygotes in testcrosses. For tall heterozygotes with antennae, the offspring are tall antennae, 46; dwarf antennae, 7; dwarf no antennae, 42; tall no antennae, 5. For heterozygotes with antennae and an upturned snout, the offspring are antennae upturned snout, 47; antennae downturned snout, 2; no antennae downturned snout, 48; no antennae upturned snout, 3. Calculate the recombination frequencies for both experiments.

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