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Chapter 4: Problem 23

A husband and wife have normal vision, although both of their fathers are red- green color-blind, an inherited X-linked recessive condition. What is the probability that their first child will be (a) a normal son? (b) a normal daughter? (c) a color-blind son? (d) a color- blind daughter?

Short Answer

Expert verified
Answer: The probabilities for their first child are 25% for a normal son, 25% for a normal daughter, 25% for a color-blind son, and 0% for a color-blind daughter.

Step by step solution

01

Identify the genotypes of the parents

Since color-blindness is X-linked, it means that the gene is located on the X chromosome. The normal X chromosome (with the normal vision gene) will be denoted as X^N, and the X chromosome with the recessive color-blind gene will be denoted as X^n. Males have one X and one Y chromosome, while females have two X chromosomes. The husband has normal vision, so his genotype must be X^NY. His father is color-blind (X^nY) and his mother has a normal vision (so she carries at least one X^N). The son must have inherited his Y chromosome from his father and his X^N chromosome from his mother. Thus, the father's genotype is X^NY. The wife also has normal vision, but her father is colorblind (X^nY). Since it's a recessive trait, she necessarily inherited an X^n from her father (and therefore must be a carrier of the color-blind gene), but she must have inherited an X^N from her mother to display normal vision. Therefore, the wife's genotype is X^NX^n.
02

Determine the genotypes of the potential offspring

To determine the genotypes of the possible offspring, we can create a Punnett square. The husband can contribute either an X^N or a Y, while the wife can contribute either an X^N or an X^n. This results in the following Punnett square: X^N | X^n ------+------- X^N | X^N ------+------- X^NY | X^nY
03

Calculate the probabilities of each scenario

Now we can calculate the probabilities for each scenario: (a) Normal son (X^NY): There is only one combination in the Punnett square that leads to a normal son (X^NY), out of 4 possible combinations, so the probability is 1/4 or 25%. (b) Normal daughter (X^NX^N): There is only one combination in the Punnett square that leads to a normal daughter (X^NX^N), out of 4 possible combinations, so the probability is 1/4 or 25%. (c) Color-blind son (X^nY): There is only one combination in the Punnett square that leads to a color-blind son (X^nY), out of 4 possible combinations, so the probability is 1/4 or 25%. (d) Color-blind daughter: There are no combinations in the Punnett square that lead to a color-blind daughter, as it requires two recessive X^n chromosomes, and the husband only has an X^N. Therefore, the probability is 0/4 or 0%. In conclusion, the probabilities for their first child are 25% for a normal son, 25% for a normal daughter, 25% for a color-blind son, and 0% for a color-blind daughter.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

X-linked Recessive Inheritance
Understanding X-linked recessive inheritance is key to various genetic conditions, including color blindness. This pattern of inheritance applies to genes located on the X chromosome, one of the two sex chromosomes in humans. Since males (XY) have only one X chromosome, a single copy of a recessive gene on the X chromosome will cause the condition. Conversely, females (XX) must receive two copies of the recessive gene, one on each X chromosome, for the condition to be expressed.

For the males, if they inherit the affected X (Xn), they will exhibit the condition since their only other sex chromosome is a Y, which doesn't carry a corresponding allele to counteract the recessive trait. On the other hand, females can be carriers if they inherit one affected X chromosome (XnXN) but will not express the condition unless they inherit two affected X chromosomes (XnXn). These carriers can potentially pass the condition to their sons or the carrier status to their daughters.
Punnett Square
The Punnett square is a simple graphical way to predict the genotypes and phenotypes of offspring for particular traits based on the genetic makeup of the parents. It鈥檚 a table that sets possible sperm on one side and possible eggs on the other and then combines them to see which alleles from each parent could combine in an offspring.

In the context of an X-linked recessive trait like color-blindness, a Punnett square becomes a powerful tool in visualizing how the X and Y chromosomes from each parent can combine to produce male and female offspring with different genotypes for that trait. This aids in calculating the probability for a child to be affected by, or carry, the trait.
Phenotype and Genotype
The terms phenotype and genotype are vital in genetics. The genotype refers to the genetic constitution of an individual organism - essentially, its complete set of genes. On the other hand, the phenotype is the set of observable characteristics or traits of an organism, such as height, eye color, or in this case, the presence of color vision or color blindness.

It's important to distinguish that the phenotype is the expression of the genotype in conjunction with environmental factors. For instance, having the genotype XNXn for a female means she is a carrier of color blindness but does not exhibit the condition; her phenotype is normal color vision.
Color Blindness Genetics
Color blindness, particularly red-green color blindness, is often used as an example of an X-linked recessive trait in genetics education. The genes involved in this form of color blindness are located on the X chromosome, and for someone to be color-blind, they must not have a functioning gene for normal color vision on at least one X chromosome.

For males (XY), inheriting one affected X chromosome (XnY) means they will display color blindness because they lack a second X chromosome that could carry a normal vision allele. However, females (XX) are often only carriers of the condition unless they inherit two affected X chromosomes (XnXn), one from each parent, which is far less common. Thus, understanding the genetic mechanisms behind color blindness is essential in predicting inheritance patterns and the risk of passing on the condition.

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Most popular questions from this chapter

In rats, the following genotypes of two independently assorting autosomal genes determine coat color: A third gene pair on a separate autosome determines whether or not any color will be produced. The \(C C\) and \(C c\) genotypes allow color according to the expression of the \(A\) and \(B\) alleles. However, the \(c c\) genotype results in albino rats regardless of the \(A\) and \(B\) alleles present. Determine the \(F_{1}\) phenotypic ratio of the following crosses: (a) \(A A b b C C \quad \times \quad\) aaBBcc (b) \(A a B B C C \quad \times \quad A A B b c c\) (c) \(A a B b C c \quad \times \quad\) AaBbcc (d) \(A a B B C c \quad \times \quad A a B B C c\) (e) \(A A B b C c \quad \times \quad A A B b c c\)

In Drosophila, an X-linked recessive mutation, scalloped (sd), causes irregular wing margins. Diagram the \(F_{1}\) and \(\mathrm{F}_{2}\) results if (a) a scalloped female is crossed with a normal male; (b) a scalloped male is crossed with a normal female. Compare these results with those that would be obtained if the scalloped gene were autosomal.

In rabbits, a series of multiple alleles controls coat color in the following way: \(C\) is dominant to all other alleles and causes full color. The chinchilla phenotype is due to the \(c^{\mathrm{ch}}\) allele, which is dominant to all alleles other than \(C\). The \(c^{h}\) allele, dominant only to \(c^{a}\) (albino), results in the Himalayan coat color. Thus, the order of dominance is \(C>c^{\mathrm{dh}}>c^{h}>c^{a} .\) For each of the fol- lowing three cases, the phenotypes of the \(\mathrm{P}_{1}\) generations of two crosses are shown, as well as the phenotype of one member of the \(\mathrm{F}_{1}\) generation. For each case, determine the genotypes of the \(P_{1}\) generation and the \(\mathrm{F}_{1}\) offspring, and predict the results of making each indicated cross between \(F_{1}\) individuals.

While vermilion is X-linked in Drosophila and causes the eye color to be bright red, brown is an autosomal recessive mutation that causes the eye to be brown. Flies carrying both mutations lose all pigmentation and are white-eyed. Predict the \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) results of the following crosses: (a) vermilion females \(\times\) brown males (b) brown females \(\times\) vermilion males (c) white females \(\times\) wild-type males

Two mothers give birth to sons at the same time at a busy urban hospital. The son of mother 1 is afflicted with hemophilia, a disease caused by an X-linked recessive allele. Neither parent has the disease. Mother 2 has a normal son, despite the fact that the father has hemophilia. Several years later, couple 1 sues the hospital, claiming that these two newborns were swapped in the nursery following their birth. As a genetic counselor, you are called to testify. What information can you provide the jury concerning the allegation?
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