In humans, the ABO blood type is under the control of autosomal multiple alleles. Color blindness is a recessive X-linked trait. If two parents who are both type A and have normal vision produce a son who is color-blind and is type 0 , what is the probability that their next child will be a female who has normal vision and is type \(0 ?\)

Short Answer

Expert verified
Answer: The probability is \(\frac{1}{16}\).

Step by step solution

01

Determine the genotypes of the parents

Because both parents are type A and have a child with type 0 blood, they most likely carry the O allele. The genotypes of the parents are both A/O. As for color blindness, which is an X-linked trait, both have normal vision, so the mother must carry the recessive gene for color blindness while the father has normal vision genes. The genotypes are Xm/Xn (mother), and X/Y (father).
02

Find the probability of the next child being female

The probability that their next child is female is 1/2, as the father passes on the X chromosome with probability 1/2 and the Y chromosome with probability 1/2.
03

Find the probability of the next child having normal vision

To have normal vision, the next female child must receive one X-linked gene for normal vision. Since the mother is Xm/Xn and father is X/Y, there is a 1/2 chance that the mother will pass on the Xn gene for normal vision to the child.
04

Find the probability of the next child having type 0 blood

To have type 0 blood, the next child must inherit the O allele from both parents. Since both parents are A/O, the probability that they both pass on the O allele to their child is (1/2)*(1/2)=1/4.
05

Calculate the overall probability

The probability that the next child is a female with normal vision and type 0 blood is the product of the probabilities from Steps 2, 3, and 4: (1/2)*(1/2)*(1/4) = 1/16. The probability that the next child will be a female with normal vision and type 0 blood is \(\frac{1}{16}\).

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Key Concepts

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

Autosomal Multiple Alleles
When we talk about autosomal multiple alleles, we are referring to genes that have more than two allele forms within a population. A classic example of this in humans is the ABO blood type system, which is determined by three alleles (A, B, O) that create four possible blood types (A, B, AB, O). It's crucial for students to understand that while individuals inherit only two alleles (one from each parent), multiple alleles can exist within the larger population.

In our exercise, the blood type A from both parents actually includes two alleles: A and O. The O blood type allele is recessive, so someone with an AO genotype will exhibit the A blood type. The child having type O blood means both parents contributed an O allele, revealing their genotype as AO (or what we noted as A/O).
X-linked Recessive Traits
Exploring X-linked recessive traits provides insight into how certain traits are passed on and why they might appear more frequently in one sex. Genes for these traits are located on the X chromosome. Males have one X and one Y chromosome, so they express X-linked traits if they inherit a recessive allele, since they lack a second X chromosome to potentially provide a dominant allele. Females, with two X chromosomes, must inherit two recessive alleles to express the trait.

In the given problem, color blindness is an X-linked recessive trait. The son is color-blind, which means he inherited a recessive allele from his carrier mother and only possesses one X chromosome. Knowing the probabilities of a female child receiving a particular X chromosome from each parent informs us about the likelihood of her being color-blind or not.
Genotype Determination
The process of genotype determination involves understanding the genetic makeup of an organism in terms of its alleles. In our exercise, we deduce the genotype of the parents based on their blood types and the fact that they produced an offspring with type O blood. For color blindness, knowing the mother must be a carrier (heterozygous for the trait) and the father is not color-blind allows us to determine their genotypes concerning the color vision gene.

This method of using known phenotypes and patterns of inheritance to deduce genotypes is a fundamental genetic skill. For X-linked traits, a phenotype in a male directly indicates his genotype because he has only one allele for the trait. It's important for students to follow these logic pathways when determining genotypes from given information.
Punnett Square
The Punnett square is an essential tool in genetics for predicting the outcomes of crosses. It visually demonstrates how alleles from each parent can combine in their offspring. By arranging possible gametes from one parent along the top and the other parent along the side, students can see all potential genotypes of their offspring.

In our exercise, a Punnett square helps to visualize the 1 in 2 probability of the child being female (XX or XY) and the 1 in 4 chance of the child having type O blood (OO from AO x AO). For the X-linked color blindness trait, it’s particularly helpful because it illustrates how a normal vision allele from the father and the potential of a normal or color-blind allele from the mother combine to give a 50% chance of a daughter not being color-blind. Overall, this method simplifies the calculation of combined probabilities to determine the likelihood of specific genetic outcomes.

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Most popular questions from this chapter

Three gene pairs located on separate autosomes determine flower color and shape as well as plant height. The first pair exhibits incomplete dominance, where the color can be red, pink (the heterozygote), or white. The second pair leads to personate (dominant) or peloric (recessive) flower shape, while the third gene pair produces either the dominant tall trait or the recessive dwarf trait. Homozygous plants that are red, personate, and tall are crossed to those that are white, peloric, and dwarf. Determine the \(F_{1}\) genotype(s) and phenotype(s). If the \(\mathrm{F}_{1}\) plants are interbred, what proportion of the offspring will exhibit the same phenotype as the \(F_{1}\) plants?

Discuss the topic of phenotypic expression and the many factors that impinge on it.

Horses can be cremello (a light cream color), chestnut (a brownish color), or palomino (a golden color with white in the horse's tail and mane). Of these phenotypes, only palominos never breed true. \(\begin{array}{ll}\text { cremello } \times \text { palomino } & \longrightarrow \begin{array}{l}1 / 2 \text { cremello } \\ 1 / 2 \text { palomino }\end{array} \\ \text { chestnut } \times \text { palomino } \longrightarrow & \begin{array}{l}1 / 2 \text { chestnut } \\ 1 / 2 \text { palomino }\end{array} \\ \text { palomino } \times \text { palomino } \longrightarrow & \begin{array}{l}1 / 4 \text { chestnut } \\ 1 / 2 \text { palomino }\end{array} \\ & 1 / 4 \text { cremello }\end{array}\) (a) From the results given above, determine the mode of inheritance by assigning gene symbols and indicating which genotypes yield which phenotypes. (b) Predict the \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) results of many initial matings between cremello and chestnut horses.

Contrast penetrance and expressivity as the terms relate to phenotypic expression.

In Drosophila , the X-linked recessive mutation vermilion (v) causes bright red eyes, in contrast to the brick-red eyes of wild type. A separate autosomal recessive mutation, suppressor of vermilion \((s u-v),\) causes flies homozygous or hemizygous for \(v\) to have wildtype eyes. In the absence of vermilion alleles, su-v has no effect on eye color. Determine the \(F_{1}\) and \(F_{2}\) phenotypic ratios from a cross between a female with wild-type alleles at the vermilion locus, but who is homozygous for \(s u\) -v, with a vermilion male who has wildtype alleles at the su-v locus.

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