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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 each specific characteristic are as follows: (a) 25% chance of having a normal son, (b) 25% chance of having a normal daughter, (c) 25% chance of having a color-blind son, (d) 0% chance of having a color-blind daughter.

Step by step solution

01

Determine the genotypes of the parents

Since red-green color blindness is X-linked, we will label the normal X chromosome as XN and the color-blind X chromosome as Xc. The father has normal vision, so he must have one normal X chromosome (XN) from his mother and a Y chromosome. The mother also has normal vision, but her father is color-blind, so she must have one normal X chromosome (XN) and one color-blind X chromosome (Xc). Thus, the genotypes are Father: XN Y and Mother: XN Xc.
02

Create a Punnett square

Set up a Punnett square to find the combinations of X and Y chromosomes that can result from the parents' genotypes: | XN Mother | Xc Mother -------|-----------|---------- XN Dad | XN XN | XN Xc Y Dad| XN Y | Xc Y
03

Identify and calculate probabilities for each child's characteristics

Now, using the Punnett square, we will identify the probabilities of each child's characteristics: (a) Normal son: XN Y - This combination is present in 1 of the 4 cells, so the probability is 1/4 or 25%. (b) Normal daughter: XN XN - This combination is present in 1 of the 4 cells, so the probability is 1/4 or 25%. (c) Color-blind son: Xc Y - This combination is present in 1 of the 4 cells, so the probability is 1/4 or 25%. (d) Color-blind daughter: There is no Xc Xc combination in the Punnett square, thus the probability is 0/4 or 0%. In conclusion, the probabilities for the couple's first child are: (a) 25% chance of having a normal son, (b) 25% chance of having a normal daughter, (c) 25% chance of having a color-blind son, (d) 0% chance of having a color-blind daughter.

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Key Concepts

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

Punnett Square
The Punnett square is a simple illustrative tool used in genetics to predict the possible genotypes of offspring from a particular cross. It's a grid that allows us to combine the genetic contributions of each parent in a visual format. For X-linked recessive conditions like red-green color blindness, where the gene is located on the X chromosome, the Punnett square helps us understand how the condition is inherited.

Parents' chromosomes are placed on the top and side of the square, and the potential combinations they can produce are shown in the grids. Since males possess one X and one Y chromosome (XY), and females have two X chromosomes (XX), the Punnett square accounts for each possible combination ensuring a clear, visual representation of genetic transmission.
Genotype
Genotype refers to the specific genetic makeup of an individual, in other words, the set of genes that the individual carries. For X-linked traits, the genotype typically involves the alleles on the X chromosome. In the case of red-green color blindness, which is a recessive condition, the genotype will manifest as the condition only if two recessive alleles are present in females (XX) or one in males (XY) because males lack a second X chromosome.

Understanding the parental genotypes is crucial in genetics problems as it helps us predict the possible genotypes (and therefore phenotypes) of their offspring. The father's genotype is XY and the mother's is XX, but since she is a carrier for color blindness, it is denoted as XN (normal) Xc (carrying color-blind allele).
Red-Green Color Blindness
Red-green color blindness is an X-linked recessive condition meaning it's carried on the X chromosome and a male would only need one affected copy of the gene to express the condition since they have only one X chromosome (XY). Females, having two X chromosomes (XX), would need two affected copies of the gene to exhibit red-green color blindness.

Due to this mode of inheritance, it is much more common in males than females. Being recessive, a female can be a carrier of the condition without expressing it, which is pivotal in understanding the inheritance pattern. A carrier female can pass on the trait to her offspring, potentially affecting her male progeny with higher probability.
Probability in Genetics
Probability in genetics refers to the likelihood of a particular genetic outcome occurring. It is grounded in the Law of Segregation and Independent Assortment, which are fundamental principles of inheritance. The Punnett square informs us about the probability of each genotype arising from a cross.

For example, by calculating the frequencies of particular genotypes or phenotypes appearing in the Punnett square grid, we can articulate the probability of an offspring with either characteristic. In the case of red-green color blindness, probability helps quantify the chances of children inheriting the condition from carrier parents.

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Most popular questions from this chapter

List all possible genotypes for the \(\mathrm{A}, \mathrm{B}, \mathrm{AB},\) and \(\mathrm{O}\) phenotypes. Is the mode of inheritance of the \(\mathrm{ABO}\) blood types representative of dominance, recessiveness, or codominance?

Pigment in mouse fur is only produced when the \(C\) allele is present. Individuals of the \(c c\) genotype are white. If color is present, it may be determined by the \(A, a\) alleles. \(A A\) or \(A a\) results in agouti color, while a results in black coats. (a) What \(F_{1}\) and \(F_{2}\) genotypic and phenotypic ratios are obtained from a cross between \(A A C C\) and aacc mice? (b) In three crosses between agouti females whose genotypes were unknown and males of the aacc genotype, the following phenotypic ratios were obtained:

Students taking a genetics exam were expected to answer the following question by converting data to a "meaningful ratio" and then solving the problem. The instructor assumed that the final ratio would reflect two gene pairs, and most correct answers did. Here is the exam question: "Flowers may be white, orange, or brown. When plants with white flowers are crossed with plants with brown flowers, all the \(F_{1}\) flowers are white. For \(F_{2}\) flowers, the following data were obtained: Convert the \(F_{2}\) data to a meaningful ratio that allows you to explain the inheritance of color. Determine the number of genes involved and the genotypes that yield each phenotype." (a) Solve the problem for two gene pairs. What is the final \(\mathrm{F}_{2}\) ratio? (b) A number of students failed to reduce the ratio for two gene pairs as described above and solved the problem using three gene pairs. When examined carefully, their solution was deemed a valid response by the instructor. Solve the problem using three gene pairs. (c) We now have a dilemma. The data are consistent with two alternative mechanisms of inheritance. Propose an experiment that executes crosses involving the original parents that would distinguish between the two solutions proposed by the students. Explain how this experiment would resolve the dilemma.

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 Cc 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) \(A a B B C c \quad \times \quad\) AaBBCc (e) \(A A B b C c \quad \times \quad\) AABbcc

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 \(C>c^{\kappa h}>c^{h}>c^{a} \cdot\) For each of the following three cases, the phenotypes of the \(P_{1}\) generations of two crosses are shown, as well as the phenotype of one member of the \(F_{1}\) generation.

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