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Chapter 4: Problem 12

In goats, development of the beard is due to a recessive gene. The following cross involving true-breeding goats was made and carried to the \(\mathrm{F}_{2}\) generation: \(P_{1}:\) bearded female \(\times\) beardless male \(\mathrm{F}_{1}:\) all bearded males and beardless females \\[ \mathrm{P}_{1} \times \mathrm{F}_{1} \longrightarrow\left\\{\begin{array}{l} 1 / 8 \text { beardless males } \\ 3 / 8 \text { bearded males } \\ 3 / 8 \text { beardless females } \\ 1 / 8 \text { bearded females }\end{array}\right.\\] Offer an explanation for the inheritance and expression of this trait, diagramming the cross. Propose one or more crosses to test your hypothesis.

Short Answer

Expert verified
To summarize, in this problem involving inheritance patterns of the beard gene in goats, we analyzed given information and used Punnett squares to predict genotypes and phenotypes in the F1 and F2 generations. Based on the results, the phenotypic ratio revealed that beard development in males is due to a recessive gene. To test this hypothesis, we proposed a cross between a beardless male (BB) and a bearded female (bb) in the F2 generation, which is expected to yield all bearded males and beardless females in a 1:1 ratio, thereby confirming the recessive nature of the gene responsible for beard development in goats.

Step by step solution

01

Analysis of Cross

Given information: - Beard development is due to a recessive gene - The cross: bearded female (bb) x beardless male (BB) - F1 generation: all bearded males (Bb) and beardless females (Bb) Here, we see that the bearded female goat represents the recessive homozygous condition (bb), and the beardless male goat represents the dominant homozygous condition (BB).
02

Punnett Square for P1 Cross

Create a Punnett square to analyze the P1 cross and predict the genotype and phenotype of F1 generation. | - | B | B | |:--:|:--:|:--:| |**b**| Bb | Bb | |**b**| Bb | Bb | All offspring in the F1 generation will have the heterozygous genotype (Bb), leading to all bearded males and beardless females.
03

Punnett Square for F1 Cross

Now, we will create a Punnett square to analyze the cross between F1 generation goats (Bb x Bb) to predict the genotype and phenotype of the F2 generation. | - | B | b | |:--:|:--:|:--:| |**B**| BB | Bb | |**b**| Bb | bb | The genotypic ratio for the F2 generation is: 1 BB : 2 Bb : 1 bb The phenotypic ratio for the F2 generation is: 3 bearded males & beardless females : 1 beardless male & bearded female This phenotypic ratio matches the one stated in the problem: 1/8 beardless males, 3/8 bearded males, 3/8 beardless females, and 1/8 bearded females.
04

Hypothesis Testing Cross

One possible cross to test the hypothesis is to cross two F2 generation goats with known genotypes. For example, cross a beardless male (BB) with a bearded female (bb). Prediction of the offspring genotypes and phenotypes: - All males offspring will develop a beard (Bb). - All female offspring will be beardless (Bb). This cross will indeed confirm that beard development is due to a recessive gene, and that its expression depends on the gender of the offspring. Bearded males and beardless females will be produced in a 1:1 ratio.

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Most popular questions from this chapter

With regard to the ABO blood types in humans, determine the genotypes of the male parent and female parent: Male parent: blood type B whose mother was type O Female parent: blood type A whose father was type B Predict the blood types of the offspring that this couple may have and the expected ratio of each.

Two mothers give birth to sons at the same time at a busy urban hospital. The son of mother 1 has hemophilia, a disease caused by an X-linked recessive allele. Neither parent has the disease. Mother 2 has a son without hemophilia, 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?

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

In cattle, coats may be solid white, solid black, or black-and-white spotted. When true-breeding solid whites are mated with truebreeding solid blacks, the \(\mathrm{F}_{1}\) generation consists of all solid white individuals. After many \(\mathrm{F}_{1} \times \mathrm{F}_{1}\) matings, the following ratio was observed in the \(\mathrm{F}_{2}\) generation: \(12 / 16\) solid white \(3 / 16\) black-and-white spotted \(1 / 16\) solid black Explain the mode of inheritance governing coat color by determining how many gene pairs are involved and which genotypes yield which phenotypes. Is it possible to isolate a true-breeding strain of black-and-white spotted cattle? If so, what genotype would they have? If not, explain why not.

Horses can be cremello (a light cream color), chestnut (a reddish brown color), or palomino (a golden color with white in the horse's tail and mane). Of these phenotypes, only palominos never breed true. The following results have been observed: cremello \(\times\) palomino \(\longrightarrow 1 / 2\) cremello \(1 / 2\) palomino chestnut \(\times\) palomino \(\longrightarrow 1 / 2\) chestnut \(1 / 2\) palomino palomino \(\times\) palomino \(\longrightarrow 1 / 4\) chestnut \(1 / 2\) palomino \(1 / 4\) cremello (a) From these results, determine the mode of inheritance by assigning sene 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.
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