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Chapter 3: Problem 35

Assume that Migaloo's albinism is caused by a rare recessive gene. (a) In a mating of two heterozygous, normally pigmented whales, what is the probability that the first three offspring will all have normal pigmentation? (b) What is the probability that the first female offspring is normally pigmented? (c) What is the probability that the first offspring is a normally pigmented female?

Short Answer

Expert verified
a) Three normally pigmented offspring in a row. b) The first female offspring being normally pigmented. c) The first offspring being a normally pigmented female. Answer: a) The probability of three normally pigmented offspring in a row is 27/64. b) The probability of the first female offspring being normally pigmented is 3/8. c) The probability of the first offspring being a normally pigmented female is 3/8.

Step by step solution

01

(a) Probability of three normally pigmented offspring

To calculate this probability, we need to first understand the probability of obtaining a single normally pigmented offspring. Since both parents are heterozygous (one dominant gene and one recessive gene), we can apply the rules of the Punnett square: Heterozygous Parent 1 (Aa) x Heterozygous Parent 2 (Aa): AA Aa Aa aa There is a \(\frac{3}{4}\) chance for each offspring to be normally pigmented (having either AA or Aa genotype) and a \(\frac{1}{4}\) chance to have albinism (having the aa genotype). Now, we can calculate the probability of obtaining three normally pigmented offspring in a row: P(normally pigmented offspring) = \(\left(\frac{3}{4}\right)^3 = \frac{27}{64}\) So, the probability of the first three offspring all having normal pigmentation is \(\frac{27}{64}\).
02

(b) Probability of the first female offspring being normally pigmented

To find this probability, we will first look at the probability of a female offspring. Since the sex of the offspring is determined by the X or Y chromosomes (XX for female and XY for male), and the male contributes either X or Y, we find that the likelihood of having a female offspring is \(\frac{1}{2}\). Now, let's combine this with the probability of obtaining a normally pigmented offspring, which is \(\frac{3}{4}\). P(normally pigmented female offspring) = \(\frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}\) So, the probability of the first female offspring being normally pigmented is \(\frac{3}{8}\).
03

(c) Probability of the first offspring being a normally pigmented female

This is a combination of the probabilities calculated in parts (a) and (b). However, since we are looking specifically for first offspring with both traits (female and normally pigmented), we can use the probability calculated in part (b). P(normally pigmented female offspring) = \(\frac{3}{8}\) Thus, the probability that the first offspring is a normally pigmented female is \(\frac{3}{8}\).

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Key Concepts

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

Understanding the Punnett Square
The Punnett square is a visual tool used in genetics to predict the genotypes of offspring from a particular cross. By arranging the alleles from each parent, we can show all the possible combinations for their offspring's genotypes.

The square is comprised of a grid where the alleles from one parent are placed on the top row and those from the second parent are placed on the side column. The intersection of these rows and columns indicates the possible genotypes for the offspring.

For example, when two heterozygous whales (Aa) mate, each can provide allele A or allele a. The Punnett square would look like this:
  • Top: A a
  • Side: A a

If we fill out the squares, we end up with the following genotypic combinations:
  • AA (homozygous dominant)
  • Aa (heterozygous)
  • Aa (heterozygous)
  • aa (homozygous recessive)

This simple grid helps students understand how genotypes can be inherited and the likelihood of each, thereby providing the basic knowledge needed to answer genetics probability questions.
Homozygous and Heterozygous Genotypes Explained
Genotypes represent the genetic make-up of an organism, particularly regarding a specific characteristic. A homozygous genotype has two identical alleles for a particular gene, either both dominant (AA) or both recessive (aa). In contrast, a heterozygous genotype (Aa) has one dominant and one recessive allele.

If an allele is dominant, it can mask the expression of the other, which means that the traits associated with the dominant allele will appear in the organism, even if only one dominant allele is present. However, if an organism is homozygous recessive (aa), the recessive trait will only appear when no dominant allele is present.

In the whale example, AA and Aa genotypes will both result in normally pigmented whales because pigment is the dominant trait. Only the aa genotype gives the recessive trait of albinism. Understanding these concepts helps students solve probability questions involving genetic crosses.
Recessive Genes Inheritance Patterns
Recessive genes require two copies for the trait they represent to be expressed. This means an organism must have a homozygous recessive (aa) genotype to show a recessive trait.

The inheritance of recessive genes follows particular patterns. For a child to express a recessive trait like albinism in whales, both parents must carry at least one recessive allele (Aa or aa). If one parent is homozygous dominant (AA), none of the offspring will show the recessive trait, though they can still be carriers (Aa).

Understanding the inheritance patterns of recessive genes not only applies to scenarios like Migaloo's albinism but also to many human genetic conditions. Knowing how these genes are passed down can inform students on the probability of inheriting certain disorders or traits.

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Most popular questions from this chapter

In a study of black guinea pigs and white guinea pigs, 100 black animals were crossed with 100 white animals, and each cross was carried to an \(\mathrm{F}_{2}\) generation. In 94 of the crosses, all the \(\mathrm{F}_{1}\) offspring were black and an \(\mathrm{F}_{2}\) ratio of 3 black: 1 white was obtained. In the other 6 cases, half of the \(\mathrm{F}_{1}\) animals were black and the other half were white. Why? Predict the results of crossing the black and white \(\mathrm{F}_{1}\) guinea pigs from the 6 exceptional cases.

A certain type of congenital deafness in humans is caused by a rare autosomal (not X-linked) dominant gene. (a) In a mating involving a deaf man and a deaf woman (both heterozygous), would you expect all the children to be deaf? Explain your answer. (b) In a mating involving a deaf man and a deaf woman (both heterozygous), could all the children have normal hearing? Explain your answer. (c) Another form of deafness is caused by a rare autosomal recessive gene. In a mating involving a deaf man and a deaf woman, could some of the children have normal hearing? Explain your answer.

Albinism in humans is inherited as a simple recessive trait. For the following families, determine the genotypes of the parents and offspring. (When two alternative genotypes are possible, list both.) (a) Two normal parents have five children, four normal and one albino. (b) A normal male and an albino female have six children, all normal. (c) A normal male and an albino female have six children, three normal and three albino. (d) Construct a pedigree of the families in (b) and (c). Assume that one of the normal children in (b) and one of the albino children in (c) become the parents of eight children. Add these children to the pedigree, predicting their phenotypes (normal or albino).

The autosomal (not X-linked) gene for brachydactyly, short fingers, is dominant to normal finger length. Assume that a female with brachydactyly in the heterozygous condition is married to a man with normal fingers. What is the probability that (a) their first child will have brachydactyly? (b) their first two children will have brachydactyly? (c) their first child will be a brachydactylous girl?

To assess Mendel's law of segregation using tomatoes, a truebreeding tall variety (SS) is crossed with a true-breeding short variety \((s s) .\) The heterozygous \(F_{1}\) tall plants \((S s)\) were crossed to produce two sets of \(\mathrm{F}_{2}\) data, as follows. \(\begin{array}{cc}\text { Set I } & \text { Set II } \\ 30 \text { tall } & 300 \text { tall } \\ 5 \text { short } & 50 \text { short }\end{array}\) (a) Using the \(\chi^{2}\) test, analyze the results for both datasets. Calculate \(\chi^{2}\) values and estimate the \(p\) values in both cases. (b) From the above analysis, what can you conclude about the importance of generating large datasets in experimental conditions?
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