Genetics: Chapter 3 Homework

Phenylketonuria (PKU) is a disease that results from a recessive gene. Two normal parents produce a child with PKU. - What is the probability that a sperm from the father will contain the PKU allele? - What is the probability that an egg from the mother will contain the PKU allele? - What is the probability that their next child will have PKU? - What is the probability that their next child will be heterozygous for the PKU gene?

- 1/2 - 1/2 - 1/4 - 1/2

At the end of your biology class, your professor asks you to develop a project to determine the genotype of a plant with red flowers. Red petal color (R) is dominant to pink flower color (r). To accomplish this task, you cross the plant with the unknown genotype with heterozygous red‑flowered plants. A partially filled Punnett square is provided. Which ratios are valid predictions of flower colors in the offspring?

- all red flowers - 3 red : 1 pink flowers

Which of the statements can be concluded from Gregor Mendel's experiments with pea plants?

- allelic combinations for different genes may differ between parents and their offspring - only one allele determines the phenotype in heterozygous individuals

Use the interactive Punnett Squares, Level 2 to answer the questions. Suppose the parents indicated in the interactive produced 224 peas . Determine the expected number of yellow and wrinkled offspring. Now, suppose one of the pea plants is homozygous recessive for both genes, resulting in the cross Yy Rr × yy rr. Calculate the percentage of yellow and wrinkled offspring. Suppose one of the pea plants is homozygous recessive for seed color, resulting in the cross Yy Rr × yy Rr. Select the correct phenotypic ratio for the offspring.

- expected number of yellow and wrinkled offspring: 42 - yellow and wrinkled offspring: 25 - 6 yellow, round : 2 yellow, wrinkled : 6 green, round : 2 green, wrinkled

Huntington's disease is an inherited autosomal dominant disorder that can affect both men and women. Imagine a couple has had seven children, and later in life, the husband develops Huntington's disease. He is tested and it is discovered he is heterozygous for the disease allele, Hh. The wife is also genetically tested for the Huntington's disease allele, and her test results show she is unaffected, hh. What is the percent probability that the first child of this couple will have Huntington's disease? What is the probability that any two of the seven children will have Huntington's disease?

- first child: 50% - any two: 16.41%

Suppose two parents, a father with the genotype AaBbCcDdee and a mother with the genotype aaBbCCDdEe, want to have children. Assume each locus follows Mendelian inheritance patterns for dominance. What proportion of the offspring will have each of the specified characteristics? Round your answers to two decimal places.

- same genotype as the father: 0.031 - same genotype as the mother: 0.031 - phenotypically resemble the father: 0.14 - phenotypically resemble the mother: 0.14 - phenotypically resemble neither parent: 0.72

Use the image to observe the results of a cross between a tall pea plant and a short pea plant. What phenotypes and proportions will be produced for the two crosses?

- tall F1 progeny backcrossed to the short parent: one-half tall and one-half short - tall F1 progeny backcrossed to the tall parent: three fourths tall and one-fourth short

White (w) coat color in guinea pigs is recessive to black (W). In 1909, W. E. Castle and J. C. Phillips transplanted an ovary from a black guinea pig into a white female whose ovaries had been removed. They then mated this white female with a white male. All the offspring from the mating were black in color (W. E. Castle and J. C. Phillips, 1909, Science 30:312-313). Select the conclusions that are indicated by this experiment regarding the validity of the pangenesis and the germ-plasm theories discussed in Chapter 1.

- the production of black guinea pig offspring suggest that the allele for black coat was passed along to the offspring from the transplanted ovary, thus supporting the germ-plasm theory - because no white guinea pigs were produced, no white coat alleles traveled to the ovary and into the gametes of the white female, thus indicating pangenesis did not occur

In cats, curled ears result from an allele, Cu, that is dominant over an allele, cu, for normal ears. Black color results from an independently assorting allele, G, that is dominant over an allele for gray, g. A gray cat homozygous for curled ears is mated with a homozygous black cat with normal ears. All the F1 cats are black and have curled ears. What phenotypes and proportions are expected from the two crosses?

- two of the F1 cats mate: 9/16 black cats, curled ears; 3/16 black cats, normal ears; 3/16 gray cats, curled ears; and 1/16 gray cats, normal ears - An F1 cat water with a stray cat that is gray and possesses normal ears: 1/4 black cats, curled ears; 1/4 black cats, normal ears; 1/4 gray cats, curled ears; and 1/4 gray cats, normal ears

Suppose a man is heterozygous for heterochromia, an autosomal dominant disorder which causes two different‑colored eyes in an individual, produced 25‑offspring with his normal‑eyed wife. Of their children, 16 were heterochromatic and 9were‑normal. Calculate the chi‑square value for this observation. Identify the statement that best interprets the results of the chi‑square analysis. Refer to the chi‑square distribution table to identify the statement that best interprets the chi‑square results.

- x^2 = 1.96 - it is not unusual that a heterozygous man produced 16 out of 25 offspring with heterochromia

In pea plants, plant height is controlled by a single autosomal dominant gene. Tall plants (H) are dominant to short plants (h). Ina cross of two tall heterozygous plants, which phenotype ratio is expected from the resulting offspring?

3:1

Pink eye and albinism are two recessive traits found in the deer mouse, Peromyscus maniculatus. In mice with pink eye, the eye is devoid of color and appears pink from the blood vessels within it. Albino mice are completely lacking color both in their fur and in their eyes. F. H. Clark crossed pink‑eyed mice with albino mice; the resulting F1 had normal coloration in their fur and eyes. He then crossed these F1 mice with mice that were pink‑eyed and albino and obtained the mice shown in the table below. It is very hard to distinguish between mice that are albino and mice that are both pink‑eyed and albino, so he combined these two phenotypic classes (F. H. Clark, 1936, Journal of Heredity 27:259−260). Match the expected numbers of progeny with each phenotype if the genes for pink‑eye and albinism assort independently. Use a chi-square test to determine if the observed numbers of progeny fit the number expected with independent assortment. Does this chi-square value fit the number expected with independent assortment?

- x^2 = 33 - no

Suppose that goats have one gene that codes for color, where A is brown and a is white. The goats also have another gene that codes for height, where B is tall and b is short. If these two genes are unlinked, what is the probability that a cross between Aa Bb × Aa bb parents will produce three out of five offspring that are white and tall? Enter your answer as a decimal, rather than as a percentage.

0.016

Imagine that two unlinked autosomal genes with simple dominance code in goats for size, where T is tall and t is short, and for color, where R is red and r is tan. If a short, tan male goat mates with a tall, red female goat of an unknown genotype, what is the probability that they would produce short, tan offspring?

0.0625

In mice, an allele for apricot eyes (a) is recessive to an allele for brown eyes (a+). At an independently assorting locus, an allele for tan (t) coat color is recessive to an allele for black (t+) coat color. A mouse that is homozygous for brown eyes and black coat color is crossed with a mouse having apricot eyes and a tan coat. The resulting F1 are intercrossed to produce the F2. In a litter of eight F2 mice, what is the probability that exactly two will have apricot eyes and tan coats? Use three decimal places for the answer.

0.0743

Suppose that two parents are both heterozygous for sickle cell anemia, which is an autosomal recessive disease. They have seven children. Use the binomial theorem to determine the probability that three of the children have sickle cell anemia and four of the children are healthy. Round your answer to the nearest tenth.

0.2

In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2. If an F1 plant is backcrossed with the sweet, nonspotted parent, what phenotypes and proportions are expected in the offspring?

1/4 bitter fruit, yellow spots; 1/4 bitter fruit, no spots; 1/4 sweet fruit, yellow spots; and 1/4 sweet fruit, no spots

In pea plants, the allele for round seed shape, R, is completely dominant to the allele for wrinkled seed shape, r. Complete the Punnett square showing the genotypes possible among the offspring when two heterozygous individuals are crossed. Use the information from the Punnett square to answer the second question. In this cross between two heterozygous pea plants, what are the chances that an offspring with wrinkled seeds will be produced?

25%

In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2. What will be the phenotypic ratio in the F2?

9/16 bitter fruit, yellow spots; 3/16 bitter fruit, no spots; 3/16sweet fruit, yellow spots; and 1/16 sweet fruit, no spots

In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2. If an F1 plant is backcrossed with the bitter, yellow‑spotted parent, what phenotypes and proportions are expected in the offspring?

All bitter fruit with yellow spots

Of the following ideas postulated by Gregor Mendel, which one requires at least two genes to be demonstrated?

Genes assort independently in diploids

In fruit flies, gray bodies (G) are dominant over black bodies (g), and brown pigments (N) are dominant over yellow pigments (n). Each individual possesses two alleles for each trait. If a fly that is homozygous dominant for both traits is crossed with a fly that is homozygous recessive for both traits, what is the predicted genotype of the offspring?

GgNn

White (w) coat color in guinea pigs is recessive to black (W). In 1909, W. E. Castle and J. C. Phillips transplanted an ovary from a black guinea pig into a white female whose ovaries had been removed. They then mated this white female with a white male. All the offspring from the mating were black in color (W. E. Castle and J. C. Phillips, 1909, Science 30:312-313). Which statement explains the results of this cross?

the color of the offspring was determined by genes in the transplanted ovary, not the genes of the female who gave birth

How is a true breeding purple‑flowered pea plant different from a hybrid purple‑flowered pea plant?

They have the same phenotype but different genotypes

White (w) coat color in guinea pigs is recessive to black (W). In 1909, W. E. Castle and J. C. Phillips transplanted an ovary from a black guinea pig into a white female whose ovaries had been removed. They then mated this white female with a white male. All the offspring from the mating were black in color (W. E. Castle and J. C. Phillips, 1909, Science 30:312-313). Select the genotype of the offspring from this cross.

Ww

Suppose there is a vial containing a single generation of flies from a cross. There is an interesting phenotype where many individuals have abnormally long hairlike bristles, sensory organs extending from the dorsal thorax, as opposed to the short wirelike wild‑type bristles among the other siblings. References state that this mutant has a dominant mutation called Suave(Su) and that the phenotype of flies that are heterozygous or homozygous for Su appear phenotypically identical. Which fly should be crossed to a Suave male from this vial in order to generate progeny that help determine the male's genotype?

a wild-type female sibling

Alleles A and a are located on a pair of metacentric chromosomes. Alleles B and b are located on a pair of acrocentric chromosomes. A cross is made between individuals having the genotypes Aa Bb and aa bb. Which gametes can the AaBb parent generate?

aB, AB, Ab, ab

In humans, oculocutaneous (OCA) albinism is a collection of autosomal recessive disorders characterized by an absence of the pigment melanin in skin, hair, and eyes. That is, normal pigmentation (A) is dominant over albinism (a). For this question, assume it is a single gene with two alleles. If both parents display the albino phenotype, what possible phenotypes may be observed in their offspring?

albinism only

How did Mendel use self‑pollination and cross‑pollination techniques in his experiments with flower color to observe the basic patterns of inheritance?

by cross-pollinating a parental generation of plants with different-colored flowers and allowing the F1 generation to self-pollinate, Mendel observed the basic patterns of inheritances in the F2 generation

A geneticist discovers an obese mouse in his laboratory colony. He breeds this obese mouse with a normal mouse. All the F1 mice from this cross are normal in size. When he interbreeds two F1 mice, eight of the F2 mice are normal in size and two are obese.The geneticist then intercrosses two of his obese mice, and he finds that all of the progeny from this cross are obese. These results lead the geneticist to conclude that obesity in mice results from a recessive allele. A second geneticist at a different university also discovers an obese mouse in her laboratory colony. She carries out the same crosses as the first geneticist and obtains the same results. She also concludes that obesity in mice results from a recessive allele. One day, the two geneticists meet at a genetics conference, learn of each others experiments, and decide to exchange mice. They both find that, when they cross two obese mice from the different laboratories, all the offspring are normal. However, when they cross two obese mice from the same laboratory, all the offspring are obese. Which option best explains their results?

the two obesity alleles are recessive to the wild type alleles but are located at different loci

Imagine that a scientist studies two traits in peas. The scientist noticed that round is dominant over wrinkled with regard to pea shape. Additionally, yellow is dominant over green with regard to pea color. To determine if these traits are linked, two individuals that are heterozygous for both traits were crossed. The data in the table represent the number of offspring produced by this dybrid cross. Phenotypic ratios represents the predicted proportion of offspring with each set of traits that would be produced if the traits independently assort. What can be determined about these traits based on Chi-square analysis?

these traits assort independently