BIO 335 Ch 3 HW

- i forgot to add the ? lol pt 2

use product rule

- i forgot to add the ? lol

use sum rule

Suppose Maria wants to breed guinea pigs that are black with short fur. However, when she bred several guinea pigs that were black with short fur, she found that there were four different offspring phenotypes. Based on the offspring data provided, determine the number of genes and alleles that control these phenotypes. *Phenotype*: short black fur, long black fur, short brown fur, long brown fur *Number*: 18, 6, 6, 2 Number of genes and alleles - 4 genes; 1 allele each - 2 genes; 2 alleles each - 1 gene; 4 alleles Which phenotypes are coded for by the same gene? Different genes are separated by semicolons. - black and brown; short fur and long - each by a separate gene - all by the same gene - black and short fur; brown and long

- 2 genes; 2 alleles each - black and brown; short fur and long

In pea plants, plant height is controlled by a single autosomal dominant gene. Tall plants (H) are dominant to short plants (h). In a cross of two tall heterozygous plants, which phenotype ratio is expected from the resulting offspring? - 1:2:1 - 9:3:3:1 - 1:1 - 3:1

- 3:1

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? - All sweet fruit with no spots - 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 - 9/16 bitter fruit, yellow spots; 3/16 bitter fruit, no spots; 3/16 sweet fruit, yellow spots; and 1/16 sweet fruit, no spots - All bitter fruit with yellow spots - 1/2 bitter fruit, yellow spots and 1/2 sweet fruit, no spots If an F1 plant is backcrossed with the bitter, yellow‑spotted parent, what phenotypes and proportions are expected in the offspring? - All sweet fruit with no spots - All bitter fruit with yellow spots - 1/2 bitter fruit, yellow spots and 1/2 sweet fruit, no spots - 9/16 bitter fruit, yellow spots; 3/16 bitter fruit, no spots; 3/16 sweet fruit, yellow spots; and 1/16 sweet fruit, no spots - 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 If an F1 plant is backcrossed with the sweet, nonspotted parent, what phenotypes and proportions are expected in the offspring? - All sweet fruit with no spots - 9/16 bitter fruit, yellow spots; 3/16 bitter fruit, no spots; 3/16 sweet fruit, yellow spots; and 1/16 sweet fruit, no spots - 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 - All bitter fruit with yellow spots - 1/2 bitter fruit, yellow spots and 1/2 sweet fruit, no spots

- 9/16 bitter fruit, yellow spots; 3/16 bitter fruit, no spots; 3/16 sweet fruit, yellow spots; and 1/16 sweet fruit, no spots - All bitter fruit with yellow spots - 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 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)(A) is dominant over albino characteristics (a)(a) . For this question, assume the phenotype is determined by a single gene with two alleles. If both parents have normal pigmentation, what are all of the possible genotypes that may be observed in their offspring? - aa only - AA only - AA, Aa, and aa - AA and Aa

- AA, Aa, and aa

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)(A) is dominant over albino characteristics (a)(a) . For this question, assume the phenotype is determined by a single gene with two alleles. If both parents display the albino phenotype, what are all of the possible genotypes that may be observed in their offspring? - aa only - AA only - AA, Aa, and aa - AA and Aa

- aa only

At times in the past, red hair in humans was thought to be a recessive trait, and at other times it was thought to be a dominant trait. What feature of inheritance would red hair be expected to exhibit if it were a recessive trait? - two red‑haired parents will only have children with red hair - two red‑haired parents could have children without red hair - affects males and females differently What feature of inheritance would red hair be expected to exhibit if it were a dominant trait? - two red‑haired parents could have children without red hair - affects males and females differently - two red‑haired parents will only have children with red hair

- two red‑haired parents will only have children with red hair - two red‑haired parents could have children without red hair

Of the following ideas postulated by Gregor Mendel, which one requires at least two genes to be demonstrated? -Some alleles are recessive and are masked by dominant alleles. -Genes assort independently in humans. -Traits are controlled by discrete units. -One of two alleles from each parent is randomly transmitted to offspring. -Traits are not determined by blending.

-Genes assort independently in humans.

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? 1/3 3/4 1/4 2/3 1/2 What is the probability that an egg from the mother will contain the PKU allele? 3/4 1/2 1/4 2/3 1/3 What is the probability that their next child will have PKU? 1/4 3/4 1/2 1/3 2/3 What is the probability that their next child will be heterozygous for the PKU gene? 1/4 1/3 1/2 3/4 2/3

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

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. [make punnet square w/ Rr and Rr] In this cross between two heterozygous pea plants, what are the chances that an offspring with wrinkled seeds will be produced? - 50% - 1% - 25% - 75%

25%

Which gametes can the Aa Bb parent generate? (select all that apply) A B A b a B a b

A B A b a B a b

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 w/ different-colored flowers and allowing the F1 gen to self-pollinate. Mendel observed the basic patterns of inheritance in the F2 gen.

The figure represents a pair of homologous chromosomes. Assign the appropriate term to each component.

[basically know how to label the loci of a chromosome]

How is a true breeding yellow‑seeded pea plant different from a hybrid yellow‑seeded pea plant? a) They have the same phenotype but different genotypes. b) They have the same genotype but different phenotypes. c) They have a different genotype and phenotype. d) They have the same genotype and phenotype.

a) They have the same phenotype but different genotypes.