Chapter 14 - Mendel and the Gene Idea

Question 1

Fill in this diagram of a cross of round- and wrinkled-seeded pea plants. The round allele (R) is dominant and the wrinkled allel (r) is recessive.

Answer 1

a. R g. r
b. r h. Rr (round)
c. F₁ generation i. Rr (round)
d. Rr j. rr (wrinkled)
e. F₂ generation k. 3 round:1 wrinkled
f. R l. 1 RR:2Rr:1rr

Question 2

A tall pea plant is crossed with a recessive dwarf pea plant. What will the phenotypic and genotypic ratio of offspring be

a. if the tall plant was TT ?
b. if the tall plant was Tt ?

Answer 2

a. all tall (Tt ) pltans
b. 1;1 tall (Tt ) to dwarf (tt )

You can use a Punnett square to determine the expected outcome of a test cross, but a shortcut is to use only 1 column for the recessive individual’s gametes, since they produce only one type of gamete.

Question 3

A true-breeding tall, purple-flowered pea plant (TTPP ) is crossed with a true-breeding dwarf, white-flowered plant (ttpp ).

a. What is the phenotype of the F₁ generation?
b. What is the genotype of the F₁ generation?
c. What four types of gametes are formed by F₁ plants?
d. Use the picture of the Punnett square to show the offspring of the F₂ generation. Shade each phenotype a different color so yuo can see the ratio of offspring.
e. List the phenotypes and ratios found in the F₂ generation.
f. What is the ratio of tall to dwarf plants?

Of purple- to white-flowered plants?

(Note that the alleles for each individual character segregate as in a monohybrid cross.)

Answer 3

a. all tall purple plants
b. TtPp
c. TP, Tp, tP, tp

d, see picture

e. 9 tall purple:3 tall white:3 dwarf purple:1 dwarf white
f. 12:4 or 3:1 tall to dwarf
12: 4 or 3:1 purple to white

Question 4

Apply the multiplication rule to a dihybrid cross. How would you determine the probability of geting an F₂ offspring that is homozygous recessive for both traits?

Answer 4

Dihybrids produce four types of gametes. The probability of getting both recessive alleles in a gamete is 1/4, and for two such gametes to join is ¼ x ¼, or ⅟_16.

Question 5

a. In the following cross, what is the probability of obtaining offspring that show all three dominant traits?
* AaBbcc* x AabbCC

probability of offspring that are A_B_C_ = _____

( _ indicates that the second allele can be either dominant or recessive without affecting the phenotype determined by the first dominant allele.)

b. What is the probability that the offspring of this cross will show at least two dominant traits?
c. What is the probability of offspring that show only one dominant trait?

Answer 5

a. Consider the outcome for each gene as amonohybrid cross. The probability that a cross of Aa x Aa will produce an A_ offspring is ¾. The probability that a cross of Bb x bb will produce a B_ offspring is ½. The probability that a cross of cc x CC will produce a C_ offspring is 1.

To have all of these events occur simultaneously, multiply their probabilities: ¾ x ½ x 1 = ⅜.

b. Offspring could be A_bbC_, aaB_C_ , or A_B_C_. The genotype A_B_cc is not possible. Can you see why?

Probability of A_bbC_ = ¾ x ½ x 1 = ⅜

Probability of aaB_C_ = ¼ x ½ x 1 = ⅛

Probability of A_B_C_ = ¾ x ½ x 1 = ⅜

Probability of offspring showing at least two dominant traits is the sum of these independent probabilities, or ⅞.

c. There is only one type of offspring that can show only one dominant trait (aabbCc ). Can you see why? Its probability is ¼ x ½ x 1 = ⅛.

Question 6

List the possible genotypes for the following blood groups.

a. A _____
b. B _____
c. AB _____
d. O _____

Answer 6

a. I ^AI ^A and I ^Ai
b. I ^BI ^B and I ^Bi
c. I ^AI ^B
d. ii

Question 7

A dominant allele M is necessary for the production of the black pigment melanin; mm individuals are white. A dominant allele B results in the deposition of a lot of pigment in an animal’s hair, producing a black color. The genotype bb produces brown hair. Two black animals heterzygous for both genes are bred. Fill in the table for the offspring of this cross.

Answer 7

The ratio of offspring from this MmBb x MmBb cross would be 9:3:4, a common ratio when one gene is epistatic to another. All epistatic ratios are modified versons of 9:3:3:1.

Question 8

The height of spike week is a result of polygenic inheritance involving three genes, each of which can contribute an additional 5 cm to the base height of the plant, which is 10 cm. The tallest plant (AABBCC ) can reach a height of 40 cm.

a. If a tall plant (AABBCC ) is crossed with a base-height plant (aabbcc ), what is the height of the F₁ plants?
b. How many phenotypic classes will there be in the F₂?

Answer 8

a. The parental cross produced 25-cm tall F₁ plants, all AaBbCc plants with 3 units of 5 cm added to the base height of 10 cm.
b. As a general rule in the polygenic inheritance of a quantitative character, the number of phenotypic classes resulting from a cross of heterozygotes equals the number of alleles involved plus one. In this case, 6 alleles (AaBbCc ) + 1 = 7. So, there will be 7 different phenotypic classes in the F₂ among the 64 possible combinations of the 8 types of F₁ gametes. These 7 classes will go from 6 dominant alleles (40 cm), 5 dominant (35 cm), 4 dominant (30 cm), and so on, to all 6 recessive alleles (10 cm).

Question 9

Consider this pedigree for the trait of albinism (lack of skin pigmentation) in three generations of a family. (Solid symbols represent individuals who are albinos.) From your knowledge of Mendelian inheritance, answer the following questions.

a. Is this trait caused by a dominant or recessive allele? How can you tell?
b. Determine the genotypes of the parents in the first generation. (Let A and a represent the alleles.) Genotype of father _____ ; of mother _____.
c. Determine the probable genotypes of the mates of the albino offspring in the second generation and the grandson 4 in the third generation. Genotypes: mate 1 _____ ; mate 2 _____ ; grandson 4 _____.
d. Can you determine the genotype of son 3 in the second generation? Why or why not?

Answer 9

a. This trait is recessive. If it were dominant, then albinism would be present in every generation, and it would be impossible to have albino children with two nonalbino (homozygous recessive) parents.
b. father Aa; mother Aa, because neither parent is albino and they have albino offspring (aa )
c. mate 1 AA (probably); mate 2 Aa; grandson 4 Aa
d. The genotype of son 3 could be AA or Aa. If his wife is AA, then he could be Aa (both his parents are carriers) and the recessive allele never would be epressed in his offspring. Even if he and his wife were both carriers (heterozygotes), there would be a ²⁴³/₁₀₂₄ (¾ x ¾ x ¾ x ¾ x ¾) or 24% chance that all five children would be normally pigmented.

Question 10

a. What is the probability that a mating between two carriers will produce an offspring with a recessively inherited disorder?
b. What is the probability that a phenotypically normal child produced by a mating of two heterozygotes will be a carrier?

Answer 10

a. ¼
b. ²/₃. Of offspring with a normal phenotype, ²/₃ would be predicted to be heterozygotes and, thus, carriers of the recessive allele.

Question 11

If two prospective parents both have siblings who have a recessive genetic disorder, what is the chance that they would have a child who inherits the disorder?

Answer 11

Both sets of prospective grandparents must have been carriers. The prospective parents do not have the disorder so they are not homozygous recessive. Thus, each has a ²/₃ chance of being a heterozygote carrier.

The probability that both parents are carriers is ²/₃ x ²/₃ = ⁴/₉ ; the chance that two heterozygotes will have a recessive homozygous child is ¼.

The overall chance that a child will inherit the disease is ⁴/₉ x ¹/₄ = ¹/₉.

Should this couple have a baby that has the disease, this would establish that they are both carriers, and the chance that a subsequent child would have the disease is ¼.

Question 12

After obtaining two heads from two tosses of a coin, the probability of tossing the coin and obtaining a head is

a. 1/2
b. 1/4
c. 1/8
d. 3/8
e. 3/4

Answer 12

a. 1/2

Question 13

The probability of tossing three coins simultaneously and obtaining three heads is

a. 1/2
b. 1/4
c. 1/8
d. 3/8
e. 3/4

Answer 13

c. 1/8

Question 14

The probability of tossing three coins simultaneoulsy and obtaining two heads and one tail is

a. 1/2
b. 1/4
c. 1/8
d. 3/8
e. 3/4

Answer 14

d. 3/8

There are three different ways to get this outcome: HHT, HTH, THH. Each outcome has a probability of 1/8.

Question 15

The F₂ generation

a. has a phenotypic ratio of 3:1.
b. is the result of the self-fertilization or crossing of F₁ individuals.
c. can be used to determine the genotype of individuals with the dominant phenotype.

d, has a phenotypic ratio that equals its genotypic ratio.

e. has 16 different genotypic possibilities.

Answer 15

b. is the result of the self-fertilization or crossing of F₁ individuals.

Question 16

The base height of the dingdong plant is 10 cm. Four genes contribute to the height of the plant, and each dominant allele contributes 3 cm to height. If you cross a 10-cm plant (quadruply homozygous recessive) with a 34-cm plant, how many phenotypic classes will there be in the F₂?

a. 4
b. 5
c. 8
d. 9
e. 64

Answer 16

d. 9

Question 17

According to Mendel’s law of segregation,

a. there is a 50% probability that a gamete will get a dominant allele.
b. gene pairs segregate independently of other genes in gamete formation.
c. allele pairs separate in gamete formation.
d. the laws of probability determine gamete formation.
e. there is a 3:1 ratio in the F₂ generation.

Answer 17

c. allele pairs separate in gamete formation.

Question 18

A 1:1 phenotypic ratio in a testcross indicates that

a. the alleles are dominant.
b. one parent must have been homozygous dominant.
c. the dominant phenotype parent was a heterozygote.
d. the alleles segregated independently.
e. the alleles are codominant.

Answer 18

c. the dominant phenotype parent was a heterozygote.

Question 19

Carriers of a genetic disorder

a. are indicated by solid symbols on a family pedigree.
b. are involved in consanguineous matings.
c. will produce children with the disease.
d. are heterzygotes for the gene that can cause the disorder.
e. have a homozygous recessive genotype.

Answer 19

d. are heterzygotes for the gene that can cause the disorder.

Question 20

If both parents are carriers of a lethal recessive gene, the probability that their child will inherit and express the disorder is

a. 1/8
b. 1/4
c. 1/2
d. 1/2 x 1/2 x 1/4, or 1/16
e. 2/3 x 2/3 x 1/4, or 1/9

Answer 20

b. 1/4

Question 21

Which phase of meiosis is most directly related to the law of independent assortment?

a. prophase I
b. prophase II
c. metaphase I
d. metaphase II
e. anaphase II

Answer 21

c. metaphase I

Question 22

You think that two alleles for coat color in mice show incomplete dominance. What is the best and simplest cross to perform in order to support your hypothesis?

a. a testcross of a homozygous recessive mouse with a mouse of unknown genotype
b. a cross of F₁ mice to look for a 1:2:1 ratio in the offspring
c. a reciprocal cross in which the sex of the mice of each coat color is reversed
d. a cross of two true-breeding mice of different colors to look for an intermediate phenotype in the F₁
e. a cross of F₁ mice to look for a 9:7 ratio in the offspring

Answer 22

d. a cross of two true-breeding mice of different colors to look for an intermediate phenotype in the F₁

Question 23

Which of the following human diseases is inherited as a simple recessive trait?

a. Tay-Sachs disease
b. cancer
c. diabetes
d. Alzheimer’s disease
e. cardiovascular disease

Answer 23

a. Tay-Sachs disease

Question 24

A multifactorial disorder

a. can usually be traced to consanguineous matings.
b. is caused by recessively inherited lethal genes.
c. has both genetic and environmental causes.
d. has a collection of symptoms traceable to an epistatic gene.
e. is usually associated with quantitative traits.

Answer 24

d. has a collection of symptoms traceable to an epistatic gene.

Question 25

A mother with type B blood has two children, one with type A blood and one with type O blood. Her husband has type O blood. Which of the following could you conclude from this information?

a. The husband could not have fathered either child.
b. The husband could have fathered both children.
c. The husband must be the father of the child with type O blood and could be the father of the type A child.
d. The husband could be the father of the child with type O blood, but not the type A child.
e. Neither the mother nor the husband could be the biological parent of the type A child.

Answer 25

d. The husband could be the father of the child with type O blood, but not the type A child.

Question 26

In guinea pigs, the brown coat color allele (B) is dominant over red (b), and the solid color allele (S) is dominant over spotted (s). The F₁ offspring of a cross between true-breeding brown, solid-colored guinea pigs and red, spotted pigs are crossed. What proportion of their offspring (F₂) would be expected to be red and solid colored?

a. 1/9
b. 1/16
c. 3/16
d. 9/16
e. 3/4

Answer 26

c. 3/16

Question 27

A dominant allele P causes the production of purple pigment, pp individuals are white. A dominant allele C is also required for color production; cc individuals are white. What proportion of offspring will be purple from a ppCc x PpCc cross?

a. 1/8
b. 3/8
c. 1/2
d. 3/4
e. None

Answer 27

b. 3/8