Chapter 2 Practice Problems

Question 1

observable characteristics

Answer 1

phenotype

Question 2

alternate forms of a gene

Answer 2

alleles

Question 3

alleles of a gene separate into gametes randomly with respect to alleles of other genes

Answer 3

independent assortment

Question 4

reproductive cells containing only one copy of each gene

Answer 4

gametes

Question 5

the heritability entity that determines a characteristic

Answer 5

gene

Question 6

the separation of the two alleles of a gene into different gametes

Answer 6

segregation

Question 7

an individual with two different alleles of a gene

Answer 7

heterozygote

Question 8

the allele expressed in the phenotype of the heterozygote

Answer 8

dominant

Question 9

offspring of the P generation

Answer 9

F1

Question 10

the cross of an individual of ambiguous genotype with a. homozygous recessive individual

Answer 10

testcross

Question 11

the alleles an individual has

Answer 11

genotype

Question 12

the allele that does not contribute to the phenotype of the heterozygote

Answer 12

recessive

Question 13

a cross between individuals both heterozygous for two genes

Answer 13

dihybrid cross

Question 14

having two identical alleles of a given gene

Answer 14

homozygote

Question 15

an albino corn snake is crossed with a normal colored corn snake. The offspring are all normal-colored. when the first generation progeny snakes are crossed among themselves, they produce 32 normal colored snakes and 10 albino snakes.

a. How do you know that only a single gene is responsible for the color differences between these snakes?
b. which of these phenotypes is controlled by the dominant allele?
c. a normal colored female snake is involved in a. test cross. this cross produces 10 normal colored and 11 albina offspring. what are the genotypes of the parents?

Answer 15

a. Two phenotypes are seen in the second generation of this cross: normal and albino. Thus, only one gene with two alleles is needed to control the phenotypes observed. The 3:1 ratio of these phenotypes in the F2 generation will be seen only if a single gene is involved.
b. the allele controlling the normal phenotype (A) is dominant to the allele controlling the albino phenotype (a).
c. the male parent’s genotype is aa. The normally colored offspring must receive an A allele from the mother, so the genotype of the normal offspring of the testcross is Aa. The albino offspring must receive an a allele from the mother, so the genotype of the albino offspring of the testcross is aa. Thus, the female parent must be heterozygous Aa.

Question 16

two short haired cats mate and produce 6 short haired cats and two long haired kittens. What does this information suffuse about hair length is inherited.

Answer 16

Long hair is recessive and the two parents are heterozygous for the trait.

Question 17

piebald spotting is a condition found in humans in which there are patches of skin that lack pigmentation. The condition results from the inability of pigment-producing cells to migrate properly during development. Two adults with piebald spotting have one child who has this trait and a second child with normal skin pigmentation.

a. is the piebald spotting trait dominant or recessive? what info led you to this answer?
b. what are the genotypes of the parents?

Answer 17

a. Two affected individuals have an affected child and a normal child. This outcome is not possible if the affected individuals were homozygous for a recessive allele conferring piebald spotting, and if the trait is controlled by a single gene. Therefore, the piebald trait must be the dominant phenotype.
b. they are both heterozygous for the trait

Question 18

As a Drosophila research geneticist, you keep stocks of flies of specific genotypes. You have a fly that has normal wings (dominant phenotype). Flies with short wings are homozygous recessive allele of the wide length gene. You need to know if this fly with normal wings is pure breeding or heterozygous for the wing length trait. what cross would you do to determine the genotype, and what results would you expect for each possible genotype?

Answer 18

You would conduct a testcross between your normal-winged fly (W–) and a short-winged fly that must be homozygous recessive (ww). if the normal-winged fly is a homozygote, all the progeny should have normal wings, but if the normal-winged fly is a heterozygote, half the progeny should have normal wings and the other half should have short wings.

Question 19

a mutant cucumber plant has flowers that fail to open when mature. Crosses can be done with this plant by manually opening and pollinating the flowers with pollen from another plant. When closed X open crosses were done, all the F1 progeny were open. The F2 plants were 145 open and 50 closed. A cross of closed xF1 gave 81 open and 77 closed. How is this trait inherited? what evidence led you to your conclusion?

Answer 19

The results of the crosses fit the pattern of inheritance of a single gene, with the open trait being dominant and the closed trait recessive. The phenotype of the F1 plants is open, indicating that open is dominant. The closed parent must be homozygous for the recessive allele. Because only one phenotype is seen among the F1 plants, the open parent must be homozygous for the dominant allele. The 3:1 ratio in the F2 shows that a single gene controls the phenotypes and that the F1 plants are all hybrids (that is, they are heterozygotes).

Question 20

In a particle population of mice, certain individuals display a phenotype called short tail, which is inherited as dominant trait. Some individuals display a recessive trait called dilute, which affects coat color. Which of these traits would be easier to eliminate from the population by selective breeding? why?

Answer 20

The dominant trait (short tail) is easier to eliminate from the population by selective breeding. The reason is you can recognize every animal that has inherited the short tail allele, because only one such dominant allele is needed to see the phenotype. If you prevent all the short-tailed animals from mating, then the allele would become extinct.

Question 21

If you roll a die, what is the probability you will roll: a) a 6? b) an even number? c) a number divisible by 3? d) if you Roll a pair of dice, what is the probability that you will roll 2 6’s? e) an even number on one and an odd number on the other? f) matching numbers g) two numbers both over 4?

Answer 21

a) 1/6
b) 1/2
c) 1/3
d) 1/36
e) 1/2
f) 1/6
g) 1/9

Question 22

In a standard deck of playing cards, four suits exist. each suit has 13 cards. In a single draw, what is the probability that you will draw a face card? a red card? a red face card?

Answer 22

12/52; 1/2; 6/52

Question 23

how many genetically different eggs could be formed by women with the following genotypes?

a. Aa bb CC DD
b. AA Bb Cc dd
c. Aa Bb cc Dd
d. Aa Bb Cc Dd

Answer 23

a. two different types, AbCD or abCD
b. 4 different types
c. 8 different types
d. 16 different types

Question 24

What is the probability of producing a child that will phenotypically resemble either one of the two parent s in the following four crosses? How many phenotypically different kinds of progeny could potentially result from each of the four crosses?

a. Aa Bb Cc Dd x aa bb cc dd
b. aa bb cc dd x AA BB CC DD
c. Aa Bb Cc Dd x Aa Bb Cc Dd
d. aa bb cc dd x aa bb cc dd

Answer 24

a. 1/8; 16 potential phenotypes
b. 0; 1
c. 81/256; 16
d. all; 1

Question 25

a mouse sperm of genotype a B C D E fertilizes an egg of genotype a b c D e. what are all the probabilities for the genotypes of a) the zygote and b) a sperm or egg produced that develops from this fertilization?

Answer 25

a. The combination of alleles in the egg and sperm allows only one genotype for the zygote: aa Bb Cc DD Ee
b. [(i) a B C D E : (ii)a B C D e : (iii) a B c D E : (iv) a B c D e : (v) a b C D E : (vi) a b C D e : (vii) a b c D E : (viii) a b c D e

Question 26

Your friend is pregnant with triplets. She thinks that it is equally likely that she will be the mother of 3 sons, 3 daughters, 2 sons and 1 daughter, or 1 son and 2 daughters. Is she correct? explain.

Answer 26

she is wrong.
probability of 3 girls is 1/8
probability of 3 boys is 1/8
probability of 2 girls and 1 boys is 3/8
probability of 2 boys and 1 girl is 3/8

Question 27

The achoo syndrome (sneezing in response to bight light) and trembling chin (triggered by anxiety) are both dominant traits in humans,

a. what is the probability that the first child of parents who are heterozygous for both the school gene and trembling chin will have have achoo syndrome but lack the trembling chin?
b. What is the probability that the first child will have neither achoo syndrome nor trembling chin?

Answer 27

a. 3/16

b. 1/16

Question 28

a young couple went to see a counselor because each had a sibling with cystic fibroses. Cystic fibrosis is a recessive disease, and neither member of the couple nor any of their four parents is affected.

a. what is the probability that the female of this couple is a carrier?
b. what are the chances that their child will have cystic fibrosis?
c. what is the probability that their child will be a carrier of the cystic fibrosis disease allele?

Answer 28

a. 2/3
b. 1/9
c. 4/9

Question 29

Huntington Disease is a rare fatal, degenerative neurological disease in which individuals start to show symptoms in their 40s. It is caused by a dominant allele. Joe, a man in his 20s, just learned that his father has Huntington disease.

a. what is the probability that Joe will also develop the disease?
b. Joe and his new wife have been eager to start a family. what is the probability that their first child will eventually develop the disease?

Answer 29

a. 1/2

b. 1/4

Question 30

explain why disease alleles for cystic fibrosis are recessive to the normal alleles, yet the disease alleles responsible for Huntington disease are dominant to the normal alleles.

Answer 30

In the case of cystic fibrosis, the alleles causing the disease do not specify active protein [in this case, the cystic fibrosis transmembrane receptor (CFTR)]. Some CF disease alleles specify defective CFTR proteins that do not allow the passage of chloride ions, while other CF disease alleles do not specify any CFTR protein at all. As you will learn in a later chapter, CF disease alleles of either type are called loss-of-function alleles. In a heterozygote, the normal CF+ allele still specifies active CFTR protein, which allows for the passage of chloride ions. Because the phenotype of heterozygotes is unaffected, these individuals must have enough active CFTR protein to allow passage of enough chloride ions for the cells to function normally. Most loss-of-function alleles of most genes are recessive to normal alleles for similar reasons. But it is imperative to realize that important exceptions exist in which loss-of-function mutations are actually dominant to normal alleles; you will see examples of these exceptions later in the book.
In the case of Huntington disease, the disease-causing allele is dominant to the normal allele. The reason is that the mutant huntingtin protein specified by the HD disease allele has, in addition to its normal function (which is not entirely understood), a second function that is toxic to nerve cells. This fact makes the HD disease allele a gain-of-function allele. The reason HD is dominant to HD+ is that the protein specified by the disease allele will be toxic to cells even if the cells have normal huntingtin protein specified by the normal allele. Most (but again not all) gain-of-function mutations are dominant for similar reasons.