A particular corn strain is
*
Wx/wx*
,
*
C/C*
, and
*
S/s*
. This describes the ___ of the plant.

genotype

phenotype

karyotype

genotype and phenotype

Answer: A

A young woman has blond hair, has blue eyes, and is lactose intolerant. This describes the girl's

genotype.

phenotype.

karyotype.

proteotype.

Answer: B

An albino woman marries a phenotypically normal man with no family history of albinism. What percentage of their children is likely to be albino?

0

25

50

100

Answer: A

An individual with the dominant phenotype but an unknown genotype is testcrossed, and all of many generated offspring exhibit the dominant phenotype. The genotype of the unknown individual is

homozygous dominant.

heterozygous dominant.

homozygous recessive.

heterozygous recessive.

Answer: A

For any given gene, the principle of segregation predicts that each gamete produced as a result of meiosis will contain

both alleles of a gene pair.

a single allele of a gene pair.

half of one allele of a gene pair.

half of both alleles of the gene pair.

Answer: B

The number of genotypic classes produced in a monohybrid F 1 cross is 4, the number produced in a dihybrid F 1 cross is 16, and the number corresponding to a trihybrid F 1 cross is 64. There is a simple mathematical pattern here, where the number of classes produced can be expressed as

½ n , where n is the number of genes.

4 n , where n is the number of genes.

N 4 , where n is the number of genes.

(1 + n) 4 , where n is the number of genes.

Answer: B

In a cross between a homozygous dominant parent and a homozygous recessive parent, the number of phenotypic classes observed in the offspring is

1.

2.

3.

4.

Answer: A

Two parents, one normal and one albino, have several children, all but one of whom are normal (one is albino). What is the probability that the normal children carry the albinism allele?

2/3

1/2

1

1/4

Answer: C

In a cross between two true-breeding parents that both exhibit the dominant phenotype, ___ percent of the offspring will exhibit the recessive phenotype.

0

25

50

100

Answer: A

In a cross between two true-breeding parents, one showing the dominant phenotype and the other the recessive phenotype, the phenotypic ratio in the F 2 generation is expected to be

1:2:1.

1:1.

3:1.

2:1.

Answer: C

In a particular type of melon, orange fruit is dominant over green fruit, and round shape is dominant to oval. If a plant that was heterozygous for both traits were crossed with a plant that was homozygous recessive for both traits, what proportion of the offspring would be heterozygous for both traits?

1/4

1/2

3/4

9/16

Answer: A

In a particular type of melon, orange fruit is dominant over green fruit, and round shape is dominant to oval. If a melon plant heterozygous for both traits were crossed with a homozygous recessive plant (for both traits), what proportion of the offspring would be heterozygous for fruit color, and homozygous recessive for fruit shape?

1/2

1/4

1/8

1/16

Answer: B

In a particular type of tomato plant, red fruit color is dominant over yellow. In a cross between a heterozygous plant and a homozygous recessive plant, what percentage of the offspring would have yellow fruit?

0

25

50

100

Answer: C

In a particular type of tomato plant, red fruit is dominant over yellow fruit, and dwarf height is a recessive trait. If two plants that were heterozygous for both traits were crossed, what proportion of the offspring would show the recessive phenotype for both traits?

1/2

1/4

1/8

1/16

Answer: D

In a particular type of tomato plant, red fruit is dominant over yellow fruit, and dwarf height is a recessive trait. If two heterozygous plants were crossed, what proportion of the offspring would show the dominant phenotype for both traits?

1/4

2/3

3/8

9/16

Answer: D

In a testcross, an individual of unknown genotype is crossed with a ___ individual to determine the unknown genotype.

homozygous dominant

heterozygous

homozygous recessive

hemizygous

Answer: C

The first generation produced in a series of monohybrid crosses is referred to as the ___ generation.

A

P

F 1

F 2

Answer: C

The use of the product rule in computing the expected proportion of Mendelian genotypes or phenotypes assumes __________ of the genes involved.

combination

independence

dominance

segregation

Answer: B

Two phenotypically normal people have four children. Three are phenotypically normal like their parents, but one is an albino. What are the probable genotypes of the parents?

Both parents are homozygous dominant.

Both parents are homozygous recessive.

One parent is homozygous dominant and one is homozygous recessive.

Both parents are heterozygous.

Answer: D

What percentage of the offspring of a woman with type O blood will carry the O allele? (Note that since you do not know the genotype or phenotype of the father of the children, you can answer this question only if you make an assumption about whether the O allele is dominant or recessive.)

0

25

50

100

Answer: D

In rabbits, agouti coat color is conditioned by the
*
A*
gene, solid coat color is conditioned by the
*
a*
gene, black coat color is conditioned by
*
B*
gene, and blue coat color is produced by the
*
b*
gene. Thus, in a cross between rabbits heterozygous at
the A and B loci, one would expect the 9:3:3:1 ratio. In a sample of
32 progeny rabbits obtained from the initial 7 litters of such
crosses, the following results were expected and observed.

**Expected**

**Observed**

Black agouti (A-B-)

18

17

Blue agouti (or opal) (A-bb)

6

9

Solid black (aaB-)

6

5

Solid blue (aabb)

2

4

**Total**

32

32

What is the probability that the observed results would be obtained by chance alone if the 9:3:3:1 model is correct?

(Table 11.5 in your textbook contains chi-square values used in this calculation.)

The probability that these data fit a 9:3:3:1 ratio and that the difference between the observed and expected is due to chance is less than 1%.

The probability that these data fit a 9:3:3:1 ratio and that the difference between the observed and expected is due to chance is between 50 and 70%.

The chi-square value for these data is 4.55. With 3 degrees of freedom, we expect this much variation from the expected values between 20 and 30% of the time. From a chi-square table, we can learn that these observed values support hypothesis of a 9:3:3:1 ratio greater than 5% of the time.

The probability that these data fit a 9:3:3:1 ratio and that the difference between the observed and expected is due to chance cannot be determined from the dataset given.

Answer: C

In rabbits, agouti coat color is conditioned by the
*
A*
gene, solid coat color is conditioned by the
*
a*
gene, black coat color is conditioned by
*
B*
gene, and blue coat color is produced by the
*
b*
gene. Thus, in a cross between rabbits heterozygous at
the A and B loci, one would expect the 9:3:3:1 ratio. In a sample of
320 progeny rabbits obtained from 62 litters from such crosses, the
following results were expected and observed.

**Expected**

**Observed**

Black agouti (A-B-)

180

165

Blue agouti (or opal) (A-bb)

60

74

Solid black (aaB-)

60

52

Solid blue (aabb)

20

29

**Total**

320

320

What is the probability that the observed results would be obtained by chance alone if the 9:3:3:1 model is correct?

(Table 11.5 in your textbook contains chi-square values used in this calculation.)

The probability that these data fit a 9:3:3:1 ratio and that the difference between the observed and expected is due to chance is between 1 and 5%.

The probability that these data fit a 9:3:3:1 ratio and that the difference between the observed and expected is due to chance is less than 95%.

The probability that these data fit a 9:3:3:1 ratio and that the difference between the observed and expected is due to chance is between 20 and 30%.

The probability that these data fit a 9:3:3:1 ratio and that the difference between the observed and expected is due to chance cannot be determined from the dataset given.

Answer: A

In the preceding two questions, the initial litters of rabbits observed (7 litters with 32 individuals) appeared to produce progeny that fit the 9:3:3:1 ratio. However, when more litters were observed (62 litters with 320 individuals), those data do not fit the 9:3:3:1 ratio. Which of these observations is most trustworthy, and why?

The smaller sample is more trustworthy because it has a better chi-square value.

The larger sample is more trustworthy because a larger number of litters produces a greater number of degrees of freedom in calculating chi-square.

The larger sample is more trustworthy because random chance cannot influence the larger numbers as easily as smaller numbers.

Both samples are equally reliable.

Answer: C

Which of Mendel's laws of inheritance apply only to pea plants and not other organisms?

the law of segregation

the law of independent assortment

Both the law of segregation and the law of independent assortment apply to organisms that reproduce sexually.

Both the law of segregation and the law of independent assortment apply only to peas, because that was all Mendel studied.

Answer: C

Mendel worked with smooth (S-) and wrinkled (ss) peas and with yellow (Y-) and green (yy) peas. By crossing smooth yellow heterozygotes and examining the progeny of such crosses, Mendel reasoned that if these 2 genes assorted independently, they would give the 9:3:3:1 phenotypic ratio, e.g., 90 - S-Y- seeds : 30 - S-yy seeds : 30 ssY- seeds : 10 - ssyy seeds. What would we now conclude if Mendel had observed 120 - S-Y- seeds : 0 - S-yy seeds : 0 ssY- seeds : 40 - ssyy seeds instead? Note that Mendel had no basis for the interpretation of his data, but today we do have an interpretation that would explain these alternative observations.

By random chance there were great discrepancies in the dataset. Statistics tells us that such things would happen with a low frequency.

We would conclude that the gene loci for smooth/wrinkled and yellow/green seeds do not assort independently and are therefore likely both on the same chromosome linked closely together.

Smooth/wrinkled phenotype and yellow/green phenotype are produced by the same gene. Thus, one gene cannot assort independently.

Both S-yy and ssY- genotypes are lethal and do not survive.

Answer: B