Activities Quiz
1.

Based on the results of this cross, you determine that _____. (Activity 15A)

question #1

round eyes are dominant to vertical eyes, and the absence of a tooth is dominant to the presence of a tooth
round eyes are dominant to vertical eyes, and the presence of a tooth is dominant to the absence of a tooth
the allele for round eyes is linked to the allele for no tooth
vertical eyes are dominant to round eyes, and the absence of a tooth is dominant to the presence of a tooth
vertical eyes are dominant to round eyes, and the presence of a tooth is dominant to the absence of a tooth


2.

The results of the following cross indicate that the _____. (Activity 15A)

question #2

absence of a tooth is dominant to the presence of a tooth
two genes are linked
MendAlien species is polyploid
absence of a tooth is dominant to vertical eyes
two genes assort independently


3.

An F1 individual can produce _____ different gametes when both eye and tooth genes are considered. (Activity 15A)

question #3

1
2
3
4
5


4.

Given these chromosomes, which of the choices represents the possible recombinant gametes? (Activity 15A)

question #4

answer #1
 answer #2
 answer #3
 answer #4
 answer #5


5.

The results of a F1 testcross are: 250 bald head, four ears : 247 hairy head, six ears : 21 bald head, six ears : 19 hairy head, four ears. How many map units apart are the head and ear genes? (Activity 15A)

3.5
3.9
7.5
50.5
92.6


6.

Which of the individuals is homozygous recessive for both of the gene pairs? (Activity 15A)

question #6

the female parent
the male parent
the F1 generation
both the male and female parents
both the male parent and the F1 generation


7.

The results of the following cross indicates that the _____. (Activity 15A)

question #7

absence of a tooth is dominant to the presence of a tooth
two genes are linked
MendAlien species is polyploid
absence of a tooth is dominant to vertical eyes
two genes assort independently


8.

The recombination frequency between gene A and gene B is 8.4%, the recombination frequency between gene A and gene C is 6.8%, and the recombination frequency between gene B and gene C is 15.2%. Which of these is the correct arrangement of these genes? (Activity 15A)

ABC
ACB
BCA
CAB
CBA


9.

A color-blind woman mates with a male with normal color vision. Which of these results would indicate that color blindness is caused by an X-linked recessive allele? (Activity 15B)

Half of the sons and half of the daughters are color-blind.
All of the daughters, and none of the sons, are color-blind.
All of the sons, and none of the daughters, are color-blind.
The offspring occur in a ratio of 3 color-blind : 1 normal vision.
The offspring occur in a ratio of 9 normal vision males : 3 color-blind vision males : 3 normal vision females : 1 color-blind female.


10.

Color blindness is an X-linked recessive trait. A color-blind man has a daughter with normal color vision. What is the genotype of the daughter? (Activity 15B)

XCXc
XCXC
XcXc
XCY
XcY


11.

Color blindness is an X-linked recessive trait. A color-blind man has a daughter with normal color vision. She mates with a male who has normal color vision. What is the expected phenotypic ratio of their offspring? (Activity 15B)

1 normal vision female : 1 color-blind female : 1 normal vision male : 1 color-blind male
All the offspring have normal color vision.
2normal vision females : 1 normal vision male : 1 color-blind male
3 normal vision female : 1 color-blind male
1 normal vision female : 1 color-blind male


12.

Color blindness is an X-linked recessive trait. A color-blind man has a daughter with normal color vision. She mates with a color-blind male. What is the expected phenotypic ratio of their offspring? (Activity 15B)

All the offspring have normal color vision.
2 normal vision females : 1 normal vision male : 1 color-blind male
3 normal vision female : 1 color-blind male
1normal vision female : 1 color-blind female : 1 normal male : 1 color-blind male
1 normal vision female : 1 color-blind male


13.

Color blindness is an X-linked recessive trait. A woman who is homozygous for normal color vision mates with a color-blind male. What is the expected phenotypic ratio of their offspring? (Activity 15B)

2 normal vision females : 1 normal vision male : 1 color-blind male
3 normal vision female : 1 color-blind male
All the offspring have normal color vision.
1 normal vision female : 1 color-blind female : 1 normal male : 1 color-blind male
1 normal vision female : 1 color-blind male


14.

Color blindness is an X-linked recessive trait. A color-blind woman mates with a male with normal color vision. What is the expected phenotypic ratio of their offspring? (Activity 15B)

1 normal vision female : 1 color-blind female : 1 normal male : 1 color-blind male
1normal vision daughter : 1 color-blind son
2 normal vision females : 1 normal-vision male : 1 color-blind male
3 normal vision female : 1 color-blind male
All the offspring have normal color vision.


15.

Color blindness is an X-linked recessive trait. Under what conditions can an unaffected male have a color-blind daughter? (Activity 15B)

His mate is color-blind.
He can't.
He is heterozygous for color vision.
His father is color-blind.
His mother is color-blind.


16.

Hypophosphatemia (vitamin D-resistant rickets) is inherited as an X-linked dominant. An unaffected woman mates with a male with hypophosphatemia. What is the expected phenotypic ratio of their offspring? (Activity 15B)

1 normal female : 1 female with hypophosphatemia : 1 normal male : 1 male with hypophosphatemia
1 normal daughter : 1 son with hypophosphatemia
1daughter with hypophosphatemia : 1 normal son
2 normal females : 1 normal male : 1 male with hypophosphatemia
3 normal female : 1 male with hypophosphatemia


17.

Hypophosphatemia (vitamin D-resistant rickets) is inherited as an X-linked dominant. A woman without hypophosphatemia and a man with hypophosphatemia have a daughter. The daughter mates with a male without hypophosphatemia. What is the expected phenotypic ratio of their offspring? (Activity 15B)

2 unaffected females : 1 unaffected male : 1 male with hypophosphatemia
1unaffected female : 1 female with hypophosphatemia : 1 unaffected male : 1 male with hypophosphatemia
1 unaffected daughter : 1 son with hypophosphatemia
1 daughter with hypophosphatemia : 1 unaffected son
3 unaffected females : 1 male with hypophosphatemia


18.

Suppose that having three nostrils is a Y-linked character. A woman with two nostrils mates with a man with three nostrils. What is the expected phenotypic ratio of their offspring? (Activity 15B)

1daughter with two nostrils : 1 son with three nostrils
1 daughter with two nostrils: 1 daughter with three nostrils: 1 son with two nostrils : 1 son with three nostrils
2 daughters with two nostrils: 1 son with two nostrils: 1 son with three nostrils
2 daughters with three nostrils: 1 son with two nostrils: 1 son with three nostrils
2 sons with two nostrils: 1 daughter with two nostrils: 1 daughter with three nostrils


19.

Humans are diploid and have 46 chromosomes (or two sets). How many sets of chromosomes are found in each human gamete? (Activity 15C)

1
2
3
4
5


20.

Humans are diploid and have 46 chromosomes. How many chromosomes are found in each human gamete? (Activity 15C)

12
23
36
45
92


21.

_____ is the process by which haploid gametes form a diploid zygote. (Activity 15C)

Embryogenesis
Meiosis
Gastrulation
Fertilization
Mitosis


22.

A particular diploid plant species has 48 chromosomes, or two sets. A mutation occurs and gametes with 48 chromosomes are produced. If self-fertilization occurs, the gametes will have _____ set(s) of chromosomes. (Activity 15C)

1
2
3
4
5


23.

Which of these terms applies to an organism with extra sets of chromosomes? (Activity 15C)

monosomy
haploid
trisomy
polyploid
diploid


24.

Mutant tetraploid plants _____. (Activity 15C)

are usually sickly
are able to interbreed with their parents
have an odd number of chromosomes
are unable to interbreed with a diploid plant
unable to self-fertilize


25.

Most polyploid plants arise as a result of _____. (Activity 15C)

self-fertilization
a mutation of gamete formation
meiosis
mitosis
hybridization



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