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AP Biology - Chapter 12

Total questions: 58

Worksheet time: 40mins

Name
Class
Date
1.
One of the main reasons genes assort independent of one another  is that 
a)
they produce unrelated traits
b)
they produce related traits
c)
they are on the same chromosome
d)
they are on different chromosomes.
2.
In humans, hemophilia is an X-linked recessive trait. If a man and a woman have a son who is affected with hemophilia, which of the following is definitely true?
a)
The mother carries an allele for hemophilia
b)
The father carries an allele for hemophilia
c)
The father is afflicted with hemophilia
d)
both parents carry an allele for hemophilia
3.

A researcher crossed a male Drosophila melanogaster having a grey body and long wings with a female D. melanogaster having a black body and apterous wings. The following distribution of traits was observed in the offspring.

What conclusion is supported by the data?

a)

The alleles for gray body and long wings are dominant.

b)

The alleles for gray body and long wings are recessive.

c)

Genes for the two traits are located on two different chromosomes, and independent assortment occurred.

d)

Genes for the two traits are located close together on the same chromosome, and crossing over occurred between the two gene loci.

4.

A male fruit fly (Drosophila melanogaster) with red eyes and long wings was mated with a female with purple eyes and vestigial wings. All of the offspring in the F1 generation had red eyes and long wings. These F1 flies were test crossed with purple-eyed, vestigial-winged flies. Their offspring, the F2 generation, appeared as indicated below.


If in the F1 and F2 generations the same characteristics appeared in both males and females, it would be safe to assume that these traits for eye color and wing length are...

a)

are sex-linked

b)

are sex-influenced characteristics

c)

are autosomal characteristics

d)

follow the Mendelian rule of independent assortment

5.

A male fruit fly (Drosophila melanogaster) with red eyes and long wings was mated with a female with purple eyes and vestigial wings. All of the offspring in the F1 generation had red eyes and long wings. These F1 flies were test crossed with purple-eyed, vestigial-winged flies. Their offspring, the F2 generation, appeared as indicated below.


Based on the F1 generation, what was the phenotype of the male fruit fly?

a)

RRLL

b)

RrLl

c)

rrll

d)

RRll

6.

The pedigree of a family with a history of a particular genetic disease is shown below. Squares represent males and circles represent females. Shaded symbols represent those who have the disease.


If Individual 2 were to marry a woman with no family history of the disease, which of the following would most likely be true of their children?

a)

All of the children would have the disease.

b)

None of the children would have the disease

c)

Only the sons would have the disease

d)

All of the sons would be carriers of the disease

7.

The process depicted in the image above is best summarized by which of the following descriptions?

a)

During the synthesis phase of the cell cycle, DNA molecules replicate to generate identical daughter cells

b)

Centromeres align specific gene sequences of homologous chromosomes during mitotic divisions.

c)

The spindle apparatus attaches at chiasma during metaphase of mitosis.

d)

During meiosis, crossing over leads to recombination of alleles between homologous chromosomes

8.
Which genes would show the highest frequency of crossing over?
a)
A-B
b)
A-D
c)
D-C
d)
B-C
9.
Fruit fly body cells have 8 chromosomes. After mitosis, you would expect a resulting fruit fly daughter cell to have ...
a)
16 chromosomes.
b)
46 chromosomes.
c)
8 chromosomes.
d)
4 chromosomes.
10.
X-linked conditions are more common in men than in women because
a)
The genes associated with the  X linked conditions are linked to the X chromosome, which determines maleness.
b)
Men need to inherit only one copy of the recessive allele for the condition to be fully expressed
c)
Women simply do not develop the disease regardless of their genetic composition
d)
The sex chromosomes are more active in men than in women
11.
What type of heredity is shown in the pedigree?
(hint: check your notes - "modes of inheritance")
a)
Sex-Linked Dominant
b)
Sex-Linked Recessive
c)
Autosomal Dominant
d)
Autosomal Recessive 
12.
In Drosophila, the genes for eye colour (pr), wing shape (vg), and body colour (eb) are all found on the same chromosome. The following crossover frequencies for these genes were determined by experimentation. Determine the sequence of genes on the chromosome.
a)
eb - vg - pr
b)
vg - eb - pr
c)
vg - pr - eb
d)
pr- vg - eb
13.
Tendency for alleles of genes on the same chromosome to be inherited together
a)
genetic linkage
b)
genetics
c)
Punnett square
d)
gene locus
14.
Hemophilia is a recessive x-linked disorder.
Which genotype represents a male with hemophilia?
a)
XHXh
b)
XhXh
c)
XHY
d)
XhY
15.
A person with Turner syndrome has only one X chromosome. This means one of their gametes was missing a chromosome.
Which of the following is why gametes sometimes lack a complete chromosome?
a)
Incomplete dominance
b)
Nondisjunction
c)
Inversion mutation
d)
Substitution mutation
16.
The red and blue pair of chromosomes in the bottom row are
a)
genetically identical
b)
carrying the same  genes, but different alleles
c)
carrying completely different genes
d)
carrying the same genes and the same alleles
17.
Which genetic abnormality does this baby have?
a)
Partial Deletion
b)
Partial Addition
c)
Trisomy
d)
Monosomy
18.
In humans, hemophilia is an X-linked recessive trait. If a man and a woman have a son who is affected with hemophilia, which of the following is definitely true?
a)
The mother carries an allele for hemophilia
b)
The father carries an allele for hemophilia
c)
The father is afflicted with hemophilia
d)
both parents carry an allele for hemophilia
19.

If a particular pedigree follows a sex-linked recessive trait, you would expect to see

a)

more affected females

b)

all carrier males

c)

more affected males

d)

an equal number of affected males and females

20.
Chromosomes that do NOT play a role in determining the sex of an individual are called
a)
sex chromosoems
b)
X and Y
c)
autosomes
d)
A and B are correct
21.
In humans, if a non-disjunction event leads to an individual that is XXY, they would
a)
be female because they have 2 X chromosomes
b)
be male because they have a Y chromosome
c)
not survive
d)
have both male and female characteristics
22.
Unlike Mendel's experiments, when a very tall person and a very short person mate, the children are variable in height. Why is one trait (tall or short) not dominant over the other?
a)
the gene for height in humans has incomplete dominance
b)
the gene for height has multiple co-dominant alleles
c)
height in not controlled by genes in humans
d)
human height is controlled by multiple genes
23.

Describe why a male is more likely to be affected by a sex-linked trait?

a)

Males have only 1 Y chromosome

b)

Females have only 1 Y chromosome

c)

Males have only 1 X chromosome

d)

Females have only 1 X chromosomes

24.

Which of the following is the predicted ratio of a testcross with a dihybrid?

a)

9:3:3:1

b)

9:6:1

c)

1:2:1

d)

1:1:1:1

25.

Which of the following is a predicted ratio of a dihybrid cross?

a)

9:3:3:1

b)

9:3:4

c)

1:2:1

d)

1:1:1:1

26.

Snowshoe hares are brown in summer and white in winter, describe this event.

a)

Hares undergo phenotypic plasticity & environment influences gene express

b)

Hares are white in winter due to the snow binding to hair follicles

c)

Hares are brown from exposed dirt in summer

d)

Hares remain hidden in the winter months inhibiting fur darken from sun

27.

Identify the type of inheritance of the pedigree.

a)

Sex-Linked

b)

Mitochondrial

c)

Autosomal

d)

Random

28.

In animals, traits determine by the mitochondrial DNA are inherited from the _______________.

a)

mother

b)

brother

c)

father

d)

sister

29.

Describe the cause of a cross resulting in greater than 50% parental phenotypes and less then 50% recombinant phenotypes

a)

Genes are linked on the same chromosome

b)

Genes are located on different arms of the same chromosome

c)

Gene are on different chromosomes

d)

Gene are found in different gametes

30.

Which of the following best describes the phenotypic ratio of 6:3:3:2:1:1?

a)

Incomplete dominance & complete dominance on dihybrid cross

b)

Incomplete dominance on dihybrid cross

c)

Complete dominance on dihybrid cross

d)

Epistasis on dihybrid cross

31.

Which of the following best describes the inheritance pattern of the shown pedigree?

a)

Autosomal Dominant

b)

Sex-Linked Dominant

c)

Autosomal Recessive

d)

Sex-Linked Recessive

32.

What is a trisomy?

a)

A chromosomal mutation were a section of DNA was copied multiple times

b)

A condition where an individual is missing a chromosome

c)

A mutation where a section of a chromosome has removed

d)

A mutation resulting from nondisjunction where an individual has an extra copy of a chromosome

33.

If a nondisjunction occurs during meiosis II, how many gametes are affected

a)

All 4

b)

Just 1

c)

2 of the 4

d)

None of them

34.

How is it possible that genetically identical organisms can express different phenotypes

a)

Genotypes do not play a role in determining phenotypes

b)

Due to epigenetic factors that can affect gene expression

c)

This is impossible

d)

This can only result from a mutation

35.

Map the following genes: A – B = 20% A - C = 10% B – C = 10% A – D = 15%, D – C = 5%

a)

BADC

b)

CADB

c)

ACBD

d)

ACDB

36.

Why are fathers not able to pass on X-linked recessive traits to their sons?

a)

This is untrue, males can pass on x-linked traits to their sons

b)

Males will only pass on dominant X-linked traits

c)

Males cannot carry recessive x-linked traits

d)

Males do not pass on X chromosomes to their sons

37.

How can you determine if genes are linked or not by looking at the phenotypes of parents and their offspring?

a)

The genes will be inherited separately majority of the time.

b)

The genes will be inherited together greater than 50% of the time.

c)

This cannot be determined by observation

38.

When will you reject the null hypothesis in a Chi-square test?

a)

When you found no correlation between the variables

b)

When the test statistic is greater than the value on the Chi-square table

c)

When the test statistic is less than the value on the Chi-square table

d)

When the observed values are very close to the expected values

39.

If two individuals are unaffected by a recessive disorder, yet they had a child with the disorder, what must be true of the parents?

a)

The parents gametes contained a mutation

b)

The parents are both heterozygous

c)

One parent was a carrier for the disorder

d)

One parent had to be homozygous

40.

One true-breeding line of mice is obese and dark and another is lean and light. Dark is dominant to light, but obese and lean exhibit incomplete dominance. What proportion of offspring from a dihybrid cross should be dark and obese?

a)

1/16

b)

3/16

c)

1/4

d)

3/8

41.
In a sex-linked inheritance pedigree you would expect to see
a)
more affected females
b)
all carrier males
c)
more affected males
d)
an equal number of affected males and females
42.
Hemophilia is a sex-linked recessive disorder.  If a hemophiliac male mates with a normal female, how many of their daughters will be carriers?
a)
0
b)
25 %
c)
50 %
d)
100 %
43.
Hemophilia is a sex-linked recessive disorder.  If a hemophiliac male mates with a normal female, how many of their sons will be hemophiliac?
a)
0
b)
1/4
c)
1/2
d)
3/4
44.

Australian shepherd dogs can have a solid coat color (mm) or a mixed pattern coat color called merle (Mm). The homozygous dominant coat color is called a lethal white (MM) which produces pups that are deaf and blind. The Australian Shepherds of America Club discourages from mating two merles. Which of the following best explains why merle-to-merle matings are undesirable?

a)

The cross has the probability of producing litters with 50% solid coat color pups.

b)

The cross produces all homozygous recessive pups

c)

The cross has the probability of producing litters with 25% merle coat pups.

d)

The cross has a 25% chance of producing homozygous dominant pups.

45.

Australian Shepherds are a breed of dogs whose coat color is directly impacted by two different genes. The gene that determines basic coat color exhibits a dominant allele (B) for black coat color and a recessive allele (b) for red coat color. Additionally, these dogs can have a solid coat color (mm) or a mixed pattern coat color called merle (Mm). The homozygous dominant coat color is called a lethal white (MM) which produces pups that are deaf and blind. What is the probability that two red Australian shepherds will produce a black pup?

a)

0

b)

1/4

c)

1/2

d)

3/4

46.

Experiment 1:Pure white eyed male Drosophilia x pure wildtype females

f1 were wildtype

Exp 2: pure wildtype males x pure white eyed females

f1: males white eyed ; females are wildtype


Question: what can be determined by experiment 1 & 2

a)

Eye color is autosomal

b)

Eye color is sex linked

47.

Dark spots are dominant to light; Flat leaves are dominant to thick.

If there was a Ddff x ddFf

what will the probability be of a dark thick leaved plant?

a)

1/4

b)

1/2

c)

0

d)

1

48.

Experiment 1: pure vestigal wing, wildtype body males x pure wildtype wings, no bristle body females


offspring 200 wildtpye wing & wildtype body males

190 wildtype wing & wildtype body females


Experiment 2 wild type males above x vestigal wing, no bristle body female

offspring:

A- 150 Wildtype wing & body males & females

B- 24 wildtype wing, no bristle body males and females

C- 146 Vestigal wing, no bristle body males & females

D- 30 vestigal wing, wildtype body males & females


What caused the recombinant offspring

a)

crossing over of linked genes

b)

Independent assortment of these un-linked genes

49.

Morgan used fruit flies (Drosophila) in his studies for all the following reasons except

a)

The are r selective

b)

only have 8 chromosomes

c)

can easily be self or cross pollinated

d)

distinct traits & wild type (common traits)

50.

Experiment 1: pure veinless wing, wildtype body males x pure wildtype wings, dark colored body females


offspring 500 wildtpye wing & wildtype body males

490 wildtype wing & wildtype body females


Experiment 2: wild type females above x veinless wing brown body

offspring:

A- 150 Wildtype wing & body males & females

B- 153 wildtype wing, brown body males and females

C- 146 Veinless wing, brown body males & females

D- 135 Veinless wing, wildtype body males and females


What caused the Recombinant offspring

a)

Crossing over of these linked genes during in meiosis in the parents

b)

Independent assortment of theses unlinked genes (separate chromosomes) in meiosis in the parents

51.

Experiment 1: pure veinless wing, wildtype body males x pure wildtype wings, dark colored body females


offspring 500 wildtpye wing & wildtype body males

490 wildtype wing & wildtype body females


Experiment 2: wild type females above x veinless wing brown body

offspring:

A- 150 Wildtype wing & body males & females

B- 153 wildtype wing, brown body males and females

C- 146 Veinless wing, brown body males & females

D- 135 Veinless wing, wildtype body males and females


Which is the Recombinant offspring

a)

A & B

b)

A &C

c)

B & D

52.

Experiment 1: pure veinless wing, wildtype body males x pure wildtype wings, dark colored body females


offspring 500 wildtpye wing & wildtype body males

490 wildtype wing & wildtype body females


Experiment 2: wild type females above x veinless wing brown body

offspring:

A- 150 Wildtype wing & body males & females

B- 159 wildtype wing, brown body males and females

C- 146 Veinless wing, brown body males & females

D- 145 Veinless wing, wildtype body males and females


What is the map distance between these 2 genes

a)

51 map Units

b)

These genes are not linked, they are sorting independently (1:1:1:1 ratio)

53.

Experiment 1: pure vestigal wing, wildtype body males x pure wildtype wings, no bristle body females


offspring 200 wildtpye wing & wildtype body males

190 wildtype wing & wildtype body females


Experiment 2 wild type males above x vestigal wing, no bristle body female

offspring:

A- 150 Wildtype wing & body males & females

B- 24 wildtype wing, no bristle body males and females

C- 146 Vestigal wing, no bristle body males & females

D- 30 vestigal wing, wildtype body males & females


What is the map distance between these 2 genes

a)

15.4 map units

b)

84.5 map units

c)

48.6 map units

54.

Experiment 1: pure vestigal wing, wildtype body males x pure wildtype wings, no bristle body females


offspring 200 wildtpye wing & wildtype body males

190 wildtype wing & wildtype body females


Experiment 2 wild type males above x vestigal wing, no bristle body female

offspring:

A- 150 Wildtype wing & body males & females

B- 24 wildtype wing, no bristle body males and females

C- 146 Vestigal wing, no bristle body males & females

D- 30 vestigal wing, wildtype body males & females


Question which apply

a)

A testcross was done

b)

A reciprocal cross was done

c)

A & D are recombinant offspring

d)

Wildtype wing and wildtype body are Autosomal dominant

e)

These genes are not on the same chromosomes & sorting independently

55.

Experiment 1: Pure tall legs,wrinkled skin males X Pure Short legs, smooth skin females ( mole rates)

F1: Tall smooth males and female


Which genetic concepts apply

a)

polygenic

b)

autosomal domiance & recessive traits

c)

Sex-linked traits

d)

Dihybrid cross

e)

Sorting Independently

56.

Experiment 1: Pure tall stem,wrinkled seeds X Pure Short stem, round seeds

F1: Tall Round

Experiment 2: F1 xF1

off springs were counted , but their phenotypes were not noted; assuming that the genes are sorting independently, which of these offspring were Tall ROUND

A- 44

B-135

C-138

D-400

a)

B or C

b)

A

c)

D

57.

Experiment 1: Pure tall stem,wrinkled seeds X Pure Short stem, round seeds

F1: Tall Round

Experiment 2: F1 xF1

off springs were counted , but their phenotypes were not noted; assuming that the genes are sorting independently, which of these offspring were SHORT WRINKLED

A- 44

B-135

C-138

D-400

a)

B or C

b)

A

c)

D

58.

Experiment 1: Pure tall stem,wrinkled seeds X Pure Short stem, round seeds

F1: Tall Round

Experiment 2: F1 xF1

off springs were counted , but their phenotypes were not noted; assuming that the genes are sorting independently, which of these offspring were TALL WRINKLED

A- 44

B-135

C-138

D-400

a)

B or C

b)

A

c)

D