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Unit 6 Review

Total questions: 27

Worksheet time: 31mins

Name
Class
Date
1.

The sex chromosomes of a normal female are:

a)

XXX

b)

XYY

c)

XX

d)

XY

2.

A gene that is located on one of the sex chromosomes is said to be:

a)

multiple

b)

recessive

c)

dominant

d)

sex-linked

3.

The sex chromosomes of a normal male are:

a)

XXX

b)

XXY

c)

XX

d)

XY

4.

How many pairs of chromosomes are found in a normal human cell?

a)

46

b)

23

c)

1

d)

2

5.

Blood type alleles may be A, B, or O. Which of the following describes this?

a)

translocation

b)

trisomy

c)

multiple-allele gene

d)

two-gene system

6.

What would be the genotypes of the parents of the following offspring: Tt, tt, Tt, tt? (Hint, draw your punnett square and work backwards).

a)

tt on the top; tt along the side

b)

Tt on the top; tt along the side

c)

Tt on the top; Tt along the side

d)

TT on the top; TT along the side

7.

If the cross from question 7 produced 240 offspring, how many of the offspring would be expected to have the homozygous recessive alleles. (Hint: look at your punnett square, determine how many are homozygous recessive then divide the total number of offspring by that number. 240 divided by ?)

a)

0

b)

60

c)

120

d)

180

8.

If a man with bushy eyebrows with the genotype of Bb married a woman with fine eyebrows with the genotype, bb, what are the ratios/percents of the possible genotypes of the children?

a)

all are bb (100%)

b)

all are Bb (100%)

c)

1BB, 2Bb, 1bb (25%, 50%, 25%)

d)

1/2 Bb, 1/2 bb (50%, 50%)

9.

Use the following Punnett Square, which shows the alleles for black (B) and white (b) fur in Guinea pigs and short (S) and long (s) fur, to answer the next three questions (12-14). The alleles for black fur and short fur are dominant to the alleles for white fur and long fur. We crossed two guinea pigs that were heterozygous for both traits.


What percentage of the offspring have white fur?

a)

50% (8/16)

b)

75% (12/16)

c)

25% (4/16)

d)

100% (16/16)

10.

Use the following Punnett Square, which shows the alleles for black (B) and white (b) fur in Guinea pigs and short (S) and long (s) fur, to answer the next three questions (12-14). The alleles for black fur and short fur are dominant to the alleles for white fur and long fur. We crossed two guinea pigs that were heterozygous for both traits.


What is the genotype of each parent? (hint: use the gametes provided to do the foil method BACKWARDS)

a)

BBSS;Bbss

b)

BBss;bbSS

c)

BbSs;BbSs

d)

bbSS;BBSS

11.

Use the following Punnett Square, which shows the alleles for black (B) and white (b) fur in Guinea pigs and short (S) and long (s) fur, to answer the next three questions (12-14). The alleles for black fur and short fur are dominant to the alleles for white fur and long fur. We crossed two guinea pigs that were heterozygous for both traits.


What are the phenotyopic ratios of this cross? (Hint: count how many times each genotype occurs)

a)

16:0

b)

8:8

c)

4:4:4:4

d)

9:3:3:1

12.

At the end of the process, which of the following produces 4 cells from 1 cell?

a)

meiosis

b)

mitosis

c)

replication

d)

fertilization

13.

At the end of the process, which of the following produces 2 cells from 1?

a)

meiosis

b)

mitosis

c)

replication

d)

fertilization

14.

Which sex chromosome(s) can an egg cell possibly contain?

a)

X only

b)

Y only

c)

X and Y

d)

AB

15.

Which sex chromosome(s) can a sperm cell possibly contain?

a)

X only

b)

Y only

c)

X and Y

d)

AB

16.

If a sperm cell (XY) and an egg cell (XX) were crossed, what is the probability of having a male offspring? (Make a punnett square)

a)

10%

b)

25%

c)

75%

d)

50%

17.

A diagram of photographed chromosomes organized in numerical order is called a:

a)

pedigree

b)

punnett square

c)

karyotype

d)

gene map

18.

What is the disorder shown in the diagram?

a)

Monosomy 21

b)

Monosomy 10 (Neiman Pic disease)

c)

Trisomy 21 (down syndrome)

d)

Quadrosomy 6

19.

What is the sex of the person in the karyotype shown?

a)

male

b)

female

c)

both

d)

neither

20.

At the end of the process, which of these produces diploid cells from a diploid cell?

a)

meiosis

b)

mitosis

c)

replication

d)

osmosis

21.

At the end of the process, which of the following produces haploid cells from a diploid cell?

a)

meiosis

b)

mitosis

c)

replication

d)

osmosis

22.

In some types of dogs, the allele for wire hair (W) is dominant to the allele for smooth hair (w). A dog breeder mated two dogs. The genotypes for the two dogs are: Ww x Ww. What is the probability that a dog with wire hair will be produced from this mating? (Hint: draw a punnett square)

a)

25% (1/4)

b)

50% (1/2)

c)

75% (3/4)

d)

100% (4/4)

23.

In humans tongue rolling (T) is dominant over non-rolling (t) and right-handedness (R) is dominant over left-handedness (r). A person who is heterozygous for both traits could produce which of the following sets of gametes. (Hint: use the foil method)

a)

TT, RR, tt, rr

b)

TR, TR, Tr, Tr

c)

TT, RR, tr, tr

d)

TR, Tr, tR, tr

24.

In carnations, red flower color is incompletely dominant over white flowers. What phenotype would a heterozygous carnation display? (Hint: incomletely dominant traits are shown as blended traits)

a)

red and white speckled

b)

pink

c)

red

d)

white

25.

In rats, black fur (B) is incompletely dominant over white fur (b) and the heterozygote displays gray fur. If two gray rats mate, what percentage of their offspring would be expected to have white fur?

a)

25%

b)

50%

c)

75%

d)

100%

26.

In certain chicken breeds, black (B) and white (W) feather color are co-dominant. What is the genotype and phenotype of a heterozygous chick?

a)

BW, black and white speckled

b)

BB, black

c)

WW, white

d)

BW, gray

27.

Hemophilia is a sex-linked disease where the blood won’t clot normally. Normal (X) is dominant over hemophilia (XH). A hemophiliac male is crossed with a carrier female. What will be the phenotypes of the offspring? (Hint: draw your punnett square. For females BOTH Xs have to have the allele, for males only one X has to have the allele)

a)

2 females with hemophilia, 2 boys with hemophilia

b)

2 females with hemophilia, 1 male with hemophilia, 1 normal male

c)

1 hemophilic female, 1 carrier female, 2 males with hemophilia

d)

1 hemophilic female, 1 carrier female, 1 hemophilic male, 1 normal male