Wayground logo

Free Printable Worksheets

Font size

S
M
L
XL
Worksheets

Genetics Review

Total questions: 54

Worksheet time: 1hrs 7mins

Name
Class
Date
1.

Oompahs can have red, blue, or purple hair. The allele that controls this trait is incompletely dominant. Use R for red and B for blue.

Orville Oompah has Purple hair and is married to Opal Oompah who brags that she has the

bluest hair in the valley.


What are the possible Genotypes for the offspring?

a)

RR, RB, BB

b)

RR, RB

c)

BB, RB

d)

RB

2.

Oompahs can have red, blue, or purple hair. The allele that controls this trait is incompletely dominant. Use R for red and B for blue.

Orville Oompah has Purple hair and is married to Opal Oompah who brags that she has the

bluest hair in the valley.

What are the possible phenotypes for the offspring?

a)

Red, Purple, Blue

b)

Red, Purple

c)

Blue, Purple

d)

Purple

3.

Oompahs can have red, blue, or purple hair. The allele that controls this trait is incompletely dominant. Use R for red and B for blue.

Orville Oompah has Purple hair and is married to Opal Oompah who brags that she has the

bluest hair in the valley.

What is the probability a child would have red hair?

a)

0%

b)

25%

c)

50%

d)

75%

e)

100%

4.

Oompahs can have red, blue, or purple hair. The allele that controls this trait is incompletely dominant. Use R for red and B for blue.

Orville Oompah has Purple hair and is married to Opal Oompah who brags that she has the

bluest hair in the valley.

What is the probability a child would have Blue hair?

a)

0%

b)

25%

c)

50%

d)

75%

e)

100%

5.

Oompahs can have red, blue, or purple hair. The allele that controls this trait is incompletely dominant. Use R for red and B for blue.

Orville Oompah has Purple hair and is married to Opal Oompah who brags that she has the

bluest hair in the valley.

What is the probability a child would have Purple hair?

a)

0%

b)

25%

c)

50%

d)

75%

e)

100%

6.

In some chickens the gene for feather color is controlled by codominance. The allele for Black is B and the allele for white is W. The heterozygous phenotype is known as erminette (Black and white spotter)

Two Erminete chickens are crossed with each other. 


What is the Genotype for box #1?

a)

BB

b)

WW

c)

BW

7.

In some chickens the gene for feather color is controlled by codominance. The allele for Black is B and the allele for white is W. The heterozygous phenotype is known as erminette (Black and white spotter)

Two Erminete chickens are crossed with each other. 


What is the phenotype for box #4?

a)

Black

b)

White

c)

Erminette

8.

In some chickens the gene for feather color is controlled by codominance. The allele for Black is B and the allele for white is W. The heterozygous phenotype is known as erminette (Black and white spotter)

Two Erminete chickens are crossed with each other. 


What are the probability the offspring would be Erminete?

a)

0%

b)

25%

c)

50%

d)

75%

e)

100%

9.

In some chickens the gene for feather color is controlled by codominance. The allele for Black is B and the allele for white is W. The heterozygous phenotype is known as erminette (Black and white spotter)

Two Erminete chickens are crossed with each other. 


What are the probability the offspring would be Black?

a)

0%

b)

25%

c)

50%

d)

75%

e)

100%

10.

In humans hemophilia is a recessive (h) X-Linked trait.

A male offspring was born with hemophilia, but both parents are normal.


What would the parents Genotypes have to be for this to happen?

a)

XHXH, XHY

b)

XHXh, XHY

c)

XHXH, XhY

d)

XhXh, XHY

11.

In humans hemophilia is a recessive (h) X-Linked trait.

Cross a mom who is a carrier for hemophilia with a dad who has the disease.

How many children would have the disease?

a)

0%

b)

25%

c)

50%

d)

75%

e)

100%

12.

In humans hemophilia is a recessive (h) X-Linked trait.

Cross a mom who is a carrier for hemophilia with a dad who has the disease.

How many children would have the disease?

a)

0%

b)

25%

c)

50%

d)

75%

e)

100%

13.

In humans hemophilia is a recessive (h) X-Linked trait.

Cross a mom who is a carrier for hemophilia with a dad who has the disease.

How many would be male with the disease?

a)

0%

b)

25%

c)

50%

d)

75%

e)

100%

14.

In humans hemophilia is a recessive (h) X-Linked trait.

Cross a mom who is a carrier for hemophilia with a dad who has the disease.

How many would be female with the disease?

a)

0%

b)

25%

c)

50%

d)

75%

e)

100%

15.

"Bent" is a dominant X-linked gene in mice. It results in a short, crooked tail. A recessive allele produces normal tails.

If a normal-tailed female is mated to a bent-tailed male, what are the chances of a male having a bent tail?

a)

0%

b)

25%

c)

50%

d)

75%

e)

100%

16.

"Bent" is a dominant X-linked gene in mice. It results in a short, crooked tail. A recessive allele produces normal tails.

If a normal-tailed female is mated to a bent-tailed male, what are the chances of a male having a normal tail?

a)

0%

b)

25%

c)

50%

d)

75%

e)

100%

17.

Mrs. Canal is type A and Mr. Canal is type O. They have three children named Greg, Rosalind and Biff. Greg is type 0, Rosalind is type A and Biff is type AB...

Can BIFF be there son? Explain

a)

Yes because he has a portion of A in his blood like his mom

b)

Yes because his dad is type O and that is a universal donor

c)

No because his sister is type A blood

d)

No because neither parent has the B allele

18.

Mrs. Canal is type A and Mr. Canal is type O. They have three children named Greg, Rosalind and Biff. Greg is type 0, Rosalind is type A and Biff is type AB...

What is Mrs. Canal's genotype? How do you know?

a)

IᴬIᴬ , because Rosalind is A

b)

Iᴮi because Biff is AB

c)

Iᴬi because Greg is type O

d)

ii because it is a universal donor

19.

Mrs. Canal is type A and Mr. Canal is type O. They have three children named Greg, Rosalind and Biff. Greg is type 0, Rosalind is type A and Biff is type AB...

What is Mr. Canal's Genotype?

a)

IᴬIᴬ

b)

IᴮIᴮ

c)

IᴬIᴮ

d)

ii

20.

Punnett Square: Type AB father and type O mother. What are the percentages of each offspring?

a)

Type A: 50%, Type B 0%, Type AB 50%, Type 0 0%

b)

Type A: 50%, Type B 50%, Type AB 0%, Type 0 0%

c)

Type A: 25%, Type B 25%, Type AB 25%, Type 0 25%

d)

Type A: 0%, Type B 0%, Type AB 50%, Type 0 50%

21.

What is the Genotype for Type A that had a type O parent

a)

IᴬIᴮ

b)

Iᴬi

c)

IᴬIᴬ

d)

ii

22.

What is the Genotype for Type 0

a)

IᴬIᴮ

b)

Iᴬi

c)

IᴬIᴬ

d)

ii

23.

What is the Genotype for Heterozygous Type A Blood

a)

IᴬIᴮ

b)

Iᴬi

c)

IᴬIᴬ

d)

IᴮIᴮ

24.

What is the Genotype for Homozygous "B" allele

a)

IᴬIᴮ

b)

Iᴮi

c)

IᴬIᴬ

d)

IᴮIᴮ

25.

Two parents think their baby was switched at the hospital. The mother has blood type O, the father has blood type AB and the baby has blood type B.

Complete a punnett square and then answer the question. What possible Genotypes COULD the baby have inherited from the parents? (what genotypes could the parents produce)

a)

Iᴬi

b)

IᴮIᴮ

c)

Iᴮi

d)

ii

26.

Two parents think their baby was switched at the hospital. The mother has blood type O, the father has blood type AB and the baby has blood type B.

Complete a punnett square and then answer the question. What is the mother's genotype? What is the father's genotype?

a)

Mom genotype is IᴬIᴬ, Dad genotype is IᴬIᴮ

b)

Mom genotype is ii, Dad genotype is IᴮIᴮ

c)

Mom genotype is ii Dad genotype is IᴬIᴮ

d)

Mom genotype is ii Dad genotype is ii

27.

Two parents think their baby was switched at the hospital. The mother has blood type O, the father has blood type AB and the baby has blood type B.

Complete a punnett square and then answer the question. Was the baby switched at birth? (Do they have a baby that is not theirs? )

a)

No, the baby is type B which is possible when Dad is AB

b)

No, the baby is theirs - it has type B just like the mom

c)

Yes - neither parent is type B so it must have been switched

28.

The blending of two traits is known as ---

a)

incomplete dominance

b)

codominance

29.

A condition in which both alleles for a gene are expressed so that both traits are visible is known as ---

a)

Codominance

b)

Incomplete dominance

30.

What type of inheritance pattern is shown here?

a)

Codominance

b)

Incomplete dominance

31.

Some flowers are controlled by codominance. Red flowers (FRFR), blue flowers (FBFB) and Speckled flowers (FRFB).


Cross a red flower with a speckled flower. What is the phenotypic ratio?

a)

50% red, 50% speckled

b)

25% red, 25% blue

c)

100% blue

32.

A red flowered plant (RR) is crossed with a white flowered plant (WW). The gene for petal color in these plants expresses incomplete dominance. What percentage of the offspring will have red (RR) flowers?

a)

0%

b)

25%

c)

50%

d)

100%

33.

In fruit flies, eye color is a sex linked trait. Red (R) is dominant to white (r). What is the sex and eye color of flies with the following genotype: X R X r ?

a)

male with white eyes

b)

female with red eyes

c)

male with red eyes

d)

female with white eyes

34.

Look at this cross: XB XB x Xb Y

What proportion/percent of the male children are colorblind (b)?

a)

100%

b)

75%

c)

25%

d)

0%

35.

Colorblindness is a recessive X-linked disorder.

Which genotype represents a male with normal vision?

a)

XN YN

b)

Xn Yn

c)

XN Y

d)

Xn Y

36.

A genotype where only dominant alleles were passed on

from both parents for one trait

BB AA TT LL

a)

Homozygous Recessive

b)

Homozygous Dominant

c)

Dihybrid

d)

Heredity

37.
Dominant alleles are represented by a(n)
a)
phenotype
b)
lower case letter
c)
upper case letter
d)
punnett square 
38.

A genotype where one dominant and one recessive allele was passed on from both parents for one trait.

Bb Aa Tt Ll

a)

Allele

b)

Homozygous Dominant

c)

Homozygous Recessive

d)

Heterozygous

39.

A genotype where only recessive alleles

were passed on from both parents for one trait

bb aa tt ll

a)

Heredity

b)

Homozygous Dominant

c)

Homozygous Recessive

d)

Phenotype

40.
Green peas are dominant (G) to yellow peas (g). What is the genotype for a hetereozygous dominant offspring?
a)
Gg
b)
gg
c)
GG
41.
What is the percentage of homozygous recessive offspring?
a)
0%
b)
25%
c)
50%
d)
75%
42.
Round (R) seeds are dominant to wrinkled (r) seeds.
What is the genotype for a homozygous dominant offspring?
a)
RR
b)
Rr
c)
rr
43.
What do you call the physical expression of a gene?
a)
genotype
b)
dominant
c)
phenotype
d)
allele
44.
Two brown eyed parents (Bb) have a baby. What is the chance the baby is blue eyed?
a)
0 %
b)
25%
c)
50%
d)
75%
45.
this image is an example of
a)
codominance
b)
dominance
c)
incomplete dominance
46.
A breed of chicken shows codominance for feather color. One allele codes for black feathers, another codes for white feathers. The feathers of heterozygous chickens of this breed will be:
a)
all black
b)
all white
c)
all gray
d)
speckled black and white
47.
If you have a parent who has GgBb as their genotype, what are the possible alleles they could donate to their gametes?
a)
GB, Gb, gB, gb
b)
GB, GB, gb, gb
c)
Gb, Gb, Gb, GB
d)
GG, BB, gg, bb
48.
T4 - What genotype is missing from this Punnett Square?
a)
RrYy
b)
RRYY
c)
rryy
d)
RrYY
49.
T4 - Based off this Punnett Square, what fraction of the offspring will have wrinkled, yellow seeds?
a)
9/16
b)
3/16
c)
1/16
d)
16/16
50.
In cacti, long arms (A) are dominant to short arms (a).  Suppose 2 heterozygous cacti are crossed.  What percentage of their offspring are expected to have short arms?
a)
0%
b)
25%
c)
75%
d)
100%
51.
In watermelons, green rinds (G) are dominant to striped rinds (g).  What is the genotype of a heterozygous green watermelon?
a)
GG
b)
Gg
c)
gg
d)
green
52.
T4 - What fraction of offspring from the cross AaBB x Aabb will be heterozygous for both traits (ex. AaBb) ?
a)
1/16
b)
1/4
c)
1/2
d)
3/16
53.

In flies, red eyes (R) are dominant to brown eyes (r) and brown bodies (B) are dominant to yellow bodies (b). A fly with genotype (RRbb) is mated with a fly that is (rrBb). What fraction of the offspring will have red eyes and yellow bodies?

a)

1/4

b)

2/4

c)

3/4

d)

1/2

54.
In a dihybrid cross
a)
one trait is crossed.
b)
two traits are crossed.
c)
four boxes are needed for the punnett square.
d)
mom's alleles are dominant.