# Genetics Homozygous Quiz Questions

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There are two types of Allele that a person can have and they are either dominant or recessive. Over the past week in genetics class. The quiz below is made up on a series of questions that are designed to test your understanding when it comes to genetics homozygous. Do you feel like you can tackle it with ease? Give it a shot and see how much you understood.

• 1.

### In  cats, the Manx (M) allele results in cats without tails, while the recessive allele (m) allows for tail growth. Homozygous M/M cats dies before birth. If 2 Manx cats mate, and they have 3 offspring, what is the chance that at least one has a tail?

• A.

2/3

• B.

1/3

• C.

8/27

• D.

19/27

D. 19/27
Explanation
The chance that none of the Manx cats has a tail is 2/3 X 2/3 X 2/3 =8/27
The chance that at least one has a tail is 27/27-8/27=19/27

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• 2.

### Red hair is recessive to dark hair. Jenny has red hair and has 2  children with Tony who does not have red hair, but his mother does. What is the chance that both of their kids have red hair?

• A.

1/4

• B.

1/2

• C.

1/8

• D.

3/4

A. 1/4
Explanation
R= Dark hair r=red hair. Jenny is r/r Tony must be R/r since his mother has red hair but he does not. If they have 2 kids, the chance they both have red hair is 1/2 X 1/2 =1/4

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• 3.

### The A and B blood types differ by a carbohydrate at the end of a chain of carbohydrates.  In type O blood neither type of carbohydrate  is added to the chain. There is a condition called The Bombay phenotype in which the carbohydrate which normally has the A or B carbohydrates attached is missing. This is a recessive condition on another chromosome. Individuals who have this condition cannot have blood type A, B or AB even if their ABO alleles give that blood type.    We will use M to indicate the dominant allele and m to indicate the recessive allele.  Someone who is type O; +/- ; M/m has children with someone who is type AB; +/-; M/m. What is the chance that they have a type  B positive child?   Note that the M gene has no effect on the +/- gene.

• A.

3/4

• B.

9/32

• C.

9/64

• D.

1/32

B. 9/32
Explanation
There is a 3/4 chance of not having the Bombay phenotype. There is a 1/2 chance of being B and there is a 3/4 chance of being positive. 3/ X 1/2 X 3?4 =9/32

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• 4.

### Billy has the following genotype IA IB ; +/-  Penny has the following genotype IA IA; +/-What is the chance that they have a child who is type  A negative?

• A.

None

• B.

1/4

• C.

1/8

• D.

1/16

C. 1/8
Explanation
There is a 1/2 chance that the child is type A and a 1/4 chance that the child is -. 1/2 X 1/2 =1/4

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• 5.

### Billy has the following genotype IA IB ; +/-  Penny has the following genotype IA IA; +/- Which one of the following blood types could be found among the offspring

• A.

Type O positive

• B.

Type AB negative

• C.

Type B positive

• D.

All of the above could be found

B. Type AB negative
Explanation
Neither parent has an i allele, so the offspring can not be type 0. All offspring get at least one A allele, so the offspring can not be type B.

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• 6.

### This disease is lethal and strikes early in childhood.

• A.

True

• B.

False

B. False
Explanation
Since parents pass it on to their offspring, it is unlikely to have a child hood onset.

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• 7.

### The pedigree below is

• A.

Autosomal Recessive

• B.

Autosomal Dominant

• C.

• D.

B. Autosomal Dominant
Explanation
Every individual who has it has at least one parent who has it. Males pass it to both sons and daughters so it can not be sex linked dominant.

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• 8.

### The pedigree below is

• A.

Autosomal Recessive

• B.

Autosomal Dominant

• C.

• D.

Explanation
It skips generations so it can not be dominant. It only affects males, making it likely to be sex linked.

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• 9.

### The genotype off the individual indicated by the arrow is

• A.

XA/Y (A should be superscript)

• B.

X/A/Xa (A and a should be superscript)

• C.

XA/XA (A should be superscript)

• D.

A/a

B. X/A/Xa (A and a should be superscript)
Explanation
Since it is sex linked, the X designation is used. Since the indicated female gives rise to both affected and unaffected males, she must be heterozygous. (Also see that her father has it and she does not. This also means she must be heterozygous

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• Mar 21, 2023
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• Feb 04, 2014
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