AP Biology : Understanding Autosomal and Sex-Linked Inheritance

Study concepts, example questions & explanations for AP Biology

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Example Questions

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Example Question #21 : Understanding Autosomal And Sex Linked Inheritance

A certain mutation is a sex-linked recessive. If a father is not affected and the mother is affected, what is the probability that the couple will have an affected son?

Possible Answers:

Correct answer:

Explanation:

If a disease is sex linked recessive, the disease resides on the x-chromosome (unless otherwise stated). The question asks about males, who genetically have one x-chromosome and one y-chromosome. The easiest way to solve this problem is to create a punnet square (shown below).  

  Punnet square  3

As you can see, there is a 100% chance that the males will be affected. Similarly, there is a 100% chance that the daughters will be carriers of the disease. As a side note, the daughters will not be affected because they have one "good" x-chromosome to compensate for the "bad" x-chromosome that carries the mutation. 

Example Question #22 : Understanding Autosomal And Sex Linked Inheritance

Which of the following is not an example of trisomy? 

Possible Answers:

Kleinfelters

Patau

Downs

Turners

Edwards

Correct answer:

Turners

Explanation:

Trisomy is a condition of an extra chromosome copy, such as in Patau (trisomy 13), Edwards (trisomy 18), Kleinfelters (XXY), and Downs (trisomy 21). Turners syndrome is a condition of monosomy (X).

Example Question #23 : Understanding Autosomal And Sex Linked Inheritance

Using the genotype below, and assuming there is no gene linkage present, answer the following question. 

 

AABbCcDdEeFf

How many different possible gametes could be formed? 

Possible Answers:

Correct answer:

Explanation:

Each homozygous pair of alleles will result in 100% identical dominant or recessive alleles, therefore cross multiple "1" for each homozygous pair and multiple "2" for each heterozygous pair. 

Therefore, the appropriate multiplication for determining the number of possible gene combinations is 

 possible combinations. 

Example Question #24 : Understanding Autosomal And Sex Linked Inheritance

Which of the following does not result from nondisjunction of homologous chromosomes during meiosis?

Possible Answers:

Trisomy 18 (Edward's Syndrome)

All of these result from nondisjunction

 Trisomy XXY (Klinefelter's)

Trisomy 21 (Down's Syndrome)

Monosomy X (Turner's Syndrome) 

Correct answer:

All of these result from nondisjunction

Explanation:

All of the following are results of a nondisjunction event. Trisomy and monosomy disorders may occur as a result of improper separation in diploid human meiosis events.

Example Question #25 : Understanding Autosomal And Sex Linked Inheritance

Disease A is inherited through an autosomal recessive process. What is the probability that two unaffected carriers of this disease will produce an offspring that inherits disease A?

Possible Answers:

75%

0%

100%

25%

50%

Correct answer:

25%

Explanation:

The correct answer is 25% because each parent has a 50% chance of giving up their recessive allele to the offspring. There are two parents so the probability will be:

Example Question #26 : Understanding Autosomal And Sex Linked Inheritance

A man with a sex-linked disorder for color-blindness and a woman with a normal genotype have two children. What is the probability that both of them will be carriers?

Possible Answers:

100%

75%

50%

0%

25%

Correct answer:

25%

Explanation:

The man's genotype can be written as , where  represents the allele with the disorder. The woman, who has a normal genotype, is therefore .

Their possible children are represented by the Punnett square shown here:

                   

          

               

Any sons will inherit the chromosome from the father and an unaffected from their moth. Any daughters will inherit chromosomes from both parents. By necessity, all daughters will inherit an normal from the mother and an affected  from the father. All of the couple's daughters will be carriers, while all of their sons will be phenotypically normal; therefore, there is a 50% chance that their child will be a carrier.

For two children, the probability is .

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