Mutations (College Board AP® Biology)

Exam Questions

28 mins9 questions
1a3 marks

Information flow in cells can be regulated by various mechanisms.

Describe the role of THREE of the following in the regulation of protein synthesis:

  • RNA splicing

  • repressor proteins

  • methylation

  • siRNA 

1b4 marks

Information flow can be altered by mutation. Describe THREE different types of mutations and their effect on protein synthesis.

1c4 marks

Identify TWO environmental factors that increase the mutation rate in an organism, and discuss their effect on the genome of the organism.  

1d1 mark

Epigenetics is the study of heritable changes in the phenotype caused by mechanisms other than changes in the DNA sequence. Describe ONE example of epigenetic inheritance. 

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2a2 marks

The table below shows the amino acid sequence of the carboxyl-terminal segment of a conserved polypeptide from four different, but related, species. Each amino acid is represented by a three-letter abbreviation, and the amino acid residues in the polypeptide chains are numbered from the amino end to the carboxyl end. Empty cells indicate no amino acid is present. 

 

Relative Amino Acid Position

Species 

1

2

3

4

5

6

7

8

9

10

I

Val

His

Leu

Val

Glu

Glu

His

Val

Glu

His

II

Val

His

Leu

Lys

Glu

Glu

His

Val

Glu

His

III

Val

His

Leu

Val

Glu

Glu

His

Val

 

 

IV

Val

His

Leu

Val

Arg

Trp

Ala

Cys

Met

Asp

Assuming that species I is the ancestral species of the group, explain the most likely genetic change that produced the polypeptide in species II and the most likely genetic change that produced the polypeptide in species III. 

2b2 marks

Predict the effects of the mutation on the structure and function of the resulting protein in species IV. Justify your prediction.

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3a2 marks

Gibberellin is the primary plant hormone that promotes stem elongation. GA 3-beta-hydroxylase (GA3H) is the enzyme that catalyzes the reaction that converts a precursor of gibberellin to the active form of gibberellin. A mutation in the GA3H gene results in a short plant phenotype. When a pure-breeding tall plant is crossed with a pure-breeding short plant, all offspring in the F1 generation are tall. When the F1 plants are crossed with each other, 75 percent of the plants in the F2 generation are tall and 25 percent of the plants are short.

qp3-2017-frq-ap-biology

Figure 1. The universal genetic code

The wild-type allele encodes a GA3H enzyme with alanine (Ala), a nonpolar amino acid, at position 229. The mutant allele encodes a GA3H enzyme with threonine (Thr), a polar amino acid, at position 229. Describe the effect of the mutation on the enzyme and provide reasoning to support how this mutation results in a short plant phenotype in homozygous recessive plants.

3b1 mark

Using the codon chart provided, predict the change in the codon sequence that resulted in the substitution of alanine for threonine at amino acid position 229.

3c1 mark

Describe how individuals with one (heterozygous) or two (homozygous) copies of the wild-type GA3H allele can have the same phenotype.

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4a1 mark

Table 1. The data show the growth of haploid Saccharomyces cerevisiae yeast strains on media that differ in amino acid content. A plus sign (+) indicates that the yeast strains grow, and a minus sign (—) indicates that the strains do not grow.

 

 

STRAINS

 

MEDIUM

Wild Type

Mutant 1

Mutant 2

Treatment I

All amino acids present

+

+

+

Treatment II

No amino acids present

+

Treatment III

All amino acids present EXCEPT methionine

+

+

Treatment IV

All amino acids present EXCEPT leucine

+

+

The yeast Saccharomyces cerevisiae is a single-celled organism. Amino acid synthesis in yeast cells occurs through metabolic pathways, and enzymes in the synthesis pathways are encoded by different genes. The synthesis of a particular amino acid can be prevented by mutation of a gene encoding an enzyme in the required pathway.

A researcher conducted an experiment to determine the ability of yeast to grow on media that differ in amino acid content. Yeast can grow as both haploid and diploid cells. The researcher tested two different haploid yeast strains (Mutant I and Mutant 2), each of which has a single recessive mutation, and a haploid wild-type strain.

The resulting data are shown in Table 1.

Identify the role of treatment I in the experiment.

4b1 mark

Provide reasoning to explain how Mutant I can grow on treatment I medium but cannot grow on treatment III medium.

4c1 mark

Yeast mate by fusing two haploid cells to make a diploid cell. In a second experiment, the researcher mates the Mutant 1 and Mutant 2 haploid strains to produce diploid cells. Using the table provided, predict whether the diploid cells will grow on each of the four media. Use a plus sign (+) to indicate growth and a minus sign (−) to indicate no growth. 

 

 

STRAINS

 

MEDIUM

Wild Type

Mutant 1

Mutant 2

Diploid Cells
Produced by
Mating
Mutant 1 and
Mutant 2

Treatment I

All amino acids present

+

+

+

 

Treatment II

No amino acids present

+

 

Treatment III

All amino acids present EXCEPT methionine

+

+

 

Treatment IV

All amino acids present EXCEPT leucine

+

+

 

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