High School Biology : Genetics Principles

Study concepts, example questions & explanations for High School Biology

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Example Questions

Example Question #1 : Hardy Weinberg Equilibrium

In an isolated population of ants in the Amazon rainforest, color is determined by a single gene. Ants that are homozygous dominant (BB) and heterozygous (Bb) for this gene are black, but ants that are homozygous recessive (bb) for this gene are brown. 

If the phenotype frequencies are 0.75 black and 0.25 brown, what are the allele frequencies within this population?

Possible Answers:

Correct answer:

Explanation:

We can use the Hardy-Weinberg equations to solve this question:

We know that the black allele (B) is dominant and the brown allele (b) is recessive.

Using these two equations, we know that 0.25 are homozygous recessive (brown). This value will be equal to .

We can then solve as follows:

Therefore, we know that the frequency of each allele is equal to 0.5.

Example Question #1 : Understanding Hardy Weinberg Calculations

In a population of rabbits, the dominant B allele produces brown fur and the recessive b allele produces white fur. There are 544 brown rabbits and 306 white rabbits in the population. 

How many of the brown rabbits would you expect to be homozygous dominant (BB)? 

Possible Answers:

Cannot be determined from the information given

Correct answer:

Explanation:

In the Hardy-Weinberg equation, the value of the term  is the percentage of homozygous dominant (BB) individuals in a population. Thus, to answer this question we will need to solve for .

With the information we are given, we can calculate , which is the percentage of the population that is homozygous recessive (bb). We know the total number of homozygous recessive rabbits (306 white rabbits) and the total number of rabbits in the population (306 + 544 = 850). We can thus divide the number of white rabbits by the population total to find :

36% of the population is homozygous recessive. We then take the square root of  to find :

This value tells us that 60% of all the alleles in the population are recessive (b).  

The Hardy-Weinberg equation is based off the idea that the total number of alleles in a population is the sum of the dominant and recessive alleles, as shown by this equation:

We can use this equation to solve for  using :

This value tells us that 40% of all the alleles in the population are dominant (B).

We can now square  to obtain . Recall that this will be the frequency of homozygous dominant individuals in the population:

The result is 16%, meaning 16% of the rabbits in the population are homozygous dominant. We can then multiply 16% by the total population to find the number of rabbits who are homozygous dominant:

Thus, 136 homozygous dominant rabbits would be expected in this population.

Example Question #2 : Understanding Hardy Weinberg Calculations

A population of frogs, which is at equilibrium, carries a dominant allele for black spots (B) and a recessive allele for no spots (b). 76 frogs have no spots, and 399 frogs have black spots.

How many of the frogs are heterozygous (Bb) for the black spot trait?

Possible Answers:

Correct answer:

Explanation:

In the Hardy-Weinberg equation, the value of the term  is the percentage of heterozygous (Bb) individuals in a population. Thus, to answer this question we will need to solve for .

With the information we are given, we can calculate , which is the percentage of the population that is homozygous recessive (bb). We know the total number of homozygous recessive rabbits (76 without spots) and the total number of rabbits in the population (76+ 399= 475). We can thus divide the number of frogs without spots by the population total to find :

16% of the population is homozygous recessive. We then take the square root of  to find :

This value tells us that 40% of all the alleles in the population are recessive (b).  

The Hardy-Weinberg equation is based off the idea that the total number of alleles in a population is the sum of the dominant and recessive alleles, as shown by this equation:

We can use this equation to solve for  using :

This value tells us that 60% of all the alleles in the population are dominant (B).

Now that we have  and , we can calculate . Recall that this will be the frequency of heterozygous individuals in the population:

The result is 48%, meaning 48% of the frogs in the population are heterozygous.  We can then multiply 48% by the total population to find the number of frogs who are heterozygous::

Thus, 228 heterozygous frogs would be expected in this population.

Example Question #1 : Understanding Hardy Weinberg Calculations

For a particular gene, the allele  is dominant to the allele . If in a population the allele frequency for  is 0.85, what is the frequency of individuals that are heterozygotes? Assume Hardy-Weinberg equilibrium conditions are met.

Possible Answers:

Correct answer:

Explanation:

There are two equations for Hardy-Weinberg equilibrium:

Using the first equation, we can substitute in 0.85 for  and solve to get .

We can then use the second equation to find the frequencies of each genotype.

 = frequency of  genotype

 = frequency of  genotype

(Why do we multiply by 2? Because we must count both  and  as heterozygotes.)

 = frequency of  genotype

Thus we get .

Example Question #1 : Understanding Hardy Weinberg Calculations

If p=0.9 and q=0.1 what will be the final genotypic frequencies of a population be under Hardy-Weinberg equilibrium? 

Possible Answers:

Correct answer:

Explanation:

Recall that under Hardy-Weinberg conditions, the allele frequencies remain constant. The formulas for Hardy-Weinberg equilibrium are:

and

Here,  is the frequency of the homozygous dominant genotype,  is the frequency of the heterozygous genotype, and  is the frequency of the homozygous recessive genotype. Since we have all the variables we need, and we want to find the genotypic frequencies (not the phenotypic) we plug into the second equation.

Be sure to check to make sure these frequencies add up to 1.

Example Question #1 : Understanding Hardy Weinberg Calculations

If p=0.3 and q=0.7, what will be the final genotypic frequencies of a population be under Hardy-Weinberg equilibrium?     

Possible Answers:

Correct answer:

Explanation:

Recall that under Hardy-Weinberg conditions, the allele frequencies remain constant. The formulas for Hardy-Weinberg equilibrium are:

and

Here,  is the frequency of the homozygous dominant genotype,  is the frequency of the heterozygous genotype, and  is the frequency of the homozygous recessive genotype. Since we have all the variables we need, and we want to find the genotypic frequencies (not the phenotypic) we plug into the second equation.

Be sure to check to make sure these frequencies add up to 1.

Example Question #1 : Understanding Hardy Weinberg Calculations

Tail length in a population of aardvarks is determined by one gene, where L=long tails and l=short tails. If the frequency of L in the population is 0.4, determine the expected frequencies of each possible genotype: homozygous dominant (LL), heterozygous (Ll), and homozygous recessive (ll). 

Possible Answers:

LL = 0.16

Ll = 0.68

ll = 0.16

LL = 0.16

Ll = 0.48

ll = 0.36 

LL = 0.4

Ll = 0.0

ll = 0.6 

LL= 0.24

Ll= 0.58

ll= 0.18

There is not enough information given to determine an answer. 

Correct answer:

LL = 0.16

Ll = 0.48

ll = 0.36 

Explanation:

To begin, we must know that we are working with the Hardy-Weinberg equations:

We can designate the dominant allele (L) as  and the recessive allele (l) as

Since the frequency of L was given (0.4) we know the the frequency of l must be:

Now we have both allele frequencies and can plug them into the equation:

So for , LL (homozygous dominant), is equal to 0.16

for , Ll (heterozygous), is equal to 0.48

for , ll (homozygous recessive), is equal to 0.36

Example Question #1 : Understanding Types Of Mutation

Which of the following describes a frameshift mutation?

Possible Answers:

Insertion or deletion of nucleotides that results in a premature stop codon

Insertion or deletion of a group of nucleotides that is a multiple of three

Substitution of one nucleotide for another

Insertion or deletion of a group of nucleotides that is not a multiple of three

Correct answer:

Insertion or deletion of a group of nucleotides that is not a multiple of three

Explanation:

A frameshift mutation indicates that the reading frame of the sequence in altered, resulting in production of different codons downstream of the mutation. Because codons are encoded by groups of three nucleotides, a frameshift mutation results from the insertion or deletion of a number of nucleotides that is not a multiple of three.

For example:

ATG-CGT

Add one nucleotide: ATT-GCG-T

Add two nucleotides: ATT-CGC-GT

Add three nucleotides: ATT-CAG-CGT

In the first two additions, there are unpaired nucleotides that will shift the reading frame. In the final addition (three nucleotides), the final group is complete; thus, any codons after the mutation (downstream) will not be affected.

A nonsense involves the addition of a premature stop codon. A missense mutation results in a different codon, and changes the primary structure of the protein. A silent mutation alters the DNA sequence without altering the amino acid result.

Example Question #1 : Understanding Types Of Mutation

Which of the following mutations will not result in a change to the amino acid sequence of a protein?

Possible Answers:

Nonsense mutation

Silent mutation

Neutral mutation

Frameshift mutation

Correct answer:

Silent mutation

Explanation:

If a mutation does not alter the amino acid sequence of a protein, it is considered a silent mutation. A neutral mutation changes the amino acid, but not the function of the protein. Both frameshift and nonsense mutations can severely affect the function and structure of a protein.

Example Question #1 : Mutation

Which class of DNA mutation results in the substitution of one amino acid for another in the protein product?

Possible Answers:

Frameshift

Nonsense

Missense

Silent

Correct answer:

Missense

Explanation:

A missense mutation results in the presence of a different amino acid than was encoded by the parental sequence. This type of mutation can have a drastic effect or no effect at all depending on the importance of the amino acid and the type of amino acid that replaces it. Some amino acids are structurally similar and may be able to act as viable substitutes for each other. For example, changing one acidic amino acid to another may ot affect the final protein, but changing a polar amino acid to a nonpolar amino acid will likely disrupt the structure.

A nonsense mutation results in the addition of a premature stop codon, creating a truncated protein product. A silent mutation is a mutation that occurs within the DNA sequence, but does not alter the amino acid sequence. Silent mutations can occur in introns, which are spliced out before translation. Finally, a frameshift mutation is an insertion or deletion of a nucleotide sequence that alters the reading frame of the gene.

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