BIOL 210 | BIOL210 Final Exam: Genetics Updated
and Latest Questions and Correct Answers with
Rationale - Portage Learning
1. In a population of 1000 individuals, 160 show a recessive phenotype. Assuming Hardy-
Weinberg equilibrium, calculate the frequency of the dominant allele (p).
A. 0.4
B. 0.16
C. 0.6
D. 0.84
Correct Answer: C
Explanation: The frequency of the recessive phenotype is q squared, which is 0.16 in this
scenario. Taking the square root gives a recessive allele frequency (q) of 0.4. Since p plus q
equals one, the dominant allele frequency p must be 0.6. Option A is incorrect because it
represents the frequency of the recessive allele q. This calculation is a fundamental
application of the Hardy-Weinberg principle in population genetics.
2. Which enzyme is responsible for unwinding the DNA double helix during the initiation of
replication?
A. Helicase
B. DNA Ligase
C. DNA Polymerase III
D. Primase
Correct Answer: A
Explanation: Helicase is the specific enzyme that breaks hydrogen bonds to unzip the DNA
strands. Option A is incorrect because DNA Polymerase III synthesizes the new DNA strand.
Option B is wrong as DNA Ligase joins Okazaki fragments on the lagging strand. Option D is
incorrect because Primase synthesizes the RNA primer needed for polymerization. Proper
unwinding is essential for the replication fork to progress.
3. In Mendel’s dihybrid cross of RrYy x RrYy, what is the expected phenotypic ratio of the
offspring?
A. 9:3:3:1
B. 1:2:1
C. 3:1
,D. 1:1:1:1
Correct Answer: A
Explanation: The 9:3:3:1 ratio is the classic result of a dihybrid cross involving two
independent traits. Option A represents the phenotypic ratio of a standard monohybrid
cross. Option B is incorrect as it typically describes the genotypic ratio of a monohybrid
cross. Option D is wrong because it represents the result of a dihybrid test cross with a
homozygous recessive. This ratio demonstrates the law of independent assortment in
action.
4. A woman who is a carrier for color blindness (X-linked recessive) marries a man with
normal vision. What is the probability that their son will be color blind?
A. 0%
B. 50%
C. 25%
D. 100%
Correct Answer: B
Explanation: Since the mother is a carrier (XcX), she has a 50 percent chance of passing
the recessive allele to her son. The son receives his Y chromosome from the father, making
his phenotype dependent solely on the mother’s X. Option A is incorrect because sons can
indeed inherit the trait from a carrier mother. Option B is incorrect because 25 percent
describes the probability for any child regardless of sex. X-linked inheritance patterns are
critical for tracking genetic disorders in pedigrees.
5. What occurs during the process of alternative splicing in eukaryotic cells?
A. Removal of exons and joining of introns
B. Addition of a 5’ cap to the mRNA transcript
C. Joining of different exons from the same pre-mRNA
D. Attachment of a poly-A tail to the 3’ end
Correct Answer: C
Explanation: Alternative splicing allows a single gene to code for multiple proteins by
varying which exons are included. Option A is incorrect because introns are removed and
exons are joined, not the reverse. Option C and D are incorrect because they describe
different types of mRNA processing like capping and polyadenylation. This mechanism
significantly increases the proteomic diversity of eukaryotic organisms. It explains why
humans have more proteins than genes.
, 6. If the recombination frequency between two genes is 15%, how many map units (cM) apart
are they on the chromosome?
A. 7.5 cM
B. 30 cM
C. 15 cM
D. 150 cM
Correct Answer: C
Explanation: By definition, one percent recombination frequency is equal to one map unit
or centimorgan. Therefore, a 15 percent frequency translates directly to 15 map units.
Option A is incorrect because the frequency is not halved for mapping. Option C and D are
incorrect as they utilize wrong conversion factors. Linkage mapping relies on the linear
relationship between distance and crossing over probability.
7. Which type of mutation results in a premature stop codon in the mRNA sequence?
A. Nonsense mutation
B. Missense mutation
C. Silent mutation
D. Frameshift mutation
Correct Answer: A
Explanation: A nonsense mutation changes a functional codon into a stop codon, halting
translation early. Option A is incorrect because silent mutations do not change the amino
acid sequence. Option B is wrong as missense mutations replace one amino acid with
another. Option D is incorrect because, while it changes the reading frame, it doesn’t
specifically define a premature stop. Nonsense mutations usually result in nonfunctional,
truncated proteins.
8. In the lac operon, what happens to the repressor protein in the presence of lactose?
A. Allolactose binds to it, causing it to release from the operator
B. It is degraded by proteasomes
C. It binds to the operator to block transcription
D. It binds to the promoter to recruit RNA polymerase
Correct Answer: A
Explanation: Lactose is converted to allolactose, which acts as an inducer by changing the
repressor’s shape. This change prevents the repressor from binding to the operator,
allowing transcription to proceed. Option A describes the state when lactose is absent.
Option B is incorrect because the protein is not destroyed but regulated. Option D is
and Latest Questions and Correct Answers with
Rationale - Portage Learning
1. In a population of 1000 individuals, 160 show a recessive phenotype. Assuming Hardy-
Weinberg equilibrium, calculate the frequency of the dominant allele (p).
A. 0.4
B. 0.16
C. 0.6
D. 0.84
Correct Answer: C
Explanation: The frequency of the recessive phenotype is q squared, which is 0.16 in this
scenario. Taking the square root gives a recessive allele frequency (q) of 0.4. Since p plus q
equals one, the dominant allele frequency p must be 0.6. Option A is incorrect because it
represents the frequency of the recessive allele q. This calculation is a fundamental
application of the Hardy-Weinberg principle in population genetics.
2. Which enzyme is responsible for unwinding the DNA double helix during the initiation of
replication?
A. Helicase
B. DNA Ligase
C. DNA Polymerase III
D. Primase
Correct Answer: A
Explanation: Helicase is the specific enzyme that breaks hydrogen bonds to unzip the DNA
strands. Option A is incorrect because DNA Polymerase III synthesizes the new DNA strand.
Option B is wrong as DNA Ligase joins Okazaki fragments on the lagging strand. Option D is
incorrect because Primase synthesizes the RNA primer needed for polymerization. Proper
unwinding is essential for the replication fork to progress.
3. In Mendel’s dihybrid cross of RrYy x RrYy, what is the expected phenotypic ratio of the
offspring?
A. 9:3:3:1
B. 1:2:1
C. 3:1
,D. 1:1:1:1
Correct Answer: A
Explanation: The 9:3:3:1 ratio is the classic result of a dihybrid cross involving two
independent traits. Option A represents the phenotypic ratio of a standard monohybrid
cross. Option B is incorrect as it typically describes the genotypic ratio of a monohybrid
cross. Option D is wrong because it represents the result of a dihybrid test cross with a
homozygous recessive. This ratio demonstrates the law of independent assortment in
action.
4. A woman who is a carrier for color blindness (X-linked recessive) marries a man with
normal vision. What is the probability that their son will be color blind?
A. 0%
B. 50%
C. 25%
D. 100%
Correct Answer: B
Explanation: Since the mother is a carrier (XcX), she has a 50 percent chance of passing
the recessive allele to her son. The son receives his Y chromosome from the father, making
his phenotype dependent solely on the mother’s X. Option A is incorrect because sons can
indeed inherit the trait from a carrier mother. Option B is incorrect because 25 percent
describes the probability for any child regardless of sex. X-linked inheritance patterns are
critical for tracking genetic disorders in pedigrees.
5. What occurs during the process of alternative splicing in eukaryotic cells?
A. Removal of exons and joining of introns
B. Addition of a 5’ cap to the mRNA transcript
C. Joining of different exons from the same pre-mRNA
D. Attachment of a poly-A tail to the 3’ end
Correct Answer: C
Explanation: Alternative splicing allows a single gene to code for multiple proteins by
varying which exons are included. Option A is incorrect because introns are removed and
exons are joined, not the reverse. Option C and D are incorrect because they describe
different types of mRNA processing like capping and polyadenylation. This mechanism
significantly increases the proteomic diversity of eukaryotic organisms. It explains why
humans have more proteins than genes.
, 6. If the recombination frequency between two genes is 15%, how many map units (cM) apart
are they on the chromosome?
A. 7.5 cM
B. 30 cM
C. 15 cM
D. 150 cM
Correct Answer: C
Explanation: By definition, one percent recombination frequency is equal to one map unit
or centimorgan. Therefore, a 15 percent frequency translates directly to 15 map units.
Option A is incorrect because the frequency is not halved for mapping. Option C and D are
incorrect as they utilize wrong conversion factors. Linkage mapping relies on the linear
relationship between distance and crossing over probability.
7. Which type of mutation results in a premature stop codon in the mRNA sequence?
A. Nonsense mutation
B. Missense mutation
C. Silent mutation
D. Frameshift mutation
Correct Answer: A
Explanation: A nonsense mutation changes a functional codon into a stop codon, halting
translation early. Option A is incorrect because silent mutations do not change the amino
acid sequence. Option B is wrong as missense mutations replace one amino acid with
another. Option D is incorrect because, while it changes the reading frame, it doesn’t
specifically define a premature stop. Nonsense mutations usually result in nonfunctional,
truncated proteins.
8. In the lac operon, what happens to the repressor protein in the presence of lactose?
A. Allolactose binds to it, causing it to release from the operator
B. It is degraded by proteasomes
C. It binds to the operator to block transcription
D. It binds to the promoter to recruit RNA polymerase
Correct Answer: A
Explanation: Lactose is converted to allolactose, which acts as an inducer by changing the
repressor’s shape. This change prevents the repressor from binding to the operator,
allowing transcription to proceed. Option A describes the state when lactose is absent.
Option B is incorrect because the protein is not destroyed but regulated. Option D is
