BIOL 235 Exam 2 With Complete Solution(100% Verified) 200+ Questions with
100% Correct Answers With well detailed Rationales/Grade A+ Assured
Question 1
How is recombination frequency fundamentally used by geneticists to form a map of a
chromosome indicating the relative distances between specific genes?
A) By measuring the exact number of nucleotides between two loci.
B) By calculating the ratio of dominant to recessive phenotypes in an F1 generation.
C) By observing that the frequency is proportional to the physical distance; a lower frequency
indicates genes are closer together.
D) By determining the rate of DNA methylation on the promoter regions.
E) By counting the number of spindle fibers attached to the centromere during Metaphase II.
Correct Answer: C) By observing that the frequency is proportional to the physical distance;
a lower frequency indicates genes are closer together.
Rationale: Linkage maps are based on the principle that the further apart two genes are on
a single chromosome, the more likely a crossover event will occur between them.
Recombination frequency (RF) serves as an estimate of this distance. When RF is low, it
suggests the genes are physically close (linked), as there is less "room" for a chiasma to
form between them during Prophase I of meiosis.
Question 2
What is the maximum recombination frequency that can be observed between two genes located
on the same chromosome, and what is the biological reason for this limit?
A) 100%; because crossover can happen at every single nucleotide.
B) 25%; because only one sister chromatid out of four is involved in any single crossover.
C) 50%; because past a certain distance, multiple crossover events (like double crossovers) make
the genes behave as if they are independently assorting.
D) 75%; because three-quarters of the gametes will always be recombinant in large
chromosomes.
E) 10%; because chromosomal interference prevents high rates of recombination.
Correct Answer: C) 50%; because past a certain distance, multiple crossover events (like
double crossovers) make the genes behave as if they are independently assorting.
Rationale: Even if genes are very far apart on the same chromosome, the observed
recombination frequency cannot exceed 50%. At this limit, the genes appear unlinked and
follow Mendel's Law of Independent Assortment. This happens because, at great distances,
the probability of a crossover is so high that it occurs virtually every time, and multiple
crossovers (like double crossovers) can "cancel each other out," returning the alleles to a
parental configuration and underestimating the true physical distance.
Question 3
If two genes show a recombination frequency of exactly 15% in a test cross, how are they
classified, and if they show a frequency of 50%, how are they classified?
, 2
A) Both are unlinked.
B) 15% are unlinked, 50% are linked.
C) 15% are linked, 50% are unlinked.
D) Both are linked.
E) 15% are epistatic, 50% are pleiotropic.
Correct Answer: C) 15% are linked, 50% are unlinked.
Rationale: Any recombination frequency significantly lower than 50% indicates that the
genes are linked, meaning they reside on the same chromosome and do not assort
independently. A frequency of 50% indicates that the genes are either on different
chromosomes or are so far apart on the same chromosome that they appear to assort
independently.
Question 4
When two genes are confirmed to be linked, they are physically located:
A) On non-homologous chromosomes.
B) At the same locus on the same chromosome.
C) Close to each other on the same chromosome.
D) On the mitochondrial DNA only.
E) In different nuclei within the same cell.
Correct Answer: C) Close to each other on the same chromosome.
Rationale: Linkage is defined by the physical proximity of two or more genes on the same
DNA molecule (chromosome). Because they are physically tethered together, they tend to be
inherited as a unit unless separated by a recombination event during meiosis.
Question 5
In a dihybrid F2 test cross involving linked genes, which phenotypic/genotypic classes are
expected to be found in the highest proportion among the offspring?
A) Recombinant classes.
B) The homozygous recessive class only.
C) Parental classes.
D) All classes will be in a 1:1:1:1 ratio.
E) Only the double mutant class.
Correct Answer: C) Parental classes.
Rationale: In cases of linkage, the majority of gametes produced by the dihybrid parent will
maintain the original allele combinations found on its own chromosomes (parental types).
Recombination is a relatively rare event compared to the lack of a crossover, so the
parental offspring will always outnumber the recombinant offspring in a test cross of
linked genes.
, 3
Question 6
A map distance of 15.5 map units (centiMorgans) between Gene A and Gene B indicates which
of the following?
A) The genes are 15.5 million base pairs apart.
B) 15.5% of the dihybrid test cross offspring will show a recombinant phenotype.
C) The genes are on different chromosomes but interact 15.5% of the time.
D) There is a 15.5% chance of the chromosome breaking at that specific spot.
E) The genes are unlinked and follow a 9:3:3:1 ratio.
Correct Answer: B) 15.5% of the dihybrid test cross offspring will show a recombinant
phenotype.
Rationale: By definition, one map unit (or centiMorgan) is equal to a 1% recombination
frequency. Therefore, a distance of 15.5 map units directly translates to a 15.5%
probability that a crossover will occur between these two loci during gamete formation.
Question 7
In a test cross of Ab/aB x ab/ab, where genes A and B are 14.5 map units apart, what specific
proportion of the offspring is expected to have the AB/ab genotype?
A) 14.5%
B) 29%
C) 7.25%
D) 42.75%
E) 85.5%
Correct Answer: C) 7.25%
Rationale: The total recombination frequency is 14.5%. In this specific cross (Ab/aB), the
recombinant gametes are AB and ab. Since the total percentage of recombinants is 14.5%,
and there are two types of recombinants (AB and ab), we divide the total by 2 to find the
frequency of a single recombinant class. 14. = 7.25%.
Question 8
During a dihybrid test cross, you observe 1000 offspring with parental genotypes and 500
offspring with recombinant genotypes. What is the calculated genetic distance between these two
genes?
A) 50 map units
B) 33.3 map units
C) 66.6 map units
D) 25 map units
E) 150 map units
Correct Answer: B) 33.3 map units
Rationale: Recombination frequency is calculated as (Number of Recombinants / Total
, 4
Number of Offspring) x 100. In this case: () x 100 = 33.3%. Since 1%
recombination equals 1 map unit, the distance is 33.3 map units.
Question 9
Are the genes from the previous question (Question 8) considered linked?
A) No, because the frequency is too high.
B) Yes, because the recombination frequency is less than 50%.
C) No, because they must be 10 map units or less to be linked.
D) Only if they are on the X chromosome.
E) Yes, because all genes on the same chromosome are linked by definition, regardless of
distance.
Correct Answer: B) Yes, because the recombination frequency is less than 50%.
Rationale: Linkage is operationally defined by any recombination frequency less than the
50% expected under independent assortment. Since 33.3% is significantly lower than 50%,
the genes are demonstrating linkage.
Question 10
A cross was performed between AB/ab x ab/ab. The offspring counts were: 800 AB/ab, 830
ab/ab, 100 aB/ab, and 180 Ab/ab. Which genotypes represent the recombinant classes?
A) AB/ab and ab/ab
B) AB/ab and aB/ab
C) aB/ab and Ab/ab
D) ab/ab and Ab/ab
E) All are parental.
Correct Answer: C) aB/ab and Ab/ab
Rationale: In the heterozygous parent (AB/ab), the alleles A and B are on one chromosome,
and a and b are on the homologous chromosome. These are the parental configurations.
Any offspring inheriting a combination different from these (like aB or Ab) must have
resulted from a crossover event in the heterozygous parent.
Question 11
Why is the statement "Genetic distances determined from recombination frequency are always
perfectly additive" considered FALSE?
A) Because recombination only happens in females.
B) Because map units and centiMorgans are different scales.
C) Because as distance increases, the probability of double crossovers increases, which masks
recombination events and leads to an underestimation of distance.
D) Because genes are moving around the genome via transposable elements.
E) Because independent assortment happens 100% of the time.
100% Correct Answers With well detailed Rationales/Grade A+ Assured
Question 1
How is recombination frequency fundamentally used by geneticists to form a map of a
chromosome indicating the relative distances between specific genes?
A) By measuring the exact number of nucleotides between two loci.
B) By calculating the ratio of dominant to recessive phenotypes in an F1 generation.
C) By observing that the frequency is proportional to the physical distance; a lower frequency
indicates genes are closer together.
D) By determining the rate of DNA methylation on the promoter regions.
E) By counting the number of spindle fibers attached to the centromere during Metaphase II.
Correct Answer: C) By observing that the frequency is proportional to the physical distance;
a lower frequency indicates genes are closer together.
Rationale: Linkage maps are based on the principle that the further apart two genes are on
a single chromosome, the more likely a crossover event will occur between them.
Recombination frequency (RF) serves as an estimate of this distance. When RF is low, it
suggests the genes are physically close (linked), as there is less "room" for a chiasma to
form between them during Prophase I of meiosis.
Question 2
What is the maximum recombination frequency that can be observed between two genes located
on the same chromosome, and what is the biological reason for this limit?
A) 100%; because crossover can happen at every single nucleotide.
B) 25%; because only one sister chromatid out of four is involved in any single crossover.
C) 50%; because past a certain distance, multiple crossover events (like double crossovers) make
the genes behave as if they are independently assorting.
D) 75%; because three-quarters of the gametes will always be recombinant in large
chromosomes.
E) 10%; because chromosomal interference prevents high rates of recombination.
Correct Answer: C) 50%; because past a certain distance, multiple crossover events (like
double crossovers) make the genes behave as if they are independently assorting.
Rationale: Even if genes are very far apart on the same chromosome, the observed
recombination frequency cannot exceed 50%. At this limit, the genes appear unlinked and
follow Mendel's Law of Independent Assortment. This happens because, at great distances,
the probability of a crossover is so high that it occurs virtually every time, and multiple
crossovers (like double crossovers) can "cancel each other out," returning the alleles to a
parental configuration and underestimating the true physical distance.
Question 3
If two genes show a recombination frequency of exactly 15% in a test cross, how are they
classified, and if they show a frequency of 50%, how are they classified?
, 2
A) Both are unlinked.
B) 15% are unlinked, 50% are linked.
C) 15% are linked, 50% are unlinked.
D) Both are linked.
E) 15% are epistatic, 50% are pleiotropic.
Correct Answer: C) 15% are linked, 50% are unlinked.
Rationale: Any recombination frequency significantly lower than 50% indicates that the
genes are linked, meaning they reside on the same chromosome and do not assort
independently. A frequency of 50% indicates that the genes are either on different
chromosomes or are so far apart on the same chromosome that they appear to assort
independently.
Question 4
When two genes are confirmed to be linked, they are physically located:
A) On non-homologous chromosomes.
B) At the same locus on the same chromosome.
C) Close to each other on the same chromosome.
D) On the mitochondrial DNA only.
E) In different nuclei within the same cell.
Correct Answer: C) Close to each other on the same chromosome.
Rationale: Linkage is defined by the physical proximity of two or more genes on the same
DNA molecule (chromosome). Because they are physically tethered together, they tend to be
inherited as a unit unless separated by a recombination event during meiosis.
Question 5
In a dihybrid F2 test cross involving linked genes, which phenotypic/genotypic classes are
expected to be found in the highest proportion among the offspring?
A) Recombinant classes.
B) The homozygous recessive class only.
C) Parental classes.
D) All classes will be in a 1:1:1:1 ratio.
E) Only the double mutant class.
Correct Answer: C) Parental classes.
Rationale: In cases of linkage, the majority of gametes produced by the dihybrid parent will
maintain the original allele combinations found on its own chromosomes (parental types).
Recombination is a relatively rare event compared to the lack of a crossover, so the
parental offspring will always outnumber the recombinant offspring in a test cross of
linked genes.
, 3
Question 6
A map distance of 15.5 map units (centiMorgans) between Gene A and Gene B indicates which
of the following?
A) The genes are 15.5 million base pairs apart.
B) 15.5% of the dihybrid test cross offspring will show a recombinant phenotype.
C) The genes are on different chromosomes but interact 15.5% of the time.
D) There is a 15.5% chance of the chromosome breaking at that specific spot.
E) The genes are unlinked and follow a 9:3:3:1 ratio.
Correct Answer: B) 15.5% of the dihybrid test cross offspring will show a recombinant
phenotype.
Rationale: By definition, one map unit (or centiMorgan) is equal to a 1% recombination
frequency. Therefore, a distance of 15.5 map units directly translates to a 15.5%
probability that a crossover will occur between these two loci during gamete formation.
Question 7
In a test cross of Ab/aB x ab/ab, where genes A and B are 14.5 map units apart, what specific
proportion of the offspring is expected to have the AB/ab genotype?
A) 14.5%
B) 29%
C) 7.25%
D) 42.75%
E) 85.5%
Correct Answer: C) 7.25%
Rationale: The total recombination frequency is 14.5%. In this specific cross (Ab/aB), the
recombinant gametes are AB and ab. Since the total percentage of recombinants is 14.5%,
and there are two types of recombinants (AB and ab), we divide the total by 2 to find the
frequency of a single recombinant class. 14. = 7.25%.
Question 8
During a dihybrid test cross, you observe 1000 offspring with parental genotypes and 500
offspring with recombinant genotypes. What is the calculated genetic distance between these two
genes?
A) 50 map units
B) 33.3 map units
C) 66.6 map units
D) 25 map units
E) 150 map units
Correct Answer: B) 33.3 map units
Rationale: Recombination frequency is calculated as (Number of Recombinants / Total
, 4
Number of Offspring) x 100. In this case: () x 100 = 33.3%. Since 1%
recombination equals 1 map unit, the distance is 33.3 map units.
Question 9
Are the genes from the previous question (Question 8) considered linked?
A) No, because the frequency is too high.
B) Yes, because the recombination frequency is less than 50%.
C) No, because they must be 10 map units or less to be linked.
D) Only if they are on the X chromosome.
E) Yes, because all genes on the same chromosome are linked by definition, regardless of
distance.
Correct Answer: B) Yes, because the recombination frequency is less than 50%.
Rationale: Linkage is operationally defined by any recombination frequency less than the
50% expected under independent assortment. Since 33.3% is significantly lower than 50%,
the genes are demonstrating linkage.
Question 10
A cross was performed between AB/ab x ab/ab. The offspring counts were: 800 AB/ab, 830
ab/ab, 100 aB/ab, and 180 Ab/ab. Which genotypes represent the recombinant classes?
A) AB/ab and ab/ab
B) AB/ab and aB/ab
C) aB/ab and Ab/ab
D) ab/ab and Ab/ab
E) All are parental.
Correct Answer: C) aB/ab and Ab/ab
Rationale: In the heterozygous parent (AB/ab), the alleles A and B are on one chromosome,
and a and b are on the homologous chromosome. These are the parental configurations.
Any offspring inheriting a combination different from these (like aB or Ab) must have
resulted from a crossover event in the heterozygous parent.
Question 11
Why is the statement "Genetic distances determined from recombination frequency are always
perfectly additive" considered FALSE?
A) Because recombination only happens in females.
B) Because map units and centiMorgans are different scales.
C) Because as distance increases, the probability of double crossovers increases, which masks
recombination events and leads to an underestimation of distance.
D) Because genes are moving around the genome via transposable elements.
E) Because independent assortment happens 100% of the time.
