BIOD 210 Genetics Modules 1-7 Exam Bundle Q&A | 2026/2027
Update - 80 Questions and Answers Already Graded A+
Premium Exam Tested And Verified
Subject Area Genetics
Description This exam covers key concepts from Modules 1-7 of BIOD 210 Genetics,
including molecular genetics, Mendelian and non-Mendelian inheritance,
population genetics, genomics, and genetic technologies. Questions require
synthesis of multiple principles and application to complex scenarios.
Expected Grade A+
Total Questions 80
Duration 3 hours
Learning Outcomes 1. Analyze complex inheritance patterns and calculate genetic risks
2. Evaluate molecular mechanisms of gene expression and regulation
3. Apply population genetics principles to predict allele frequency changes
4. Interpret genomic data and assess implications of genetic technologies
5. Integrate knowledge of epigenetics, gene editing, and evolutionary genetics
Accreditation Meets standards for R1 university genetics coursework, aligned with ASHG and
ABMG competencies.
Page 1
,1. In a population of diploid organisms, a locus has two alleles, A and a, with
frequencies p and q. Under Hardy-Weinberg equilibrium, which of the following
conditions would cause the frequency of heterozygotes to be higher than 2pq after
one generation?
A. Non-random mating with inbreeding
B. Overdominance with selection favoring heterozygotes
C. Gene flow from a population with different allele frequencies
D. Recurrent mutation from A to a at a high rate
Answer: B. Overdominance with selection favoring heterozygotes
Overdominance (heterozygote advantage) increases heterozygote frequency above
Hardy-Weinberg expectations because selection favors heterozygotes. Inbreeding
reduces heterozygotes. Gene flow and mutation can change allele frequencies but do not
necessarily increase heterozygosity above 2pq.
2. A researcher performs a three-point testcross in Drosophila involving genes w (eye
color), m (wing size), and b (body color). The progeny counts show that the double
crossover class is underrepresented. Which phenomenon most likely explains this
observation?
A. Positive interference
B. Negative interference
C. Complete linkage
D. Gene conversion
Answer: A. Positive interference
Positive interference reduces the frequency of double crossovers compared to the
product of single crossover frequencies. Negative interference increases double
crossovers. Complete linkage would yield no recombinants. Gene conversion involves
non-reciprocal transfer, not crossover interference.
Page 2
,3. In a eukaryotic cell, a mutation in the spliceosome component U1 snRNA prevents
base pairing with the 5' splice site. What is the most likely immediate consequence
for pre-mRNA processing?
A. Intron retention and translation of aberrant protein
B. Use of cryptic splice sites leading to exon skipping
C. Failure of 5' cap addition at the transcription start site
D. Polyadenylation at alternative sites within the intron
Answer: A. Intron retention and translation of aberrant protein
U1 snRNA recognizes the 5' splice site; its mutation prevents initial spliceosome
assembly, leading to intron retention. Cryptic splice sites may be used but are not
immediate; exon skipping is a later outcome. Cap addition is independent of splicing.
Polyadenylation is not directly affected.
4. A pedigree shows a disorder affecting both males and females, with affected
individuals having at least one affected parent. However, some affected males pass
the trait to all daughters but no sons, and affected females pass the trait to half of
offspring regardless of sex. What is the mode of inheritance?
A. Autosomal dominant with incomplete penetrance
B. X-linked dominant
C. Y-linked
D. Mitochondrial
Answer: B. X-linked dominant
X-linked dominant inheritance: affected males pass the trait to all daughters (who
inherit their X) and no sons (who inherit Y). Affected females (heterozygous) have a
50% chance of passing to each child. Autosomal dominant would show male-to-male
transmission. Y-linked only affects males. Mitochondrial is maternal only.
Page 3
, 5. In a genome-wide association study (GWAS) for type 2 diabetes, a SNP with
p-value 5×10 is identified in a non-coding region near a gene involved in insulin
secretion. Which of the following is the most appropriate next step to establish
causality?
A. Calculate the odds ratio and population attributable risk
B. Perform fine-mapping and functional assays in relevant cell types
C. Conduct a replication study in an independent cohort
D. Sequence the entire gene in all cases and controls
Answer: B. Perform fine-mapping and functional assays in relevant cell types
GWAS identifies associations, not causality. Fine-mapping narrows the causal variant,
and functional assays (e.g., CRISPR editing in beta cells) test biological effect.
Replication confirms association but not causality. Odds ratio measures effect size.
Sequencing may miss regulatory variants.
6. A researcher introduces a plasmid containing a gene of interest into E. coli and
selects for antibiotic resistance. After transformation, some colonies grow on
selective plates but do not express the gene. Which of the following is the most likely
explanation?
A. The gene is integrated into the bacterial chromosome via homologous recombination
B. The plasmid has a mutation in the promoter of the resistance gene
C. The cells took up the plasmid but the gene of interest is in a different reading frame
D. The plasmid is present but the gene of interest is silenced by DNA methylation
Answer: C. The cells took up the plasmid but the gene of interest is in a different
reading frame
If the gene of interest is in a different reading frame relative to its promoter (e.g., due to
cloning error), it may not be translated properly, while the resistance gene (on same
plasmid) is expressed. Integration is rare without homology. Promoter mutation would
affect resistance. E. coli lacks extensive DNA methylation silencing.
Page 4
Update - 80 Questions and Answers Already Graded A+
Premium Exam Tested And Verified
Subject Area Genetics
Description This exam covers key concepts from Modules 1-7 of BIOD 210 Genetics,
including molecular genetics, Mendelian and non-Mendelian inheritance,
population genetics, genomics, and genetic technologies. Questions require
synthesis of multiple principles and application to complex scenarios.
Expected Grade A+
Total Questions 80
Duration 3 hours
Learning Outcomes 1. Analyze complex inheritance patterns and calculate genetic risks
2. Evaluate molecular mechanisms of gene expression and regulation
3. Apply population genetics principles to predict allele frequency changes
4. Interpret genomic data and assess implications of genetic technologies
5. Integrate knowledge of epigenetics, gene editing, and evolutionary genetics
Accreditation Meets standards for R1 university genetics coursework, aligned with ASHG and
ABMG competencies.
Page 1
,1. In a population of diploid organisms, a locus has two alleles, A and a, with
frequencies p and q. Under Hardy-Weinberg equilibrium, which of the following
conditions would cause the frequency of heterozygotes to be higher than 2pq after
one generation?
A. Non-random mating with inbreeding
B. Overdominance with selection favoring heterozygotes
C. Gene flow from a population with different allele frequencies
D. Recurrent mutation from A to a at a high rate
Answer: B. Overdominance with selection favoring heterozygotes
Overdominance (heterozygote advantage) increases heterozygote frequency above
Hardy-Weinberg expectations because selection favors heterozygotes. Inbreeding
reduces heterozygotes. Gene flow and mutation can change allele frequencies but do not
necessarily increase heterozygosity above 2pq.
2. A researcher performs a three-point testcross in Drosophila involving genes w (eye
color), m (wing size), and b (body color). The progeny counts show that the double
crossover class is underrepresented. Which phenomenon most likely explains this
observation?
A. Positive interference
B. Negative interference
C. Complete linkage
D. Gene conversion
Answer: A. Positive interference
Positive interference reduces the frequency of double crossovers compared to the
product of single crossover frequencies. Negative interference increases double
crossovers. Complete linkage would yield no recombinants. Gene conversion involves
non-reciprocal transfer, not crossover interference.
Page 2
,3. In a eukaryotic cell, a mutation in the spliceosome component U1 snRNA prevents
base pairing with the 5' splice site. What is the most likely immediate consequence
for pre-mRNA processing?
A. Intron retention and translation of aberrant protein
B. Use of cryptic splice sites leading to exon skipping
C. Failure of 5' cap addition at the transcription start site
D. Polyadenylation at alternative sites within the intron
Answer: A. Intron retention and translation of aberrant protein
U1 snRNA recognizes the 5' splice site; its mutation prevents initial spliceosome
assembly, leading to intron retention. Cryptic splice sites may be used but are not
immediate; exon skipping is a later outcome. Cap addition is independent of splicing.
Polyadenylation is not directly affected.
4. A pedigree shows a disorder affecting both males and females, with affected
individuals having at least one affected parent. However, some affected males pass
the trait to all daughters but no sons, and affected females pass the trait to half of
offspring regardless of sex. What is the mode of inheritance?
A. Autosomal dominant with incomplete penetrance
B. X-linked dominant
C. Y-linked
D. Mitochondrial
Answer: B. X-linked dominant
X-linked dominant inheritance: affected males pass the trait to all daughters (who
inherit their X) and no sons (who inherit Y). Affected females (heterozygous) have a
50% chance of passing to each child. Autosomal dominant would show male-to-male
transmission. Y-linked only affects males. Mitochondrial is maternal only.
Page 3
, 5. In a genome-wide association study (GWAS) for type 2 diabetes, a SNP with
p-value 5×10 is identified in a non-coding region near a gene involved in insulin
secretion. Which of the following is the most appropriate next step to establish
causality?
A. Calculate the odds ratio and population attributable risk
B. Perform fine-mapping and functional assays in relevant cell types
C. Conduct a replication study in an independent cohort
D. Sequence the entire gene in all cases and controls
Answer: B. Perform fine-mapping and functional assays in relevant cell types
GWAS identifies associations, not causality. Fine-mapping narrows the causal variant,
and functional assays (e.g., CRISPR editing in beta cells) test biological effect.
Replication confirms association but not causality. Odds ratio measures effect size.
Sequencing may miss regulatory variants.
6. A researcher introduces a plasmid containing a gene of interest into E. coli and
selects for antibiotic resistance. After transformation, some colonies grow on
selective plates but do not express the gene. Which of the following is the most likely
explanation?
A. The gene is integrated into the bacterial chromosome via homologous recombination
B. The plasmid has a mutation in the promoter of the resistance gene
C. The cells took up the plasmid but the gene of interest is in a different reading frame
D. The plasmid is present but the gene of interest is silenced by DNA methylation
Answer: C. The cells took up the plasmid but the gene of interest is in a different
reading frame
If the gene of interest is in a different reading frame relative to its promoter (e.g., due to
cloning error), it may not be translated properly, while the resistance gene (on same
plasmid) is expressed. Integration is rare without homology. Promoter mutation would
affect resistance. E. coli lacks extensive DNA methylation silencing.
Page 4
