Genetics Practice Test: Practice Test 1

Question 1

1. The premature stop codon destabilizes the RNA polymerase, causing it to dissociate from the DNA template.
2. The ribosome dissociates prematurely, exposing a cryptic Rho utilization (rut) site on the nascent mRNA. (correct answer)
3. The absence of a complete protein product triggers a feedback loop that transcriptionally silences the operon.
4. The short, non-functional mRNA produced is rapidly targeted for degradation by cellular ribonucleases.

Explanation: The correct answer is B. In prokaryotes, transcription and translation are tightly coupled. Ribosomes follow closely behind the RNA polymerase. Rho-dependent termination requires the Rho protein to bind to a C-rich 'rut' site on the nascent RNA and translocate towards the polymerase. Normally, ribosomes translating the mRNA cover the rut sites, preventing Rho from binding. A premature stop codon causes the ribosome to fall off the mRNA early. This exposes the previously shielded rut site, allowing Rho to bind and initiate premature transcription termination. This reduces the synthesis of the distal parts of the gene and any downstream genes in the operon. A is incorrect because the stop codon affects the ribosome, not the RNA polymerase directly. C is incorrect; while feedback loops exist, the direct mechanism of polarity is physical, involving Rho. D is incorrect because while the mRNA is indeed degraded, polarity describes the premature termination of transcription, which is the cause of the shortened mRNA, not the result of its degradation.

Question 2

1. ( \log_2(300 / 50) )
2. ( \log_2( (300/1500) / (50/1000) ) ) (correct answer)
3. ( \log_2( (300-50) / (1500-1000) ) )
4. ( \log_2(300) / \log_2(50) )

Explanation: Proper calculation of fold change requires normalization to a housekeeping gene to control for variations in initial sample amount. The expression of the target gene (MYC) should be divided by the expression of the housekeeping gene (GAPDH) for each condition separately. Then, the ratio of the normalized treated value to the normalized untreated value is calculated. The log₂ of this final ratio gives the normalized log₂(fold change).

Question 3

1. The number of biological replicates (10 plots per condition) is insufficient for a field ecology experiment.
2. The experiment lacks a positive control, such as a plot treated with a chemical known to affect flower color.
3. The treatment (pesticide) is perfectly confounded with an environmental variable (sun exposure), making it impossible to attribute causality. (correct answer)
4. The initial allele frequencies were not controlled with sufficient precision, which will invalidate the final statistical comparison.

Explanation: The correct answer is C. The most severe flaw is the confounding variable. Any observed difference in flower color frequency between the two groups could be due to the pesticide, the difference in sun exposure, or an interaction between the two. The design cannot distinguish between these possibilities. (A) While more replicates are often better, 20 total plots is a substantial number, and the confounding is a more fundamental error. (B) A positive control is not as essential as eliminating confounding variables. (D) Minor variations in starting frequencies can be accounted for; they do not invalidate the experiment in the way a major confounder does.

Question 4

1. The three genes are located very close to one another on the chromosome.
2. The observed number of double-crossover progeny is zero.
3. The coefficient of coincidence is zero.
4. The genes are on different chromosomes. (correct answer)

Explanation: Interference is a phenomenon that applies to linked genes on the same chromosome. If the genes were on different chromosomes, they would assort independently, and the concepts of linkage, recombination frequency between them, and interference would not be applicable. An interference of 1.0 (complete interference) means C=0 and no double crossovers are observed, a situation that occurs when genes are very tightly linked.

Question 5

1. The alleles for both genes will assort independently, following Mendelian predictions for a dihybrid cross.
2. The alleles for the two genes will segregate as if they were a single gene, with only parental combinations of alleles appearing in gametes. (correct answer)
3. The law of segregation no longer applies, and gametes may receive both or neither allele for a given gene.
4. Recombinant gametes will be produced at a frequency of 50%, mimicking the results of independent assortment.

Explanation: Linkage means the genes are physically located on the same chromosome. If crossing over does not occur, there is no mechanism to separate the alleles on a given parental chromosome. Therefore, the alleles that started together on one chromosome (e.g., AB) will end up in gametes together, and the alleles on the homologous chromosome (e.g., ab) will also be inherited as a unit. This is complete linkage, and the genes segregate as a single unit. Independent assortment (A) and 50% recombination (D) only occur for unlinked genes (or very distant linked genes). The law of segregation (C) still applies; each gamete will get one homologous chromosome and thus one allele for each gene, just not in new combinations.

Question 6

1. P - R - S (correct answer)
2. R - S - P
3. S - P - R
4. The order cannot be determined from this information.

Explanation: To determine the gene order, we compare the allele combination on a parental chromosome with the combination on a double-crossover (DCO) chromosome. The parental chromosomes are P r S and p R s. The DCO chromosomes are p r s and P R S. Let's compare a parental, P r S, with a DCO, P R S. The alleles for P and S are identical between the two, but the allele for R is different (r vs. R). The gene that is exchanged between the parental and DCO classes is the middle gene. Therefore, R is the middle gene, and the order is P-R-S.

Question 7

1. Gene A
2. Gene B (correct answer)
3. Gene C
4. Gene D

Explanation: A volcano plot displays statistical significance (-log10(p-value) or adjusted p-value) on the y-axis versus magnitude of change (log2(Fold Change)) on the x-axis. 'Strongly upregulated' means a high positive log2(Fold Change), placing it far to the right. 'High variance between replicates' leads to a low statistical significance, meaning a low -log10(p-value). Gene B fits these criteria, as it has a high log2(Fold Change) but is below the typical significance threshold (dashed line). Gene A is significantly upregulated. Gene C is significantly downregulated. Gene D is not differentially expressed.

Question 8

1. Differences in the general transcription machinery, such as RNA polymerase II, between humans and chimps.
2. A frameshift mutation in the protein-coding sequence of the human FOXP2 gene.
3. Sequence differences in non-coding cis-regulatory elements like enhancers that control the FOXP2 gene. (correct answer)
4. Differences in the amino acid sequences of the trans-acting transcription factors that bind to the FOXP2 promoter in both species.

Explanation: When the protein product of a gene is conserved but its expression pattern has diverged, the most likely cause is evolution of the cis-regulatory elements (e.g., enhancers, promoters, silencers). These non-coding sequences control when, where, and how much a gene is transcribed. Changes in these elements can alter TF binding and lead to new expression patterns without changing the protein itself. Changes in trans-factors or general machinery would likely affect many genes, not just one.

Question 9

1. High in erythroid cells; high in myeloid cells.
2. Low in erythroid cells; low or absent in myeloid cells.
3. Low or absent in erythroid cells; high in myeloid cells.
4. High in erythroid cells; low or absent in myeloid cells. (correct answer)

Explanation: When analyzing gene regulation questions, focus on how enhancers and silencers work together to control transcription. Enhancers increase gene expression when bound by activators, while silencers decrease expression when bound by repressors. Let's trace through each cell type. In erythroid precursor cells, the activator Act-E binds to Element 1 (the strong enhancer), promoting transcription. Crucially, the repressor Rep-S is absent, so Element 2 (the silencer) remains unoccupied and cannot inhibit transcription. This results in high Glo1 expression. In myeloid precursor cells, Act-E still binds Element 1, providing the same enhancing effect. However, Rep-S now binds to Element 2, activating the silencer function. The silencer's repressive effect counteracts or overrides the enhancer's activation, resulting in low or absent gene expression. Answer choice A incorrectly suggests high expression in both cell types, ignoring the silencer's effect in myeloid cells. Choice B wrongly predicts low expression in erythroid cells, missing that the enhancer works unopposed when the repressor is absent. Choice C completely reverses the scenario, suggesting the silencer somehow increases expression in myeloid cells. Choice D correctly captures that erythroid cells have high expression (enhancer active, silencer inactive) while myeloid cells have low/absent expression (enhancer active but overruled by silencer). Study tip: Remember that silencers typically dominate over enhancers when both are active. Always track which regulatory elements are occupied in each cell type and consider their combined effect on transcription.

Question 10

1. An affected male's phenotypically normal sister has an affected son.
2. An affected male is confirmed to be the biological father of an unaffected son. (correct answer)
3. The founding male in the pedigree is affected, but his brother is unaffected.
4. The trait is observed in only two of the five branches of the extended family.

Explanation: Y-linked (holandric) inheritance involves the transmission of traits on the Y chromosome. Therefore, a father passes his Y chromosome, and thus the Y-linked trait, to all of his sons. The observation of an affected male having a biologically confirmed, unaffected son directly contradicts this mode of inheritance. A) is impossible for Y-linkage as the trait cannot pass through a female. B) is a direct contradiction. C) is irrelevant as the brother represents a different lineage. D) is expected, as the trait only follows male lines.

Question 11

1. 40%
2. 34%
3. 68%
4. 62% (correct answer)

Explanation: When you encounter linked gene problems with interference, you need to account for how crossovers in one region affect crossovers in adjacent regions. This requires calculating both single and double crossover frequencies. Start by finding the recombination frequencies: R-S distance is 20 cM (20% recombination) and S-T distance is 20 cM (20% recombination). If genes were unlinked, you'd expect 4% double crossovers (0.20 × 0.20 = 0.04). However, interference = 0.5 means only half the expected double crossovers actually occur, so the actual double crossover frequency is 2%. Now calculate the gamete frequencies:

• Single crossover between R-S: 20% - 2% = 18%
• Single crossover between S-T: 20% - 2% = 18%
• Double crossovers: 2%
• Parental types: 100% - 18% - 18% - 2% = 62%

Question 12

1. The heights of individuals in the study population form a bell-shaped, continuous distribution.
2. Adult offspring's height is positively correlated with the average height of their parents.
3. The population can be sorted into exactly five distinct, non-overlapping height categories. (correct answer)
4. Genome-wide association studies (GWAS) identify over 700 different genetic loci associated with height.

Explanation: The correct answer is C. A primary characteristic of a polygenic trait is continuous variation, where phenotypes fall along a spectrum rather than in discrete categories. Finding a small number of distinct, non-overlapping height classes would be highly unexpected and would suggest a much simpler mode of inheritance (e.g., Mendelian inheritance with one or two genes), contradicting the known polygenic nature of height.

• A is incorrect. A normal (bell-shaped) distribution is the expected pattern for a quantitative, polygenic trait in a large population.
• B is incorrect. Because height is heritable, a correlation between parental and offspring height is expected.
• D is incorrect. The finding that hundreds or even thousands of genetic loci are associated with height is a key piece of modern evidence confirming its highly polygenic nature. This would be expected, not surprising.

Question 13

1. Somatic mutations are acquired during an individual's lifetime and are not passed to offspring, whereas MELAS is caused by a germline mutation from the mother. (correct answer)
2. Age-related mutations only occur in nuclear genes that support mitochondria, while MELAS is a primary mtDNA mutation.
3. Inherited mitochondrial diseases like MELAS are typically homoplasmic, whereas age-related changes result in heteroplasmy.
4. Somatic mutations are efficiently repaired by mitochondrial proofreading, while inherited germline mutations bypass these repair mechanisms.

Explanation: The key distinction is between somatic and germline mutations. Age-related accumulation of mtDNA mutations occurs in somatic cells and is not heritable. In contrast, classic mitochondrial diseases are caused by pathogenic mutations present in the mother's germline (oocytes) and are therefore passed down to all offspring, establishing the mutation in all their cells from the zygote stage.

Question 14

1. Because double-crossover events result in non-viable gametes, reducing their observed frequency.
2. Because the probability of two independent crossover events occurring between three linked genes is the product of their individual probabilities. (correct answer)
3. Because double-crossover events can only occur in repulsion phase heterozygotes, which are less common.
4. Because positive interference always enhances the rate of double crossovers above the expected frequency.

Explanation: Crossover events in adjacent regions of a chromosome are largely independent. The probability of a double crossover is the probability of a crossover happening in the first interval multiplied by the probability of a crossover happening in the second interval. Since these individual probabilities (recombination frequencies) are always less than 0.5, their product will be a much smaller number than either individual probability. For example, if P(crossover 1) = 0.2 and P(crossover 2) = 0.1, then P(double crossover) = 0.2 * 0.1 = 0.02. This low joint probability makes the double-crossover class the rarest among the progeny.

Question 15

1. 1/6 (correct answer)
2. 1/4
3. 1/3
4. 1/2

Explanation: First, determine the genotypes of the parents (III-1's parents). Individual II-4 has Type B blood and a Type O ((ii)) mother, so her genotype must be (I^B i). Individual II-2 has Type A blood. His parents (I-1 and I-2) are Type A and Type B, respectively, but they have a Type O child (II-1, genotype (ii)). This means the grandparents' genotypes must be (I^A i) and (I^B i). The possible Type A offspring from this cross are (I^A I^A) and (I^A i) in a 1:2 ratio. So, there is a 1/3 probability that II-2 is (I^A I^A) and a 2/3 probability he is (I^A i). Now, calculate the probability of a Type B child (genotype (I^B i)). Case 1: II-2 is (I^A I^A) (Prob = 1/3). Cross: (I^A I^A \times I^B i). The probability of a (I^B i) child is 0. Case 2: II-2 is (I^A i) (Prob = 2/3). Cross: (I^A i \times I^B i). The probability of a (I^B i) child is 1/4. The total probability is the sum of probabilities of each case: (1/3 * 0) + (2/3 * 1/4) = 0 + 2/12 = 1/6.

Question 16

1. The endogenous miRNA pathway leading primarily to translational repression.
2. The RNA interference (RNAi) pathway, where RISC mediates target mRNA cleavage. (correct answer)
3. Nonsense-mediated decay (NMD) triggered by the introduction of a premature stop codon.
4. Transcriptional gene silencing through methylation of the Gene X promoter.

Explanation: Short, synthetic dsRNAs that are perfectly complementary to a target mRNA are the hallmark of RNA interference (RNAi), a pathway that utilizes siRNAs (short interfering RNAs). In mammals, perfect complementarity typically directs the Argonaute protein within the RISC complex to cleave the target mRNA, leading to its rapid degradation. (A) is less likely; while there is overlap with the miRNA pathway, perfect complementarity and significant mRNA reduction point strongly to cleavage-based RNAi rather than translational repression. (C) is incorrect as NMD is triggered by premature stop codons, not dsRNA. (D) describes transcriptional silencing, a different mechanism that would not cause a rapid reduction in existing mRNA levels.

Question 17

1. A larger deviation between the observed and expected phenotypic counts. (correct answer)
2. An increase in the p-value threshold for significance from 0.05 to 0.10.
3. An increase in the number of phenotypic categories being analyzed.
4. A closer agreement between the observed and expected phenotypic counts.

Explanation: Chi-square analysis tests whether observed data significantly deviates from expected patterns under a null hypothesis. The chi-square statistic is calculated as $\chi^2 = \sum \frac{(observed - expected)^2}{expected}$ for each category. Looking at this formula reveals what increases the chi-square value. The numerator contains the squared difference between observed and expected counts. When this difference gets larger, the chi-square value increases proportionally. Choice A correctly identifies that larger deviations between observed and expected phenotypic counts will increase the calculated chi-square value. Choice B incorrectly suggests that changing the p-value threshold affects the calculated chi-square. The p-value threshold (0.05 vs 0.10) only determines how we interpret the chi-square value for statistical significance—it doesn't change the actual calculated number. Choice C is wrong because adding more phenotypic categories doesn't automatically increase chi-square; it depends on how well each category fits expectations. More categories could actually decrease chi-square if the additional data fits the expected pattern well. Choice D describes the opposite scenario—closer agreement between observed and expected counts means smaller deviations, which would decrease the chi-square value. Remember that chi-square measures "goodness of fit"—how well your data matches expectations. Larger deviations from expected patterns always increase the chi-square statistic, while better fits decrease it. Focus on the mathematical relationship in the formula rather than getting distracted by significance thresholds or experimental design elements.

Question 18

1. The mRNA level might not correlate with the protein level for VEGF.
2. An mRNA level of 150 units is generally considered too low to be biologically active.
3. Angiogenesis is controlled by a network of genes, not just VEGF.
4. There is no comparison to VEGF expression in a non-tumor control tissue. (correct answer)

Explanation: The term 'upregulated' is inherently comparative. It means the expression is higher relative to something else. Without a control sample, such as adjacent normal tissue from the same patient, it is impossible to determine if the measured value of 150 units represents an increase, a decrease, or no change. The conclusion is therefore unfounded based on the data presented.

Question 19

1. Cross the mutant (m/m) to a wild-type strain and test cross the F1 to each deletion strain, looking for linkage.
2. Cross different deletion strains to each other and screen for the production of non-viable progeny.
3. Cross each deletion strain to a wild-type strain and look for any F1 progeny that display the mutant phenotype.
4. Cross the mutant (m/m) to each deletion strain and look for F1 progeny that display the mutant phenotype. (correct answer)

Explanation: When you encounter deletion mapping problems, remember that deletions act as null alleles—they completely remove genetic material from a chromosome. This creates a powerful tool for quickly localizing mutations. The correct approach (D) exploits a key principle: if you cross a recessive mutant (m/m) with a deletion strain that has deleted the chromosomal region containing the m gene, the F1 offspring will be m/deletion. Since the deletion provides no functional copy of the gene, these offspring will express the recessive mutant phenotype despite being technically heterozygous. When the deletion doesn't include the m locus, F1 offspring will be m/+ and show the wild-type phenotype. Option A is unnecessarily complex and time-consuming. You'd first need to generate F1 progeny, then perform test crosses with each deletion strain, then analyze linkage patterns—a multi-generation approach when a single cross suffices. Option B makes no biological sense. Crossing deletion strains to each other won't tell you anything about where your specific mutation m is located, and the viability of progeny depends on which essential genes are deleted, not on the location of m. Option C has the logic backwards. Crossing deletion strains to wild-type produces F1 that are heterozygous (deletion/+), which will show wild-type phenotypes regardless of which chromosomal region is deleted, since they retain one functional copy of all genes. Remember: deletion mapping works because deletions unmask recessive mutations when both are present in the same individual. Look for crosses that directly combine your mutation with each deletion.

Question 20

1. Cross the jerker male with a female known to be homozygous dominant (JJ).
2. Cross the jerker male with one of his daughters produced from a cross with a non-jerker female.
3. Cross the jerker male with a female that exhibits the recessive, non-jerker phenotype (jj). (correct answer)
4. Cross the jerker male with a female that also has the jerker phenotype but is of unknown genotype.

Explanation: A test cross is specifically designed to determine an unknown genotype of a dominant-phenotype individual by crossing it with an individual that is homozygous recessive for the trait. In this case, crossing the jerker male (J_) with a non-jerker female (jj) will reveal if the male carries the recessive allele. If any offspring are non-jerkers (jj), the male must be heterozygous (Jj).

Question 21

1. The genes F and G are tightly linked, with a map distance close to 0 cM.
2. The genes F and G are linked and are approximately 25 cM apart.
3. The observed data are not significantly different from the ratio expected for unlinked genes. (correct answer)
4. The chi-square test is invalid because the sample size was too small.

Explanation: A 1:1:1:1 phenotypic ratio is the classic expectation for a test cross involving two unlinked genes that assort independently. A high p-value (typically > 0.05) from a chi-square test indicates that there is no statistically significant difference between the observed data and the expected results under the null hypothesis. In this case, the null hypothesis is independent assortment. Therefore, the data strongly support the conclusion that the genes are unlinked (or are so far apart on the same chromosome that they assort independently), and the observed ratio is consistent with this hypothesis.

Question 22

1. 10.0 cM (correct answer)
2. 45.0 cM
3. 50.0 cM
4. 90.0 cM

Explanation: First, determine the genotype and linkage phase of the F1 dihybrid. The parents were tall-pubescent (DDpp) and dwarf-smooth (ddPP). The F1 plant is thus DdPp, with the alleles in repulsion phase (genotype Dp/dP). Second, identify the parental and recombinant progeny from the test cross (Dp/dP x dp/dp). Parental progeny from the Dp/dP F1 are tall-pubescent (445) and dwarf-smooth (455). Recombinant progeny result from crossover gametes (DP and dp). These are tall-smooth (52) and dwarf-pubescent (48). Third, calculate the map distance. Total progeny = 445 + 455 + 52 + 48 = 1000. Total recombinants = 52 + 48 = 100. Recombination Frequency (RF) = (100/1000) × 100 = 10%. The map distance in centiMorgans equals the recombination frequency, thus 10.0 cM.

Question 23

1. 8%
2. 16% (correct answer)
3. 42%
4. 84%

Explanation: The parent is in repulsion phase: dpy + / + unc. With 16% recombination, parental gametes (dpy + and + unc) each have frequency 42%, while recombinant gametes (dpy unc and + +) each have frequency 8%. Progeny with the Dumpy phenotype come from two sources: dpy + gametes (42%) giving Dumpy-wild type, and dpy unc gametes (8%) giving Dumpy-Unc. Total Dumpy progeny = 42% + 8% = 50%. Among Dumpy progeny, those that are also Unc = 8%/(42%+8%) = 8%/50% = 16%.

Question 24

1. 1/4
2. 1/2
3. 2/3 (correct answer)
4. 1

Explanation: Since individuals I-1 and I-2 are unaffected but have an affected child (II-1), they must both be heterozygous carriers (Aa). The possible genotypes for their offspring are AA, Aa, and aa in a 1:2:1 ratio. Individual II-3 is phenotypically normal, so her genotype cannot be aa. The remaining possible genotypes are AA and Aa. There is one AA genotype and two Aa genotypes in this subset of outcomes. Therefore, the probability that II-3 is a carrier (Aa) is 2/3.

Question 25

1. The ASO uses the SMN2 transcript as a template to repair the SMN1 gene via reverse transcription.
2. The ASO sterically blocks a splicing repressor from binding to the ISS, thereby promoting the inclusion of exon 7. (correct answer)
3. The ASO binds to and activates the mutated ESE on exon 7, overriding the effect of the silencer.
4. The ASO directly binds to the faulty SMN protein produced from SMN2, restoring its normal function.

Explanation: Nusinersen is designed to interfere with the splicing regulation of SMN2. By binding to the ISS in intron 7, the ASO physically prevents repressor proteins (like hnRNPs) from binding to this silencing element. Removing this repressive signal tips the balance of splicing regulation in favor of exon 7 inclusion. This leads to an increased production of full-length, functional SMN protein from the SMN2 gene, compensating for the lack of SMN protein from the missing SMN1 gene.