Designer Genes Practice Test — Hard Level

A free 10-question Designer Genes practice test for Science Olympiad, written at national level. Every question and answer choice is listed below — correct answers and scoring are revealed when you take the timed test.

10 questions15 min timedHard levelAI-generated — verify answers against reliable sources

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1. A researcher sequences a gene and finds the following change in the coding (sense) DNA strand near the 5' end of the gene: Wild type: 5'-ATG GAG TTT CCA GCT ...-3' Mutant: 5'-ATG GAG TAT TCC AGC T...-3' Assume standard splicing and that transcription and translation initiate normally at the first ATG. Which classification best describes this mutation at the protein level? (Use the standard genetic code: GAG → Glu, TTT → Phe, TAT → Tyr, TCC → Ser, AGC → Ser.)

A.A) Silent mutation, because the amino acid sequence is unchanged
B.B) Missense mutation, because one amino acid is replaced by another
C.C) Nonsense mutation, because a premature stop codon is introduced
D.D) Frameshift mutation, because all codons downstream are altered

2. You perform a testcross AaBb × aabb to determine whether genes A and B assort independently in a particular plant species. You score 1,000 offspring and obtain the following phenotypes: - AB (dominant for both traits): 410 - Ab: 96 - aB: 104 - ab (recessive for both traits): 390 Assume that a single crossover between the loci is the only source of recombinants and that there is no differential viability. Which conclusion is best supported by these data? (Useful information: For a 1:1:1:1 ratio with 3 degrees of freedom, χ² critical at p = 0.05 is 7.815.)

A.A) The genes assort independently; deviations from 1:1:1:1 are due to random sampling error.
B.B) The genes are linked at approximately 10 map units apart, and the null hypothesis of independent assortment should be rejected.
C.C) The genes are linked at approximately 20 map units apart, and the null hypothesis of independent assortment should be rejected.
D.D) The genes are completely linked (0 map units), and recombinants arise from rare mutations rather than crossing over.

3. An X-linked recessive disorder occurs in 1 out of every 2,500 males in a large population that is in Hardy–Weinberg equilibrium. A city has 1,000,000 inhabitants with an equal number of males and females, and the population can be treated as a random sample from this larger population. Assuming Hardy–Weinberg conditions hold and ignoring the extremely small number of affected females, approximately how many females in the city are expected to be heterozygous carriers for the disorder?

A.A) 40
B.B) 200
C.C) 400
D.D) 800

4. In an E. coli cell containing a functional lac operon with its native promoter, operator, and regulatory genes intact, under which condition will transcription of the lactose-metabolizing structural genes (lacZYA) be at its highest level? Assume normal regulation by both the lac repressor and the CAP–cAMP complex.

A.A) High glucose, no lactose present
B.B) High glucose, high lactose present
C.C) Low glucose, no lactose present
D.D) Low glucose, high lactose present

5. In a malaria-endemic region, the frequency of individuals with sickle-cell disease (homozygous recessive for the Hb^S allele) is 1 in 400. Assume Hardy–Weinberg equilibrium and that only heterozygotes (Hb^A Hb^S) have increased malaria resistance. Approximately what proportion of the population is expected to have increased resistance to malaria due to this allele?

A.A) 0.25%
B.B) 2.5%
C.C) 5.0%
D.D) 9.5%

6. A heterozygous fly with genotype AaBbCc is testcrossed to a fully recessive fly aabbcc. The three genes assort as a linked group. The phenotypes of 1,000 offspring (corresponding directly to the gametes produced by the AaBbCc parent) are recorded as follows: - ABC: 410 - abc: 400 - Abc: 60 - aBC: 52 - ABc: 30 - abC: 28 - AbC: 10 - aBc: 10 Assuming all deviations are due to crossing over and that these are the only three loci considered, what is the map distance between genes A and B (in centimorgans, cM)?

A.A) 7.8 cM
B.B) 9.0 cM
C.C) 13.2 cM
D.D) 19.0 cM

7. A trait in a large human pedigree shows the following characteristics: • Affected individuals appear in every generation. • Approximately equal numbers of affected males and females are observed. • Each affected individual has at least one affected parent. • Two unaffected parents never have an affected child. • Transmission from father to son is common. Which mode of inheritance best fits this pattern?

A.A) Autosomal dominant
B.B) Autosomal recessive
C.C) X-linked recessive
D.D) Mitochondrial

8. Researchers measure β-galactosidase activity in two E. coli strains using a chromogenic substrate. Activities are reported in arbitrary units. Strain W (wild-type): - No glucose, no lactose: 5 - No glucose, +lactose: 500 - +glucose, no lactose: 2 - +glucose, +lactose: 50 Strain M (mutant): - No glucose, no lactose: 450 - No glucose, +lactose: 480 - +glucose, no lactose: 40 - +glucose, +lactose: 45 Which single genetic change in the lac regulatory system most plausibly explains the behavior of Strain M?

A.A) Loss-of-function mutation in lacI, producing a repressor that cannot bind the operator in the absence of inducer (constitutive expression, still sensitive to glucose via CAP–cAMP).
B.B) Loss-of-function mutation in the CAP (CRP) protein, so the operon cannot respond to cAMP levels and is expressed only in the presence of lactose.
C.C) Mutation in the lac operator that abolishes RNA polymerase binding but leaves repressor binding intact, preventing transcription under all conditions.
D.D) Nonsense mutation early in lacZ, eliminating functional β-galactosidase while leaving transcriptional regulation of the operon unchanged.

9. In a very large, randomly mating population at Hardy–Weinberg equilibrium for a single gene with two alleles, A and a, the frequency of individuals with genotype aa (homozygous recessive) is 1 in 1,600. Which expression correctly gives the expected frequency of heterozygous carriers (Aa) in this population, assuming Hardy–Weinberg conditions hold? Let f(aa) denote the observed frequency of the homozygous recessive genotype.

A.A) f(Aa) = [f(aa)]²
B.B) f(Aa) = 2 × f(aa)
C.C) f(Aa) = 2 × √f(aa) × (1 − √f(aa))
D.D) f(Aa) = 1 − f(aa)

10. A research lab uses CRISPR–Cas9 to introduce a 1-bp insertion that disrupts an EcoRI site (GAATTC) within a human gene. The edited and unedited alleles are indistinguishable by size alone, so the lab designs a PCR–restriction assay: - PCR amplifies a 1,000 bp fragment spanning the target site. - In the wild-type allele, EcoRI cuts this fragment into 600 bp and 400 bp pieces. - In the edited allele, the 1-bp insertion destroys the EcoRI site; the 1,000 bp fragment remains uncut. After PCR and EcoRI digestion, DNA from three clonal cell lines (derived from single cells of a diploid cell line) is run on an agarose gel alongside a wild-type control: - Lane WT (control): 600 bp and 400 bp bands - Lane 1: 600 bp and 400 bp bands - Lane 2: single 1,000 bp band - Lane 3: 1,000 bp, 600 bp, and 400 bp bands Which interpretation of the clonal lines is most accurate?

A.A) Line 1 is biallelically edited; Lines 2 and 3 are unedited.
B.B) Line 2 is biallelically edited; Line 1 is unedited; Line 3 is heterozygous (one edited, one wild-type allele).
C.C) Line 3 is biallelically edited; Line 1 is heterozygous; Line 2 is unedited.
D.D) All three lines are mixtures of edited and unedited cells; none are true clones.