Recombinant DNA and Biotechnology (1B)
Help Questions
MCAT Biological and Biochemical Foundations of Living Systems › Recombinant DNA and Biotechnology (1B)
A study uses CRISPR-Cas9 to knock out a DNA repair gene in a human cell line to test sensitivity to a chemotherapy agent. Two clonal lines are isolated: Clone X and Clone Y. Sequencing of the target locus shows that Clone X has a frameshift insertion near the start codon, while Clone Y has an in-frame 3-bp deletion. After drug treatment, Clone X shows markedly reduced survival compared with the parental line; Clone Y shows survival similar to parental.
Based on the scenario, which conclusion is most supported by the use of CRISPR-Cas9 editing in this context?
Clone X cannot be compared to parental because CRISPR edits only RNA, not DNA.
Both clones must have identical phenotypes because CRISPR always produces complete gene deletion.
The reduced survival in Clone X proves the chemotherapy agent directly cuts DNA at the CRISPR target site.
Clone Y likely retains substantial gene function because the in-frame deletion may preserve the reading frame.
Explanation
The skill being tested is CRISPR-Cas9 genome editing in recombinant DNA and biotechnology for generating gene knockouts. CRISPR-Cas9 introduces targeted double-strand breaks, leading to indels via non-homologous end joining, which can disrupt gene function if frameshifts occur. In this scenario, Clone X has a frameshift insertion, showing reduced survival after chemotherapy, while Clone Y has an in-frame deletion with survival similar to parental cells. Choice D is correct because the in-frame deletion in Clone Y likely preserves some gene function, explaining the parental-like phenotype. Choice C is incorrect because CRISPR does not always produce complete deletions; outcomes vary by repair type. For similar questions, sequence edits to correlate genotypes with phenotypes. This principle aids in interpreting editing efficiency and functional impacts in biotechnological applications.
A researcher uses CRISPR-Cas9 to edit a single base in a gene associated with an inherited disease, aiming to restore the wild-type sequence in patient-derived cells. After editing, sequencing shows both corrected alleles and unedited alleles in the population.
Which result would be expected if the edited cells are not clonally isolated before analysis?
Sequencing will show only the corrected sequence because CRISPR edits all cells uniformly.
The presence of unedited alleles proves the guide RNA was amplified by PCR in the cells.
Bulk sequencing can show a mixture of corrected and uncorrected sequences due to a heterogeneous cell population.
Only RNA sequencing can detect CRISPR edits because Cas9 modifies transcripts, not DNA.
Explanation
The skill being tested is analyzing CRISPR-Cas9 editing in heterogeneous populations for biotechnology applications. Bulk sequencing reveals mixed alleles if editing is incomplete, showing both edited and unedited sequences. In this scenario, sequencing post-editing shows corrected and unedited alleles without clonal isolation. Choice D is correct because non-clonal populations yield mixed signals. Choice B is incorrect because editing is not always uniform. For similar questions, perform clonal expansion for pure genotypes. This ensures precise evaluation of base editing outcomes.
A lab clones a therapeutic peptide gene into a plasmid designed to secrete the peptide into the bacterial periplasm using a signal sequence. After induction, most peptide remains in the cytosolic fraction and little is detected in the periplasm. Sequencing confirms the peptide coding region is correct.
Which outcome is most consistent with gene cloning and expression in this context?
Periplasmic secretion would require CRISPR-mediated integration into the bacterial chromosome.
Cytosolic localization indicates the plasmid lacks an origin of replication, preventing expression.
A mutation or mismatch affecting the signal sequence or secretion context could impair targeting despite correct peptide coding sequence.
Periplasmic secretion fails because PCR cannot amplify genes encoding secreted proteins.
Explanation
The skill being tested is directing recombinant protein localization in bacterial expression systems using biotechnology. Signal sequences target proteins to periplasm, but mutations can impair secretion despite correct coding. In this scenario, peptide remains cytosolic despite signal sequence, with correct coding confirmed. Choice A is correct because issues in secretion signals can cause retention. Choice B is incorrect because PCR amplifies secreted protein genes effectively. For similar questions, validate localization by fractionation. This optimizes yields of functional secreted proteins.
A researcher clones a gene into a plasmid downstream of a promoter, but the insert is accidentally ligated in the reverse orientation. The plasmid is transformed into bacteria and induced, yet no target protein is detected.
Which outcome is most consistent with gene cloning in this context?
No protein is detected because CRISPR-Cas9 excised the insert during bacterial growth.
Reverse orientation can prevent correct transcription of the coding sequence from the promoter, reducing expression.
Reverse orientation increases expression because RNA polymerase reads both DNA strands equally well.
Orientation does not matter because translation begins at any AUG in the plasmid backbone.
Explanation
The skill being tested is the importance of insert orientation in plasmid cloning for recombinant DNA expression. Correct orientation ensures transcription from the promoter produces sense mRNA; reverse yields antisense. In this scenario, reverse ligation leads to no protein detection post-induction. Choice A is correct because reverse orientation prevents proper transcription. Choice B is incorrect because RNA polymerase is directional, not reading both strands equally. For similar questions, verify orientation by restriction mapping. This prevents expression failures in cloning experiments.
A CRISPR-Cas9 study introduces a premature stop codon into a cytokine gene to reduce secretion. ELISA of culture supernatant shows decreased cytokine levels, but qPCR of cytokine mRNA shows similar transcript abundance in edited and control cells.
Which result would be expected if the CRISPR edit primarily affects protein production rather than transcription?
qPCR cannot measure mRNA because PCR only amplifies proteins after translation.
mRNA levels can remain similar while secreted protein decreases due to truncated, unstable, or nonsecreted protein.
mRNA levels must increase because stop codons stimulate RNA polymerase activity.
Secreted cytokine must increase because CRISPR always creates gain-of-function mutations.
Explanation
The skill being tested is the impact of CRISPR-Cas9 edits on gene expression at RNA and protein levels in biotechnology. Premature stop codons can lead to nonsense-mediated decay or truncated proteins, affecting protein but not necessarily mRNA levels. In this scenario, ELISA shows decreased cytokine, but qPCR shows similar mRNA. Choice D is correct because the edit likely produces unstable proteins without altering transcript abundance. Choice B is incorrect because stop codons do not stimulate transcription. For similar questions, measure both RNA and protein. This distinguishes transcriptional from post-transcriptional effects.
A forensic lab uses PCR to amplify short tandem repeat (STR) loci from a low-quantity DNA sample. The electropherogram shows allelic dropout at one locus (only one allele detected) while other loci show two alleles. The lab suspects stochastic effects from low template input.
Which conclusion is most supported by the use of PCR in this context?
Allelic dropout indicates the primers annealed to RNA rather than DNA, creating a single allele.
PCR cannot amplify repetitive DNA such as STRs, so the result must be from sequencing.
Low template amounts can cause preferential amplification, producing apparent homozygosity at a locus.
Allelic dropout proves the individual has only one chromosome at that locus.
Explanation
The skill being tested is recognizing artifacts in PCR amplification from low-template DNA in forensic biotechnology. Low template amounts can cause allelic dropout due to stochastic amplification, leading to apparent homozygosity. In this scenario, one STR locus shows only one allele, while others show two, attributed to low input. Choice A is correct because preferential amplification mimics homozygosity in low-template PCR. Choice B is incorrect because dropout does not indicate monosomy; it's a technical artifact. For similar questions, use replicate amplifications to confirm. This principle improves reliability in forensic genotyping.
A lab clones a gene into a plasmid with a strong promoter and ribosome binding site to overexpress the protein in bacteria. After induction, the target protein accumulates mostly in insoluble fractions, while little is found in the soluble lysate.
Which outcome is most consistent with gene cloning for bacterial overexpression in this context?
Overexpression can lead to inclusion body formation, reducing soluble protein despite high total expression.
The protein is insoluble because PCR amplification adds introns that bacteria cannot remove.
The insoluble fraction indicates the gene failed to ligate into the plasmid.
Insoluble protein proves the plasmid integrated into the bacterial genome at multiple sites.
Explanation
The skill being tested is bacterial overexpression of recombinant proteins using plasmids in biotechnology. High expression can lead to insoluble inclusion bodies, sequestering protein from soluble fractions. In this scenario, the protein is mostly insoluble post-induction despite strong promoter use. Choice D is correct because overexpression often causes aggregation into inclusion bodies. Choice B is incorrect because insolubility does not indicate ligation failure; expression occurred. For similar questions, optimize induction conditions to enhance solubility. Tags or chaperones can aid in recovering functional protein.
A CRISPR-Cas9 knockout is performed to evaluate whether a transporter is required for uptake of a fluorescent drug analog. After editing, cells show reduced fluorescence compared with control cells. However, a subset of edited cells still shows high fluorescence. The lab suspects incomplete knockout in the population.
Which result would be expected if the observed heterogeneity is due to incomplete editing by CRISPR-Cas9?
Single-cell cloning would yield some clones with wild-type transporter sequence and others with disruptive indels.
Transporter uptake would increase because CRISPR activates transcription at the target site.
PCR of the transporter locus would be impossible because Cas9 permanently blocks DNA polymerase.
All cells would show identical fluorescence because CRISPR edits are always 100% efficient.
Explanation
The skill being tested is assessing CRISPR-Cas9 editing efficiency and heterogeneity in biotechnology. Incomplete editing results in mixed populations with varying genotypes, detectable by functional assays or sequencing. In this scenario, edited cells show overall reduced fluorescence, but some retain high levels, suggesting incomplete knockout. Choice D is correct because single-cell cloning would reveal wild-type and edited clones, explaining heterogeneity. Choice B is incorrect because CRISPR efficiency is not always 100%; variability occurs. For similar questions, quantify editing rates via sequencing. This ensures accurate interpretation of phenotypic screens.
A researcher clones a human membrane receptor cDNA into a bacterial plasmid to produce protein for structural studies. After induction, little to no full-length receptor is detected, and bacterial growth is impaired. The researcher suspects toxicity and poor expression of membrane proteins in bacteria.
Which outcome is most consistent with the limitations of gene cloning and expression in this context?
Bacteria may express the receptor inefficiently or in a misfolded form, reducing yield and stressing cells.
The receptor can only be produced by PCR amplification of the plasmid, not by bacterial translation.
Induction prevents transcription of plasmid genes, so receptor expression should decrease upon induction.
Bacteria will splice introns from the receptor pre-mRNA, improving expression of full-length protein.
Explanation
The skill being tested is limitations of bacterial expression systems for eukaryotic proteins in recombinant DNA technology. Bacteria often misfold or inefficiently express membrane proteins, leading to toxicity and low yields. In this scenario, induction yields little full-length receptor and impairs bacterial growth, suspected due to toxicity. Choice A is correct because poor expression and misfolding of eukaryotic membrane proteins in bacteria can stress cells. Choice B is incorrect because bacteria lack splicing machinery for introns. For similar questions, consider host-specific challenges. Switching to eukaryotic systems can improve outcomes for complex proteins.
A researcher clones a bacterial toxin gene into a plasmid that includes an antibiotic resistance marker. After ligation and transformation, bacteria are plated on antibiotic-containing media. Hundreds of colonies grow. However, sequencing reveals that many colonies contain plasmid without the toxin insert.
Which outcome is most consistent with this gene cloning result?
Ligation cannot produce plasmids without inserts because restriction enzymes prevent vector self-ligation.
Antibiotic selection confirms plasmid uptake but does not guarantee insertion of the toxin gene.
The presence of empty plasmid indicates that PCR was not performed on the colonies.
Antibiotic selection ensures only bacteria with chromosomal integration of the toxin gene survive.
Explanation
The skill being tested is plasmid-based gene cloning and selection in recombinant DNA techniques. Antibiotic selection ensures plasmid uptake but does not confirm the presence of the desired insert, as empty vectors can also confer resistance. In this scenario, many colonies grow on antibiotic media, but sequencing shows plasmids without the toxin gene insert. Choice C is correct because selection verifies transformation but not successful ligation of the insert. Choice B is incorrect because plasmids typically do not integrate into the chromosome; they replicate episomally. For similar questions, screen colonies by PCR or sequencing post-selection. This step optimizes cloning efficiency in biotechnology workflows.