Mechanisms of Cell Differentiation, Development (2C)

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MCAT Biological and Biochemical Foundations of Living Systems › Mechanisms of Cell Differentiation, Development (2C)

Questions 1 - 10
1

Researchers observe that two daughter cells produced by an asymmetric division show different levels of a fate determinant protein (Det-1). The daughter with higher Det-1 later expresses a lineage marker; the daughter with lower Det-1 does not. Blocking polarized localization of Det-1 before division makes both daughters similar and reduces marker expression overall. Cellular principle assessed: asymmetric segregation of cell fate determinants. Which mechanism best explains the development of different cell types?

Unequal inheritance of Det-1 biases gene expression programs in daughter cells, promoting divergent fates

Blocking Det-1 localization increases genetic recombination during mitosis, reducing lineage markers

Both daughters become identical because asymmetric division normally changes DNA sequence in one daughter

Det-1 is a mitochondrial protein, so unequal inheritance changes ATP levels and randomly alters fate

Explanation

This question assesses asymmetric segregation of cell fate determinants, leading to divergent daughter fates. Cell differentiation can occur via asymmetric division, where determinants are unequally distributed, biasing gene expression in progeny. Here, unequal Det-1 inheritance correlates with marker expression, and blocking it equalizes daughters and reduces markers. Choice A is correct because unequal Det-1 biases programs for divergence. Choice D is incorrect as asymmetric division does not change DNA sequence. Verify asymmetry by tracking determinant localization and progeny fates. A strategy is to perturb segregation and observe effects on differentiation outcomes.

2

A differentiation factor (DF) is expressed at similar mRNA levels in two cell types, but DF protein is high only in Cell Type A. Polysome profiling shows DF mRNA is efficiently translated in Cell Type A but not in Cell Type B. Cellular principle assessed: translational control contributing to cell-type-specific protein expression. Which mechanism is most consistent with these data?

DF protein is high in Cell Type A because DF mRNA is alternatively spliced into an unrelated enzyme

Cell Type A has reduced transcription of DF, which increases DF protein through feedback

Cell Type B contains a repressor that reduces translation initiation on DF mRNA, limiting DF protein

Cell Type B has increased DF gene copy number, which lowers DF protein by dilution

Explanation

This question assesses translational control contributing to cell-type-specific protein expression, beyond mRNA levels. Cell differentiation can regulate proteins translationally, with cell-specific factors affecting ribosome association and synthesis rates. Here, DF mRNA is similar but protein and translation efficiency differ between cell types. Choice A is correct because a repressor in Cell Type B limits translation. Choice B is incorrect as increased copy number would not lower protein by dilution. Verify translational control via polysome profiling. A transferable check is to compare mRNA and protein levels for discrepancies indicating post-transcriptional regulation.

3

A lineage-specific gene (Gene L) is induced during differentiation only when cells are plated at high density. Conditioned medium from high-density cultures partially rescues Gene L induction in low-density cultures. Cellular principle assessed: paracrine signaling influencing differentiation. Which process is most likely involved in the scenario described?

Cells secrete a diffusible factor at high density that promotes Gene L expression in nearby cells

High density increases DNA replication errors, generating mutants that overexpress Gene L

High density forces cells into G0, and Gene L is transcribed only during S phase

Conditioned medium supplies nucleotides that directly activate the Gene L promoter by base pairing

Explanation

This question assesses paracrine signaling influencing differentiation, where secreted factors affect neighbors. Cell differentiation can depend on density via paracrine factors that promote gene expression in nearby cells. Here, high density induces Gene L, partially rescued by conditioned medium. Choice A is correct because diffusible factors from dense cultures promote induction. Choice B is incorrect as no replication errors are implied. Test paracrine effects with conditioned media experiments. A reasoning tool is to distinguish autocrine/paracrine by density and media transfer outcomes.

4

A lab observes that a differentiation marker is expressed in a patchy pattern across a tissue even though all cells are genetically identical and exposed to the same external conditions. Time-lapse imaging shows that once a cell turns the marker on, it tends to stay on through subsequent divisions. Cellular principle assessed: heritable gene expression states without DNA sequence change. Which mechanism best explains the stable patchy pattern?

Transient fluctuations in ATP levels that reset completely at each cell division

Unequal cytokinesis permanently changes chromosome number in marker-positive clones

Epigenetic inheritance of chromatin states that maintain marker gene expression across cell divisions

Somatic recombination in each cell generating unique marker gene alleles that are inherited

Explanation

This question assesses heritable gene expression states without DNA sequence change, via epigenetics. Cell differentiation maintains stable, heritable patterns through chromatin states propagated across divisions. Here, patchy marker expression persists through divisions in identical cells. Choice C is correct because epigenetic inheritance sustains expression. Choice B is incorrect as no recombination generates alleles. Confirm heritability by tracking expression stability over divisions. A useful check is to exclude genetic changes when patterns are non-uniform in uniform conditions.

5

Investigators tracked a developmental gene (Gene R) during early differentiation. In progenitors, the Gene R promoter was heavily DNA-methylated and Gene R mRNA was low. After exposure to a differentiation signal, methylation at the promoter decreased and Gene R mRNA increased ~20-fold without changes in Gene R copy number. Cellular principle assessed: epigenetic regulation of transcription. Which process is most likely involved in the scenario described?

Homologous recombination inserting additional Gene R exons to boost mRNA output

Increased methylation at the Gene R promoter recruiting RNA polymerase II more efficiently

Ribosomal frameshifting that increases translation of Gene R without altering mRNA abundance

Demethylation at the Gene R promoter enabling transcription factor binding and increased transcription

Explanation

This question assesses epigenetic regulation of transcription, focusing on how modifications like DNA methylation influence gene expression during development. Cell differentiation relies on epigenetic changes, such as promoter demethylation, which can relieve repression and allow transcription factor binding to activate genes. In this case, decreased methylation at the Gene R promoter coincides with a 20-fold increase in mRNA without copy number changes, indicating epigenetic derepression. Choice D is correct because demethylation enables transcription factor binding and boosts transcription. Choice B is incorrect as increased methylation typically represses, not enhances, transcription by hindering polymerase recruitment. To confirm epigenetic involvement, look for expression changes without genetic alterations. A key strategy is to differentiate between transcriptional and post-transcriptional regulation by measuring mRNA levels.

6

In a developmental model, a morphogen-like signal was applied uniformly to a population of progenitor cells, but only a subset differentiated into Cell Type T. Single-cell analysis showed that cells that became Cell Type T had higher baseline expression of receptor R before morphogen exposure. When receptor R was experimentally overexpressed in all cells, the fraction differentiating into Cell Type T increased. Cellular principle assessed: differential responsiveness to the same extracellular signal can arise from differences in receptor expression, leading to distinct cell fates. Which outcome would be expected during differentiation under uniform morphogen exposure?

Receptor R overexpression changes the DNA sequence of lineage genes, making differentiation into Cell Type T heritable across generations

Cells with higher receptor R levels are more likely to activate downstream transcriptional programs that specify Cell Type T

Cells with higher receptor R levels are less likely to respond to morphogen because receptors sequester ligand away from signaling

All cells will adopt Cell Type T because uniform morphogen exposure eliminates any need for receptor-mediated signaling

Explanation

This question tests understanding of how differential receptor expression creates heterogeneous responses to uniform signals during differentiation. Cell fate decisions often depend on the cell's ability to respond to extracellular signals, which is determined by receptor expression levels. Cells with higher receptor R expression have greater capacity to transduce the morphogen signal, making them more likely to activate downstream transcriptional programs specifying Cell Type T fate. The correct answer (D) logically explains how receptor expression differences lead to different outcomes despite uniform signal exposure. Answer B is incorrect because higher receptor levels increase, not decrease, signal responsiveness - receptors transduce signals rather than sequestering ligands away from signaling. To identify receptor-dependent differentiation heterogeneity, correlate initial receptor expression levels with final cell fate outcomes under uniform signaling conditions.

7

Two differentiated cell types were generated from the same stem cell line. RNA-seq showed Gene S was highly expressed in Cell Type 1 and nearly absent in Cell Type 2. ATAC-seq indicated that the promoter region of Gene S was accessible in Cell Type 1 but not in Cell Type 2. No differences were detected in Gene S coding sequence. Cellular principle assessed: chromatin accessibility influences transcription and can differ between cell types. Which mechanism best explains the observed Gene S expression pattern?

Gene S is transcribed equally in both cell types, but Cell Type 2 lacks ribosomes, eliminating detectable mRNA

Reduced promoter accessibility in Cell Type 2 limits transcription factor binding and decreases transcription of Gene S

Gene S is absent in Cell Type 2 due to loss of the entire chromosome during asymmetric cell division

Increased promoter accessibility in Cell Type 2 prevents RNA polymerase II recruitment, lowering Gene S expression

Explanation

This question tests understanding of how chromatin accessibility regulates cell-type-specific gene expression. Chromatin accessibility determines whether transcription factors and RNA polymerase can access gene regulatory regions to initiate transcription. ATAC-seq reveals that Gene S promoter is in open chromatin in Cell Type 1, allowing transcription factor binding and high expression, while closed chromatin in Cell Type 2 prevents access and results in low expression despite identical DNA sequences. The correct answer (A) correctly identifies that reduced accessibility limits transcription factor binding and decreases transcription. Answer B is incorrect because increased accessibility would enhance, not prevent, RNA polymerase recruitment. To determine if chromatin state regulates gene expression, compare chromatin accessibility data (ATAC-seq) with expression data (RNA-seq) - genes with accessible promoters in specific cell types should show higher expression in those cells.

8

During differentiation of mesenchymal stem cells, investigators observed that microRNA miR-21 levels increased early (day 1), while protein levels of transcriptional repressor REP decreased by day 2. REP mRNA levels remained unchanged. By day 4, a differentiation marker Gene Z was strongly expressed. Introducing an miR-21 inhibitor prevented the drop in REP protein and reduced Gene Z expression. Cellular principle assessed: post-transcriptional regulation can influence differentiation by altering protein levels without changing mRNA abundance. Which mechanism best explains the development of different cell types in this scenario?

miR-21 increases REP transcription, which then activates Gene Z as a downstream target

miR-21 binds REP mRNA and decreases REP translation, relieving repression of Gene Z expression

miR-21 methylates the REP promoter, increasing REP protein stability and driving Gene Z expression

miR-21 is translated into a small protein that directly binds the Gene Z promoter to initiate transcription

Explanation

This question tests understanding of post-transcriptional regulation by microRNAs during cell differentiation. MicroRNAs regulate gene expression by binding to target mRNAs and either degrading them or inhibiting their translation, thereby reducing protein levels without affecting mRNA abundance. miR-21 binds to REP mRNA and inhibits its translation, reducing REP protein levels while REP mRNA remains constant; since REP is a transcriptional repressor of Gene Z, decreased REP protein relieves repression and allows Gene Z expression. The correct answer (D) correctly describes how miR-21 decreases REP translation to indirectly activate Gene Z. Answer B is incorrect because microRNAs typically decrease, not increase, target gene expression, and the data shows REP mRNA levels unchanged. To identify microRNA-mediated regulation, look for changes in protein levels without corresponding mRNA changes, and test whether microRNA inhibitors reverse the phenotype.

9

Neural progenitor cells were induced to differentiate in vitro. At day 0, both Gene P and Gene Q were transcribed at low levels. After 3 days in Differentiation Condition 1, Gene P mRNA increased 20-fold while Gene Q remained low. In Differentiation Condition 2, Gene Q increased 20-fold while Gene P remained low. A translation inhibitor added only on day 0 prevented the later rise of either Gene P or Gene Q, even after the inhibitor was removed. Cellular principle assessed: early gene expression can produce regulatory proteins required for later lineage-specific transcriptional programs. Based on the information, which outcome would be expected during differentiation?

The inhibitor prevents differentiation by forcing cells to enter meiosis, which suppresses somatic gene expression

Day-0 translation is required because mRNA cannot be transcribed unless ribosomes are actively translating

Day-0 translation produces regulatory proteins that are required to activate later transcription of Gene P or Gene Q

Blocking translation on day 0 permanently mutates the promoters of Gene P and Gene Q, preventing transcription

Explanation

This question tests understanding of how early gene expression establishes regulatory cascades necessary for later differentiation programs. Cell differentiation often involves sequential waves of gene expression where early proteins regulate later genes. The day-0 translation produces regulatory proteins (likely transcription factors or chromatin modifiers) that are essential for activating the condition-specific programs leading to either Gene P or Gene Q expression by day 3. The correct answer (D) logically explains why blocking early translation prevents later transcriptional changes - without the initial regulatory proteins, the downstream cascade cannot proceed. Answer B is incorrect because transcription and translation are independent processes; mRNA synthesis doesn't require active translation. When analyzing differentiation timecourses, identify early translation-dependent steps by testing whether protein synthesis inhibitors at specific timepoints block later gene expression programs.

10

A lab compared two cell populations derived from the same hematopoietic stem cell: Population 1 expressed high levels of erythroid marker E, while Population 2 expressed high levels of myeloid marker M. Chromatin immunoprecipitation showed that transcription factor GATA bound near the enhancer of marker E only in Population 1, while transcription factor PU.1 bound near the enhancer of marker M only in Population 2. Cellular principle assessed: lineage-specific transcription factor binding at enhancers directs differential gene expression. Which process is most likely involved in the scenario described?

Enhancer binding causes deletion of the marker genes in the alternative lineage, preventing any future expression

Transcription factor binding primarily changes the amino acid sequence of E and M proteins by editing mRNA codons

Enhancer binding by lineage-specific transcription factors increases transcription of target genes in a cell-type-specific manner

Transcription factor binding converts enhancer DNA into mRNA, which then acts as the marker protein

Explanation

This question tests understanding of how lineage-specific transcription factors direct cell fate through enhancer binding. Cell differentiation involves master transcription factors that bind to enhancers of lineage-specific genes, recruiting transcriptional machinery and increasing gene expression. GATA binding at erythroid enhancers drives erythroid differentiation, while PU.1 binding at myeloid enhancers drives myeloid differentiation, creating mutually exclusive cell fates from the same progenitor. The correct answer (A) accurately describes how transcription factor-enhancer interactions increase target gene transcription in a cell-type-specific manner. Answer B is incorrect because transcription factors regulate expression, not delete DNA sequences - both populations retain all genes but express different subsets. To identify transcription factor-driven differentiation, look for correlations between specific transcription factor binding at enhancers and expression of nearby lineage markers.

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