Lectures 2-6 Practice Problems

Transcription is about to begin at the transcription start site (+1) of a eukaryotic gene. Which statement is correct:

A. RNA is synthesised 5′→3′, and the template strand is read 3′→5′

B. RNA is synthesised 5′→3′, and the template strand is also read 5′→3′

C. RNA is synthesised 3′→5′, and the template strand is read 5′→3′

D. Both DNA strands are transcribed outward from the +1 site

A. RNA is synthesised 5′→3′, and the template strand is read 3′→5′.

Explanation: RNA always grows 5′→3′, so the template has to be read in the opposite direction, 3′→5′. Only one strand is used.

Which of the following correctly lists the four levels at which eukaryotic gene expression is regulated:

A. Transcriptional, replicational, translational, post-translational

B. Pre-transcriptional, transcriptional, translational, post-translational

C. Transcriptional, post-transcriptional, translational, post-translational

D. Chromosomal, transcriptional, translational, degradational

C. Transcriptional, post-transcriptional, translational, post-translational

Explanation: The four levels are transcriptional, post-transcriptional, translational and post-translational. Control of synthesis, processing and degradation all matter, but replication is not one of them.

Which protein recognizes the core promoter and nucleates assembly of the RNA polymerase II pre-initiation complex:

A. TFIIH

B. Mediator

C. A TBP-associated factor (TAF) acting on its own

D. The TATA-binding protein (TBP), as a component of TFIID

D. The TATA-binding protein (TBP), as a component of TFIID

Explanation: TBP binds the TATA box as part of TFIID and unwinds the DNA locally. TAFs recognise other elements such as INR and DPE, but not on their own. TFIIH arrives later and catalyses ATP-powered unwinding.

Which statement about enhancers is correct:

A. They are found only in bacteria

B. They can lie thousands of base pairs away, upstream or downstream, and act by DNA looping

C. They must lie within about 100 bp of the transcription start site

D. They are bound directly by RNA polymerase II, without any regulatory proteins

B. They can lie thousands of base pairs away, upstream or downstream, and act by DNA looping

Explanation: Enhancers act at a distance and in either orientation. Looping brings the regulatory proteins bound at the enhancer into contact with the promoter.

In the torpedo model of transcription termination, what causes RNA polymerase II to leave the DNA:

A. A hairpin forms in the nascent RNA and pulls the polymerase off the template

B. The polymerase changes conformation after it transcribes the poly(A) site

C. A ribonuclease degrades the excess RNA, catches up with the polymerase, and displaces it

D. TFIID is released from the TATA box and the pre-initiation complex collapses

C. A ribonuclease degrades the excess RNA, catches up with the polymerase, and displaces it

Explanation: Option A is the allosteric model — the other of the two. In the torpedo model an RNase chases the transcript down and knocks the polymerase off.

A nucleosome is being assembled on newly replicated DNA. Which histones associate with the DNA first:

A. The two H3–H4 dimers

B. The linker histone H1

C. All eight core histones bind simultaneously

D. The two H2A–H2B dimers

A. The two H3–H4 dimers

Explanation: The two H3–H4 dimers associate with DNA first, and the two H2A–H2B dimers then join to complete the octamer. About 146 bp of DNA wraps the octamer in 1.65 turns. H1 is a linker histone and is not part of the core octamer.

Chromatin remodeling complexes bind DNA and nucleosomes with no sequence specificity of their own. Where does the specificity for particular genes come from?

A. The ATPase (DNA translocase) domain of the remodeler

B. Sequence-specific DNA-binding transcription factors that recruit the remodeler

C. The linker histone H1

D. The dyad axis of the nucleosome

B. Sequence-specific DNA-binding transcription factors that recruit the remodeler

Explanation: Remodelers bind nucleosomes and DNA without reading sequence. Specificity is supplied by sequence-specific transcription factors that recruit them, and by histone tail marks read through bromodomains and chromodomains.

After DNA replication, CG and CHG methylation can be restored by copying from the parent strand, but CHH methylation cannot. Why not?

A. CHH sites occur only in mammals

B. CHH sites lie exclusively within CpG islands

C. DNMT1 cannot recognize hemimethylated DNA

D. CHH is asymmetric, so replication leaves no methylated partner strand to copy from

D. CHH is asymmetric, so replication leaves no methylated partner strand to copy from

Explanation: CG and CHG are symmetric, so the same context exists on both strands and the parent strand can template the daughter. CHH is asymmetric, so the mark has to be re-established each cycle, in plants by RNA-directed DNA methylation.

Which statement about 5-hydroxymethylcytosine (5hmC) is correct:

A. It is produced from 5mC by TET enzymes, and because DNMT1 reads it poorly it also drives passive loss of methylation

B. It is added directly to cytosine by DNMT3A

C. It is recognised efficiently by DNMT1 and is therefore faithfully maintained through replication

D. It is found only in plants

A. It is produced from 5mC by TET enzymes, and because DNMT1 reads it poorly it also drives passive loss of methylation

Explanation: TET enzymes oxidise the methyl group of 5mC to give 5hmC, and further to 5fC and 5caC. DNMT1 is a poor reader of 5hmC, so oxidation before maintenance means the mark is simply not copied. That is why TET feeds both the active and the passive route.

Primordial germ cells are remethylated in a sex-specific manner. Which statement is correct:

A. In males remethylation is complete by birth; in females it occurs after birth, during oocyte growth

B. Both sexes complete remethylation before birth

C. Neither sex remethylates until after fertilization

D. In females remethylation is complete by birth; in males it occurs after puberty

A. In males remethylation is complete by birth; in females it occurs after birth, during oocyte growth

Explanation: Male PGC methylation is fully established by birth and is then maintained through many rounds of mitosis, which is why the male germline accumulates more maintenance errors. Female remethylation happens after birth, during oocyte growth.