Haploid genome size varies widely among animals and does not track organismal complexity; some salamanders, for example, have many times more DNA per haploid genome than humans. This C-value paradox is explained mainly by differences in

the amount of non-coding DNA, especially repeats and transposon-derived sequences

the number of protein-coding genes, which rises roughly in step with genome size

the amount of non-coding DNA, especially repeats and transposon-derived sequences

the number of chromosomes, since a larger genome is split among more chromosomes

the length of each codon, since large-genome species read their mRNA in longer units

A sequence in an animal genome closely matches the mRNA of a functional gene elsewhere: it lacks all of that gene's introns, ends in a run of A residues, is flanked by short direct repeats and has no promoter. It most probably arose by

reverse transcription of a processed mRNA and insertion of the cDNA at a new site

unequal crossing over between tandem copies of the gene, followed by point mutations

tandem duplication of the whole gene followed by precise deletion of each intron

reverse transcription of a processed mRNA and insertion of the cDNA at a new site

cut-and-paste transposition of the gene by a transposase acting on inverted repeats

Which of the following statements about chromatin-level regulation in mammalian cells are correct?

Acetylation of lysines in histone tails neutralises their positive charge and loosens chromatin

H3K27 trimethylation laid down by Polycomb repressive complex 2 keeps target genes such as Hox genes silent

Acetylation of lysines in histone tails neutralises their positive charge and loosens chromatin

DNMT1 is the main enzyme that methylates previously unmethylated CpGs de novo in the early embryo

H3K27 trimethylation laid down by Polycomb repressive complex 2 keeps target genes such as Hox genes silent

Treating cells with an inhibitor of histone deacetylases lowers the overall level of histone acetylation

In mice the growth-factor gene Igf2 is imprinted so that only the paternally inherited allele is expressed. A male heterozygous for a null Igf2 allele is crossed with a wild-type female; separately, a female heterozygous for the same null allele is crossed with a wild-type male. Which outcome is expected?

Half the offspring of the heterozygous father are growth-deficient, and all offspring of the heterozygous mother grow normally

Half the offspring of the heterozygous mother are growth-deficient, and all offspring of the heterozygous father grow normally

Half the offspring in each cross are growth-deficient, because one null allele halves the amount of Igf2 made

No offspring in either cross are affected, because the wild-type allele always compensates for the null allele

Half the offspring of the heterozygous father are growth-deficient, and all offspring of the heterozygous mother grow normally

In Drosophila, a chromosomal inversion moves the white⁺ gene from its normal euchromatic site to a position next to centromeric heterochromatin. Flies carrying the inversion have eyes with patches of red and white facets. The best explanation is that

heterochromatin silences white⁺ in some eye precursor cells and not others, and daughter cells inherit that state

the inversion breakpoint mutated white⁺ in some eye cells only, while the germ line kept the normal allele

the relocated gene has become imprinted, so it is expressed only in those eye cells that inherited the paternal chromosome

the inverted chromosome is lost by non-disjunction in about half of the dividing eye cells of every fly

A 300 bp DNA element placed 8 kb downstream of a reporter gene raises its transcription 50-fold, and it works equally well when its orientation is inverted. The element is best described as

an enhancer, because it binds activators and acts at a distance in either orientation

a core promoter, because it determines where RNA polymerase II starts transcription

an insulator, because it shields the reporter gene from the surrounding chromatin

an enhancer, because it binds activators and acts at a distance in either orientation

a silencer, because regulatory elements lying downstream of a gene usually repress it

Factor X binds the DNA sequence site-X but has lost its activation domain through a deletion. Factor Y has a strong activation domain and binds site-Y. A fusion of X's DNA-binding domain to Y's activation domain is expressed in cells carrying two reporter genes, one driven by site-X and the other by site-Y, with no intact Y present. What is expected?

Only the site-X reporter is activated, because the DNA-binding domain chooses the target and the activation domain works wherever it is brought

Only the site-Y reporter is activated, because the activation domain decides which promoter is used

Only the site-X reporter is activated, because the DNA-binding domain chooses the target and the activation domain works wherever it is brought

Both reporters are activated, because the fusion carries working parts of both factors and so binds both kinds of site

Neither reporter is activated, because domains taken from different factors cannot fold and act together in one protein

Which of the following statements about transcription by RNA polymerase II are correct?

TBP, a subunit of TFIID, binds the TATA box and bends the DNA

TFIIH unwinds promoter DNA and phosphorylates serine 5 of the Pol II C-terminal domain

Pol II recognises the TATA box on its own and needs general transcription factors only during elongation

TBP, a subunit of TFIID, binds the TATA box and bends the DNA

TFIIH unwinds promoter DNA and phosphorylates serine 5 of the Pol II C-terminal domain

An enhancer works only when it lies within 100 bp upstream of the TATA box in its native orientation

Liver and small intestine both express the single apolipoprotein B gene, yet the liver secretes apoB-100 while the intestine secretes apoB-48, whose sequence matches the N-terminal 48% of apoB-100. The intestinal protein arises because

a cytidine deaminase changes one C to U in the intestinal mRNA, turning a CAA codon into UAA

an intestine-specific promoter inside the gene starts transcription halfway along the coding region

a protease in the intestinal ER cleaves apoB-100 in two and secretes only the N-terminal half

alternative splicing in the intestine skips the exons that encode the C-terminal half of the protein

a cytidine deaminase changes one C to U in the intestinal mRNA, turning a CAA codon into UAA

A cell line expresses a mutant iron-regulatory protein that binds iron-responsive elements tightly whatever the iron level. Grown in iron-rich medium, compared with normal cells in the same medium these cells will show

low ferritin synthesis and high transferrin receptor levels, the pattern normal cells show when iron is scarce

high ferritin synthesis and low transferrin receptor levels, the normal response of cells to abundant iron

high ferritin synthesis and high transferrin receptor levels, because IRP binding stimulates the use of both mRNAs

low ferritin synthesis and low transferrin receptor levels, because IRP binding blocks the translation of both mRNAs

In the animal microRNA pathway, the enzyme that cuts the exported pre-miRNA hairpin in the cytoplasm to release a duplex of about 22 nucleotides is

Dicer, an RNase III enzyme that measures and cuts the double-stranded stem

Drosha, which processes the primary transcript inside the nucleus

Argonaute, which cleaves target mRNAs that pair fully with the guide

Dicer, an RNase III enzyme that measures and cuts the double-stranded stem

Exportin-5, which trims the hairpin while it carries it out through the nuclear pore

Plasma C-peptide is measured to judge how much insulin a diabetic patient's own pancreas still secretes, even while the patient injects insulin. This works because C-peptide

is cut from proinsulin in β-cell granules and released in equal amounts with insulin, but is absent from injected insulin

is the signal peptide removed in the ER, so its level tracks how fast the insulin gene is transcribed

is cut from proinsulin in β-cell granules and released in equal amounts with insulin, but is absent from injected insulin

is released when the liver degrades circulating insulin, so it reflects the total turnover of injected and natural insulin

is the phosphorylated, active form of insulin that only the pancreas can produce from the precursor

Which of the following statements about transposable elements in animals are correct?

LTR retrotransposons move through an RNA intermediate that is reverse-transcribed into DNA before it integrates

In the germ line, piRNAs bound to Piwi proteins silence transposons

DNA transposons such as the P element move by cut and paste, using a transposase that acts at terminal inverted repeats

Alu elements encode their own reverse transcriptase and endonuclease and so move without help

LTR retrotransposons move through an RNA intermediate that is reverse-transcribed into DNA before it integrates

In the germ line, piRNAs bound to Piwi proteins silence transposons

DNA transposons such as the P element move by cut and paste, using a transposase that acts at terminal inverted repeats

In Drosophila, crossing males of a P strain (whose genomes carry P elements) with females of an M strain (which lack them) gives progeny with sterile gonads and frequent germ-line mutations, while the reciprocal cross, M males with P females, gives normal progeny. The best explanation is that

P-strain mothers load P-element piRNAs into the egg that silence transposition, and M-strain eggs lack them

P elements are carried only on the Y chromosome, so that only the sons of P-strain fathers ever inherit them

the P-element transposase is encoded in mitochondrial DNA, which the progeny inherit from the mother

M-strain eggs carry a transposase inhibitor, so P elements can move only when the mother is a P female

P-strain mothers load P-element piRNAs into the egg that silence transposition, and M-strain eggs lack them

Gene Expression in Eukaryotes: Genome Organisation, Levels of Regulation and Transposable Elements

This chapter answers Section XL4.6 of XL4 Zoology, one of the five optional sections of the Life Sciences (XL) paper: “Eukaryotic genome organization and regulation of gene expression, transposable elements.” It first lays out an animal genome: genome size and the C-value paradox, split genes, gene families and pseudogenes, repetitive DNA and the mitochondrial genome. It then follows a gene through every level at which its output can be controlled, from chromatin and transcription through RNA processing, export and decay to translation and the protein itself, and ends with transposable elements: how they move, what they do to genomes and how hosts restrain them. The numericals are the coding fraction of a genome, protein length from exon structure, isoform counts from alternative splicing and target-site duplication arithmetic.

1. Organisation of the eukaryotic genome

Genome size is measured as the C-value, the DNA of one haploid set, in picograms or base pairs (1 pg is just under 10⁹ bp). It varies enormously and does not follow organismal complexity: some salamanders and lungfishes carry many times the roughly 3.1 × 10⁹ bp of the human genome. This C-value paradox is explained by non-coding DNA, mostly repeats, not by extra genes. The human genome has about 20,000 protein-coding genes, yet coding exons make up only about 1 to 2% of it; the rest is introns, regulatory sequence and repeats, and nearly half derives from transposable elements.

Most animal genes are split: exons are interrupted by introns that are transcribed and then spliced out, and introns are often far longer than exons, so a 100 kb gene may yield an mRNA of a few kilobases. Many genes belong to gene families made by duplication and divergence, like the α-like and β-like globin clusters whose genes are switched on in sequence from embryo to adult. Duplicates that have lost function are pseudogenes; processed pseudogenes lack introns, end in an A-rich tract and are flanked by short direct repeats, the marks of an mRNA that was reverse-transcribed and inserted. Genes needed in bulk, such as rRNA genes, sit in tandem arrays.

Classes of DNA in an animal genome
DNA class	Unit and arrangement	Examples
Single-copy	One or a few copies per haploid genome	Most protein-coding genes
Satellite DNA	Units repeated in long tandem arrays, often of distinct buoyant density	Centromeric alpha satellite (171 bp unit) of primates
Minisatellites (VNTRs)	Units of about 10 to 100 bp; copy number differs between individuals	Classical DNA fingerprinting
Microsatellites (STRs)	Units of 1 to 6 bp	(CA)n markers; CAG expansion in Huntington disease
Interspersed repeats	Copies scattered across the genome	LINEs (L1), SINEs (Alu), LTR elements, DNA transposons

In renaturation (Cot) analysis, sequences present in many copies find partners fastest, so highly repetitive DNA reanneals at the lowest Cot values, moderately repetitive DNA next and single-copy DNA last. Animal mitochondrial DNA is a small circle, about 16.6 kb in humans, with 37 genes (13 respiratory-chain polypeptides, 2 rRNAs, 22 tRNAs), no introns and little non-coding DNA beyond the control region; it is present in many copies, maternally inherited and read with a slightly different genetic code.

2. Regulation at the level of chromatin

Nucleosomes restrict access to DNA, so chromatin is the first level of control. Histone acetyltransferases (HATs), many recruited by activators, acetylate lysines in histone tails; this neutralises their positive charge, loosens histone–DNA and nucleosome–nucleosome contacts and creates docking sites for bromodomain proteins. Histone deacetylases (HDACs), recruited by repressors, reverse it. Histone methylation activates or represses depending on the residue: H3K4me3 marks active promoters; H3K9me3 is bound by the chromodomain of HP1 to build heterochromatin; and H3K27me3, laid down by Polycomb repressive complex 2, keeps developmental genes such as Hox genes off where they do not belong. ATP-dependent remodellers such as SWI/SNF slide or evict nucleosomes to expose promoters.

In mammals DNA methylation adds a methyl group to cytosine in CpG dinucleotides. DNMT3A and DNMT3B methylate de novo; DNMT1 recognises hemimethylated CpGs after replication and copies the pattern onto the new strand, so it is inherited through cell divisions. Methylated CpG islands at promoters are silent, partly because methyl-CpG-binding proteins such as MeCP2 recruit HDACs. Methylation underlies genomic imprinting, in which one parental allele is silenced: mouse Igf2 is expressed only from the paternal allele and the neighbouring H19 only from the maternal one. In X inactivation Xist RNA coats one X in female mammals and recruits silencing complexes, turning it into facultative heterochromatin, the Barr body. Position matters too: in Drosophila a gene moved next to heterochromatin is silenced in some cells and not others, so relocating white⁺ gives mottled eyes, position-effect variegation.

3. Regulation of transcription

Most eukaryotic genes are regulated chiefly at transcription initiation. A Pol II core promoter often carries a TATA box about 25 to 30 bp upstream of the start site and an initiator element at the start; proximal elements such as CAAT and GC boxes lie within about 200 bp. Pol II cannot find a promoter on its own: general transcription factors assemble first. The TATA-binding protein (TBP) of TFIID binds and bends the TATA box; TFIIB, Pol II with TFIIF, TFIIE and TFIIH follow. TFIIH unwinds the promoter with its helicase activity and phosphorylates serine 5 of the C-terminal domain (CTD) of Pol II, releasing it into elongation; the phosphorylated CTD also recruits the capping and splicing machinery.

Enhancers bind activators and work at a distance, upstream, downstream or inside introns, in either orientation; DNA looping, aided by the Mediator complex and cohesin, brings them to the promoter. Silencers bind repressors, and insulators bound by CTCF stop an enhancer from acting on genes beyond them. Transcription factors are modular: a DNA-binding domain (zinc finger, homeodomain or other helix-turn-helix, basic leucine zipper, basic helix-loop-helix) picks the site, and a separate activation domain recruits Mediator, HATs and remodellers. Domain swaps prove it: the yeast Gal4 DNA-binding domain fused to the activation domain of the herpesvirus protein VP16 activates any gene carrying Gal4 sites, the logic behind the yeast two-hybrid assay. Control is combinatorial, a limited set of factors in different combinations giving each cell type its own pattern, and signal-dependent: steroid receptors are activated by their hormones, and CREB is phosphorylated by protein kinase A when cAMP rises.

4. RNA processing, export and decay

A pre-mRNA is processed while it is being made. A 5′ cap, 7-methylguanosine joined 5′ to 5′ through a triphosphate, is added after the first 20 to 30 nucleotides; it protects the RNA from 5′ exonucleases and is bound by cap-binding proteins for export and translation. Splicing removes introns, which almost always begin with GU and end with AG, with a branch-point A upstream of the 3′ splice site. The spliceosome (snRNPs U1, U2, U4, U5 and U6 with many proteins) carries out two transesterifications: the 2′-OH of the branch A attacks the 5′ splice site, forming a lariat, and the freed 3′-OH of the upstream exon then attacks the 3′ splice site, joining the exons. U1 pairs with the 5′ splice site and U2 with the branch point. At the 3′ end the AAUAAA signal and a downstream GU-rich element direct cleavage and polyadenylation: poly(A) polymerase adds a tail of about 200 A without a template.

Alternative splicing lets one gene make several proteins: exons can be skipped (cassette exons) or chosen one of a pair (mutually exclusive exons), splice sites can shift, and alternative promoters and poly(A) sites change the ends; most human multi-exon genes are alternatively spliced. In Drosophila, sex is set by a splicing cascade: Sex-lethal protein, made only in females, causes transformer pre-mRNA to be spliced into a functional form, and Transformer with Transformer-2 makes doublesex splice into its female isoform, while males make the default male isoforms. RNA editing changes the sequence after transcription: in the human small intestine a cytidine deaminase converts one C to U in apolipoprotein B mRNA, creating a UAA stop codon, so the intestine makes the short apoB-48 and the liver the full apoB-100 from the same gene; ADAR enzymes convert A to inosine, read as G, in many neural transcripts.

🧠 Codon arithmetic

Protein length from an mRNA: count the nucleotides from the first base of AUG to the last base of the stop codon, divide by 3 and subtract 1 for the stop codon. Example: a mature mRNA of 750 nt (cap and tail excluded) with a 75 nt 5′ UTR and a 135 nt 3′ UTR has 540 nt from AUG through the stop, 180 codons and 179 amino acids. Skipping an exon whose length is a multiple of 3 keeps the frame; any other length shifts it.

Only fully processed mRNAs, marked by the cap, the exon-junction complexes left by splicing and the poly(A) tail, are exported through nuclear pores. In the cytoplasm their lifetime is regulated: AU-rich elements in 3′ UTRs speed deadenylation and decay of short-lived messages such as cytokine mRNAs, and nonsense-mediated decay destroys an mRNA whose stop codon lies well upstream of an exon junction, sparing the cell truncated proteins.

5. Translational and post-translational control

Global control acts on initiation factors. Phosphorylation of eIF2α by stress-activated kinases blocks the recycling of eIF2·GDP and slows almost all initiation; in reticulocytes the haem-regulated inhibitor does this when haem is scarce, so globin is not made without haem. Growth and nutrient signals act through mTOR, which phosphorylates 4E-BPs so that they release eIF4E for cap-dependent translation. Specific control uses sequences in UTRs, and iron is the classic case. When iron is low, iron-regulatory proteins (IRPs) bind iron-responsive elements (IREs), stem-loops in mRNA. An IRE in the 5′ UTR of ferritin mRNA, once bound, blocks ribosome scanning, so the storage protein is not made; IREs in the 3′ UTR of transferrin receptor mRNA, once bound, protect it from degradation, so more receptor imports iron. When iron is high the IRPs let go and the opposite follows. Oocytes store masked maternal mRNAs with short poly(A) tails that are translated only after cytoplasmic polyadenylation at maturation or fertilisation.

MicroRNAs are about 22 nucleotides long. A Pol II primary transcript (pri-miRNA) is cut in the nucleus by Drosha into a hairpin pre-miRNA, exported by exportin-5 and cut in the cytoplasm by Dicer; one strand is loaded onto an Argonaute protein to form the RNA-induced silencing complex (RISC). Animal miRNAs pair imperfectly with 3′ UTRs, chiefly through a seed of nucleotides 2 to 8, repressing translation and promoting deadenylation and decay; one miRNA can regulate hundreds of mRNAs. siRNAs, cut by Dicer from long double-stranded RNA, pair perfectly and make Argonaute cleave the target, the basis of experimental RNA interference.

Control continues after translation. Proteolytic processing activates many proteins: preproinsulin loses its signal peptide in the ER, and proinsulin is cut in secretory granules into insulin and C-peptide, which are released in equal amounts. Covalent modifications such as phosphorylation, acetylation and glycosylation switch activity or location, and regulated degradation ends a protein's life: chains of ubiquitin linked through lysine 48 send it to the 26S proteasome, which is how cyclins, securin and p53 are removed on time.

6. Transposable elements

Transposable elements are DNA sequences that move to new sites in the genome. Barbara McClintock discovered them in maize in the 1940s through the Ac/Ds system, in which the autonomous Activator element supplies the transposase that lets the non-autonomous Dissociation element move. Elements are classified by mechanism.

The two classes of transposable element
Class	Mechanism	Structure	Animal examples
Class I: LTR retrotransposons	Copy and paste through RNA: reverse transcriptase makes cDNA, integrase inserts it	Long terminal repeats at both ends; gag and pol genes, like a retrovirus without env	Drosophila copia; endogenous retroviruses of mammals
Class I: LINEs	RNA reverse-transcribed directly at the nicked target site by the element's own endonuclease and reverse transcriptase	About 6 kb when full length, 3′ poly(A); most copies 5′-truncated and inactive	L1, about 17% of the human genome
Class I: SINEs	Non-autonomous; borrow LINE enzymes	About 300 bp, derived from 7SL RNA or tRNA genes	Alu, over a million copies, about 10% of the human genome
Class II: DNA transposons	Cut and paste: transposase excises the element at its ends and inserts it elsewhere	Terminal inverted repeats flanking a transposase gene	Drosophila P element, mariner, Tc1 of C. elegans

Insertions are flanked by short target-site duplications. The transposase or integrase cuts the two target strands a few base pairs apart, the element is joined to the overhanging single-stranded ends, and repair fills the gaps, so the short stretch between the cuts now appears twice, as direct repeats on either side of the element. Each element makes duplications of a characteristic length, for example 8 bp for the P element. Excision of a DNA transposon often leaves a small footprint, so even an element that has left can leave a mutation behind.

Transposons are mutagens and engines of genome change. A new L1 insertion into the factor VIII gene was found in a patient with haemophilia A; recombination between dispersed copies causes deletions and rearrangements; insertions can donate promoters and enhancers to nearby genes; and some transposon enzymes have been domesticated, such as the RAG1 recombinase of vertebrate V(D)J recombination. Hosts restrain them. In the animal germ line piRNAs loaded onto Piwi proteins cleave transposon RNAs and direct repressive chromatin to their copies, and DNA methylation keeps most copies silent in somatic cells. Hybrid dysgenesis in Drosophila shows the defence failing: males of P strains (carrying P elements) crossed with females of M strains (lacking them) give sterile progeny with many germ-line mutations, because the M egg supplies no P-element piRNAs, whereas the reciprocal cross is normal. Tamed, the P element became the standard vector for making transgenic flies.

Key takeaways

Genome size does not track complexity (C-value paradox) because of non-coding DNA: the human genome of about 3.1 × 10⁹ bp has about 20,000 protein-coding genes, only 1 to 2% coding sequence and nearly half transposon-derived DNA. Satellites cluster at centromeres, VNTRs and STRs vary between individuals, and processed pseudogenes lack introns.
Acetylation (HATs) opens chromatin and HDACs close it; H3K4me3 marks active promoters, H3K9me3 (HP1) and H3K27me3 (Polycomb) repress. CpG methylation, set by DNMT3 and maintained by DNMT1, silences promoters and underlies imprinting (Igf2 paternal, H19 maternal) and X inactivation (Xist).
Pol II needs general factors: TBP of TFIID binds the TATA box and TFIIH unwinds DNA and phosphorylates CTD serine 5. Enhancers act at a distance in either orientation by looping through Mediator; transcription factors are modular (DNA-binding plus activation domain) and act combinatorially.
Processing adds a 5′ cap, removes GU…AG introns through a lariat and adds a poly(A) tail after AAUAAA. Alternative splicing (Drosophila Sxl → tra → dsx) and editing (intestinal apoB-48) multiply products; amino acids = (nucleotides from AUG through stop)/3 − 1.
eIF2α phosphorylation and mTOR–4E-BP set global translation; IRP binding to IREs blocks ferritin translation (5′ UTR) and stabilises transferrin receptor mRNA (3′ UTR) when iron is low. miRNAs (Drosha, Dicer, Argonaute, seed 2–8) repress many mRNAs; proteolytic processing and K48-ubiquitin degradation control proteins after translation.
Class I retrotransposons copy and paste through RNA (LTR elements, LINE L1 about 17%, SINE Alu about 10% of the human genome); Class II DNA transposons cut and paste with a transposase acting at terminal inverted repeats. Insertions carry target-site duplications; piRNAs guard the germ line, and their absence from M-strain eggs causes hybrid dysgenesis.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Haploid genome size varies widely among animals and does not track organismal complexity; some salamanders, for example, have many times more DNA per haploid genome than humans. This C-value paradox is explained mainly by differences in
1. the number of protein-coding genes, which rises roughly in step with genome size
2. the amount of non-coding DNA, especially repeats and transposon-derived sequences
3. the number of chromosomes, since a larger genome is split among more chromosomes
4. the length of each codon, since large-genome species read their mRNA in longer units
Show answer
Answer: B — the amount of non-coding DNA, especially repeats and transposon-derived sequences
Large genomes are large mainly because they carry more non-coding DNA: introns, intergenic sequence and above all repeats, many derived from transposable elements. Gene number varies far less (a few tens of thousands across vertebrates), chromosome number does not track DNA content, and the triplet code is the same in all these animals.
A haploid animal genome of 3.0 × 10⁹ bp contains 20,000 protein-coding genes whose coding sequences (start codon through stop codon) average 1,350 bp. The percentage of the genome that is protein-coding sequence is ____ % (to one decimal place).
Numerical answer — type the value.
Show answer
Answer: 0.9
Total coding sequence = 20,000 × 1,350 = 2.7 × 10⁷ bp. Fraction = 2.7 × 10⁷/3.0 × 10⁹ = 0.009 = 0.9%. The stem gives the coding length directly, so no correction for introns or UTRs is needed; the point is that over 99% of such a genome does not code for protein. Forgetting to convert the fraction to a percentage gives 0.009.
A sequence in an animal genome closely matches the mRNA of a functional gene elsewhere: it lacks all of that gene's introns, ends in a run of A residues, is flanked by short direct repeats and has no promoter. It most probably arose by
1. unequal crossing over between tandem copies of the gene, followed by point mutations
2. tandem duplication of the whole gene followed by precise deletion of each intron
3. reverse transcription of a processed mRNA and insertion of the cDNA at a new site
4. cut-and-paste transposition of the gene by a transposase acting on inverted repeats
Show answer
Answer: C — reverse transcription of a processed mRNA and insertion of the cDNA at a new site
Every feature points to an mRNA copied back into DNA: introns already spliced out, a poly(A) remnant, no promoter (promoters are not transcribed into mRNA) and target-site duplications from insertion, usually by LINE-encoded reverse transcriptase. This is a processed pseudogene. Unequal crossing over and tandem duplication would keep the introns and promoter and place the copy next to the original; DNA-transposon movement would leave terminal inverted repeats, not a poly(A) tract.
Which of the following statements about chromatin-level regulation in mammalian cells are correct?
1. Acetylation of lysines in histone tails neutralises their positive charge and loosens chromatin
2. DNMT1 is the main enzyme that methylates previously unmethylated CpGs de novo in the early embryo
3. H3K27 trimethylation laid down by Polycomb repressive complex 2 keeps target genes such as Hox genes silent
4. Treating cells with an inhibitor of histone deacetylases lowers the overall level of histone acetylation
Show answer
Answer: A — Acetylation of lysines in histone tails neutralises their positive charge and loosens chromatin; C — H3K27 trimethylation laid down by Polycomb repressive complex 2 keeps target genes such as Hox genes silent
Lysine acetylation removes the positive charge that holds tails on DNA and recruits bromodomain proteins, and H3K27me3 written by PRC2 is the Polycomb mark that keeps developmental genes off. DNMT1 is the maintenance methyltransferase that copies existing patterns onto hemimethylated DNA after replication; de novo methylation is done by DNMT3A and DNMT3B. Blocking HDACs leaves acetyl groups in place, so overall acetylation rises, not falls.
In mice the growth-factor gene Igf2 is imprinted so that only the paternally inherited allele is expressed. A male heterozygous for a null Igf2 allele is crossed with a wild-type female; separately, a female heterozygous for the same null allele is crossed with a wild-type male. Which outcome is expected?
1. Half the offspring of the heterozygous mother are growth-deficient, and all offspring of the heterozygous father grow normally
2. Half the offspring in each cross are growth-deficient, because one null allele halves the amount of Igf2 made
3. No offspring in either cross are affected, because the wild-type allele always compensates for the null allele
4. Half the offspring of the heterozygous father are growth-deficient, and all offspring of the heterozygous mother grow normally
Show answer
Answer: D — Half the offspring of the heterozygous father are growth-deficient, and all offspring of the heterozygous mother grow normally
Each pup expresses only the Igf2 allele it received from its father. From the heterozygous father, half the pups receive the null allele as their only active copy and are growth-deficient. From the heterozygous mother, the null allele, if inherited, sits on the silenced maternal copy, while every pup gets an active wild-type allele from its wild-type father, so all grow normally. The reversed pattern would fit a maternally expressed gene such as H19; a simple dosage or full-compensation model ignores imprinting.
In Drosophila, a chromosomal inversion moves the white⁺ gene from its normal euchromatic site to a position next to centromeric heterochromatin. Flies carrying the inversion have eyes with patches of red and white facets. The best explanation is that
1. heterochromatin silences white⁺ in some eye precursor cells and not others, and daughter cells inherit that state
2. the inversion breakpoint mutated white⁺ in some eye cells only, while the germ line kept the normal allele
3. the relocated gene has become imprinted, so it is expressed only in those eye cells that inherited the paternal chromosome
4. the inverted chromosome is lost by non-disjunction in about half of the dividing eye cells of every fly
Show answer
Answer: A — heterochromatin silences white⁺ in some eye precursor cells and not others, and daughter cells inherit that state
This is position-effect variegation. Heterochromatin spreads into the relocated gene to a variable extent; once a cell's choice is made it is inherited, so clones of silenced (white) and expressing (red) cells form patches. All cells of the fly carry the same chromosomes, so imprinting by parental origin cannot make a mosaic, and a mutation caused by the breakpoint would be present in every cell. Chromosome loss in half the cells is not what happens in these flies and would not depend on the gene's new position.
A 300 bp DNA element placed 8 kb downstream of a reporter gene raises its transcription 50-fold, and it works equally well when its orientation is inverted. The element is best described as
1. a core promoter, because it determines where RNA polymerase II starts transcription
2. an insulator, because it shields the reporter gene from the surrounding chromatin
3. an enhancer, because it binds activators and acts at a distance in either orientation
4. a silencer, because regulatory elements lying downstream of a gene usually repress it
Show answer
Answer: C — an enhancer, because it binds activators and acts at a distance in either orientation
Activation from kilobases away, independent of orientation and position relative to the gene, is the defining behaviour of an enhancer, which binds activators and contacts the promoter by DNA looping through Mediator. A core promoter must sit at the start site in a fixed orientation. An insulator blocks enhancer action or heterochromatin spread rather than raising transcription 50-fold, and a silencer represses; position downstream does not make an element a silencer.
Factor X binds the DNA sequence site-X but has lost its activation domain through a deletion. Factor Y has a strong activation domain and binds site-Y. A fusion of X's DNA-binding domain to Y's activation domain is expressed in cells carrying two reporter genes, one driven by site-X and the other by site-Y, with no intact Y present. What is expected?
1. Only the site-Y reporter is activated, because the activation domain decides which promoter is used
2. Only the site-X reporter is activated, because the DNA-binding domain chooses the target and the activation domain works wherever it is brought
3. Both reporters are activated, because the fusion carries working parts of both factors and so binds both kinds of site
4. Neither reporter is activated, because domains taken from different factors cannot fold and act together in one protein
Show answer
Answer: B — Only the site-X reporter is activated, because the DNA-binding domain chooses the target and the activation domain works wherever it is brought
Transcription factors are modular. The fusion binds DNA only through X's DNA-binding domain, so it reaches site-X alone, and Y's activation domain, once tethered there, recruits Mediator and coactivators to switch the gene on. It cannot bind site-Y, because it carries no Y DNA-binding domain. Domain-swap fusions such as Gal4–VP16 work well, which rules out the claim that foreign domains cannot cooperate.
Which of the following statements about transcription by RNA polymerase II are correct?
1. Pol II recognises the TATA box on its own and needs general transcription factors only during elongation
2. TBP, a subunit of TFIID, binds the TATA box and bends the DNA
3. TFIIH unwinds promoter DNA and phosphorylates serine 5 of the Pol II C-terminal domain
4. An enhancer works only when it lies within 100 bp upstream of the TATA box in its native orientation
Show answer
Answer: B — TBP, a subunit of TFIID, binds the TATA box and bends the DNA; C — TFIIH unwinds promoter DNA and phosphorylates serine 5 of the Pol II C-terminal domain
Promoter recognition starts with TBP of TFIID binding and sharply bending the TATA box, and TFIIH supplies both the helicase that opens the promoter and the kinase that phosphorylates CTD serine 5 for promoter escape. Pol II cannot recognise a promoter by itself; the general factors are needed for initiation. Enhancers act from kilobases away, upstream or downstream and in either orientation.
A gene has three exons whose lengths in the mature mRNA are 250, 180 and 300 nucleotides (exons 1, 2 and 3; cap and poly(A) tail not counted). The AUG start codon begins at nucleotide 101 of exon 1, and the stop codon ends 120 nucleotides before the end of exon 3. In one tissue exon 2 is skipped; this keeps the reading frame and creates no new stop codon. Counting the initiator methionine, the number of amino acids in the protein translated from the exon-2-skipped mRNA is ____.
Numerical answer — type the value.
Show answer
Answer: 109
Without exon 2 the mRNA is 250 + 300 = 550 nt. Removing the 100 nt 5′ UTR and the 120 nt 3′ UTR leaves 330 nt from the A of AUG through the stop codon, which is 110 codons; the stop codon adds no amino acid, so the protein has 109. The full-length mRNA (730 nt) gives 510/3 − 1 = 169, and skipping 180 nt removes exactly 60 amino acids. Forgetting to subtract the stop codon gives 110.
A gene contains one pair of mutually exclusive exons (every mature mRNA includes exactly one of the two) and, elsewhere, four cassette exons, each of which is included or skipped independently of all the others. All remaining exons are constitutive. The number of distinct mature mRNA isoforms the gene can produce is ____.
Numerical answer — type the value.
Show answer
Answer: 32
The mutually exclusive pair gives 2 choices, and each of the four cassette exons gives 2 (in or out), so the number of isoforms is 2 × 2⁴ = 2 × 16 = 32. Treating the mutually exclusive pair as two more independent cassettes would give 2⁶ = 64, which wrongly includes mRNAs with both or neither of the pair.
Liver and small intestine both express the single apolipoprotein B gene, yet the liver secretes apoB-100 while the intestine secretes apoB-48, whose sequence matches the N-terminal 48% of apoB-100. The intestinal protein arises because
1. an intestine-specific promoter inside the gene starts transcription halfway along the coding region
2. a protease in the intestinal ER cleaves apoB-100 in two and secretes only the N-terminal half
3. alternative splicing in the intestine skips the exons that encode the C-terminal half of the protein
4. a cytidine deaminase changes one C to U in the intestinal mRNA, turning a CAA codon into UAA
Show answer
Answer: D — a cytidine deaminase changes one C to U in the intestinal mRNA, turning a CAA codon into UAA
ApoB-48 is the classic case of mRNA editing. An intestinal cytidine deaminase deaminates a single C in the apoB mRNA, converting a glutamine codon (CAA) into a stop codon (UAA), so translation ends about halfway. An internal promoter would give a C-terminal fragment, not an N-terminal one, and the protein is not made by cleaving apoB-100. Exon skipping is a plausible-sounding mechanism, but the intestinal and hepatic mRNAs have the same exons and differ only at the edited base.
A cell line expresses a mutant iron-regulatory protein that binds iron-responsive elements tightly whatever the iron level. Grown in iron-rich medium, compared with normal cells in the same medium these cells will show
1. low ferritin synthesis and high transferrin receptor levels, the pattern normal cells show when iron is scarce
2. high ferritin synthesis and low transferrin receptor levels, the normal response of cells to abundant iron
3. high ferritin synthesis and high transferrin receptor levels, because IRP binding stimulates the use of both mRNAs
4. low ferritin synthesis and low transferrin receptor levels, because IRP binding blocks the translation of both mRNAs
Show answer
Answer: A — low ferritin synthesis and high transferrin receptor levels, the pattern normal cells show when iron is scarce
The effect of IRP binding depends on where the IRE lies. Bound to the 5′ UTR IRE of ferritin mRNA, IRP blocks scanning ribosomes, so ferritin synthesis falls; bound to the 3′ UTR IREs of transferrin receptor mRNA, it protects the mRNA from degradation, so receptor levels rise. A mutant that binds regardless of iron locks the cell into the iron-starved pattern even in iron-rich medium. The other options either describe the normal response to iron or wrongly give IRP the same effect on both mRNAs.
In the animal microRNA pathway, the enzyme that cuts the exported pre-miRNA hairpin in the cytoplasm to release a duplex of about 22 nucleotides is
1. Drosha, which processes the primary transcript inside the nucleus
2. Argonaute, which cleaves target mRNAs that pair fully with the guide
3. Dicer, an RNase III enzyme that measures and cuts the double-stranded stem
4. Exportin-5, which trims the hairpin while it carries it out through the nuclear pore
Show answer
Answer: C — Dicer, an RNase III enzyme that measures and cuts the double-stranded stem
Dicer cuts the pre-miRNA hairpin in the cytoplasm into the roughly 22 nt duplex, one strand of which is loaded onto Argonaute. Drosha acts earlier, in the nucleus, on the pri-miRNA; exportin-5 only transports the pre-miRNA and does not cut it; Argonaute carries the mature guide and may cleave targets but does not process the hairpin.
Plasma C-peptide is measured to judge how much insulin a diabetic patient's own pancreas still secretes, even while the patient injects insulin. This works because C-peptide
1. is the signal peptide removed in the ER, so its level tracks how fast the insulin gene is transcribed
2. is cut from proinsulin in β-cell granules and released in equal amounts with insulin, but is absent from injected insulin
3. is released when the liver degrades circulating insulin, so it reflects the total turnover of injected and natural insulin
4. is the phosphorylated, active form of insulin that only the pancreas can produce from the precursor
Show answer
Answer: B — is cut from proinsulin in β-cell granules and released in equal amounts with insulin, but is absent from injected insulin
Proinsulin folds with its A and B chains linked by the C-peptide; prohormone convertases in the secretory granules cut the C-peptide out, and insulin and C-peptide are secreted together, mole for mole. Therapeutic insulin contains no C-peptide, so plasma C-peptide reports only the patient's own secretion. The signal peptide is removed in the ER and destroyed, C-peptide is not a breakdown product of circulating insulin, and it is a separate peptide, not a modified insulin.
A DNA transposon 1,250 bp long, including its terminal inverted repeats, is inserted into a chromosome by a transposase that cuts the two strands of the target 9 bp apart. After the single-stranded gaps are filled and sealed, the chromosome is longer than before by ____ bp.
Numerical answer — type the value.
Show answer
Answer: 1259
The staggered cuts leave 9-nt single-stranded overhangs on each side of the inserted element. Filling them copies the 9 bp target sequence once more, so it now appears twice, once on each side, as a target-site duplication. The net gain is the element plus one extra copy of the target stretch: 1,250 + 9 = 1,259 bp. Adding 2 × 9 = 18 (1,268) double-counts, because one of the two flanking copies was already there; 1,250 ignores the duplication.
Which of the following statements about transposable elements in animals are correct?
1. Alu elements encode their own reverse transcriptase and endonuclease and so move without help
2. LTR retrotransposons move through an RNA intermediate that is reverse-transcribed into DNA before it integrates
3. In the germ line, piRNAs bound to Piwi proteins silence transposons
4. DNA transposons such as the P element move by cut and paste, using a transposase that acts at terminal inverted repeats
Show answer
Answer: B — LTR retrotransposons move through an RNA intermediate that is reverse-transcribed into DNA before it integrates; C — In the germ line, piRNAs bound to Piwi proteins silence transposons; D — DNA transposons such as the P element move by cut and paste, using a transposase that acts at terminal inverted repeats
LTR retrotransposons resemble retroviruses: their RNA is copied into cDNA by reverse transcriptase and integrated. piRNAs loaded onto Piwi-family Argonautes cleave transposon transcripts and direct repressive chromatin in the germ line. P elements and other Class II elements are excised and reinserted by transposase acting at their terminal inverted repeats. Alu is a non-autonomous SINE about 300 bp long that encodes no enzymes and is mobilised by the L1 machinery.
In Drosophila, crossing males of a P strain (whose genomes carry P elements) with females of an M strain (which lack them) gives progeny with sterile gonads and frequent germ-line mutations, while the reciprocal cross, M males with P females, gives normal progeny. The best explanation is that
1. P elements are carried only on the Y chromosome, so that only the sons of P-strain fathers ever inherit them
2. the P-element transposase is encoded in mitochondrial DNA, which the progeny inherit from the mother
3. M-strain eggs carry a transposase inhibitor, so P elements can move only when the mother is a P female
4. P-strain mothers load P-element piRNAs into the egg that silence transposition, and M-strain eggs lack them
Show answer
Answer: D — P-strain mothers load P-element piRNAs into the egg that silence transposition, and M-strain eggs lack them
When P elements enter through the sperm into an M egg, the egg has no maternally deposited piRNAs against them, so the elements transpose freely in the germ line, causing sterility, mutations and chromosome breaks. A P-strain egg carries those piRNAs and keeps the elements silent, so the reciprocal cross is normal. P elements are spread over the autosomes and the X, and daughters of the dysgenic cross are affected too; transposase is encoded by the element itself in the nucleus; and an inhibitor in M eggs would protect, not harm, the dysgenic cross.