BIOL1007 MCQ Practice

[L2] Which list contains only properties of life? A) Reproduction, motility, mutation, diffusion B) Order, combustion, homeostasis, crystallisation C) Order, homeostasis, adaptation, evolution D) Growth, photosynthesis, mitosis, adaptation

C) Order, homeostasis, adaptation, evolution — The 8 properties: order, energy processing, response to stimuli, reproduction, growth & development, homeostasis, adaptation, evolution.

[L2] Homeostasis is best defined as: A) Adaptation of a population over generations B) A feature that improves survival in an environment C) The organisation of molecules into cells D) Maintaining stable internal conditions despite a changing environment

D) Maintaining stable internal conditions despite a changing environment — Homeostasis = stable internal conditions; evolution and adaptation are separate properties.

[L2] The three domains of life are: A) Plants, Animals, Fungi B) Bacteria, Archaea, Eukarya C) Bacteria, Protists, Eukarya D) Prokaryota, Eukarya, Viruses

B) Bacteria, Archaea, Eukarya — Domains are the highest rank; kingdoms divide them further.

[L2] Which four elements make up ~96% of body mass? A) H, O, Na, Cl B) C, N, P, S C) C, H, N, O D) C, H, O, P

C) C, H, N, O — C, H, N and O are the common elements; Na, Mg, K, Ca, P, S, Cl occur in smaller amounts.

[L2] Which statement is TRUE? A) O2 is a compound B) All compounds are molecules, but not all molecules are compounds C) All molecules are compounds, but not all compounds are molecules D) A compound contains only one type of atom

B) All compounds are molecules, but not all molecules are compounds — A compound needs 2+ different elements; O2 is a molecule but not a compound.

[L2] Which atom does NOT normally make a molecule polar? A) Nitrogen B) Phosphorus C) Carbon D) Oxygen

C) Carbon — C is neutral and hydrophobic; O, N, P (sometimes S) create polarity.

[L2] A molecule with both hydrophobic and hydrophilic regions is: A) Hydrophobic B) Non-polar C) Isotonic D) Amphipathic

D) Amphipathic — Phospholipids are the classic amphipathic molecule.

[L2] Why does ice float on water? A) Ice has stronger covalent bonds than liquid water B) Water expands on freezing, so ice is less dense than liquid water C) Ice contains dissolved air D) Ice has a higher specific heat than water

B) Water expands on freezing, so ice is less dense than liquid water — H-bonds lock water into an open lattice when frozen.

[L2] Water's high specific heat contributes mainly to: A) Capillary action B) Its ability to dissolve ions C) Moderation of temperature D) Ice floating

C) Moderation of temperature — High specific heat and heat of vaporisation buffer temperature changes.

[L2] Capillary action and high surface tension result from water's: A) Expansion on freezing B) Non-polar nature C) Low heat of vaporisation D) Cohesion

D) Cohesion — Cohesion = H-bonding between water molecules.

[L2] A glycosidic bond is formed by: A) Phosphorylation B) Dehydration synthesis C) Hydrolysis D) Oxidation

B) Dehydration synthesis — Dehydration removes water to join sugars; hydrolysis adds water to break them.

[L2] The storage polysaccharide of plants is: A) Chitin B) Glycogen C) Cellulose D) Starch (amylose + amylopectin)

D) Starch (amylose + amylopectin) — Glycogen is the animal store; cellulose and chitin are structural.

[L2] Compared with amylopectin, glycogen is: A) A structural polysaccharide B) Made of fructose C) More highly branched D) Unbranched

C) More highly branched — Glycogen is the animal storage form, even more branched than amylopectin.

[L2] Which is a STRUCTURAL polysaccharide found in arthropod exoskeletons? A) Starch B) Amylose C) Chitin D) Glycogen

C) Chitin — Other structural polysaccharides: cellulose (plants), peptidoglycan (bacteria).

[L2] Peptidoglycan is found in: A) Plant cell walls B) Animal cell membranes C) Arthropod exoskeletons D) Bacterial cell walls

D) Bacterial cell walls — Archaea lack peptidoglycan as a major wall component.

[L2] A fatty acid consists of: A) A carboxylic acid group and a hydrocarbon tail B) A phosphate group and a sugar C) An amino group and an R group D) Glycerol and three phosphates

A) A carboxylic acid group and a hydrocarbon tail — The –COOH head is attached to a non-polar hydrocarbon chain.

[L2] Adding more hydrogen (saturating) a fatty acid will: A) Increase its melting temperature B) Add C=C double bonds C) Decrease its melting temperature D) Make it more polar

A) Increase its melting temperature — Saturated chains are straight and pack tightly, so they melt at higher temperatures.

[L2] The main lipid used for energy storage is: A) Wax B) Triacylglycerol C) Cholesterol D) Phospholipid

B) Triacylglycerol — Steroids signal, waxes protect, phospholipids build membranes.

[L2] Which lipid class acts mainly as signalling molecules? A) Glycolipids B) Steroids C) Triacylglycerols D) Waxes

B) Steroids — e.g. hormones such as oestrogen and cortisol.

[L2] Which pair are both purines? A) Guanine and cytosine B) Cytosine and thymine C) Adenine and uracil D) Adenine and guanine

D) Adenine and guanine — Purines are double-ringed (A, G); pyrimidines are C, T, U.

[L2] Uracil replaces which base in RNA? A) Thymine B) Cytosine C) Adenine D) Guanine

A) Thymine — T is DNA-only; U is RNA-only.

[L2] The three parts of a nucleotide are: A) Amino group, carboxyl group, R group B) Phosphate, pentose sugar, nitrogenous base C) Phosphate, hexose sugar, amino acid D) Glycerol, fatty acid, phosphate

B) Phosphate, pentose sugar, nitrogenous base — The phosphate gives nucleic acids their negative charge.

[L2] Every amino acid contains: A) A hydrocarbon tail and glycerol B) A phosphate group and a nitrogenous base C) A carbonyl group and a glycosidic bond D) An amino group, a carboxylic acid group and an R group on an α-carbon

D) An amino group, a carboxylic acid group and an R group on an α-carbon — The R group gives each of the 20 amino acids its identity.

[L2] How many amino acids are commonly found in proteins and encoded by genes? A) 22 B) 20 C) 64 D) 4

B) 20 — 64 is the number of codons, not amino acids.

[L3] A phosphodiester bond links: A) A base to the 1' carbon of the sugar B) The 3' carbon of one sugar to the 5' carbon of the next via a phosphate C) Two bases across the double helix D) Two amino acids

B) The 3' carbon of one sugar to the 5' carbon of the next via a phosphate — Base–sugar is a glycosidic bond; base–base is hydrogen bonding.

[L3] Why is the DNA backbone hydrophilic? A) Its sugars are hydrophobic B) Its bases are non-polar C) It contains hydrocarbon chains D) Its phosphate groups carry negative charges

D) Its phosphate groups carry negative charges — Charged phosphates interact readily with water.

[L3] Nucleic acids are synthesised: A) Only in the 5' → 3' direction B) Only in the 3' → 5' direction C) From the C-terminus to the N-terminus D) In both directions equally

A) Only in the 5' → 3' direction — New nucleotides are always added to the free 3'-OH.

[L3] A G–C base pair is more stable than an A–T pair because it has: A) Three hydrogen bonds instead of two B) A covalent bond between bases C) Two purines D) A phosphodiester bond

A) Three hydrogen bonds instead of two — This is also why A/T-rich regions (origins, TATA) are easier to separate.

[L3] In double-stranded DNA, the ratio of purines to pyrimidines is: A) 1:2 B) 2:1 C) Variable between species D) 1:1

D) 1:1 — Every purine pairs with a pyrimidine (A–T, G–C).

[L3] RNA does not usually form a B-DNA-type double helix because: A) It contains thymine B) The extra 2'-OH on ribose prevents it C) It lacks a phosphate backbone D) It is synthesised 3' to 5'

B) The extra 2'-OH on ribose prevents it — The 2'-OH is the key difference between ribose and deoxyribose.

[L3] Which RNA acts as an adaptor that brings amino acids to the ribosome? A) siRNA B) mRNA C) rRNA D) tRNA

D) tRNA — mRNA is the template; rRNA is part of the ribosome.

[L3] The major and minor grooves of DNA are important because they: A) Allow proteins to access and interact with the bases B) Hold the two strands together covalently C) Store energy for replication D) Are where ribosomes attach

A) Allow proteins to access and interact with the bases — Transcription factors and histones interact through the grooves.

[L3] Peptide bond formation is: A) Catalysed by DNA polymerase B) Spontaneous and exergonic C) Energetically unfavourable and requires energy input D) A hydrolysis reaction

C) Energetically unfavourable and requires energy input — It is a dehydration synthesis that needs energy (driven by ribosome/ATP/GTP).

[L3] The peptide bond is flat and rigid because it: A) Is a hydrogen bond B) Contains a phosphate C) Has partial double-bond character D) Is a disulphide bridge

C) Has partial double-bond character — Resonance prevents rotation around the C–N bond.

[L3] The repeating unit of the polypeptide backbone is: A) C–O–P B) P–S–B C) N–C–C D) C–C–O

C) N–C–C — Amide N, α-carbon, carbonyl carbon.

[L3] Polypeptides are synthesised from: A) C-terminus to N-terminus B) Both ends simultaneously C) N-terminus to C-terminus D) 5' to 3'

C) N-terminus to C-terminus — Like nucleic acids, proteins have defined direction.

[L4] The central dogma of molecular biology is: A) DNA → RNA → protein B) RNA → DNA → protein C) Protein → RNA → DNA D) DNA → protein → RNA

A) DNA → RNA → protein — Transcription then translation.

[L4] Approximately what percentage of the human genome encodes proteins? A) ~50% B) ~15% C) ~98% D) ~1.5%

D) ~1.5% — Most of the genome is non-coding.

[L4] The transcriptome is: A) All the RNA in a given cell at a given time B) All the DNA in an organism C) All the proteins in a cell D) The set of all genes that encode proteins

A) All the RNA in a given cell at a given time — Genome = DNA, transcriptome = RNA, proteome = protein.

[L4] Which cell lacks a genome? A) Mature red blood cell B) Liver cell C) Skin fibroblast D) Neuron

A) Mature red blood cell — RBCs have no nucleus.

[L4] Two different cell types from the same person will usually have: A) The same genome but different proteomes B) Identical transcriptomes C) Different genomes and the same proteome D) Different genomes and different proteomes

A) The same genome but different proteomes — Each cell expresses a different subset of genes.

[L4] Which is a feature of eukaryotic genomes? A) Always smaller than 5 Mb B) No associated proteins C) Linear chromosomes wrapped around histones D) A single circular chromosome

C) Linear chromosomes wrapped around histones — Prokaryotes usually have one small circular chromosome.

[L4] Which statement about genome size is correct? A) Prokaryotes have larger genomes than eukaryotes B) More chromosomes always means a more complex organism C) Genome size does not correlate strongly with organismal complexity D) Humans have the largest known genome

C) Genome size does not correlate strongly with organismal complexity — Paris japonica has ~149 billion bp vs ~6 billion in humans.

[L4] Why must the genetic code use triplet codons? A) Ribosomes can only read three bases B) There are exactly 3 types of RNA C) Doublets only give 16 combinations – too few for 20 amino acids D) Singlets give 64 combinations

C) Doublets only give 16 combinations – too few for 20 amino acids — 4³ = 64 combinations covers 20 amino acids + stops.

[L4] The start codon is: A) AUG B) UGA C) AAA D) UAA

A) AUG — AUG codes for methionine.

[L4] Which is NOT a stop codon? A) UAG B) UAA C) UGA D) AUG

D) AUG — AUG is the start (Met) codon.

[L4] 'Non-overlapping' genetic code means: A) Each nucleotide is part of only one codon B) Each amino acid has only one codon C) Genes never overlap on a chromosome D) Codons can be read in any direction

A) Each nucleotide is part of only one codon — The reading frame divides the sequence into consecutive triplets.

[L4] Codon sequences are conventionally written: A) N → C B) 5' → 3' as found on mRNA C) As the tRNA anticodon D) 3' → 5' on the template DNA

B) 5' → 3' as found on mRNA

[L4] The fact that most amino acids have more than one codon is called: A) Degeneracy (redundancy) of the code B) Universality C) Non-overlap D) Frameshifting

A) Degeneracy (redundancy) of the code — 64 codons code for 20 amino acids.

[L5] DNA replication is described as: A) Semi-conservative, bidirectional and semi-discontinuous B) Dispersive, bidirectional and continuous C) Conservative, unidirectional and continuous D) Semi-conservative, unidirectional and discontinuous

A) Semi-conservative, bidirectional and semi-discontinuous — Confirmed by Meselson–Stahl.

[L5] In semi-conservative replication, each daughter helix contains: A) Mixed fragments of old and new DNA in each strand B) Two new strands C) Two parental strands D) One parental strand and one new strand

D) One parental strand and one new strand — Mixed fragments describes the dispersive model.

[L5] Which experiment confirmed semi-conservative replication? A) Hershey–Chase B) Pasteur's swan-neck flask C) Griffith's transformation D) Meselson–Stahl

D) Meselson–Stahl

[L5] Which is NOT required for DNA replication? A) A primer B) A promoter C) dNTPs D) A template strand

B) A promoter — Promoters are for transcription.

[L5] Which enzyme unwinds DNA by breaking hydrogen bonds between bases? A) Primase B) Ligase C) Topoisomerase D) Helicase

D) Helicase

[L5] Topoisomerase's role is to: A) Lay down RNA primers B) Seal Okazaki fragments C) Stabilise single strands D) Relieve supercoiling by cutting, unwinding and rejoining DNA

D) Relieve supercoiling by cutting, unwinding and rejoining DNA

[L5] Single-stranded binding proteins: A) Remove RNA primers B) Join DNA fragments C) Synthesise primers D) Stop separated strands re-annealing

D) Stop separated strands re-annealing

[L5] Primase is needed because DNA polymerase: A) Cannot read A/T-rich DNA B) Can only work 3' → 5' C) Requires an RNA template D) Can only add nucleotides to an existing 3'-OH

D) Can only add nucleotides to an existing 3'-OH

[L5] Which enzyme replaces RNA primers with DNA? A) DNA polymerase I B) Helicase C) Primase D) DNA polymerase III

A) DNA polymerase I — Pol III is the main synthesising enzyme.

[L5] Okazaki fragments are joined by: A) DNA ligase B) DNA polymerase III C) Primase D) Topoisomerase

A) DNA ligase

[L5] Okazaki fragments are found on the: A) Leading strand B) Template for transcription C) Lagging strand D) mRNA

C) Lagging strand — The lagging strand is made discontinuously away from the fork.

[L5] The leading strand is synthesised: A) Continuously toward the replication fork from one primer B) 3' → 5' C) Discontinuously away from the fork D) Without any primer

A) Continuously toward the replication fork from one primer

[L5] Why do eukaryotes have many origins of replication? A) They lack helicase B) Their genomes are large and linear, so one origin would be too slow C) Their DNA is circular D) Their DNA is single-stranded

B) Their genomes are large and linear, so one origin would be too slow — Humans have ~40,000–80,000 origins.

[L5] Origins of replication are typically rich in: A) Uracil B) Methylated cytosine C) A/T base pairs D) G/C base pairs

C) A/T base pairs — A–T pairs have only 2 H-bonds so separate more easily.

[L5] What provides the energy to form each phosphodiester bond during replication? A) The proton gradient B) Release and breakdown of pyrophosphate from the dNTP C) Hydrolysis of GTP by helicase D) ATP from ligase

B) Release and breakdown of pyrophosphate from the dNTP

[L5] Which is TRUE of transcription but NOT replication? A) A DNA template is used B) Synthesis is 5' → 3' C) Nucleotide triphosphates are used D) No primer is required

D) No primer is required — RNA polymerase starts de novo at the promoter.

[L5] The substrates for RNA polymerase are: A) Okazaki fragments B) NTPs (ribonucleotide triphosphates) C) Amino acids D) dNTPs

B) NTPs (ribonucleotide triphosphates)

[L5] In bacteria, the promoter is recognised by: A) Rho protein B) The sigma factor C) Ligase D) Helicase

B) The sigma factor — Sigma helps RNA polymerase bind, then leaves.

[L5] The bacterial consensus promoter regions are located at: A) −10 and −35 B) +1 and +10 C) +10 and +35 D) −100 and −200

A) −10 and −35 — Relative to the +1 transcription start site.

[L5] Why is the −10 (TATA) region important? A) It is where ribosomes bind B) It is A/T-rich and easy to melt, allowing local strand separation C) It forms the termination hairpin D) It encodes the start codon

B) It is A/T-rich and easy to melt, allowing local strand separation

[L5] Intrinsic (rho-independent) termination relies on: A) Rho binding a rut site B) A G/C-rich hairpin in the RNA followed by A/U-rich sequence C) A release factor D) A stop codon

B) A G/C-rich hairpin in the RNA followed by A/U-rich sequence — The hairpin pauses polymerase, which dissociates.

[L5] Rho-dependent termination involves Rho protein: A) Recognising the −35 region B) Adding a poly-A tail C) Binding a C-rich rut site and using helicase activity to release RNA polymerase D) Forming a G/C hairpin

C) Binding a C-rich rut site and using helicase activity to release RNA polymerase

[L5] A protein that blocks RNA polymerase binding or movement is a(n): A) Sigma factor B) Promoter C) Repressor D) Activator

C) Repressor

[L5] Which is NOT a way transcription is regulated in the lecture? A) Promoter strength B) Changing the genetic code C) Small-molecule modulators D) Activators

B) Changing the genetic code

[L5] Which is a reason cells use RNA as an intermediate? A) RNA stores the genome long-term B) RNA can be replicated without a template C) RNA is more stable than DNA D) Many disposable copies can be made, amplifying expression while DNA stays protected

D) Many disposable copies can be made, amplifying expression while DNA stays protected

[L5] During transcription, RNA polymerase copies: A) The mRNA strand B) Both strands simultaneously C) One DNA strand at a time (either strand can be the template for different genes) D) Only the strand with the promoter at its 3' end, for every gene

C) One DNA strand at a time (either strand can be the template for different genes)

[L6] Which enzyme attaches an amino acid to its tRNA? A) Aminoacyl-tRNA synthetase B) Peptidyl transferase C) RNA polymerase D) Ligase

A) Aminoacyl-tRNA synthetase — Uses ATP – the 'hidden cost' of translation.

[L6] Peptide bonds are catalysed by: A) mRNA B) Release factors C) rRNA in the large ribosomal subunit (peptidyl-transferase centre) D) tRNA synthetase

C) rRNA in the large ribosomal subunit (peptidyl-transferase centre)

[L6] The incoming charged tRNA first binds the ribosome's: A) E site B) P site C) Shine–Dalgarno site D) A site

D) A site — A → P → E.

[L6] Uncharged tRNA leaves the ribosome from the: A) P site B) E site C) Peptidyl-transferase centre D) A site

B) E site

[L6] The Shine–Dalgarno sequence is: A) A eukaryotic promoter B) The ribosome binding site on bacterial mRNA C) A stop codon D) The −10 promoter region

B) The ribosome binding site on bacterial mRNA

[L6] The first amino acid in bacterial proteins is: A) Formyl-methionine (fMet) B) Methionine C) Tryptophan D) Glycine

A) Formyl-methionine (fMet) — Eukaryotes use plain Met.

[L6] Translation terminates when: A) The ribosome reaches the poly-A tail B) A G/C hairpin forms C) Rho protein binds D) A stop codon enters the A site and release factors bind

D) A stop codon enters the A site and release factors bind

[L6] Which is a difference between prokaryotic and eukaryotic translation? A) Prokaryotes start with Met, eukaryotes with fMet B) Eukaryotes use one release factor for all three stop codons C) Only eukaryotes have elongation D) Prokaryotic ribosomes lack rRNA

B) Eukaryotes use one release factor for all three stop codons — Prokaryotes have 2–3 release factors.

[L6] Transcription and translation can be coupled in: A) Neither B) Eukaryotes, because they have a nucleus C) Prokaryotes, because they have no nucleus D) Only mitochondria-free eukaryotes

C) Prokaryotes, because they have no nucleus

[L6] α-helices are stabilised by hydrogen bonds between: A) Disulphide bridges B) R groups of adjacent amino acids C) C=O and N–H groups four residues apart D) Bases on opposite strands

C) C=O and N–H groups four residues apart

[L6] Which level of protein structure includes β-sheets? A) Tertiary B) Secondary C) Primary D) Quaternary

B) Secondary

[L6] Disulphide bridges and hydrophobic interactions between R groups mainly stabilise: A) The peptide bond B) Secondary structure only C) Tertiary structure D) Primary structure

C) Tertiary structure

[L6] Haemoglobin, made of four subunits, has: A) No tertiary structure B) Only primary structure C) Only secondary structure D) Quaternary structure

D) Quaternary structure

[L6] Denaturation is sometimes reversible because: A) Primary structure is unaffected B) Denaturation only affects DNA C) Quaternary structure is never affected D) Peptide bonds reform instantly

A) Primary structure is unaffected

[L6] Sickle cell disease results from a change in haemoglobin's: A) Quaternary structure only B) DNA backbone C) Primary structure (one amino acid substitution) D) Glycosidic bonds

C) Primary structure (one amino acid substitution) — Glu → Val in the β subunit.

[L6] In sickle cell disease, haemoglobin molecules aggregate into fibres because of: A) Disulphide bonds B) Phosphodiester bonds C) Ionic bonds with oxygen D) Hydrophobic interactions

D) Hydrophobic interactions

[L7] The first law of thermodynamics states that energy: A) Is lost forever as heat B) Cannot be created or destroyed, only transformed C) Increases in the universe over time D) Always becomes more ordered

B) Cannot be created or destroyed, only transformed

[L7] The second law of thermodynamics states that: A) Reactions with negative ΔG are slow B) Enzymes lower activation energy C) Entropy of the universe always increases D) Energy is conserved

C) Entropy of the universe always increases

[L7] Chemical energy in glucose is an example of: A) Thermal energy B) Potential energy C) Entropy D) Kinetic energy

B) Potential energy

[L7] At equilibrium: A) Forward and reverse rates are equal, so there is no net change B) Only products remain C) All reactions stop D) ΔG is strongly negative

A) Forward and reverse rates are equal, so there is no net change — Equilibrium is dynamic.

[L7] A reaction with ΔG < 0 is: A) At equilibrium B) Always fast C) Exergonic and spontaneous D) Endergonic and non-spontaneous

C) Exergonic and spontaneous

[L7] A reaction with ΔG > 0: A) Requires an input of energy B) Releases free energy C) Occurs spontaneously D) Is catalysed without enzymes

A) Requires an input of energy

[L7] ΔG tells you: A) The amount of enzyme required B) Whether a reaction can occur spontaneously, not how fast C) The activation energy D) How fast a reaction occurs

B) Whether a reaction can occur spontaneously, not how fast — Kinetics determine rate.

[L7] Sucrose breakdown is favourable but extremely slow. This shows that: A) Sucrose is at equilibrium B) Enzymes change ΔG C) ΔG for sucrose hydrolysis is positive D) Thermodynamically favourable reactions can be kinetically slow

D) Thermodynamically favourable reactions can be kinetically slow

[L7] Enzymes speed up reactions by: A) Making ΔG more negative B) Shifting the equilibrium toward products C) Supplying ATP D) Lowering the activation energy

D) Lowering the activation energy