topic 4 exam questions Flashcards
Describe how a gene is a code for the production of a polypeptide. Do not
include information about transcription or translation in your answer.
- (Because) base/nucleotide sequence;#
- (In) triplet(s);
- (Determines) order/sequence of amino acid sequence/primary
structure (in polypeptide)
The nucleus and a chloroplast of a plant cell both contain DNA.
Give three ways in which the DNA in a chloroplast is different from DNA in
the nucleus.
In chloroplasts
Must be comparative statements.
Accept alternatives in context of nuclear DNA
- DNA shorter;
Accept smaller
- Fewer genes;
- DNA circular not linear;
Accept DNA in a loop not linear
Accept no chromosomes (in chloroplast) unlike nucleus
- Not associated with protein/histones, unlike nuclear DNA;
- Introns absent but present in nuclear DNA;
Ignore references to double and single stranded DNA
Not all mutations in the nucleotide sequence of a gene cause a change in
the structure of a polypeptide.
- Triplets code for same amino acid
Accept: DNA/code/triplets are degenerate
Reject: codons (as question states within genes) - Occurs in introns /non-coding sequence;
Reject: codons (as question states within genes)
Ignore junk DNA
Reject: multiple repeats
Compare and contrast the DNA in eukaryotic cells with the DNA in
prokaryotic cells.
- Nucleotide structure is identical;
Accept labelled diagram or description of nucleotide
as phosphate, deoxyribose and base - Nucleotides joined by phosphodiester bond;
OR
Deoxyribose joined to phosphate (in sugar, phosphate backbone); - DNA in mitochondria / chloroplasts same / similar (structure) to DNA
in prokaryotes;
Accept shorter than nuclear DNA/is circular not
linear/is not associated with protein/histones unlike
nuclear DNA;
Contrasts
4. Eukaryotic DNA is longer;
5. Eukaryotic DNA contain introns, prokaryotic DNA does not;
6. Eukaryotic DNA is linear, prokaryotic DNA is circular;
7. Eukaryotic DNA is associated with / bound to protein / histones,
prokaryotic DNA is not;
Describe how mRNA is formed by transcription in eukaryotes.
- Hydrogen bonds (between DNA bases) break;
Ignore DNA helicase.
Reject hydrolysing hydrogen bonds.
- (Only) one DNA strand acts as a template;
- (Free) RNA nucleotides align by complementary base pairing;
For ‘align by complementary base pairing’, accept
‘align to complementary bases’ or ‘align by base
pairing’. - (In RNA) Uracil base pairs with adenine (on DNA)
OR
(In RNA) Uracil is used in place of thymine;
Do not credit use of letters alone for bases. - RNA polymerase joins (adjacent RNA) nucleotides;
Reject suggestions that RNA polymerase forms
hydrogen bonds or joins complementary bases. - (By) phosphodiester bonds (between adjacent nucleotides);
- Pre-mRNA is spliced (to form mRNA)
OR
Introns are removed (to form mRNA);
Give two structural differences between a molecule of messenger RNA
(mRNA) and a molecule of transfer RNA (tRNA)
- mRNA does not have hydrogen bonds / base pairing, tRNA does;
OR
mRNA is linear / straight chain, tRNA is cloverleaf; - mRNA does not have an amino acid binding site, tRNA does;
Accept mRNA cannot carry an amino acid, tRNA
can - mRNA has more nucleotides;
Accept mRNA is longer or converse - (Different) mRNAs have different lengths, all tRNAs are similar /
same length; - mRNA has codons, tRNA has an anticodon;
Statements must be comparative
Describe how mRNA is produced in a plant cell.
- The DNA strands separate by breaking the H bonds;
OR
H bonds broken between (complementary) (DNA) bases; - (Only) one of the strands/template strand is used (to make
mRNA/is transcribed); - (Complementary) base pairing so A ⟶ U, T ⟶ A, C ⟶ G, G
⟶ C; - (RNA) nucleotides joined by RNA polymerase;
- pre-mRNA formed;
- Splicing / introns removed to form mRNA;
- Ignore ‘hydrolysis’ of bonds
- Accept DNA “unzips” by breaking the H bonds
- Accept ‘non-coding’ sections for introns
Give the two types of molecule from which a ribosome is made
- One of RNA / ribonucleic acid(s) / nucleotide(s)/nucleic acid(s) /
rRNA / ribosomal RNA / ribosomal ribonucleic acid
and
one of protein(s) / polypeptide(s) / amino acid(s) / peptide(s) /
ribosomal protein
Describe the role of a ribosome in the production of a polypeptide. Do not
include transcription in your answer.
- mRNA binds to ribosome;
- Idea of two codons / binding sites;
- (Allows) tRNA with anticodons to bind / associate;
- (Catalyses) formation of peptide bond between
amino acids
(held by tRNA molecules); - Moves along (mRNA to the next codon) / translocation
described;
Define ‘gene mutation’ and explain how a gene mutation can have no negative effect
- Change in the base/nucleotide (sequence of chromosomes/DNA);
For 4 marks at least one mark must be scored in
each section of the answer.
Accept named mutation for ‘change’. - Results in the formation of new allele;
(Has no effect because) - Genetic code is degenerate (so amino acid sequence may not
change);
OR
Mutation is in an intron (so amino acid sequence may not change);
Accept description of ‘degenerate’, eg some amino
acids have more than one triplet/codon. - Does change amino acid but no effect on tertiary structure;
- (New allele) is recessive so does not influence phenotype;
(Has positive effect because) - Results in change in polypeptide that positively changes the
properties (of the protein)
OR
Results in change in polypeptide that positively changes a named
protein;
For ‘polypeptide’ accept ‘amino acid sequence’ or
‘protein’.
- May result in increased reproductive success
OR
May result in increased survival (chances);
Give two differences between mitosis and meiosis.
- One division, two divisions in meiosis;
- (Daughter) cells genetically identical,
daughter cells genetically different in
meiosis;
Reference to ‘genetically’ needed once - Two cells produced, (usually) four cells produced in meiosis;
- Diploid to diploid/haploid to haploid, diploid to haploid in
meiosis;
Accept same number chromosomes in
mitosis, but half the number in meiosis - Separation of homologous chromosomes only in meiosis;
- Crossing over only in meiosis;
- Independent segregation only in meiosis;
Crossing over greatly increases genetic diversity in this species of moss.
Describe the process of crossing over and explain how it increases genetic
diversity.
- Homologous pairs of chromosomes associate / form a bivalent;
- Chiasma(ta) form;
- (Equal) lengths of (non-sister) chromatids / alleles are
exchanged; - Producing new combinations of alleles;
- Accept descriptions of homologous pairs
- Accept descriptions of chiasma(ta) e.g.
chromatids / chromosomes entangle / twist - Neutral Crossing / cross over
- Reject genes are exchanged
- Accept lengths of DNA are exchanged
- Do not accept references to new combinations
of genes unless qualified by alleles
A student used a dilution series to investigate the number of cells present
in a liquid culture of bacteria.
Describe how he made a 1 in 10 dilution and then used this to make a 1 in
1000 dilution of the original liquid culture of bacteria.
- Add 1 part (bacteria) culture to 9 parts (sterile) liquid (to make 10–1
dilution);
Accept water / nutrient / broth for liquid
- Mix (well);
Accept stir - Repeat using 9 parts fresh (sterile) liquid and 1 part of 10–1 and 10–2
dilutions to make 10–3 dilution;
OR
Add 1 part 10–1 (suspension) to 99 parts (sterile) liquid (to make 10–3
dilution);
all taxons
domain, kingdom, phylum (plural, phyla), class, order, family, genus (plural, genera), and species.
Explain why it is more useful to calculate an index of diversity than to
record species richness.
- (Index of diversity also) measures abundance / number /
population (size) of each species;
Ignore “total number of species” unqualified
Accept: every species for each species.
AQA Biology A-Level - Biodiversity within a community MS PhysicsAndMathsTutor.com - (So useful because) may be many of some species
OR
(So useful because) may be few of other species;
