4/5/6 Mark Qs Genetic Information, Varyation & Relationships Between Organisms

Question 1

Compare and contrast the DNA in eukaryotic cells with the DNA in prokaryotic cells. (5)

Answer 1

Comparisons
1. Nucleotide structure is identical;
Accept labelled diagram or description of nucleotide as phosphate, deoxyribose and base
2. Nucleotides joined by phosphodiester bond;
OR
Deoxyribose joined to phosphate (in sugar, phosphate backbone);
3. 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;

Question 2

Compare and contrast the DNA in eukaryotic cells with the DNA in prokaryotic cells. (5)

Answer 2

Comparisons
1. Nucleotide structure is identical;
Accept labelled diagram or description of nucleotide as phosphate, deoxyribose and base
2. Nucleotides joined by phosphodiester bond;
OR
Deoxyribose joined to phosphate (in sugar, phosphate backbone);
3. 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;

Question 3

Describe how mRNA is formed by transcription in eukaryotes. (5)

Answer 3

1. Hydrogen bonds (between DNA bases) break;
   Ignore DNA helicase.
   Reject hydrolysing hydrogen bonds.
2. (Only) one DNA strand acts as a template;
3. (Free) RNA nucleotides align by complementary base pairing;
   For ‘align by complementary base pairing’, accept ‘align to complementary bases’ or ‘align by base pairing’.
4. (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.
5. RNA polymerase joins (adjacent RNA) nucleotides;
   Reject suggestions that RNA polymerase forms hydrogen bonds or joins complementary bases.
6. (By) phosphodiester bonds (between adjacent nucleotides);
7. Pre-mRNA is spliced (to form mRNA)
   OR
   Introns are removed (to form mRNA);

Question 4

Describe how a polypeptide is formed by translation of mRNA. (5)

Answer 4

. (mRNA attaches) to ribosomes
OR
(mRNA attaches) to rough endoplasmic reticulum;
2. (tRNA) anticodons (bind to) complementary (mRNA) codons;
3. tRNA brings a specific amino acid;
4. Amino acids join by peptide bonds;
5. (Amino acids join together) with the use of ATP;
6. tRNA released (after amino acid joined to polypeptide);
7. The ribosome moves along the mRNA to form the polypeptide;

Question 5

Starting with mRNA in the cytoplasm, describe how translation leads to the
production of a polypeptide.
Do not include descriptions of transcription and splicing in your answer. (5)

Answer 5

mRNA associates with a ribosome / ribosome attaches to mRNA;
Idea of association is required
2. Ribosome moves to / finds the start codon / AUG;
3. tRNA brings / carries (appropriate / specific) amino acid;
Must be explicitly stated and not inferred.
4. Anticodon (on tRNA complementary) to codon (on mRNA);
5. Ribosome moves along to next codon;
OR
Ribosome ‘fits’ around two codons / can fit two tRNAs;
Must be explicitly stated and not inferred.
6. (Process repeated and) amino acids join by peptide bonds / condensation reaction (to form polypeptide);
OR
(Process repeated and) amino acids joined using (energy from) ATP (to form polypeptide);

Question 6

Describe how mRNA is produced in a plant cell. (5)

Answer 6

. The DNA strands separate by breaking the H bonds;
OR
H bonds broken between (complementary) (DNA) bases;
2. (Only) one of the strands/template strand is used (to make mRNA/is transcribed);
3. (Complementary) base pairing so A ⟶ U, T ⟶ A, C ⟶ G, G ⟶ C;
4. (RNA) nucleotides joined by RNA polymerase;
5. pre-mRNA formed;
6. Splicing / introns removed to form mRNA;
1. Ignore ‘hydrolysis’ of bonds
1. Accept DNA “unzips” by breaking the H bonds
6. Accept ‘non-coding’ sections for introns

Question 7

Crossing over greatly increases genetic diversity in this species of moss.
Describe the process of crossing over and explain how it increases genetic diversity. (4)

Answer 7

1. Homologous pairs of chromosomes associate / form a bivalent;
2. Chiasma(ta) form;
3. (Equal) lengths of (non-sister) chromatids / alleles are exchanged;
4. Producing new combinations of alleles;
5. Accept descriptions of homologous pairs
6. Accept descriptions of chiasma(ta) e.g. chromatids / chromosomes entangle / twist
7. Neutral Crossing / cross over
8. Reject genes are exchanged
9. Accept lengths of DNA are exchanged
10. Do not accept references to new combinations of genes unless qualified by alleles