▶️ Answer/Explanation
(a)
(i) DNA can be collected from blood, hair roots, cheek cells (saliva), or semen.
(ii) The DNA is amplified using PCR (polymerase chain reaction), which uses Taq polymerase to replicate DNA at high temperatures.
(b)
The father is Male 2. This is because all the bands in the child’s DNA either match the mother’s or Male 2’s profile. Any band in the child that is not from the mother must come from the biological father, and only Male 2 shares all those matching bands.
Markscheme
(a)
(i) from blood sample/hair/cheek/saliva/semen;
(ii) PCR/polymerase chain reaction
OR
using Taq polymerase;
(b)
a. male 2;
b. each band in the child’s DNA must be the same as a band in either the mother
or the father
OR
any band in the child’s profile that is not present in the mother’s profile must be
present in the father’s;
▶️ Answer/Explanation
a.
- DNA is a double helix made of two strands.
- The strands are antiparallel – one runs 5′ to 3′, the other 3′ to 5′.
- Each strand has a sugar-phosphate backbone.
- DNA is made up of nucleotides, each consisting of a phosphate group, deoxyribose sugar, and a nitrogenous base.
- The four bases are adenine (A), thymine (T), cytosine (C), and guanine (G).
- A pairs with T and C pairs with G using hydrogen bonds.
The base pairing holds the two strands together.
b.
- Helicase – unzips the DNA strands by breaking hydrogen bonds between bases.
- DNA polymerase III – adds complementary nucleotides to the new strand in the 5′ to 3′ direction.
- Ligase – joins Okazaki fragments together on the lagging strand to create a continuous DNA strand.
c.
- All cells in the body contain the insulin gene, but it is only expressed in pancreatic β cells.
- During development, cells differentiate, meaning certain genes are turned on or off depending on cell type.
- In β cells, the insulin gene is activated, while in other cells it remains inactive.
- Transcription factors bind to regulatory regions (like enhancers/silencers) to control gene expression.
- These regulatory proteins are specific to certain genes and only found in certain cell types.
- In β cells, glucose levels trigger transcription of insulin mRNA, leading to insulin production.
- Epigenetic modifications like DNA methylation and histone acetylation also influence gene activation or silencing.
Markscheme
a
a. two stranded/double helix ✔
b. antiparallel / strands running in opposite directions
OR
one strand organized 5′ to 3′ and the other 3′ to 5′ ✔
c. sugar-phosphate backbone ✔
d. each strand formed by chains of nucleotides ✔
e. each nucleotide is formed by a phosphate, a deoxyribose and a base / annotated diagram of a nucleotide clearly indicated as a nucleotide ✔
f. the bases are adenine, guanine, cytosine and thymine ✔
g. strands held together by hydrogen bonds (between complementary base pairs)
OR
A pairs with T and C pairs with G ✔
Both helix and two strands needed for point a. Double helix is sufficient for the mark.
Points can be awarded to annotated diagrams.
For point c, the explicit label sugar phosphate backbone is required.
b
a. helicase to separate/unwind DNA strands ✔
b. gyrase/toposiomerase to relax the tension as bacterial DNA is being uncoiled / prevent supercoiling ✔
c. primase to synthesise primers ✔
d. polymerase (I) removes primers and replaces with nucleotide ✔
e. polymerase (III) adds nucleotides (in a 5′ to 3′ direction) ✔
f. ligase joins (Okazaki) fragments together ✔
Accept the enzyme name without ‘DNA’ included; e.g. ‘DNA ligase’ or ‘ligase’ can both be accepted.
c
a. insulin production is determined by a gene ✔
b. gene for insulin (is found in all cells), but only activated in (β cells of) pancreas ✔
c. stem cells differentiate into specialized cells/(into pancreatic β) ✔
d. during differentiation some genes are turned on and others off ✔
e. insulin is a hormone that regulates the amount of glucose/sugar in blood ✔
f. pancreatic β cells have sensors that detect glucose level in blood ✔
g. an increase in glucose will increase transcription of mRNA of insulin ✔
h. the site of transcription of insulin is in the pancreatic β cells ✔
i. gene transcription is regulated by proteins that bind to specific base sequence in DNA/enhancers/silencers/promoter proximal elements ✔
j. regulatory sequences/proteins are specific to the gene they regulate / insulin regulator proteins are only found in the pancreatic β cells ✔
k. (DNA) methylation (usually) inhibits gene expression / (histone) acetylation promotes gene expression / tightness of coiling of DNA around histones affects gene expression ✔
Accept sugar as equivalent to glucose.
▶️ Answer/Explanation
a.
- Organic compounds like sucrose are transported through the phloem.
- Transport occurs from sources (e.g. leaves where sugars are made) to sinks (e.g. roots, fruits, growing tissues where sugars are used or stored).
- Sieve tube elements, connected end to end with sieve plates, form the conducting tissues.
- Companion cells assist with loading sugars into phloem using active transport.
- Proton pumps move H⁺ ions out of the companion cells, creating a gradient.
- Sucrose enters the phloem through co-transport with H⁺ ions.
- The high solute concentration in the phloem causes water to enter by osmosis.
The resulting high pressure drives the flow of sap from source to sink – this is called mass flow or translocation.
b.
- Meiosis – produces haploid male (pollen) and female (ovule) gametes.
- Pollination – transfer of pollen from the anther to the stigma of a flower.
- Fertilisation – occurs when the male gamete fuses with the female gamete to form a zygote.
Seed dispersal – seeds are spread to new locations to reduce competition and allow growth in suitable environments.
c.
- Helicase unwinds the double helix and breaks hydrogen bonds between the bases.
- Topoisomerase (gyrase) relieves tension caused by the unwinding.
- Each separated strand serves as a template for the formation of a new strand – this is called semi-conservative replication.
- Primase adds a short RNA primer to allow DNA polymerase to start.
- DNA polymerase III adds complementary DNA nucleotides in a 5′ to 3′ direction.
- A pairs with T and C pairs with G – this is complementary base pairing.
- On the leading strand, replication is continuous; on the lagging strand, it’s discontinuous, forming Okazaki fragments.
- DNA polymerase I removes the RNA primers and replaces them with DNA.
DNA ligase seals the gaps between Okazaki fragments to form a continuous strand.
Markscheme
a
a. phloem transports organic compounds/sucrose
b. from sources/leaves/where produced to sinks/roots/where used
c. through sieve tubes/columns of cells with sieve plates/perforated end walls
d. loading of organic compounds/sucrose into /H+ ions out of phloem/sieve tubes by active transport/using ATP
e. high solute concentration causes water to enter by osmosis (at source)
f. high (hydrostatic) pressure causes flow (from source to sink)
g. companion cells help with loading / plasmodesmata provide a path between sieve tubes and companion cell
h. translocation/mass flow
b
a. meiosis / production of male and female gametes
b. pollination / transfer of pollen from anther to stigma
c. fertilization happens after pollination / fertilisation is joining of gametes
d. seed dispersal / spread of seeds to new locations
Reject fruit dispersal.
c
a. helicase unwinds the double helix
b. gyrase/topoisomerase relieves strains during uncoiling
c. helicase separates the two strands of DNA/breaks hydrogen bonds
Accept unzips here but not for mark point a.
d. each single strand acts as a template for a new strand / process is semi-conservative
e. DNA polymerase III can only add nucleotides to the end of an existing chain/to a primer
f. (DNA) primase adds RNA primer/short length of RNA nucleotides
g. DNA polymerase (III) adds nucleotides in a 5′ to 3′ direction
h. complementary base pairing / adenine to thymine and cytosine to guanine
Do not accept letters.
i. DNA polymerase (III) moves towards the replication fork on one strand and away from it on the other strand
j. continuous on the leading strand and discontinuous/fragments formed on the lagging strand
k. DNA polymerase I replaces primers/RNA with DNA
l. ligase joins the fragments together/seals the nicks
▶️ Answer/Explanation
a.
- Bacteriophages (viruses that infect bacteria) were used in the experiment.
- Radioactive sulfur (³⁵S) was used to label protein (since sulfur is found in proteins but not DNA).
- Radioactive phosphorus (³²P) was used to label DNA (since phosphorus is found in DNA but not in proteins).
- The labeled viruses were allowed to infect bacterial cells.
- After infection, a blender was used to separate the viral protein coats from the bacterial cells.
- Centrifugation was done to separate the heavier bacterial cells from the lighter viral coats.
- In the ³²P (DNA) experiment, radioactivity was found inside the bacteria, indicating DNA had entered.
- In the ³⁵S (protein) experiment, radioactivity remained outside, showing proteins did not enter.
This showed that DNA, not protein, carries the genetic information into the cell.
b.
Regulation of gene expression – some non-coding sequences control when and how genes are turned on or off.
Markscheme
a
a. radioactive isotopes used to label viruses/bacteriophages/phages
b. proteins labelled with radioactive sulphur/35S and DNA labelled with radioactive phosphorous/32P
c. phage infects bacterium
d. only viral DNA enters bacterium «viral coat/capsid/shell do not»
e. parts of phage remaining outside bacterial cell are removed
OR
bacteria are separated from phage parts «by centrifuge»
f. bacteria contain the labelled/radioactive DNA
b
a. regulate gene expression
b. act as promoter
c. role in chromosome pairing/crossing over/recombination
d. introns
▶️ Answer/Explanation
a.
b.
- DNA samples are collected from the crime scene, suspects, and victims.
- Polymerase Chain Reaction (PCR) is used to amplify the small DNA sample.
- Restriction enzymes cut the DNA at specific sequences.
- DNA fragments are loaded into a gel for gel electrophoresis.
- An electric current is applied to separate fragments by size.
- This produces a DNA profile—a unique pattern of bands for each individual.
- The suspect’s DNA is compared with the DNA from the crime scene.
- A match can help identify the criminal or victim.
DNA is stable, so even old or degraded samples can be used.
c.
- A gene is a segment of DNA that codes for a polypeptide (a sequence of amino acids).
- The gene is first transcribed into mRNA.
- The mRNA is then translated at the ribosome into a polypeptide.
- This concept is known as “one gene, one polypeptide.”
- However, this is an oversimplification:
- Some genes don’t code for polypeptides, but for functional RNA like tRNA or rRNA.
Alternative splicing means one gene can produce multiple polypeptides.
Regulatory sequences control expression without coding for proteins.
Some genes produce non-coding RNA with regulatory roles.
- Mutations in a gene can change the amino acid sequence of the polypeptide, affecting its function.
- Overall, the relationship is complex but fundamental to gene expression and protein synthesis.
Markscheme
a
The diagram must show four nucleotides shown with two on each side showing phosphate-sugar backbones and nitrogen base pairs bonded between them.
Award [1] for each of the following clearly drawn and correctly labelled:
• phosphate – shown connected to deoxyribose
• deoxyribose – shown connected to phosphate
• (nitrogenous) bases – shown bonded to deoxyribose
• base pairs – shown with labels adenine/A bonded to thymine/T and cytosine/C bonded to guanine/G
• hydrogen bonds – shown connecting bases
• covalent bonds – shown connecting deoxyribose to phosphates
• nucleotide – clearly identified
Award [4 max] if diagram is not shown double stranded.
b
• DNA samples are taken from crime scene, suspects and victims
• polymerase chain reaction/PCR used to increase the amount of DNA
• restriction enzymes used to cut DNA
• electrophoresis involves electric field/placing sample between electrodes
• used to separate DNA fragments according to size
• creating DNA profiles/unique patterns of bands
• comparison is made between the patterns
• criminals/victims can be identified in this way
• DNA is (quite) stable / DNA can be processed long after the crime
c
• DNA codes for a specific sequence of amino acids/polypeptide
• the DNA code for one polypeptide is a gene
• DNA is transcribed into mRNA
• mRNA moves to a ribosome
• where mRNA is translated into a polypeptide
• originally it was thought that one gene always codes for one polypeptide
• some genes do not code for a polypeptide
• some genes code for transfer RNA/tRNA/ribosomal RNA/rRNA
• some sections of DNA code for regulators that are not polypeptides
• antibody production does not follow this pattern (of simple transcription-translation)
• change in the gene/mutation will affect the primary structure of the polypeptide
▶️ Answer/Explanation
Phase A:
This is anaphase, and it happens during the middle (intermediate) stage of mitosis.
Phase B:
This is prophase, and it takes place during the early stage of mitosis.
a (i)
- The part of the chromosome in the middle (called the centromere) splits.
- The two identical halves (called sister chromatids) get pulled apart.
- These chromatids move towards opposite sides of the cell.
- This pulling is done by structures called spindle fibers, which get shorter to drag the chromatids away.
b.
If cell division isn’t properly controlled, it can lead to cancer or tumors, which are groups of cells growing in an abnormal way.
c.
- Complementary base pairing makes sure the DNA is copied correctly.
- The bases always pair up the same way: Adenine with Thymine, and Cytosine with Guanine.
- This ensures that the new DNA strands are exact copies of the original, so the new cells get the right genetic instructions.
Markscheme:
a (i)
phase A: anaphase (occurs at an) intermediate (stage); (both needed)
phase B: prophase (occurs at an) early (stage); (both needed)
a (ii)
• centromeres split/break
• (sister) chromatids/chromosomes separate
• dragged/pulled/movement to separate poles
• by shortening of spindle microtubules
Do not allow events other than those in anaphase
b
tumours / cancer
c
• conservation of the base sequence of DNA
• adenine pairs with thymine and cytosine pairs with guanine
(do not accept initials only)
• both (daughter) cells/DNA strands produced have identical genetic information
▶️ Answer/Explanation
(a)
(b)
- Y’s father is heterozygous, which means he has one allele for the disorder.
- Y’s mother is unaffected, so she doesn’t carry the allele.
- Since the father can pass on either the normal or the mutated allele, and the mother can only pass on a normal one, Y has a:
- 50% or 1 in 2 chances of inheriting the allele for the disorder.
(c)
Looking at the family in box B:
- The father is heterozygous, and the daughter is unaffected.
- If the gene were sex-linked and on the X chromosome, any daughter of a male with the allele would definitely inherit that X, and so she’d be heterozygous.
- But the daughter is completely unaffected, which shows she didn’t inherit the allele.
- So, this suggests that the gene is not on the sex chromosomes, meaning it’s not sex-linked.
(d)
- Proteins are made by reading the DNA base sequence, which tells the cell what amino acids to link together.
- A base substitution means one base is changed for example, an A becomes a G.
- This could change the triplet code, which changes the amino acid that gets added.
- Even one amino acid change can lead to a different shape or function in the final protein.
So, a single base change in the DNA might lead to a different protein being made, which could cause a condition like thrombophilia.
Markscheme
a
b
50%/0.5/1/2
c
a. if it was sex-linked it would be on the X chromosome
b. there cannot be a heterozygous male if the trait is sex-linked
c. males would pass the allele to their daughter
d. daughter is not shown as heterozygous so it is not sex-linked
d
a. sequence of DNA bases determines the amino acid sequence of a protein
b. changing one base (on the DNA) can cause the triplet/mRNA to code for a different amino acid
c. changing one base (on the DNA) causes a different protein to be made (during translation)