Revision Library

revisionlibrary.co.uk

A-Level · Biology

AB8 The Control of Gene Expression

AQA 7402/7405/7408 · Calculator allowed · about 115 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Gene mutations occur when the base sequence of DNA is changed.

(a)A single base substitution changes one DNA triplet but, because of the degenerate nature of the genetic code, the amino acid coded for does not change. Which term correctly describes this type of mutation?(1)

A) Nonsense mutation
B) Silent mutation
C) Frameshift mutation
D) Missense mutation

(b)Explain why a base deletion is more likely than a single base substitution to have a significant effect on the polypeptide produced.(3)

(c)State one reason why a mutation occurring within an intron is unlikely to affect the amino acid sequence of the final protein.(1)

(Total for Question 1 is 5 marks)

Oestrogen is a lipid-soluble hormone that can alter the transcription of target genes inside a cell.

(a)Describe how oestrogen brings about the transcription of a target gene, from entering the cell to RNA polymerase acting on the gene.(4)

(b)A mutation changes the tertiary structure of the DNA-binding region of this transcription factor. Explain the likely effect on transcription of the target gene.(3)

(c)State the term used to describe the region of DNA to which a transcription factor binds to initiate transcription.(1)

(Total for Question 2 is 8 marks)

Epigenetics refers to heritable changes in gene expression that occur without a change in the base sequence of DNA. Two epigenetic mechanisms are DNA methylation and histone modification.

(a)Define the term "epigenetics".(1)

(b)Describe how increased methylation of DNA can silence a gene.(3)

(c)Acetylation of histone proteins generally increases the rate of transcription of associated genes. Explain how this occurs.(3)

(d)During development, cells with an identical genome become specialised into different cell types (differentiation). Explain how epigenetic mechanisms allow this to happen.(2)

(Total for Question 3 is 9 marks)

Explain, with reference to proto-oncogenes, tumour suppressor genes and epigenetic control of gene expression, how abnormal DNA methylation patterns can contribute to the development of a tumour.

(Total for Question 4 is 6 marks)

This question is based on a core practical technique used in genetic profiling. Restriction enzymes were used to cut DNA samples into fragments, which were then separated using gel electrophoresis. Because the relationship between fragment size and migration distance is not linear, a calibration graph is produced by plotting log10(fragment size in base pairs, bp) against migration distance (mm) for DNA fragments of known size. Two calibration points obtained were: a 100 bp fragment migrated 52 mm, and a 1000 bp fragment migrated 18 mm.

(a)Describe how DNA fragments are separated by gel electrophoresis, referring to charge and size.(2)

(b)Show that the gradient of the calibration line, plotting log10(size/bp) against migration distance (mm), is -0.0294 (to 3 significant figures), using the two calibration points given.(2)

(c)Calculate the size, in bp, of an unknown DNA fragment that migrated 40 mm on the same gel. Give your answer to 3 significant figures.(3)

(d)Explain why, when comparing DNA fingerprints from two different samples, both samples must be cut using the same restriction enzyme before electrophoresis.(2)

(Total for Question 5 is 9 marks)

This question is based on the core practical technique of the polymerase chain reaction (PCR), which is used to amplify small quantities of DNA in vitro. A forensic scientist, Priya Nandwani, set up a PCR reaction starting with 2.50 x 10^-3 ng of target DNA. Assume 100% amplification efficiency, so the amount of DNA doubles at every cycle.

(a)State two components, other than the target DNA, that must be present in a PCR reaction mixture.(2)

(b)Show that the mass of DNA present after 20 complete cycles is approximately 2620 ng.(3)

(c)Calculate the minimum number of complete cycles needed to produce at least 500 ng of DNA from the same starting mass of 2.50 x 10^-3 ng.(3)

(d)Suggest one reason why the actual mass of DNA obtained after 20 cycles in a real PCR reaction is likely to be less than the value calculated in part (b).(1)

(Total for Question 6 is 9 marks)

A biotechnology company wants to produce recombinant bacteria that express a human gene. The gene is inserted into a bacterial plasmid that carries a gene for ampicillin resistance and a separate gene for tetracycline resistance.

(a)Name the type of enzyme used to cut both the human gene and the plasmid, and describe the type of cut it produces.(2)

(b)Explain why the same restriction enzyme must be used to cut both the human gene fragment and the plasmid vector.(2)

(c)Name the enzyme used to permanently join the human gene into the cut plasmid, and state the type of bond it forms.(2)

(d)The human gene is inserted into the middle of the tetracycline resistance gene, disrupting it. Explain how replica plating on ampicillin-containing and tetracycline-containing agar plates could be used to identify bacterial colonies that have taken up a recombinant plasmid containing the human gene.(4)

(Total for Question 7 is 10 marks)

Small interfering RNA (siRNA) is a short, double-stranded RNA molecule that can be used to control the expression of specific genes.

(a)State what is meant by "gene silencing".(1)

(b)Describe how siRNA brings about the silencing of a target gene after transcription.(4)

(c)Suggest how siRNA technology could, in principle, be used to treat a cancer caused by overexpression of an oncogene.(2)

(Total for Question 8 is 7 marks)

Grace Adeyemi has a family history of breast cancer and undergoes genetic screening using a labelled DNA probe designed to be complementary to a known mutant allele of the BRCA1 gene, a tumour suppressor gene.

(a)Describe how the labelled DNA probe could be used to determine whether Grace's DNA sample contains the mutant BRCA1 allele.(4)

(b)Explain why the DNA probe must be single-stranded.(1)

(c)Grace is found to carry one copy of the mutant BRCA1 allele. Using the "two-hit hypothesis" of tumour suppressor gene function, explain why this greatly increases, but does not guarantee, her risk of developing breast cancer.(3)

(Total for Question 9 is 8 marks)

The table shows the percentage of tumour samples found to contain a mutation in the p53 tumour suppressor gene, for four types of cancer.
Cancer type: percentage of tumours with p53 mutation
Colorectal: 50%
Lung: 65%
Breast: 30%
Ovarian: 80%

(a)Identify the cancer type with the lowest proportion of tumours showing a p53 gene mutation, and suggest one other way, apart from mutation of the p53 gene itself, by which loss of p53 function could occur in these tumours.(2)

(b)Explain why a mutation that inactivates p53 is more likely to contribute to cancer if it occurs in a body cell that goes on to divide many times, compared with an identical mutation in a cell that rarely divides.(2)

(c)The p53 protein acts as a transcription factor. Explain how a mutation affecting only the DNA-binding domain of p53 (leaving its activation domain unaffected) could still result in complete loss of its tumour-suppressing function.(2)

(Total for Question 10 is 6 marks)

The human genome contains approximately 3.2 x 10⁹ base pairs but only about 20 000 protein-coding genes, whereas the human proteome is estimated to contain over 100 000 different proteins.

(a)Protein-coding sequences make up approximately 1.5% of the human genome. Assuming this coding DNA is divided evenly between the 20 000 protein-coding genes, calculate the average number of base pairs of coding DNA per gene.(3)

(b)Using the information given, explain how the number of different proteins in the human proteome (over 100 000) can be greater than the number of protein-coding genes (20 000).(3)

(Total for Question 11 is 6 marks)

Somatic (non-germ-line) gene therapy has been trialled as a treatment for cystic fibrosis, an inherited genetic disorder affecting the lungs. Discuss why gene therapy of this type may not provide a permanent cure for the disorder.

(Total for Question 12 is 6 marks)

Mark scheme · AB8 The Control of Gene Expression

Question 1

(a) B1 B cao
(a) Answer: B) Silent mutation
(b) B1 a deletion causes a frameshift: the reading frame (grouping into triplets/codons) is shifted from the point of the mutation onwards oe
(b) B1 every subsequent codon is changed, which may alter every amino acid after the mutation and/or create a premature stop codon oe
(b) B1 a substitution only ever affects one triplet/codon, so at most only one amino acid changes, and due to the degenerate code it may not change the amino acid (and therefore the protein) at all oe
(b) Answer: A deletion shifts the reading frame so that every codon after the mutation is altered, whereas a substitution changes at most one codon/amino acid and, because the code is degenerate, may have no effect at all.
(c) B1 introns do not code for amino acids and are removed (spliced out) from the pre-mRNA before translation, so a mutation there does not appear in the mature mRNA/protein oe
(c) Answer: Introns are removed by splicing before translation, so a mutation within an intron is not present in the mature mRNA and does not affect the protein.

Question 2

(a) B1 oestrogen is lipid-soluble/small enough to diffuse directly through the phospholipid bilayer of the cell membrane oe
(a) B1 oestrogen binds to a (specific) transcription factor/receptor protein in the cytoplasm, forming an oestrogen-transcription factor complex oe
(a) B1 the (activated) complex moves into the nucleus (through a nuclear pore) oe
(a) B1 the complex binds to a specific base sequence in the DNA (a promoter/binding site associated with the target gene), stimulating (or inhibiting) the transcription of that gene by RNA polymerase oe
(a) Answer: Oestrogen diffuses across the cell membrane, binds to a transcription factor to form a complex, which moves into the nucleus and binds to a specific DNA sequence to stimulate transcription by RNA polymerase.
(b) M1 the mutation changes the primary structure (amino acid sequence), altering the bonds (e.g. hydrogen/ionic/disulfide) that determine tertiary structure, changing the shape of the DNA-binding site oe
(b) M1 the transcription factor complex can no longer bind (or binds less effectively) to its specific/complementary DNA sequence oe
(b) A1 transcription of the target gene is not stimulated (or is reduced), so the gene is not expressed / the protein is not produced (or produced at a lower rate) ft
(b) Answer: The transcription factor can no longer bind to its target DNA sequence, so transcription of the gene is not stimulated and the gene product is not made (or is made at a reduced rate).
(c) B1 promoter (region) oe
(c) Answer: Promoter

Question 3

(a) B1 heritable changes in gene function/expression that do not involve a change to the base/nucleotide sequence of DNA oe
(a) Answer: Epigenetics is the study of heritable changes in gene expression that occur without any change to the base sequence of DNA.
(b) B1 methyl groups are added to cytosine bases in the DNA, often in the promoter region of a gene oe
(b) B1 increased methylation physically prevents transcription factors and/or RNA polymerase from binding to the DNA, and/or attracts proteins that cause the chromatin to condense (heterochromatin formation) oe
(b) B1 transcription of the gene is inhibited/prevented, so no mRNA (and therefore no protein) is produced from that gene oe
(b) Answer: Methyl groups added to cytosine bases (often in the promoter) block transcription factor/RNA polymerase binding and can condense chromatin structure, preventing transcription of the gene.
(c) M1 acetyl groups added to histone proteins reduce the overall positive charge on the histones oe
(c) M1 this reduces the strength of the (electrostatic/ionic) association between the (negatively charged) DNA and the histones oe
(c) A1 chromatin becomes less condensed (more open/euchromatin), making the DNA more accessible to transcription factors/RNA polymerase, increasing the rate of transcription oe
(c) Answer: Acetylation reduces the positive charge on histones, loosening the DNA-histone association so chromatin becomes less condensed; this makes the gene more accessible to transcription factors/RNA polymerase, increasing transcription.
(d) B1 different patterns of DNA methylation and/or histone modification are established in different cell lineages during development, permanently switching off genes not needed in that cell type oe
(d) B1 only the genes needed for a particular cell's specialised function remain able to be transcribed/expressed, so cells with the same genome develop different structures and functions oe
(d) Answer: Different epigenetic marks (methylation/histone modification patterns) are established in different cell lineages, permanently silencing genes not needed for that cell's role, so only a subset of genes remains expressed, producing specialised cells from an identical genome.

Question 4

Level 3 (5-6): A detailed and coherent explanation that correctly links hypomethylation of a proto-oncogene (or its promoter) to increased expression and uncontrolled cell division, AND hypermethylation of a tumour suppressor gene promoter to loss of its inhibitory/apoptotic function, with a clear overall link to tumour formation. Scientific terminology is used accurately throughout.
Level 2 (3-4): The answer explains how altered methylation affects transcription/gene expression and refers to either proto-oncogenes or tumour suppressor genes (or both, with limited detail), with some link to uncontrolled cell division, but the explanation of one or both mechanisms is incomplete.
Level 1 (1-2): Basic or isolated statements, such as "methylation switches genes off" or "mutations cause cancer", with little or no explanation of the mechanism linking methylation to gene expression or to cancer.
Level 0 (0): No relevant content.
Indicative content:
- Proto-oncogenes normally code for proteins that stimulate cell division (e.g. growth factors or their receptors), and are usually expressed at controlled/low levels.
- Hypomethylation (reduced methylation) of a proto-oncogene's promoter increases the rate of transcription, so more growth-stimulating protein is produced.
- This leads to an increased rate of cell division, contributing to a tumour if uncontrolled (the proto-oncogene is effectively behaving like an oncogene).
- Tumour suppressor genes normally code for proteins that inhibit cell division, repair damaged DNA, or trigger apoptosis (e.g. p53).
- Hypermethylation (increased methylation) of a tumour suppressor gene's promoter silences the gene, reducing or preventing its transcription.
- Less (or no) functional tumour suppressor protein is produced, so cell division is not inhibited and/or damaged cells are not directed to undergo apoptosis.
- The combined effect of proto-oncogene activation and tumour suppressor gene silencing removes the normal controls on the cell cycle, allowing uncontrolled mitosis/cell division and the growth of a tumour.
- Because methylation patterns do not change the DNA base sequence, they can be copied to daughter cells during DNA replication/mitosis, so the abnormal gene expression pattern (and increased division rate) can persist and spread through a cell population.

Question 5

(a) B1 DNA fragments are negatively charged (due to phosphate groups) and are loaded into wells in an agarose gel; a (direct) electric current is applied across the gel oe
(a) B1 fragments move towards the positive electrode (anode); smaller fragments move further/faster through the pores of the gel than larger fragments in a given time, separating the fragments by size oe
(a) Answer: DNA fragments, being negatively charged, migrate towards the anode when a current is applied; smaller fragments move further through the gel than larger fragments in the same time, separating them by size.
(b) M1 correct substitution: gradient = (log10(100) - log10(1000)) / (52 - 18) = (2 - 3) / 34 shown
(b) A1 gradient = -1/34 = -0.0294 (3 sf) cso
(b) Answer: -0.0294 log10(bp) per mm
(c) M1 correct intercept found: c = 3 - (-0.0294 x 18) = 3.53 (3 sf) oe (or equivalent method using either calibration point)
(c) M1 log10(size) = (-0.0294 x 40) + 3.53 = 2.35 (3 sf) shown
(c) A1 size = 10^2.35 = 225 bp (awrt 220-230 bp) cao
(c) Answer: Approximately 225 bp
(d) B1 the same restriction enzyme cuts DNA at the same specific (recognition) base sequence in both samples oe
(d) B1 this produces comparable/corresponding fragments in each sample, so differences in the resulting banding patterns reflect real differences in the DNA sequence between the samples, allowing valid comparison oe
(d) Answer: Using the same restriction enzyme ensures both samples are cut at the same recognition sequences, producing comparable fragments so that any differences in the banding pattern are due to genuine sequence differences between the samples.

Question 6

(a) B1 any two from: (heat-tolerant) DNA polymerase (e.g. Taq polymerase); (two) primers; free (deoxyribonucleotide) nucleotides; a thermocycler to control temperature (max 2 marks) oe
(a) B1 second correct component from the list above
(a) Answer: Any two of: heat-tolerant (Taq) DNA polymerase; primers; free nucleotides; a thermocycler.
(b) M1 DNA doubles each cycle, so mass after 20 cycles = starting mass x 2²⁰ oe
(b) M1 2²⁰ = 1 048 576 and 2.50 x 10^-3 x 1 048 576 shown
(b) A1 = 2621.44 ng, which rounds to approximately 2620 ng (3 sf) cso
(b) Answer: Approximately 2620 ng (2621.44 ng before rounding)
(c) M1 correct inequality set up: 2.50 x 10^-3 x 2ⁿ ≥ 500 oe
(c) M1 rearranged and solved: 2ⁿ ≥ 200 000, so n ≥ log2(200 000) = 17.6 shown
(c) A1 minimum of 18 (complete) cycles cao (must be rounded up to a whole number)
(c) Answer: 18 cycles
(d) B1 any one from: amplification efficiency falls below 100% in later cycles as primers/free nucleotides become depleted; DNA polymerase activity declines over repeated heating cycles; single strands increasingly reanneal with each other rather than with primers, reducing effective amplification oe
(d) Answer: In practice, reagents (primers/nucleotides) become depleted and enzyme activity declines over many cycles, so amplification efficiency falls below the assumed 100%, giving a lower yield than the theoretical maximum.

Question 7

(a) B1 restriction endonuclease (enzyme) oe
(a) B1 produces a staggered cut, leaving short single-stranded overhangs ("sticky ends") with exposed, unpaired bases oe
(a) Answer: A restriction endonuclease, which produces a staggered cut leaving single-stranded "sticky end" overhangs.
(b) M1 the same restriction enzyme always cuts at the same recognition sequence, producing complementary/identical sticky ends (base sequences) on both the gene and the plasmid oe
(b) A1 the exposed bases on the gene fragment can then form complementary base pairs (hydrogen bonds) with the exposed bases on the cut plasmid, allowing the gene to be inserted oe
(b) Answer: The same restriction enzyme produces complementary sticky ends on both the gene fragment and the plasmid, allowing complementary base pairing between them so the gene can be inserted into the plasmid.
(c) B1 DNA ligase
(c) B1 (covalent) phosphodiester bond(s) in the sugar-phosphate backbone oe
(c) Answer: DNA ligase, which forms covalent phosphodiester bonds in the sugar-phosphate backbone.
(d) M1 bacteria are first grown/spread on agar containing ampicillin; only bacteria that have taken up a plasmid (recombinant or non-recombinant) are resistant to ampicillin and can grow oe
(d) M1 colonies are replica plated (transferred in the same relative position) onto agar containing tetracycline oe
(d) M1 bacteria containing a recombinant plasmid have a disrupted tetracycline resistance gene, so they cannot grow/are killed on the tetracycline plate oe
(d) A1 the two plates are compared: colonies present on the ampicillin plate but absent from the corresponding position on the tetracycline plate must contain the recombinant plasmid with the human gene oe
(d) Answer: Colonies are first selected on ampicillin (any transformed bacteria grow); replica plating onto tetracycline shows that only non-recombinant transformed bacteria (with an intact tetracycline gene) grow, so colonies present on ampicillin but absent on tetracycline, in the same position, contain the recombinant plasmid with the inserted gene.

Question 8

(a) B1 the prevention/blocking of the expression of a gene, so that no functional protein is produced even though the gene is present in the genome oe
(a) Answer: Gene silencing is the prevention of expression of a gene, so no (functional) protein is produced even though the gene is present.
(b) B1 double-stranded RNA is cut into short fragments (siRNA) by an enzyme oe
(b) B1 the two siRNA strands separate; one strand associates with a protein complex (RISC) oe
(b) B1 the siRNA-protein complex binds to a complementary base sequence on the target mRNA (by complementary base pairing) oe
(b) B1 this leads to the target mRNA being cut/degraded, so it cannot be translated and no (functional) protein is produced from that gene oe
(b) Answer: Double-stranded RNA is cleaved into siRNA fragments; one strand binds a protein complex which then binds, by complementary base pairing, to matching mRNA; this causes the mRNA to be degraded so it cannot be translated, silencing the gene.
(c) M1 an siRNA could be designed with a base sequence complementary to the mRNA transcribed from the oncogene oe
(c) A1 this would lead to degradation of the oncogene mRNA, preventing/reducing translation and therefore production of the (growth-stimulating) protein, reducing the rate of (uncontrolled) cell division oe
(c) Answer: A complementary siRNA could be designed to target the oncogene's mRNA, causing its degradation so the growth-stimulating protein is not translated, reducing the rate of cell division.

Question 9

(a) B1 Grace's (double-stranded) DNA sample is denatured (e.g. by heating or using alkali) to separate it into single strands oe
(a) B1 the single-stranded, labelled probe, with a base sequence complementary to the mutant allele, is added to the sample oe
(a) B1 if the mutant allele is present, the probe binds/hybridises to it by complementary base pairing (forming hydrogen bonds) oe
(a) B1 unbound probe is washed away; the presence of the label (e.g. fluorescence, or exposure of photographic film for a radioactive label) shows that the mutant allele is present oe
(a) Answer: The DNA sample is denatured to single strands; a labelled, single-stranded probe complementary to the mutant allele is added and hybridises to it if present by complementary base pairing; excess probe is washed off and detection of the label confirms the mutant allele is present.
(b) B1 so that its bases are free/exposed to form complementary base pairs (hydrogen bonds) with the complementary single strand of the sample DNA oe
(b) Answer: The probe must be single-stranded so its bases are available to hybridise (form complementary base pairs) with a matching single strand in the sample.
(c) M1 BRCA1 is a tumour suppressor gene; normally both alleles must be inactivated ("two hits") before its function (e.g. DNA repair/inhibiting cell division) is lost oe
(c) M1 Grace has already inherited one non-functional/mutant allele in every cell (the "first hit"), so she only needs a second, somatic mutation to inactivate her one remaining normal allele oe
(c) A1 the risk of cancer is greatly increased because only one further (random) mutation is needed, but it is not certain because this second "hit" is a chance event that may never occur in a particular cell oe
(c) Answer: Because both copies of a tumour suppressor gene must normally be inactivated for its function to be lost, and Grace already carries one non-functional allele in every cell, she only needs a single further somatic mutation in the remaining normal allele; this makes cancer much more likely but not certain, since that second mutation is a random event that may or may not occur.

Question 10

(a) B1 breast (cancer), 30% cao
(a) B1 epigenetic silencing of p53, e.g. hypermethylation of the p53 gene's promoter, reducing/preventing its transcription (accept: increased degradation of the p53 protein, or mutation in a gene that regulates p53 activity) oe
(a) Answer: Breast cancer (30%); loss of function could occur through epigenetic silencing, e.g. hypermethylation of the p53 promoter reducing its transcription.
(b) M1 without functional p53, cells with damaged DNA are not repaired and/or are not directed to undergo apoptosis, so further mutations are not eliminated oe
(b) A1 a cell that divides many times has many more opportunities to accumulate the additional mutations (e.g. in oncogenes) needed for cancer to develop, whereas a cell that rarely divides has few such opportunities oe
(b) Answer: Loss of p53 means damaged cells are not repaired or removed by apoptosis, so mutations accumulate; a frequently dividing cell has many more opportunities to build up the further mutations needed for cancer than a cell that rarely divides.
(c) M1 if the tertiary structure of the DNA-binding domain is altered, p53 can no longer bind to the specific target DNA sequence(s)/promoters of the genes it normally regulates oe
(c) A1 transcription of these target genes (e.g. those causing cell cycle arrest or apoptosis) is not activated, regardless of a functional activation domain, so the tumour-suppressing effect is lost oe
(c) Answer: A changed DNA-binding domain prevents p53 from binding to its target gene promoters at all, so those genes (e.g. controlling cell cycle arrest or apoptosis) cannot be activated, even though the activation domain itself still works, resulting in complete loss of function.

Question 11

(a) M1 total coding DNA = 1.5% x 3.2 x 10⁹ = 4.8 x 10⁷ (bp) shown
(a) M1 4.8 x 10⁷ / 20 000 shown
(a) A1 = 2400 bp (per gene) cao
(a) Answer: 2400 bp per gene
(b) B1 alternative splicing of pre-mRNA means a single gene's exons can be joined in different combinations, producing more than one mRNA (and therefore more than one polypeptide) from the same gene oe
(b) B1 post-translational modification (e.g. cleavage of a polypeptide, or addition of chemical groups) of a single polypeptide can produce more than one distinct functional protein oe
(b) B1 as a result, the total number of distinct proteins in the proteome can be much greater than the number of protein-coding genes in the genome oe
(b) Answer: Alternative splicing allows one gene to produce several different mRNAs/polypeptides, and post-translational modification can convert a single polypeptide into more than one functional protein, so the proteome contains far more distinct proteins than there are protein-coding genes.

Question 12

Level 3 (5-6): A detailed, coherent discussion that identifies and explains at least two well-developed reasons why somatic gene therapy is unlikely to be permanent (e.g. finite cell lifespan requiring repeat treatment, and either delivery/immune response difficulties or the restriction to somatic rather than germ-line cells), with clear scientific reasoning throughout.
Level 2 (3-4): The answer identifies at least one valid reason (e.g. treated cells are eventually replaced, or delivery of the vector is difficult) with some explanation, but the discussion lacks detail, or only one reason is meaningfully developed.
Level 1 (1-2): Basic or isolated statements, such as "the treatment might not work" or "cells die", with little scientific explanation of why the effect is not permanent.
Level 0 (0): No relevant content.
Indicative content:
- Somatic gene therapy only alters the cells that are directly treated (e.g. lung epithelial cells); it does not affect the patient's egg or sperm cells (germ-line cells), so the effect is not passed on to any offspring and the patient's own reproductive cells still carry the faulty allele.
- Cells lining the lungs are naturally replaced on a regular basis (a limited/finite lifespan); as treated cells die and are replaced by new, untreated cells that still carry the faulty allele, the beneficial effect of the therapy is gradually lost.
- As a result, treatment would need to be repeated at intervals throughout the patient's life to maintain any benefit, rather than providing a single, permanent cure.
- It can be difficult to deliver the vector (e.g. a modified virus or a liposome) to a sufficiently large number of cells throughout the affected tissue (e.g. deep into the lungs) for the treatment to have a significant, lasting clinical effect.
- The patient's immune system may recognise and destroy the vector (particularly a viral vector) after repeated exposure, reducing the effectiveness of, or triggering an immune response to, subsequent treatments.
- There is a risk that the inserted gene integrates into the host genome at a random position, potentially disrupting another functional gene (e.g. a tumour suppressor gene or a proto-oncogene), which is a further safety concern with repeated treatment.
- Germ-line gene therapy, which could in principle produce a permanent, heritable cure, is not currently permitted in humans for ethical and safety reasons, so only the less permanent somatic approach can legally be used.

The Control of Gene Expression - Worksheets, Questions and Revision