Continuity and change
D.1.1 DNA replication
You should know:
✔ DNA replication produces exact copies of DNA with identic al base sequences.
✔ DNA replic ation is semi-conservative and depends on complementary base pairing.
✔ PCR c an be used to amplify small amounts of DNA.
✔ gel electrophoresis separates proteins or fragments of DNA according to size.
Additional higher level:
✔ DNA polymerases add the 5’ of a DNA nucleotide to the 3’ end of a strand of nucleotides.
✔ the proofreading role of DNA polymerase III.
• Replication is the semiconservative synthesis of DNA.
• Helicase breaks hydrogen bonds between strands.
You will study transcription and gene expression in more detail in subtopic D1.2.
You should be able to:
✔ explain the roles of helicase and DNA polymerase in DNA replication.
✔ explain how Taq DNA polymerase produces multiple copies of DNA by PCR.
✔ analyse DNA profiles.
Additional higher level:
✔ compare DNA replication on the leading and the lagging strand.
✔ explain DNA replication (only in prokaryotes), describing the functions of DNA primase, DNA polymerases I and III, and DNA ligase.
Enzymes in DNA replication
The central principle of genetics is that DNA forms RNA, which then makes proteins. Therefore, DNA carries all the information for inheritance. DNA replication is required for reproduction and for growth and tissue replacement in multicellular organisms. Replication of DNA is needed for the duplication of the genetic material in advance of cell division by mitosis or meiosis.
In replication, DNA polymerase uses the DNA strand as a template to link nucleotides together to form a new strand with identical base sequences as the template. Replication is semi-conservative because one strand is an original strand and the other is a new strand. DNA replication is carried out by a complex system of enzymes that allow a high degree of accuracy in copying base sequences. Helicase unwinds the DNA double helix, separating the strands by breaking hydrogen bonds between the two strands of DNA. Next, DNA polymerase starts replication by adding nucleotides, thus synthesizing the complementary strand. If in the original strand there is an adenine, the base that is joined to the new strand is a thymine.
Which statement is related to DNA replication?
A. It always requires Taq DNA polymerase to occur.
B. DNA polymerase is an enzyme that separates the DNA strands.
C. New nucleotide bases attach to the original sugar–phosphate backbone.
D. The new DNA contains one original strand and one new strand.
The correct answer is D, as replication is semiconservative. Taq DNA polymerase is a bacterial enzyme that is heat resistant, which makes it ideal for use in polymerase chain reaction (PCR). Helicase is the enzyme that separates the DNA strands. The original strand acts as a template and nucleotides attach by complementary base pairing by hydrogen bonds—they do not attach to the sugar–phosphate backbone.
In 1958, Meselson and Stahl designed an experiment to test the semiconservative replication of DNA. They grew bacteria in a medium with heavy nitrogen 15N. After a few generations they transferred the bacteria to a 14N medium, where they obtained bacteria from several generations. They separated their DNA according to size using density gradient centrifugation.
original parent molecule
Samplecontent
To identify the DNA sequence, you must find the complementary base: for A it is T, and for C it is G. Remember that in DNA you have T instead of U.
second generation daughter molecules
first generation daughter molecules
◂ Figure 1 Semi-conservative replication of DNA
Source: Meselson, M. & Stahl, F. W. (1958). The replication of DNA in Escherichia coli. PNAS, 44, 671–682.
PCR and gel electrophoresis
Polymerase chain reaction (PCR) is a technique by which small amounts of DNA can be amplified into large quantities. The DNA fragments obtained by this method can be separated by size in gel electrophoresis. Because DNA has a negative charge, the fragments move towards the positive electrode. The smaller fragments migrate further than the larger ones. A marker with known sizes of DNA fragments is used to compare sizes.
DNA
primers
primers
1. Heat DNA to 93°C. This breaks the hydrogen bonds that hold the two strands of the DNA double helix together.
2. Cool to 55°C. Primers join to the ends of the DNA strands.
the cycle can be repeated
• Polymerase chain reaction (PCR) uses Taq polymerase to amplify DNA.
3. Heat to 72°C. DNA polymerase joins new nucleotides on to the DNA strands. This gives two copies of the original DNA sequence. new strand made from new DNA nucleotides
◂ Figure 2
Polymerase chain reaction (PCR) cycle
Solution
Example 1
▼ The answer is 8. Tube 1 has no primers, so even if it had contained DNA, no bands would have been seen.
▼ A short description of at least two temperatures is expected. Heat (94°C) separates the DNA strands by breaking hydrogen bonds. Although the terminology “denature” is used for this process, the answer has not specified at which temperature this occurs. Cooling (55°C) allows complementary primers to bind to the template by annealing. Heating again (72°C) is optimum for DNA Taq polymerase to add nucleotides, extending the DNA.
▲ This answer is correct, because there is less amplification of DNA so the bands will be lighter or non-existent.
Sample student answer
PCR was performed to amplify a small amount of DNA. Eight tubes were prepared as shown in the table.
Mix of nucleotides, salts, buffer and polymerase
Control
The tubes were placed in a thermal cycler with the temperatures shown in the diagram and run for 25 cycles. one PCR cycle
The image shows the result of gel electrophoresis on the eight samples.
a) State the number of the tube used as a control without DNA. [1]
This answer could have achieved 0/1 marks:
Tube 1
b) Explain the reason for changing the temperature during each cycle. [2]
This answer could have achieved 0/2 marks:
To denature DNA.
c) Predict the result that would be obtained if fewer cycles were used in this PCR process. [1]
This answer could have achieved 1/1 marks:
Bands of DNA much lighter.
Which of the three DNA profiles indicate(s) that the alleged father could be the biological father of the child?
Samplecontent
DNA profiling is performed in paternity tests. To increase its reliability, the number of markers used are increased as this reduces the probability of a false match.
Source: www.pbs.org
A. I and II only
B. II and III only
C. I and III only
D. III only
Solution
The correct answer is C, as there is coincidence between some of the bands. In II, there are no common bands between the child and the alleged father.
Directionality of DNA polymerases
In replication, DNA primase adds an RNA primer, which is a short length of RNA. Next, DNA polymerase III starts replication by adding nucleotides at the primer. It synthesizes the complementary strand in a 5’ to 3’ direction. DNA polymerase I removes the primer by replacing RNA with DNA. DNA ligase seals the nicks linking sections of replicated DNA, or Okazaki fragments, in the lagging strand. Once replication is completed, DNA polymerase III proofreads for mistakes. It removes any nucleotide from the 3’ terminal with a mismatched base, followed by replacement with a correctly matched nucleotide.
• DNA primase adds RNA primer at the 5’ end.
• DNA polymerase III adds complementary nucleotides at the primer in the 5’–3’ direction and proofreads the new DNA.
• DNA polymerase I removes the primer.
• DNA ligase seals the nicks.
Example 2
Low Res AHL
directionofDNApolymerase
•How is genetic continuity ensured between generations?
•What biological mechanisms rely on directionality?
directionofDNApolymerase
Transcription and stability of DNA templates
Transcription is the synthesis of RNA using a DNA template, whereas translation is the synthesis of proteins from mRNA. DNA contains the blueprint for the synthesis of mRNA, which will determine the primary structure of the protein. The two DNA strands are separated due to the breaking of the hydrogen bonds between complementary nucleotides. Complementary base pairing between DNA and mRNA is the same as in the DNA strands except for adenine (A) on the DNA pairs with uracil (U) on the mRNA strand. RNA polymerase adds the free 5’ end of the RNA nucleotide to the 3’ end of the growing mRNA molecule. The temporary hydrogen bonds between DNA and RNA are broken, liberating the mRNA. The mRNA either remains in the nucleus or leaves via a nuclear pore.
• Transcription is the synthesis of mRNA copied from the DNA base sequences by RNA polymerase.
What is the reason for Okazaki fragments being formed during DNA replication?
A. To enable replication of the 3’ → 5’ (lagging) strand
B. To form the template for the RNA primers
C. To initiate replication on the 5’ → 3’ (leading) strand
D. To help the DNA helicase unwinding the DNA helix
The correct answer is A, as DNA polymerase only adds nucleotides in the 5’ to 3’ direction. The template for the primers already exists, as it is the DNA. Okazaki fragments are only needed in the lagging strand, as DNA polymerase can replicate the leading strand in the 5’ to 3’ direction. Helicase does not need help to unwind the DNA. Example 3
D1.2 Protein synthesis
You should know:
✔ mRNA is copied from the DNA base sequences by RNA polymerase in transcription.
✔ the DNA sequence must be conserved throughout the lifetime of a cell.
✔ transcription is required for the expression of genes.
✔ mRNA is translated into the amino acid sequence of polypeptides according to the genetic code.
✔ translation depends on complementary base pairing between codons on mRNA and anticodons on tRNA.
✔ the genetic code is universal and degenerate.
Additional higher level:
✔ transcription and translation occur in a 5’ to 3’ direction.
✔ transcription factors bind to the promoter to initiate transcription.
✔ regulators of gene expression, introns, telomeres, and genes for rRNA and tRNA are non-coding sequences in DNA that do not code for polypeptides.
✔ many proteins are modified into their functional state.
✔ amino acids are recycled by proteasomes.
Samplecontent
DNA is a stable molecule that can resist different environmental conditions. The sequence is maintained unchanged. In a few somatic cells that do not divide after differentiation, for example in heart muscle cells, the sequences must be conserved throughout the lifetime of the cell for the cell to perform its function.
Transcription is required for the expression of genes
Transcription is a process required for the expression of genes. These gene products are usually proteins, but can also be tRNA or small RNAs that can act as gene expression regulators. Most genes are turned off, and are therefore not being transcribed at any one time. This means that gene expression is regulated, and some genes are expressed only at certain times in certain cells of different tissues. Although the genome of an organism is the same in all its cells, only some genes are expressed in different tissues.
Example 4
You should be able to:
✔ describe the role of hydrogen bonding and complementary base pairing in transcription.
✔ explain the roles of mRNA, ribosomes and tRNA in translation.
✔ deduce the sequence of amino acids coded by an mRNA strand.
✔ describe stepwise movement of the ribosome along mRNA in elongation of the polypeptide.
✔ describe a point mutation that affects protein structure.
Additional higher level:
✔ explain post-transcriptional changes to RNA and how they affect gene expression.
✔ explain alternative splicing.
✔ describe initiation of translation at the ribosome.
Why is insulin produced only in the β cells of the pancreas?
A. Only the DNA of β cells of the pancreas contain the gene for insulin.
B. Insulin gene expression is repressed in all other cells.
C. The insulin gene is expressed in all cells, but its mRNA only survives in the β cells of the pancreas.
D. The β cells of the pancreas receive all the mRNA for insulin from the rest of the cells.
Solution
The correct answer is B, as the gene for insulin is only turned on in the β cells of the pancreas. The DNA of all cells contain the gene for insulin, but its expression is turned off. The mRNA is labile and easily degraded, so it usually remains inside the cell it is produced in.
Steps of translation
When the mRNA enters the ribosome, translation will occur. The mRNA determines the amino acid sequence of polypeptides according to the genetic code. Once transcribed, the mRNA binds to the small subunit of the ribosome and two tRNAs bind simultaneously to the large subunit. Translation depends on complementary base pairing between codons on mRNA and anticodons on tRNA. Codons of three bases on mRNA correspond to one amino acid in
• Gene expression is the process by which a gene product is made using information in the DNA.
• Translation is the synthesis of polypeptides from mRNA on ribosomes.
•The genetic code is universal because it can be found in most organisms and degenerate because more than one codon codes for one amino acid.
a polypeptide. The stepwise movement of ribosomes along the mRNA allows the entrance of one new complementary tRNA at a time. Peptide bonds are formed between the amino acids in the newly attached tRNA and the growing polypeptide chain.
Example 5
The information needed to make polypeptides is carried in mRNA from the nucleus to the ribosomes of eukaryotic cells. This information is decoded during translation. The diagram below represents the process of translation. Tyr = tyrosine Met = metionine
= leucine
= alanine
A sequence of nucleotides of mRNA is shown:
AUGUUUCCACUAAAUGUAAGCAGA
= glycine
phenylalanine
a) Annotate the diagram to show the direction in which the ribosome moves during translation.
b) State the name of the next amino acid which will attach to the polypeptide
Solution
a) From left to right (5’ → 3’)
b) Alanine (Ala)
The genetic code is universal, which means all organisms have the same code. For example, human insulin can be produced in bacteria as genes can be transferred between species. The genetic code is also degenerate. This is because more than one codon codes for one amino acid. Most polypeptides start with the amino acid methionine, so the start codon of mRNA is usually AUG. Transcription starts at TAC on the antisense DNA (or ATG in the DNA coding sequence) as this is where the polymerase starts transcription.
Samplecontent
a) (i) Identify the DNA sequence that has been transcribed to produce this mRNA.
(ii) Deduce the peptide sequence translated from this mRNA sequence.
b) (i) Explain what would have happened if instead of AGA you had UGA in the last codon of this portion of this mRNA.
(ii) Suggest a reason how the change in b) (i) could have happened.
Solution
a) (i) TAC AAA GGT GAT TTA CAT TCG TCT
To identify the DNA sequence, you must find the complementary bases: for A it is T, and for C it is G, keeping in mind that in DNA you have T instead of U.
(ii) Met Phe Pro Leu Asn Val Ser Arg
To deduce the peptide sequence, you first separate the sequence of mRNA every three nucleotides. You then look in the table for A as the first letter, U for the second letter and G for the third. This is done for each codon.
b) (i) UGA is a stop codon, so the translation would be stopped here. There is no tRNA capable of joining this codon, so the peptide leaves the ribosome. The peptide produced would be shorter.
(ii) A point mutation replacing T with A on the DNA could have caused the change of A to U in the mRNA.
Mutations that change protein structure
Some mutations cause a change in the structure of the protein produced. In sickle cell anaemia, one specific base substitution causes glutamic acid to be substituted by valine as the sixth amino acid in the haemoglobin polypeptide. As a result, the structure of the haemoglobin protein is altered and it does not carry oxygen efficiently in erythrocytes. The shape of affected erythrocytes is also altered (they are “sickle-shaped”).
•A mutation is a change to the structure of a gene c aused by the alteration of single base units in the DNA. A mutation may be inherited by subsequent generations or may affect only the individual in which it occurs.
The first, second and third positions of the three bases of the codons on messenger RNA (mRNA) are shown in this genetic code.
Example 6
• Introns are sections of mRNA that are edited out before translation.
Directionality of transcription and translation and initiation of transcription
Transcription and translation always occur in the 5’ to 3’ direction. Initiation of transcription is in the promoter that is upstream of the 3’ end of the gene in the DNA template. The first nucleotide of the transcribed mRNA is at the 5’ end, forming the codon that first enters the ribosome to be translated (usually AUG on mRNA, which codes for methionine).
a) L abel the sense and antisense strands.
b) Draw an arrow on the diagram to show where the next nucleotide will be added to the growing mRNA strand.
Solution
a) The antisense strand is the blue strand on top (the one that is copied by RNA) and the sense strand the green strand on the bottom (the one that has a very similar sequence to the RNA transcribed, with exception of the change from T in DNA to U in RNA).
b) A clearly drawn arrow pointing at the free 3’ end of the RNA strand— this is on the left of the growing mRNA. This is because the RNA grows in the 5’ to 3’ direction and therefore copies the antisense DNA strand in the 3’ to 5’ direction.
Non-coding sequences in DNA do not code for polypeptides
Non-coding regions of DNA do not code for proteins but have other important functions, such as regulators of gene expression, introns, telomeres, and genes for rRNAs and tRNAs in eukaryotes. Regulators of gene expression can be areas of the DNA such as the promoter, where the RNA polymerase joins to start transcription. Many promoters are regulated by transcription factors. Activator proteins will start transcription and repressor proteins will downregulate transcription. Introns are sections found in eukaryotic DNA that are transcribed to mRNA but are edited out before translation. Telomeres are the extremes of chromosomes and protect the DNA from damage. There are 170 to 570 genes coding for tRNAs in eukaryotes. The genes for rRNA are highly conserved.
Which are examples of non-coding regions of DNA?
I. Telomeres
II. Promoter
III. Exons
A. I only
B. I and II only
C. I and III only
D. I, II and III
Solution
The correct answer is B, as exons are the coding regions.
Post-transcriptional modification in eukaryotic cells
In eukaryotes the mRNA undergoes post-transcription modifications such as polyadenylation of the 3’ end (polyA tail), capping of the 5’ end (methyl7-guanosine cap) and splicing. The addition of caps helps to stabilize the mRNA molecule and avoid digestion by nuclease enzymes. The removal of introns by special proteins and enzymes and the splicing together of exons occurs before the mature mRNA leaves the nucleus. If alternative sequences are used as introns, different peptides can be obtained from the same gene (alternative splicing). Alternative splicing reduces the amount of DNA required to produce polypeptides.
• Post-transcription modifications in mRNA include splicing of exons and addition of 3’ polyA chains and 5’ caps.
polypeptide
▴ Figure 4 Transcription and translation in protein synthesis
Initiation
of translation
Sample student answer
COOH
Explain how information in the mRNA can be translated into a polypeptide. [8]
This answer could have achieved 8/8 marks:
Information transferred from DNA to mRNA is translated into an amino acid sequence. Synthesis of the polypeptide involves a repeated cycle of events. Initiation of translation involves assembly of the components that carry out the process. Firstly the tRNA is
The diagram shows the process of transcription.
Example 7
Example 8
▲ This is a complete answer, including all the parts of mRNA translation. Although the diagram is not necessary, in this case it includes important and clear information.
initiator tRNA in the P site
activated by phosphorylation in the cytoplasm. A tRNA-activating enzyme attaches a specific amino acid to the 3’ end of a determined tRNA, using ATP for energy. It recognizes tRNA by its shape or chemical properties. To begin the process of translation, an mRNA molecule binds to the small ribosomal subunit at an mRNA binding site. An initiator tRNA molecule carrying methionine binds at the start codon (AUG). The large subunit of the ribosomes then joins the small subunit. It has three sites, the A (aminoacyl) site, P (peptidyl) site and E (exit) site. The initiator tRNA is in the P site of the ribosome. The next codon signals another tRNA to bind at the A site. A peptide bond is formed between the amino acids in the P and A sites. The ribosome translocates three bases along the mRNA (in the 5’–3’ direction), moving the tRNA in the P site to the E site. This tRNA is now free in the cytoplasm to be activated again. The tRNA with the appropriate anticodon will bind to the next codon and occupy the vacant A site. This will continue until a stop codon (UAG, UGA or UAA) is reached. The disassembly of the components follows termination of translation.
large ribosomal subunit small ribosomal subunit 5ʹ 3ʹ 5ʹ 3ʹ mRNA binding site start codon
Modification of polypeptides into their functional state
Insulin is a peptide hormone that regulates glucose levels in blood. It is synthesized as an inactive precursor molecule, pre-proinsulin. The signal peptide (shown in red in Figure 5) is removed to produce the also inactive 110 amino acid-long protein called proinsulin. The protein forms disulfide bonds and has part of its structure removed to become an active molecule. A single protein of human insulin is composed of 51 amino acids. The proinsulin is an inactive form with long-term stability, which serves to keep the highly reactive insulin protected, yet readily available. The active molecule is a much faster-reacting hormone because diffusion rate is inversely related to particle size.
Recycling of amino acids by proteasomes
Proteasomes are protein complexes, found in the nucleus and cytoplasm of eukaryotes, which oversee protein degradation. Recycling of amino acids through protein enzymatic degradation enables the cell to maintain basal metabolic activities such as gene expression and cell cycle regulation. A functional proteome requires constant protein breakdown and synthesis.
• Proteasomes are protein complexes involved in the degradation of proteins to amino acids.
Samplecontent
▸ Figure 5 Two-stage modification of pre-proinsulin to insulin
Example 9
The degradation of casein protein by the proteasome complexes 26S and 20S was studied in test tubes in the presence or absence of ATP.
chain (21 amino acids)
B” chain (30 amino acids)
amino acids
connecting peptide
Source: Peters, J. M., et al. (1994). Distinct 19 S and 20 S subcomplexes of the 26 S proteasome and their distribution in the nucleus and the cytoplasm. Journal of Biological Chemistry, 269(10), 7709–7718.
a) Calculate the difference in percentage degradation of casein between 26S with and without ATP.
b) Deduce, with a reason, whether casein degradation in the proteasome is ATP dependent.
c) Discuss which proteasome is most likely involved in casein degradation in cells.
Solution
a) 9 − 4 = 5%
b) The 26S and 20S degraded casein in an ATPdependent way as the percentage of casein degraded increased in the presence of ATP in both cases.
c) Both show casein degradation; 20S showed more casein degradation than 26S, with and without ATP. But cells require constant breakdown and synthesis of proteins, so not always the most degradation is convenient. Also, the conditions in cells might not be the same as in the test tube.
•How does the diversity of proteins produced contribute to the functioning of a cell?
•What biological processes depend on hydrogen bonding?
D1.3 Mutation and gene editing
You should know:
✔ mutations are random structural changes to genes.
✔ gene mutation c an be caused by mutagens and by errors in DNA replication or repair.
✔ some bases have higher possibility of mutation although mutations are random.
✔ mutations in germ cells are inherited whereas mutations in somatic cells may cause cancer.
✔ gene mutations are the source for genetic variation, therefore essential for evolution by natural selection.
Additional higher level:
✔ gene knockout technology can be used to study the function of a gene.
• Gene mutations are structural changes to genes at the molecular level.
• Single-nucleotide polymorphisms (SNPs) are caused by base substitution mutations. Because of the degeneracy of the genetic code, they may or may not change a single amino acid in a polypeptide.
Gene mutation
You should be able to:
✔ distinguish between substitutions, insertions and deletions.
✔ explain the consequences of base substitutions.
✔ explain the consequences of insertions and deletions.
Additional higher level:
✔ describe an example of CRISPR sequences and the enzyme Cas9 in gene editing.
✔ discuss the hypotheses to account for conserved or highly conserved sequences in genes.
• Mutagens damage the DNA molecule.
Gene mutations are structural changes to genes at the molecular level. There are different types of gene mutations. A base substitution is when one nucleotide is replaced by another in the DNA sequence. A change in the DNA determines a change in the mRNA codon. The change in mRNA could cause a different tRNA to join by its anticodon, determining a different primary sequence of the protein. If the new codon is a stop codon, the protein will be much shorter. Because the genetic code is degenerate, if the change in nucleotide determines for a codon that codes for the same amino acid, no change will occur to the protein (neutral mutation). Single-nucleotide polymorphisms (SNPs) are a result of base substitution mutations. These SNPs determine variability in a population.
An insertion is when one or more nucleotides are added to DNA. This will cause a frameshift in the DNA reading frame. The codons will be displaced, so the final peptide produced will be completely different from the original peptide. In many cases, a new stop codon is formed, and the peptide is shorter. In the rare case where the insertion is of three nucleotides, the protein will be similar, but with an extra amino acid added. In the case of a deletion, a section of the DNA is eliminated. If the section where the insertion or deletion occurred is not a coding section and the number of nucleotides is a multiple of three so there is no frameshift, then the insertion or deletion does not affect the peptide produced.
Causes of gene mutation
Mutations can arise from exposure to radiation such as ultraviolet (UV) rays, the presence of chemical mutagens, carcinogens, papilloma virus or cigarette smoke. The most common chemical mutagens are substances that greatly resemble nucleotides (base analogues) and are incorporated into the DNA during replication. Examples of chemical mutagens are mustard gas and ethidium bromide. There are many highly reactive oxygen species generated by normal cellular processes that are also mutagenic. Mutagens are also carcinogens if the damage they produce leads to cancer. Exposure to radiation such as X-rays,
gamma rays or UV light can also cause mutagenesis. UV light induces cross-linking between DNA molecules that inhibits replication and transcription. Mutagenesis can also be caused by spontaneous mutations that result from errors in DNA replication, recombination and repair. If the error is in replication, there is usually a point mutation due to the addition of a wrong amino acid to the sequence of the peptide. The error can also be in the rearrangement of chromosomes during meiosis and mitosis and can cause a change in the number of chromosomes. Some of these changes are silent as they affect non-coding or regulating regions, or because of the degeneracy of the genetic code they form an alternative codon for the same amino acid. Some carcinogens, such as anabolic steroids, increase the chances that repair mechanisms fail, therefore causing cancer.
Samplecontent
Example 10
The ABO blood group gene locus is located on human chromosome 9, and encodes for an enzyme that mediates the expression of A and B antigens on erythrocytes. The A and B genes differ in a few singlebase substitutions. The protein sequence of the allele for blood group O differs in more amino acids from the sequences for blood groups A and B. The table shows part of a protein sequence alignment of the alleles of blood groups (an asterisk means the sequence is the same and a dash means the amino acid is not present).
Allele Amino acid
blood group A
blood group
blood
Example 11
Scientists estimated thyroid cancer risks related to the type of diagnostic X-ray procedure in 75,494 people, 251 of which were diagnosed with thyroid cancer. The risk estimates were determined as hazard ratios compared to the norm. A ratio of 2 means that the patients are two times more likely to have thyroid cancer than the norm.
X-ray procedure Hazard ratio
cervical spine radiograph 0.95
skull radiograph 0.99
other head and neck radiograph 1.02
angiogram 1.04
dental radiograph (per 10 radiographs)1.11
mammogram 0.99
chest radiograph (per 5 radiographs) 0.92
upper gastrointestinal series 0.98 barium swallow 0.94
lumbar/thoracic spine radiograph 0.99
Source: Neta, G., et al. (2013). A prospective study of medical diagnostic radiography and risk of thyroid cancer. American Journal of Epidemiology, 177(8), 800–809.
a) Identify the number of amino acids that are different in this sequence between blood group A and blood group B.
b) Suggest whether this data shows that human diversity is due to SNPs.
Solution a) 2
b) There are more SNPs between the sequence for blood group O and both A and B. There are few differences between A and B. Therefore, the SNPs make a difference in the blood group, confirming that SNPs can contribute to human diversity.
a) State the procedure that has the highest risk of producing thyroid cancer.
b) Comment on the evidence of thyroid cancer risk associated with diagnostic X-rays.
Solution
a) Dental radiographs
b) The data shows no evidence of the relationship between X-ray procedures and thyroid cancer, except for dental X-rays where there is 11% more risk than the norm and 10 radiographs need to be taken. More tests should be performed, and more patients studied.
Nature of science
Commercial genetic tests can yield information about ancestry and potential future health and disease risk using only a saliva sample. The results are usually sent back to the consumer without much explanation or expert interpretation.
Randomness and consequences of mutation in germ cells and somatic cells
Mutations can occur anywhere in the base sequences of a genome, although some bases have a higher probability of mutating than others. There is no natural mechanism for making a deliberate change to a particular base with the purpose of changing a trait. The driving force for the generation of new mutations is in the replication of DNA when germ cells are created. Germ cells are especially good at preventing or repairing DNA damage but, in the unlikely event that this mutation in the germ cell is not repaired, it will be inherited by the next generation. Instead, if the mutation is in a somatic cell, it will not be inherited by the offspring, but it may lead to cancer. Mutations are present in most or all malignant cells of a tumour and have probably been selected because they confer a proliferative advantage.
Mutation as a source of genetic variation
Most mutations either have no effect on the individual or are detrimental to health and survival. There are very few truly beneficial mutations. Nevertheless, mutations are the fuel of evolution, and are therefore beneficial to the adaptation of species to changes in their environment. Mutations function as facilitators of evolution, providing variation and enabling rapid evolution of new forms.
DNA from different cancer tumours was sequenced and compared to normal cell DNA to detect transitions (change from purine to purine or pyrimidine to pyrimidine) and transversions (change from purine to pyrimidine or pyrimidine to purine). The graph shows the mutations observed only in a tumour due to random events (blue bars) and mutations that appear due to reproduction of a mutant cell (yellow bars).
a) State the most common change in base.
b) Identify the most common transversion.
c) (i) Compare and contrast random mutant events and mutations due to reproduction of mutated cells.
(ii) Suggest one reason for the difference in number of events caused by random mutation and by mutations that appear due to reproduction of a mutant cell.
Solution
a) The transition CG to TA.
b) The transversion TA to GC.
c) (i) Random events happen much more often than mutations due to reproduction of a mutated cell. Both occur for most changes, except for TA to CG that does not appear in reproduction of a mutated cell.
CRISPR sequences and Cas9 in gene editing
CRISPR–Cas9 is a complex formed by a short guide RNA (sgRNA) sequence that will target one selected region or gene on the DNA. This system can design an RNA complementary to a given DNA. Cas9 enzyme will cut the DNA at the targeted sequence, which will be repaired either by deleting some sequence or adding some sequence. A sequence of viral or prokaryotic DNA (PAM) prevents nucleases from destroying the target sequence. Several blood disorders, including leukaemia, lymphoma and thalassemia have been treated in tissues using this technology. In 2023, CRISPR–Cas9 technology was approved to treat sickle cell disease.
Nature of science
CRISPR–Cas9 is powerful in terms of the potential to start a biotechnological revolution in the field of crop development and human pathology. However, in the wrong hands, it can lead to abuse and misuse in multiple ways, including manipulation of germline genetics. Scientists across the world are subject to different regulatory systems. Therefore, an international effort is needed to harmonize regulation of the application of genome editing technologies such as CRISPR.
Source: Bielas, J. H., et al. (2006). Human cancers express a mutator phenotype. PNAS, 103(48), 18238–18242.
(ii) Mutations appear randomly in tumours, they are not just replicated. Cancer cells have increased rates of mutagenesis.
Gene knockout technology
Gene knockout (KO) is a technique for identifying the functions of coding and non-coding genomic regions by preventing the expression of this gene. Gene function can be studied using model organisms with similar sequences by using KO technology. By causing a specific gene to be inactive in an organism and observing any differences in behaviour or physiology, researchers can infer the probable function of the gene. A library of KO organisms is available for some species to use as models in research.
Source: Choi, E. & Koo, T. (2021). CRISPR technologies for the treatment of Duchenne muscular dystrophy. Molecular Therapy, 29(11), 3179–3191.
•The CRISPR–Cas9 system allows the editing of nucleotides in DNA.
• Conserved DNA sequences (or an amino acid sequence in a protein) have remained unchanged throughout evolution.
Conserved sequences
Conserved sequences are DNA molecules (or an amino acid sequence in a protein) that have remained unchanged throughout evolution. Conserved sequences are identical or similar across a species or a group of species. Highly conserved sequences are identical or similar over long periods of evolution. An example is the cytochrome c gene and its corresponding protein, which are highly conserved in vertebrates. This protein is involved in aerobic respiration, and is therefore essential for organisms. Another example is haemoglobin, which is essential for the transport of oxygen in vertebrates. In addition to protein-coding sequence, the human genome contains a significant amount of regulatory DNA, which contains highly conserved sequences. In bacteria and archaea, the 16S and 23S ribosomal genes are the most conserved genomic regions.
One hypothesis for the mechanism that maintains conserved sequences is the functional requirements for the gene products. Because the protein is essential, if there is a mutation in the gene, the organism does not survive, and the mutation is not inherited. Another hypothesis is that these genes have slower rates of mutation and that is why they do not change much throughout evolution.
Example 12
▴ Figure 6 Mechanism of action of CRISPR–Cas9