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IB Biology D1.1: DNA replication worked solutions

Worked solutions to every practice question in D1.1: DNA replication of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

[1 mark]

Which statement about DNA replication is correct? (A) Each new molecule has the same base sequence as the original. (B) It uses nucleotides that contain uracil. (C) It happens after the cell has divided. (D) It makes one molecule with two new strands.

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A. Replication copies each template strand, so each new molecule has the base sequence of the original. DNA contains thymine, not uracil, replication happens before division, and each new molecule keeps one old strand.

Final answer

Question A2

[1 mark]

A drug stops helicase working in a dividing cell. What is the immediate result at the replication fork? (A) The strands cannot separate because hydrogen bonds stay in place. (B) New nucleotides cannot bond to one another. (C) Covalent bonds in the backbone break. (D) RNA primers are not removed.

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A. Helicase breaks the hydrogen bonds between the two strands, so without it the strands cannot separate. B describes polymerase, C a bond that no replication enzyme breaks, and D a job of DNA polymerase I in prokaryotes.

Final answer

Question A3

[1 mark]

Both strands of a DNA molecule contain only ^{15}N. It is copied twice in a medium that contains only ^{14}N. How many of the four molecules that result contain any ^{15}N? (A) 0 (B) 1 (C) 2 (D) 4

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C. After the first copying there are two hybrid molecules, each with one old ^{15}N strand. In the second copying each hybrid gives one hybrid and one light molecule. The two original strands are never destroyed, so exactly 2 of the 4 molecules contain ^{15}N.

Final answer

Question A4

[1 mark]

A forensic laboratory has only a few skin cells from a crime scene. Which order of techniques gives a DNA profile? (A) Gel electrophoresis, then PCR (B) PCR, then gel electrophoresis (C) Gel electrophoresis only (D) PCR only

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B. A trace sample must first be amplified by PCR and the amplified fragments are then separated by gel electrophoresis. A and C would try to separate too little DNA, and D never separates the fragments.

Final answer

Question A5

[1 mark]

What is the function of the primers in PCR? (A) They break the hydrogen bonds between the strands. (B) They bind to the ends of the target and give the polymerase a start. (C) They supply the energy for each cycle. (D) They join the new fragments together.

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B. Primers are short single strands that bind to the ends of the target sequence and give Taq polymerase a place to start. Heat, not primers, separates the strands, and no energy or joining of fragments is supplied by them.

Final answer

Question A6

[1 mark]

A PCR programme repeats three stages: 95~°C for 30~s, 55~°C for 30~s and 72~°C for 60~s. What happens at 55~°C? (A) The strands separate. (B) Primers bind to the template strands. (C) Taq polymerase adds nucleotides. (D) The polymerase is destroyed.

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B. Cooling to about 55 °C lets the primers bind (anneal) to the template strands. Separation is at 95 °C, and Taq adds nucleotides at 72 °C.

Final answer

Question A7

[1 mark]

In gel electrophoresis, DNA fragments move towards the positive electrode. Which property of DNA causes this? (A) The phosphate groups are negatively charged. (B) The bases are positively charged. (C) The fragments differ in length. (D) DNA forms hydrogen bonds with the gel.

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A. The phosphate groups of the backbone carry a negative charge, so every fragment is attracted to the positive electrode. Fragment length decides how far it travels, not which way.

Final answer

Question A8

[1 mark]

Four DNA fragments W, X, Y and Z were separated by gel electrophoresis. The table shows how far each moved from the well. Which fragment has the fewest base pairs? Distance moved from the well (mm): W 12, X 25, Y 31, Z 18. (A) W (B) X (C) Y (D) Z

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C. The shortest fragment travels furthest. Y moved 31 mm, further than W (12 mm), Z (18 mm) and X (25 mm).

Final answer

Question A9

[1 mark]

The table lists the DNA fragments (in bp) at one marker for a child, the mother and three men. Which man could be the father? Child 450 and 250 bp; mother 450 and 350 bp; man X 350 and 150 bp; man Y 250 and 150 bp; man Z 450 and 350 bp. (A) X (B) Y (C) Z (D) X and Z

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B. The child's 450 bp band can come from the mother, so the 250 bp band must come from the father. Only Y has a 250 bp fragment. X and Z do not have 250 bp.

Final answer

Question A10

HL[1 mark]

DNA polymerase III is extending a new strand. To which end of the new strand does it add the next nucleotide? (A) The 5′ end of the new strand (B) The 3′ end of the new strand (C) The 5′ end of the template (D) The 3′ end of the template

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B. Polymerase joins the 5′ phosphate of the incoming nucleotide to the 3′ OH at the end of the growing strand, so a new strand grows 5′ to 3′. The template is only read, never extended.

Final answer

Question A11

HL[1 mark]

In a bacterium, which enzyme takes out each RNA primer and fills the gap with DNA nucleotides? (A) DNA primase (B) DNA polymerase III (C) DNA polymerase I (D) DNA ligase

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C. DNA polymerase I removes the primer and fills the gap with DNA. Primase makes primers, polymerase III extends them, and ligase seals nicks.

Final answer

Question A12

HL[1 mark]

A modified nucleotide has a hydrogen atom, instead of a hydroxyl group, on its 3′ carbon. Why does synthesis stop after it is added to a strand? (A) It cannot base-pair with the template. (B) No 3′ OH is left for the next nucleotide to bond to. (C) It has no phosphate group. (D) It destroys the template strand.

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B. The next nucleotide must bond to a 3′ OH. With none left, no covalent bond can form and the chain cannot grow. The modified nucleotide can still pair with the template and has a phosphate group.

Final answer

Tutorial practice

Question B1

[5 marks]

Bacteria were grown for many generations in a medium containing ^{15}N and then moved to a medium containing only ^{14}N. DNA was extracted at intervals and spun in a density gradient. The table gives the position of each band as the distance from the top of the tube. Generations in 14N and band position (mm): 0, 30; 1, 24; 2, 18 and 24. (a) State the position of the band for DNA that contains only ^{14}N. [1] (b) Explain why the result after one generation rejects the conservative model. [2] (c) Predict the bands, and their relative thickness, after three generations. [2]

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(a) DNA with only ^{14}N is the lightest, so it lies highest, at 18 mm (1). (b) The conservative model predicts two bands after one generation: heavy DNA at 30 mm and light DNA at 18 mm (1). Only one band was seen, at an intermediate position (24 mm), so every molecule is a hybrid and the conservative model is rejected (1). (c) After three generations there are 8 molecules: 2 hybrid and 6 light (1). There are two bands, at 24 mm and at 18 mm, and the 18 mm band is three times as thick as the 24 mm band (1).

Final answer

(a) 18 mm. (b) A conservative model predicts a heavy band (30 mm) and a light band (18 mm); one band at 24 mm is between them, so the conservative model is rejected. (c) Two bands, at 18 mm and 24 mm; the 18 mm band is three times as thick.

Question B2

[3 marks]

A student writes: “In replication the original DNA molecule stays whole and a completely new molecule is made.'' Explain why this is not correct.

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Helicase separates the two strands and each strand acts as a template (1). Free nucleotides pair with the exposed bases by complementary base pairing, A with T and C with G, and polymerase joins them into a new strand (1). Each new molecule therefore holds one original strand and one new strand, so replication is semi-conservative and the original molecule does not stay whole (1).

Final answer

Each strand of the original molecule separates and acts as a template. Free nucleotides pair with it by complementary base pairing and polymerase joins them. Each daughter molecule therefore has one original and one new strand.

Question B3

[4 marks]

A single hair root yields only a trace of DNA. Outline how PCR could produce enough DNA to analyse.

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Any four of: DNA is mixed with primers, heat-stable (Taq) polymerase and free nucleotides (1). Heating to about 95 °C separates the strands (1). Cooling to about 55 °C lets the primers bind to the ends of the target (1). At about 72 °C Taq polymerase extends the primers into new strands (1). The cycle is repeated many times and the number of copies doubles each cycle (1).

Final answer

Mix the DNA with primers, Taq polymerase and free nucleotides. Heat to about 95 °C so the strands separate. Cool to about 55 °C so primers bind. Heat to 72 °C so Taq adds nucleotides. Repeat the cycle so the copies double each time.

Question B4

[4 marks]

Four PCR tubes were set up with the standard mixture, but with the changes shown. The products were run on a gel. Tube 1 no change, band present; tube 2 annealing at 80 °C, no band; tube 3 polymerase from E. coli, no band; tube 4 no primers, no band. (a) Explain the result for tube 2. [2] (b) Explain the result for tube 3. [2]

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(a) Annealing must be cool enough for primers to pair with the template (1). At 80 °C the hydrogen bonds between primer and template do not hold, so no primer binds and nothing is extended (1). (b) The polymerase from E. coli is not heat-stable, so it is denatured at about 95 °C in the first cycle (1) and cannot add nucleotides at any later step (1).

Final answer

(a) At 80 °C the primers cannot bind to the template, so extension cannot start. (b) E. coli polymerase is denatured at 95 °C in the first cycle, so no new strand is made.

Question B5

[3 marks]

At one marker the mother has fragments of 600 and 400 bp, and the father has fragments of 500 and 400 bp. (a) List the possible band patterns for their child. [2] (b) A child has bands at 600 and 300 bp. State, with a reason, whether the child could be theirs. [1]

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(a) A child receives one fragment from each parent. The four combinations are 600 + 500, 600 + 400, 400 + 500 and 400 + 400, which shows as one band at 400 bp (2 for all four, 1 for two or three). (b) The child cannot be theirs, because the 300 bp fragment is found in neither parent (1).

Final answer

(a) 600 + 500, 600 + 400, 400 + 500, and 400 + 400 (a single band at 400 bp). (b) No: 300 bp is in neither parent.

Question B6

[3 marks]

DNA profiling compares tandem repeats at several markers. (a) State what each band in a DNA profile represents. [1] (b) Explain why twelve markers are more reliable than one. [2]

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(a) A band is a group of DNA fragments of the same length, set by the number of repeats at that marker (1). (b) At one marker two unrelated people could share a band by chance (1). The chance of matching at every one of twelve markers is the product of the separate chances, which is very small, so a false match is far less likely (1).

Final answer

(a) A DNA fragment of one length, which depends on the number of repeats at a marker. (b) Unrelated people may share a band at one marker by chance, but the chances multiply across twelve markers, so a false match becomes very unlikely.

Question B7

[4 marks]

Describe how DNA profiling could be used to decide whether a suspect left a blood sample at a crime scene.

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DNA is extracted from the scene sample and from the suspect (1). The tandem-repeat markers are amplified by PCR using primers and Taq polymerase (1). The fragments are separated by size by gel electrophoresis (1). The band patterns are compared: matching bands at every marker are consistent with the suspect, and a clear mismatch excludes him or her (1).

Final answer

Extract DNA from the sample and from the suspect, amplify the tandem-repeat markers by PCR, separate the fragments by gel electrophoresis and compare the bands. A match at every marker is consistent with the suspect; any clear mismatch excludes the suspect.

Question B8

[3 marks]

A PCR programme repeats the three stages in the table 30 times. Stage 1 95 °C for 30 s; stage 2 55 °C for 30 s; stage 3 72 °C for 60 s. (a) State what happens to the DNA at stage 1. [1] (b) Explain why the temperature is lowered for stage 2. [1] (c) Explain why stage 3 is carried out at 72~°C. [1]

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(a) At 95 °C the hydrogen bonds between the two strands break, so the strands separate (1). (b) The temperature is lowered so that primers can pair with complementary sequences at the ends of the target (1). (c) 72 °C is the optimum for Taq polymerase, which adds nucleotides to the primers to make new strands (1).

Final answer

(a) Hydrogen bonds between the strands break and the strands separate. (b) Cooler conditions let the primers bind to the template. (c) 72 °C is the best temperature for Taq polymerase to extend the primers.

Question B9

[6 marks]

Explain the process of DNA replication in the nucleus of a cell before mitosis.

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Any six of: replication happens in the S phase of interphase, before mitosis (1); helicase unwinds the double helix and breaks the hydrogen bonds between the strands (1); each separated strand acts as a template (1); free nucleotides pair with the exposed bases by complementary base pairing, A with T and C with G (1); DNA polymerase joins the nucleotides, forming covalent bonds in the new strand (1); each new molecule has one original strand and one new strand, so replication is semi-conservative (1); the two molecules have base sequences identical to the original (1).

Final answer

It occurs in the S phase of interphase. Helicase separates the strands by breaking hydrogen bonds. Each strand is a template. Free nucleotides pair by complementary base pairing (A-T, C-G). DNA polymerase joins them into a new strand. Each molecule has one old and one new strand, and the sequences are identical.

Question B10

HL[4 marks]

Outline the roles of primase, DNA polymerase I and DNA ligase in prokaryotic DNA replication.

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Primase makes a short RNA primer (1) at the start of each new strand, so polymerase III has a 3′ end to extend (1). DNA polymerase I removes the RNA primers and replaces them with DNA (1). DNA ligase seals the nicks between fragments, joining them by a covalent bond (1).

Final answer

Primase makes short RNA primers. DNA polymerase I removes the RNA primers and replaces them with DNA. DNA ligase seals the nicks between adjacent fragments, forming covalent bonds.

Question B11

HL[4 marks]

Compare leading strand and lagging strand synthesis.

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Leading is synthesised continuously whereas lagging is synthesised discontinuously in Okazaki fragments (1). The leading strand grows towards the replication fork and the lagging strand grows away from it (1). The leading strand needs one primer, but the lagging strand needs a new primer for every fragment (1). The lagging strand fragments are joined by ligase, but the leading strand needs no joining (1). Both are made in the 5′ to 3′ direction (accept as a similarity).

Final answer

The leading strand is continuous and grows towards the fork with one primer. The lagging strand is made in short Okazaki fragments growing away from the fork, each needing a primer, and ligase joins them.

Question B12

HL[3 marks]

Explain how DNA polymerase III helps to keep the error rate of replication low.

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Complementary base pairing means most nucleotides added are the correct match (1). DNA polymerase III checks the nucleotide just added at the 3′ end of the new strand (1). If it is mismatched, polymerase III removes it and replaces it with the correct nucleotide, which lowers the mutation rate (1).

Final answer

Complementary base pairing selects the correct nucleotide. Polymerase III checks the nucleotide just added at the 3′ end, removes it if it is mismatched, and adds the correct one.

Advanced practice

Question C1

HL[6 marks]

The diagram shows the two template strands at a replication fork in a bacterium. The fork is moving to the right. Only one end is labelled. Diagram: two template strands separating at a replication fork that moves to the right; the top template is labelled 3′ at its left end. (a) Label the 5′ and 3′ ends of both template strands. [2] (b) Draw the leading strand and show its direction of growth. [2] (c) Draw two Okazaki fragments on the lagging template, with arrows, and mark one RNA primer. [2]

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(a) The two templates are antiparallel. The top template has 3′ at the left, so its right end is 5′ (1); the bottom template has 5′ at the left and 3′ at the right (1). (b) The leading strand pairs with the top template, which is read 3′ to 5′ as the polymerase moves towards the fork, so it is a continuous strand drawn with an arrow pointing right, 5′ at its left end (1) and its 3′ growing end near the fork (1). (c) The lagging strand is drawn on the bottom template as separate Okazaki fragments, each with an arrow pointing left, away from the fork (1), and an RNA primer at the 5′ (right-hand) end of at least one fragment (1).

Final answer

(a) Top template: 3′ left, 5′ right. Bottom template: 5′ left, 3′ right. (b) Leading strand along the top template, arrow pointing right (towards the fork), 5′ at its left end. (c) On the bottom template, fragments with arrows pointing left (away from the fork), the RNA primer at the right (5′) end of a fragment.

Question C2

[6 marks]

A scientist amplifies a 450~bp marker from a sample that holds 5 copies of the target. Assume every cycle doubles the number of targets. The ladder used on the gel gave bands for 400~bp at 24~mm and 500~bp at 21~mm from the well. (a) Calculate the number of copies of the target after 20 cycles. [2] (b) A second tube is stopped after 14 cycles. Explain why its band is fainter. [2] (c) Predict where the 450~bp band lies on the gel, in mm from the well. [2]

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(a) N=N_0\times2^{n}=5\times2^{20} (1) =5\,242\,880\approx5.2\times10^{6} (1). (b) After 14 cycles there are 5\times2^{14}=81\,920 copies, far fewer than after 20 cycles (1), so less DNA is in the band and it appears fainter (1). (c) A 450 bp fragment is between 400 and 500 bp, so its band lies between the 500 bp band (21 mm) and the 400 bp band (24 mm) (1), at about 22 mm (accept 22 to 23 mm, interpolating on a logarithmic scale) (1).

Final answer

(a) 5\times2^{20}=5\,242\,880\approx5.2\times10^{6}. (b) Fewer cycles make fewer copies, so the band is fainter. (c) Between 21 mm and 24 mm, about 22 mm.

Question C3

[6 marks]

A bacterium with unlabelled DNA copies its chromosome once in a medium containing radioactive thymine, and then twice in a medium without radioactivity. Each copying is one complete round of replication. (a) State the number of chromosomes after the three rounds. [1] (b) Calculate how many of these chromosomes contain a radioactive strand. [2] (c) State the fraction of all strands that are radioactive. [1] (d) Explain why no chromosome has two radioactive strands. [2]

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(a) Three rounds give 2^{3}=8 chromosomes (1). (b) After round 1 there are two molecules, each with one labelled strand. The two labelled strands are never destroyed and each ends in a different molecule, so 2 of the 8 chromosomes contain label (2). (c) The 8 chromosomes hold 16 strands, of which only the 2 original labelled strands carry label, so the fraction is 2/16=1/8 (1). (d) Label was available only in round 1, so each labelled strand was made opposite an unlabelled original strand (1). In every later round a labelled strand is copied into an unlabelled partner, because the later medium has no label, so a labelled strand always pairs with an unlabelled one (1).

Final answer

(a) 8. (b) 2. (c) 2/16=1/8. (d) Label was supplied only in round 1, so every labelled strand is paired with an unlabelled one and stays that way in later rounds.

Question C4

[6 marks]

A blood sample at a crime scene and three suspects were profiled at three markers. The table gives the fragment sizes in bp. The scene profile shows only one band at marker 2. Scene: marker 1 210 and 260; marker 2 150; marker 3 320 and 380. P: 210, 260; 150, 180; 320, 380. Q: 210, 260; 150; 320, 380. R: 210, 300; 150; 320, 380. (a) State which suspect's profile matches the scene. [1] (b) Give a reason to exclude each of the other two suspects. [2] (c) The chance that two unrelated people match at each marker is 1 in 25, 1 in 10 and 1 in 40. Calculate the chance of a random match at all three markers. [2] (d) State how adding more markers would change this chance. [1]

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(a) Suspect Q matches the scene at every marker (1). (b) P has two different fragments (150 and 180 bp) at marker 2 but the scene has a single band at 150 bp, so P is excluded (1). R has 300 bp at marker 1 where the scene has 260 bp, so R is excluded (1). (c) The chances multiply: 1/25\times1/10\times1/40=1/10\,000 (1 for the product, 1 for the answer). (d) With more markers the probability of a chance match falls further, so a match is more reliable (1).

Final answer

(a) Suspect Q. (b) P has two fragments (150, 180) at marker 2 where the scene shows only one; R has 300 instead of 260 at marker 1. (c) 1/(25\times10\times40)=1/10\,000. (d) More markers reduce the chance of a false match further.

Question C5

HL[6 marks]

The chromosome of a bacterium has 4.6\times10^{6} base pairs. Replication begins at one origin and two forks move in opposite directions, each at 800 base pairs per second. (a) Calculate the time to copy the chromosome, in minutes. [2] (b) Each Okazaki fragment is about 1500 nucleotides long. Estimate the number of Okazaki fragments made at one fork. [2] (c) State the enzyme that (i) makes the primer for each fragment and (ii) replaces the primers and seals the nicks. [2]

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(a) The two forks together copy 2\times800=1600 base pairs per second (1), so the time is 4.6\times10^{6}\div1600=2875 s =47.9 min, about 48 min (1). (b) Each fork copies half the chromosome, 2.3\times10^{6} base pairs, so the lagging strand at one fork is 2.3\times10^{6} nucleotides long (1); 2.3\times10^{6}\div1500=1533\approx1.5\times10^{3} fragments (1). (c) Primase makes the RNA primer for each fragment (1). DNA polymerase I replaces the primers with DNA and ligase seals the nicks between fragments (1).

Final answer

(a) 4.6\times10^{6}\div(2\times800)=2875 s, about 48 minutes. (b) Each fork copies 2.3\times10^{6} nucleotides on the lagging template, so 2.3\times10^{6}\div1500\approx1.5\times10^{3} fragments. (c) (i) Primase. (ii) DNA polymerase I and DNA ligase.

Question C6

HL[6 marks]

A mutant strain of bacteria has a DNA polymerase III that cannot proofread. The error rate of the normal enzyme is 1 per 10^{7} bases and of the mutant enzyme 1 per 10^{5} bases. The chromosome has 4.6\times10^{6} base pairs. (a) Calculate the expected number of errors per chromosome copied by (i) the normal and (ii) the mutant enzyme. [2] (b) Explain how the normal enzyme lowers the error rate. [2] (c) Suggest one effect of the higher error rate on the mutant population. [2]

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(a) Normal: 4.6\times10^{6}\div10^{7}=0.46 errors per chromosome (1). Mutant: 4.6\times10^{6}\div10^{5}=46 errors per chromosome (1), a hundredfold more. (b) The polymerase detects a mismatched base at the 3′ end of the new strand and removes it (1), then adds the correct complementary nucleotide (1). (c) Most mutations are harmful or neutral, so many of the mutant cells die or divide more slowly (1); the extra variation from the high mutation rate may let some cells survive a change, such as an antibiotic, so the population adapts faster (1).

Final answer

(a) Normal: 4.6\times10^{6}\div10^{7}=0.46 errors; mutant: 4.6\times10^{6}\div10^{5}=46 errors. (b) The enzyme removes a mismatched nucleotide from the 3′ end and replaces it. (c) Many mutant cells carry harmful mutations and die, but the extra variation could give some cells a new advantage.

Connected questions

Question X1

Connected questions[8 marks]

D1.1 + D2.1. A human cell in G1 has 46 chromosomes. It replicates its DNA and then divides by mitosis. Each chromosome is one DNA molecule in G1. (a) State the phase in which the DNA is replicated. [1] (b) State the number of DNA molecules in the nucleus (i) in G1 and (ii) at the start of mitosis. [2] (c) Explain why the two chromatids of one chromosome have identical base sequences. [2] (d) Name the stage in which chromatids separate and state why the daughter cells receive the same DNA. [2] (e) Explain why drugs that block DNA replication harm rapidly dividing cells most. [1]

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(a) Replication takes place in the S phase of interphase (1). (b) In G1 there are 46 DNA molecules (1). After replication each chromosome has two chromatids, so there are 92 DNA molecules at the start of mitosis (1). (c) The two chromatids are copies made from the same template, each base fixed by complementary base pairing (1), so their base sequences are identical (1). (d) Chromatids separate in anaphase (1); because they are identical, each daughter nucleus receives the same genetic information (1). (e) Rapidly dividing cells replicate DNA more often, so a drug that blocks replication affects them most (1).

Final answer

(a) S phase. (b) (i) 46 (ii) 92. (c) Each chromatid is a copy of the other, made from the same template with complementary base pairing. (d) Anaphase; the chromatids are identical, so each daughter nucleus gets the same genetic information. (e) They have to replicate DNA before each division, so a drug that blocks replication stops them first.

Question X2

Connected questions[8 marks]

D1.1 + D1.2 + D1.3. A DNA polymerase inserts a wrong base opposite a template base and the error is not corrected. The gene later codes for a polypeptide in which the codon GAG is replaced by GUG in the mRNA. (a) Explain how the error becomes a permanent mutation in later DNA. [3] (b) State the effect on the polypeptide of the change from GAG to GUG. [2] (c) State why a base substitution may leave the amino acid unchanged. [1] (d) Explain why the mutation is passed to offspring only if it arises in some cells. [2]

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(a) The wrong base is in the new strand of one daughter molecule (1). At the next replication that strand is a template and the complementary base is added opposite the wrong base (1), so one molecule now carries the changed base pair in both strands and passes it to its cells (1). (b) The codon change alters one amino acid, glutamic acid (GAG) to valine (GUG) (1), so the polypeptide has a different primary structure and may fold or work differently (1). (c) The genetic code is degenerate: some codons for the same amino acid differ in the third base (1). (d) A mutation in a somatic cell is not passed to offspring (1), but one in a cell that forms gametes can be (1).

Final answer

(a) The wrong base is in the new strand. At the next replication that strand is a template, so one daughter molecule has the changed base pair on both strands and it is inherited. (b) One amino acid changes, glutamic acid to valine. (c) Several codons code for the same amino acid. (d) Only mutations in cells that form gametes pass to offspring; somatic mutations stay in the body.

Question X3

Connected questions[8 marks]

D1.1 + D3.2. Two DNA markers lie on different chromosomes. At marker 1 the mother has fragments A and B and the father has fragments C and D. At marker 2 the mother has E and F and the father has G and H. A child receives one fragment from each parent at each marker. (a) State the number of different combinations of fragments the child could have at marker 1. [1] (b) State the probability that the child has fragments A and C at marker 1. [1] (c) Calculate the probability that the child has A and C at marker 1 and E and G at marker 2. [2] (d) Explain why the two probabilities can be multiplied. [2] (e) A man matches the child at both markers. Explain why this does not prove that he is the father. [2]

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(a) A or B from the mother and C or D from the father give AC, AD, BC and BD, which is 4 (1). (b) Each combination has the same chance, so P(\text{AC})=1/4 (1). (c) P(\text{AC and EG})=1/4\times1/4=1/16 (2). (d) Chromosomes of different pairs line up independently in meiosis (1), so the fragment inherited at marker 1 does not influence the one at marker 2, and the probabilities can be multiplied (1). (e) Other men could share the same fragments by chance (1), so a match at both markers makes paternity likely, but only more markers can lower the chance of a false match (1).

Final answer

(a) 4. (b) 1/4. (c) 1/16. (d) Markers on different chromosomes assort independently in meiosis, so the inheritance at one marker does not affect the other. (e) Other men could share the same bands by chance, so a match makes paternity likely but does not prove it.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

DNA replication from the core ideas to full exam questions, with question types, worked examples and graded practice.

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