Author: Philip Colls
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Meiosis and Genetic variation Booklet revision questions   NAME:
Q1. The figure below summarises the process of meiosis. The circles represent cells and the structures within each cell represent chromosomes.
         
(a)     Describe and explain the appearance of one of the chromosomes in cell X.
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(b)     Describe what has happened during division 1 in the figure above.
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(c)     Identify one event that occurred during division 2 but not during division 1.
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(d)     Name two ways in which meiosis produces genetic variation.
1. _________________________________________________________________
2. _______________________________________________________________(2)
(Total 8 marks)
Q2. Two pairs of alleles A and a, and B and b are found on one pair of homologous chromosomes. A person has the genotype AaBb. Figure 1 shows the chromosomes at an early stage of meiosis. The position of two of the alleles is shown.
 
Figure 1
(a)     Complete Figure 1 to show the alleles present at the other marked positions.
(1)
Crossing over occurs as shown in Figure 2.
 
Figure 2
(b)     What term is used to describe the pair of homologous chromosomes shown in Figure 2?
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(c)     From Figure 2, give the genotypes of the gametes produced containing the chromatids
(i)      that have not crossed over;
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(ii)     that have crossed over.
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(d)     Give two processes, other than crossing over, which result in genetic variation. Explain how each process contributes to genetic variation.
Process ____________________________________________________________
Explanation _________________________________________________________
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Process ____________________________________________________________
Explanation _________________________________________________________
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(Total 8 marks)
Q3. (a)    Figure 1 shows one pair of homologous chromosomes.
Figure 1
 
(i)      Name X.
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(ii)     Describe the role of X in mitosis.
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 (2)
(iii)    Homologous chromosomes carry the same genes but they are not genetically identical.Explain why.
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(1)
(b)     Figure 2 shows three pairs of homologous chromosomes in a cell at the end of cell division.
Figure 2
 
(i)      The appearance of each chromosome in Figure 2 is different from those shown in Figure 1. Explain why.
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(ii)     Complete the diagram to show the chromosomes in one cell that could be produced from the cell in Figure 2  as a result of meiosis.
 
(2)
(iii)    Other than independent segregation, give one way in which meiosis allows the production of genetically different cells.
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(1)
(Total 8 marks)
Q4.
(a)     (i)      Why is the genetic code described as being universal?
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(1)
(ii)     The genetic code uses four different DNA bases. What is the maximum number of different DNA triplets that can be made using these four bases?
 
(1)
Transcription of a gene produces pre-mRNA.
(b)     Name the process that removes base sequences from pre-mRNA to form mRNA.
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(1)
(c)     The figure below shows part of a pre-mRNA molecule. Geneticists identified two mutations that can affect this pre-mRNA, as shown in the figure.

Base sequence coding
for amino acids

Base sequence removed
from pre-mRNA

Base sequence coding
for amino acids

 

Mutation 1,
single base
deletion

Mutation 2,
single base
substitution

(i)      Mutation 1 leads to the production of a non-functional protein.
Explain why.
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(3)
(ii)     What effect might mutation 2 have on the protein produced?
Explain your answer.
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(2)
(Total 8 marks)
Q5.
Phenylketonuria is a disease caused by mutations of the gene coding for the enzyme PAH. The table shows part of the DNA base sequence coding for PAH. It also shows a mutation of this sequence which leads to the production of non-functioning PAH.
 
DNA base sequence coding for PAH | C | A | G | T | T | C | G | C | T | A | C | G
DNA base sequence coding for non-functioning PAH | C | A | G | T | T | C | C | C | T | A | C | G
(a)     (i)      What is the maximum number of amino acids for which this base sequence could code?
 
(1)
(ii)     Explain how this mutation leads to the formation of non-functioning PAH.
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(3)
PAH catalyses a reaction at the start of two enzyme-controlled pathways.
The diagram shows these pathways.
 
(b)     Use the information in the diagram to give two symptoms you might expect to be visible in a person who produces non-functioning PAH.
1. _________________________________________________________________
2. _________________________________________________________________
(2)
(c)     One mutation causing phenylketonuria was originally only found in one population in central Asia. It is now found in many different populations across Asia. Suggest how the spread of this mutation may have occurred.
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(1)
(Total 7 marks)
Q6. Figure 1 shows a faulty form of meiosis that can occur in some plants.
Figure 1

(a)     Complete Figure 2 to show the chromosome content of the cells that would result from a normal meiotic division of the diploid parent cell shown in Figure 1.
Figure 2

(2)
(b)     If two diploid (2n) gametes fuse at fertilisation, it can result in the growth of a tetraploid plant which has 4 copies of each chromosome.
Red clover is a plant grown to produce cattle feed. Tetraploid red clover plants produce a higher yield than diploid red clover plants.
Whether a red clover plant produces 2n gametes is genetically controlled.
Scientists investigated the possibility of breeding red clover plants that only produced 2n gametes.
•        In breeding cycle 0, they grew red clover plants and identified plants that produced 2n gametes.
•        In breeding cycle 1, they used the plants producing 2n gametes to produce offspring.
•        In breeding cycles 2 and 3, they identified plants producing 2n gametes and used these to produce offspring.

Their results are shown in the table.
 
| Observed | Expected
Breeding cycle | Number of plants that did not produce 2n gametes | Number of plants that did produce 2n gametes | Number of plants that did not produce 2n gametes | Number of plants that did produce 2n gametes
0 | 50 | 4 | 50 | 4
1 | 14 | 42 | |
2 | 2 | 44 | |
3 | 0 | 56 | |
The scientists used the following null hypothesis.
‘The proportion of plants that produce 2n gametes will not change from one breeding cycle to the next.’
Complete the table to show the expected number of plants that did not produce 2n gametes and the expected number of plants that did produce 2n gametes after 1 cycle. Give each answer to the nearest whole number.
(2)
(c)     The scientists tested their null hypothesis using the chi-squared statistical test.
After 1 cycle their calculated chi-squared value was 350
The critical value at P=0.05 is 3.841
What does this result suggest about the difference between the observed and expected results and what can the scientists therefore conclude?
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(d)     Use your knowledge of directional selection to explain the results shown in the table.
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 (3)  (Total 9 marks)
Q7. Figure 1 shows three cells, B, C and D, from tissues in the same organism. Each cell is in a stage of either mitosis or meiosis.
Figure 1
 
(a)     Complete the table with a tick if the cell shows the feature.
 
| Cell B | Cell C | Cell D
homologous chromosomes are present | | |
a stage of mitosis | | |
(2)
(b)     Describe and explain the appearance of chromosome K in cell C.
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(2)
(c)     Explain what is happening at point J in cell B.
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(2)

(d)     Use information from all three cells in Figure 1 to explain how the number of chromosomes in cell D was produced.
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(1)
(e)     Figure 2 shows the mass of DNA present in cells of a population of healthy cells where mitosis is occurring.
Figure 2
 
Explain why some cells contain a mass of DNA between 1 and 2 arbitrary units.
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(1)
(Total 8 marks)
Q8.
In women, the first division of meiosis produces one daughter cell that has almost all of the cytoplasm. The other daughter cell consists of a nucleus surrounded by a very small amount of cytoplasm and a cell-surface membrane. This very small daughter cell is called a polar body. Polar bodies do not usually divide. The same process occurs in the second division of meiosis, resulting in one egg cell and two polar bodies.
The diagram shows the formation of an egg cell and two polar bodies during meiosis. It also shows what happens to one pair of homologous chromosomes. This pair carries two alleles of gene A.
 
(a)     Complete the diagram by putting A or a in the boxes. One box has been completed for you with A.
(1)
(b)     Put a tick (✓) in the box next to the name of the process that produced the combination of alleles on the chromosome in the first polar body in the diagram.
 
Anaphase |
Crossing over |
Independent assortment |
Semi-conservative replication |
(1)
(c)     A scientist measured the diameter of a polar body and the diameter of the nucleus inside it. The diameter of the polar body was 10.4 μm and the diameter of the nucleus was 7.0 μm. The density of mitochondria in the cytoplasm of the polar body (outside of the nucleus) was 0.08 mitochondria per μm^3.
Calculate the number of mitochondria in the polar body. You should assume polar bodies and nuclei are spherical.
The formula for the volume of a sphere is  πr^3 where π = 3.14
Show your working.
Number of mitochondria =____________________ 
(2)
(d)     Mitochondrial diseases are caused by faulty mitochondria. All of a person’s mitochondria are inherited from their mother via the egg cell. An egg cell contains approximately 3 × 10^5 mitochondria.
One proposed treatment to prevent passing on faulty mitochondria involves
•        removing the nucleus from an egg cell donated by a woman with healthy mitochondria
•        replacing this nucleus with the contents of the polar body from a woman whose egg cells are affected by mitochondrial disease.
Suggest how this treatment prevents inheritance of mitochondrial diseases.
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(2)

(e)     If most of the mitochondria in a cell are faulty, this prevents many important enzyme-catalysed reactions taking place or slows them down.
Suggest and explain one reason why.
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(2)
(Total 8 marks)
Q9.
(a)     A mutation can lead to the production of a non-functional enzyme. Explain how.

(6)
Scientists investigated the effect of a specific antibiotic on two strains of the same species of bacterium.
•        One strain, SR, shows a stringent response in the presence of this antibiotic. Part of this response involves stopping cell division. This gives this strain a greater resistance to the effects of this antibiotic.
•        The other strain, non-SR, cannot carry out a stringent response.
The scientists grew cultures of the SR strain and the non-SR strain containing the same number of bacterial cells. They then stopped each strain from dividing and exposed them to different concentrations of the antibiotic. After a fixed time, the scientists estimated the number of living bacteria remaining in the cultures.
Figure 1 shows their results.
Figure 1
 
(b)     Describe differences in the effect of increasing the concentration of antibiotic on the SR strain and the non-SR strain.
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[Extra space] _______________________________________________________
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(2)

(c)     One way in which the stringent response gives resistance to this antibiotic is by stopping cell division.
The scientists concluded that stopping cell division is not the only way in which the stringent response gives resistance to this antibiotic.
Explain how Figure 1 supports this conclusion.
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(d)     The stringent response involves a number of enzyme-catalysed reactions.
Explain how scientists could use this knowledge to design drugs that make the treatment of infections caused by the SR strain more successful.
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The antibiotic damages the bacterium by causing the production of substances called free radicals.
The scientists exposed the SR strain and the non-SR strain to the antibiotic. They then measured the amounts of free radicals and an enzyme called catalase in both strains.
Figure 2 shows their results.
Figure 2

(e)     Use the information provided and Figure 2 to suggest an explanation for the greater resistance of the SR strain to this antibiotic.
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(3)
(Total 15 marks)

Mark schemes
Q1.
(a)     1.      Chromosome is formed of two chromatids;
2.      (Because) DNA replication (has occurred);
3.      (Sister) chromatids held together by centromere.
3
(b)     1.      Chromosomes in homologous pair;
2.      One of each into daughter cells / haploid number.
2
(c)     Separation of (sister) chromatids / division of centromere.
1
(d)     1.      Independent segregation (of homologous chromosomes);
Accept random assortment
2.      Crossing over / formation of chiasmata.
2
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Q2.
(a)     
A b | | A b | a B | | a B
1
(b)     bivalent;
1
(c)     (i)      Ab, aB;
(ii)     AB, ab;
2
(d)     mutation;
different / new allele formed / genes deleted or duplicated / sequence
of genes changed (reject genetic information);
random fusion of gametes / fertilisation;
new combination of alleles;
independent assortment (of chromosomes) (accept random);
shuffling of maternal and paternal chromosomes / new combination
of alleles;
(ignore references to stages of meiosis)
any 2 × 2
4 max
[8]
Q3.
(a)     (i)      Centromere;
Accept: if phonetically correct 
Reject: centriole
1
(ii)     1.      Holds chromatids together;
2.      Attaches (chromatids) to spindle;
3.      (Allows) chromatids to be separated / move to (opposite) poles / (centromere) divides / splits at metaphase / anaphase;
3. Q Neutral: chromosomes or chromatids split / halved / divided
3. Reject: reference to homologous chromosomes being separated
Accept ‘chromosomes’ instead of ‘chromatids’
Ignore incorrect names for X
2 max
(iii)    (Homologous chromosomes) carry different alleles;
Accept alternative descriptions for ‘alleles’ eg different forms of a gene / different base sequences
Neutral: reference to maternal and paternal chromosomes
1
(b)     (i)      (In Figure 2)
1.      Chromatids have separated (during anaphase);
1. Q Neutral: split / halved / divided
1. Reject: reference to homologous chromosomes
being separated
or
2.      Chromatids have not replicated;
1. & 2. Accept ‘chromosomes’ instead of ‘chromatids’
or
3.      Chromosomes formed from only one chromatid;
Accept converse arguments for Figure 1
Ignore references to the cell not dividing as in the question stem
Ignore: named phases
1 max
(ii)     1.      Three chromosomes;
Ignore shading
2.      One from each homologous pair;
Only one mark for three chromosomes shown as pairs of chromatids
2
(iii)    Crossing over / alleles exchanged between chromosomes or chromatids / chiasmata formation / genetic recombination;
Accept: description of crossing over eg sections of chromatids break and rejoin
Neutral: random fertilisation
Reject: reference to sister chromatids
Q Neutral: genes exchanged
Neutral: mutation
1
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Q4.
(a)     (i)      (In all organisms / DNA,) the same triplet codes for the same amino acid;
Accept codon / same three bases / nucleotides
Accept plurals if both triplets and amino acids
Reject triplets code for an amino acid
Reject reference to producing amino acid
1
(ii)     64;
1
(b)     Splicing;
Ignore deletion references
Accept RNA splicing
1
(c)     (i)      1.      (Mutation) changes triplets / codons after that point / causes frame shift;
Accept changes splicing site
Ignore changes in sequence of nucleotides / bases
2.      Changes amino acid sequence (after this) / codes for different amino acids (after this);
Accept changes primary structure
Reject changes amino acid formed / one amino acid changed
3.      Affects hydrogen / ionic / sulfur bond (not peptide bond);
4.      Changes tertiary structure of protein (so non-functional);
Neutral 3-D structure
3 max
(ii)     1.      Intron non-coding (DNA) / only exons coding;
Context is the intron
Do not mix and match from alternatives
Neutral references to introns removed during splicing
1.and 2. Ignore ref. to code degenerate and get same / different amino acid in sequence
2.      (So) not translated / no change in mRNA produced / no effect (on protein) / no effect on amino acid sequence;
Accept does not code for amino acids
OR
3.      Prevents / changes splicing;
4.      (So) faulty mRNA formed;
Accept exons not joined together / introns not removed
5.      Get different amino acid sequence;
2 max
[8]
Q5.
(a)     (i)      4;
1
(ii)     1.      Change in amino acid / (sequence of) amino acids / primary structure;
1. Reject = different amino acids are 'formed'
2.      Change in hydrogen / ionic / disulphide bonds alters tertiary structure / active site (of enzyme);
2. Alters 3D structure on its own is not enough for this marking point.
3.      Substrate not complementary / cannot bind (to enzyme / active site) / no enzyme- substrate complexes form;
3
(b)     1.      Lack of skin pigment / pale / light skin / albino;
2.      Lack of coordination / muscles action affected;
2 max
(c)     Founder effect / colonies split off / migration / interbreeding;
Allow description of interbreeding e.g. reproduction between individuals from different populations
1
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Q6.
(a)     1.      1 long and 1 short chromosome, each made up of 2 chromatids held (by centromere), in each cell of 1^st division;
2.      1 long and 1 short (separate) chromosome in each cell of 2^nd division;
Allow ECF for correct chromosomes shown in each cell from candidate’s 1^st division cells.
Ignore drawing of centromere.
2
(b)     
52 | 4
;;
Allow 1 mark for numbers totalling 56 except 14/42 - repetition of observed values.
If table is blank, award 1 mark for evidence of 56.
Both 52 and 4 required in table for two marks, do not credit 52 or 4 for one mark.
Award 1 max for answers not given as whole numbers.
2
(c)     1.      There is a less than 0.05/5% probability that the difference(s) (between observed and expected) occurred by chance;
Reject ‘results (without reference to difference) occurring by chance’. Overall max 1 with this statement.
Accept ‘there is a greater than 0.95/95% probability that the difference did not occur by chance’.
Ignore ‘difference is significant’
2.      Calculated value is greater than critical value so the null hypothesis can be rejected;
Ignore ‘difference is significant’
Do not accept ‘P value’ for ‘critical value’.
3.      (The scientists can conclude that) the proportion of plants that produce 2n gametes does change from one breeding cycle to the next;
2 max
(d)     1.      The scientists selected/used for breeding plants that produced 2n gametes;
Answer must be in context of the scientists selecting plants to breed. Accept ‘artificial selection’ or ‘selectively bred’.
2.      (So these plants) passed on their alleles (for production of 2n gametes to the next generation);
Both mark points can be awarded if one correct reference is made to alleles (in either context).
3.      The frequency of alleles for production of 2n gametes increased (in the population).
Both mark points can be awarded if one correct reference is made to alleles (in either context).
For ‘production of 2n gametes’ accept ‘abnormal meiosis’.
Do not accept ‘number’ for frequency.
Accept converse answers linked to plants that produce n gametes.
3
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Q7.
(a)
| Cell B | Cell C | Cell D
homologous chromosomes are present | ✔ | ✔ |
a stage of mitosis | | ✔ |
Mark horizontally
1 mark for each correct row
2
(b)     Mark as pairs, do not mix and match
1.      (Chromosomes consist of) two chromatids connected at centromere;
Accept: sister chromatids for two chromatids
2.       (Because) DNA has replicated;
OR
3.       K is on equator of spindle;
Ignore: ‘middle’
4.       (because) attached at centromere;
Ignore reference to meiosis / bivalents / homologous pairs
2
(c)     1.      Crossing over / exchange of alleles /lengths of DNA / recombination;
Accept: description of crossing over eg sections of chromatids break and re-join
Accept: reference to chiasma/ chiasmata
2.      Between (chromatids of) homologous chromosomes;
Accept: ‘between non-sister chromatids’
Accept: ‘bivalent’ for homologous
Ignore: genes exchanged
2
(d)     Separation/segregation of pairs/homologous chromosomes;
Accept: result of meiosis I / result of division of cell B
Accept: pulled to opposite poles for ‘separation’
Ignore ref to chromatids
1
(e)     (DNA) replication taking place/not finished;
Accept: they are cells in S phase
1
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Q8.
(a)     Lowercase a in both boxes
1
(b)     Tick in box next to ‘Crossing over’;
1
(c)     32.73 / 32.7 / 32 / 33;;
Award 1 max for either
409 (409.2) for difference in volume (but incorrect number of mitochondria);
OR
Answer of 262 (261.9) (using diameter, rather than radius);
2
(d)     1.      Egg (created) has nucleus / DNA / genes of (affected) woman / mother;
Accept ref. to zygote / embryo / child for egg
Accept genetic information
Ignore references to alleles
Reject if nucleus from wrong egg / woman
2.      It has mostly / many / lots of normal mitochondria (of unaffected woman)
OR
There are few faulty mitochondria;
Reject ref. to production of healthy mitochondria as result of treatment
2
(e)     1.      Not enough / little ATP produced;
One reason asked for, so list rule applies
Ignore ref. to no ATP produced
2.      ATP provides energy for (enzyme) reactions
OR
ATP phosphorylates substrates / enzymes, so making them (more) reactive;
Accept (leads to) lower activation energy for reaction
Reject if mention energy produced
2 max
[8]
Q9.
(a)      1.      Change / mutation in base / nucleotide sequence (of DNA / gene);
Q.
Ignore: references to changing base-pairing
Accept: affect for change, if in correct context
Accept: changes triplets / codons
2.      Change in amino acid sequence / primary structure (of enzyme);
Accept: different amino acid(s) coded for
Q Reject: different amino acids produced / formed / made
3.      Change in hydrogen / ionic / disulfide bonds;
Accept: references to sulfur bonds
4.      Change in the tertiary structure / shape;
Neutral: alters 3D structure / 3D shape
5.      Change in active site;
6.      Substrate not complementary / cannot bind (to enzyme / active site) / no enzyme-substrate complexes form.
Accept: no E S complexes form
6
(b)     1.      Non-SR strain falls more / SR strain falls less / up to 10(μg / cm−3);
Must include 10 but only required once in either MP1 or MP2
Ignore: units or absence of
This must be a comparative statement
2.      Above 10(μg / cm^−3), SR strain levels out / off and non-SR strain continues to decrease;
3.      Greater difference between strains with increasing concentration of antibiotic.
This must be a comparative statement
2 max
(c)     1.      Division stopped (of both strains by scientist);
Reject: references to mitosis stopping
2.      SR strain still more resistant / fewer die / none die (at higher concentrations of antibiotic).
Accept: SR strain and non-SR strain would be similar if resistance is due to only stopping division
Need some comparison with non-SR
2
(d)     1.      Make a competitive / non-competitive inhibitor;
Mark in pairs
either MP1 and MP2 OR MP3 and MP4
2.      Competitive competes with / blocks active site / non-competitive inhibitor affects / changes active site;
Do not mix and match
OR
3.      (Make a drug) that inhibits / denatures / destroys enzyme / stringent response;
Accept: drug that ‘knocks out’ / destroys enzyme
4.      Give at the same time as / before an antibiotic.
2 max
(e)     (SR strain)
1.      Fewer free radicals (than non-SR);
Note: has to be comparative statement
2.      Produces more catalase (than non-SR);
Accept converse statements for non-SR.
3.      Catalase (might be) linked to production of fewer free radicals / breaking down / removing free radicals.
Accept: hydrolysis of radicals by catalase.
3
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