Hodder Education



WORKBOOK ANSWERS

AQA A-level Biology

Sections 7 and 8

Genetics, populations, evolution and ecosystems ? The control of gene expression

This Answers document provides suggestions for some of the possible answers that might be given for the questions asked in the workbook. They are not exhaustive and other answers may be acceptable, but they are intended as a guide to give teachers and students feedback.

Section 7

Genetics, populations, evolution and ecosystems

Inheritance

1 Genotype is the genetic make-up of an individual; phenotype is the expression of the genotype and its interaction with the environment.

2 A gene is a sequence of DNA that codes for a polypeptide/found at a specific locus on a chromosome; alleles are different variations of the same gene.

3

|Term |Definition |

|Homozygous |An organism that has both alleles of a particular gene the same, e.g. TT, tt |

|Heterozygous |An organism that has two different alleles of a particular gene, e.g. Tt |

|Dominant |An allele that is expressed in the phenotype even when only one copy is present |

|Recessive |An allele that is only expressed in the phenotype when two copies are present |

4

|Parent phenotypes |Manx |Manx |

|Parent genotypes |Tt |Tt |

|Gametes |T t |× |T t |

| |T |t |

|T |TT |Tt |

|t |Tt |tt |

|Offspring genotypes |TT |Tt |tt |

| |(do not survive) | | |

|Offspring phenotypes |Manx |: |Normal tail |

|Ratio |2 |: |1 |

5

|Parent phenotypes |Normal |Normal |

|Parent genotypes |Ff |Ff |

|Gametes |F f |× |F f |

(We know that the parents are heterozygous as they already have a baby with CF.)

| |F |f |

|F |FF |Ff |

|f |Ff |ff |

|Offspring genotypes |FF |Ff |ff |

|Offspring phenotypes |Normal |: |CF |

|Ratio |3 |: |1 |

Probability of next child having CF = 1 in 4, 0.25 or 25%

6a Recessive (no mark unless evidence given) because 10/12 has the condition but neither parent/7 and 8 do not have the condition.

6b 3 = Aa/heterozygous

7 = Aa/heterozygous

11 = Aa/heterozygous or AA/homozygous dominant

7

|Parent phenotypes |Roan, hornless |Roan, hornless |

|Parent genotypes |CRCWHh |CRCWHh |

|Gametes |CRH CRh CWH CWh |× |CRH CRh CWH CWh |

| |CRH |CRh |CWH |CWh |

|CRH |CRCRHH |CRCRHh |CRCWHH |CRCWHh |

|CRh |CRCRHh |CR CRhh |CRCWHh |CRCWhh |

|CWH |CRCWHH |CRCWHh |CWCWHH |CWCWHh |

|CWh |CRCWHh |CRCWhh |CWCWHh |CWCWhh |

|Phenotype |Red, hornless |Red, horned |

|Parent genotypes |IAIODd |IBIOdd |

|Gametes |IAD IAd IOD IOd |× |IBd IBd IOd IOd |

(We know that both parents are heterozygous for ABO blood group because they already have a child with group O. We also know that the father is heterozygous for the Rhesus group because their child is Rhesus negative.)

| |IAD |IAd |IOD |IOd |

|IBd |IAIBDd |IAIBdd |IBIODd |IBIOdd |

|IBd |IAIBDd |IAIBdd |IBIODd |IBIOdd |

|IOd |IAIODd |IAIOdd |IOIODd |IOIOdd |

|IOd |IAIODd |IAIOdd |IOIODd |IOIOdd |

|Phenotype |A Rh+ |A Rh– |

|Parent genotypes |XGXB |XBY |

|Gametes |XG XB |× |XB Y |

| |XG |XB |

|XB |XGXB |XBXB |

|Y |XGY |XBY |

|Offspring |XGXB |

|genotypes | |

|AaBb |Purple |

|AAbb |White |

|aaBB |White |

11

|Parent phenotypes |Black |Black |

|Parent genotypes |BbEe |BbEe |

|Gametes |BE Be bE be |× |BE Be bE be |

| |BE |Be |bE |be | |

|Ratio |9 |: |

|Tall, green seeds |80 |90 |

|Tall, yellow seeds |40 |30 |

|Short, green seeds |32 |30 |

|Short, yellow seeds |8 |10 |

13b The probability of getting these results by chance is less than 0.05; therefore the null hypothesis is accepted/the results are consistent with a 9:3:3:1 ratio.

Exam-style questions

1a Epistasis

1b

|Genotype |Phenotype |

|AabbCC |Cinnabar |

|aaBBCC |Vermillion |

|AABbCc |Red |

1c

|Parent phenotypes |Normal male |Normal female |

|Parent genotypes |XGY |XGXg |

|Gametes |XG Y |× |XG Xg |

| |XG |Y |

|XG |XG XG |XG Y |

|Xg |XG Xg |Xg Y |

Ratio of phenotypes is 2 normal females : 1 normal male : 1 goggle-eyed male.

2a Two parents with achondroplasia (1 and 2) have a child (7) without achondroplasia.

2b

|Parent phenotypes |Red eyes |Red eyes |

|Parent genotypes |AaBb |AaBb |

|Gametes |AB aB Ab ab |× |AB aB Ab ab |

| |AB |aB |Ab |ab |

|AB |AABB |AaBB |AABb |AaBb |

Ratio of phenotypes is 9 red : 3 brown : 3 orange : 1 white.

Populations

1 Homozygous recessives = q2 = 1/10 000 = 0.0001

q = (0.0001 = 0.01

p + q = 1

Therefore, p = 1 – 0.01 = 0.99

Frequency of heterozygotes = 2pq = 2 × 0.99 × 0.01 = 0.0198 (or 1.98% or 198 per

10 000)

2 The population is large; there is no immigration or emigration; there is random mating between members of the population; there is no mutation; there is no selection/each genotype has an equal chance of reproductive success.

Maximum 4 marks

3 If 64% have unattached earlobes, 36% must have attached earlobes.

Therefore, q2 = 0.36

q = (0.36 = 0.6

p = 1 – q = 1 – 0.6 = 0.4

2pq = 2 × 0.6 × 0.4 = 0.48

Therefore, 48% are heterozygous.

Evolution may lead to speciation

1 Random segregation/independent assortment; chiasmata formation

2 Mutation; random fertilisation

3a Elephants with tusks are killed/tuskless elephants survive; reproduce and pass on alleles; frequency of tuskless allele increases/allele for tusks decreases.

3b Directional selection, because one extreme (of phenotype) is being selected for/tuskless selected for/tusks selected against.

4 Genetic bottleneck has occurred/description of this, e.g. great reduction in gene pool; little variation in population/have similar alleles; few with resistance to disease/most susceptible.

5 Both spotted and black have selective advantage/described; intermediate not advantage/disadvantage in both areas; intermediate less likely to survive and pass on allele.

6 Small clutch size means few offspring survive; large clutch size means many offspring will not get enough food, therefore few survive; intermediate means enough food available and more survive; pass on their alleles for intermediate clutch size.

Maximum 3 marks

7 Changes are due to chance; the smaller the population, the more susceptible it is to small (random) changes; can result in large changes in allele frequency in small populations.

Maximum 2 marks

8a Interbreed; will not produce fertile offspring (if separate species).

8b Geographical isolation; each population shows variation; different conditions in each area means different phenotypes are at an advantage/selected for; those with advantageous phenotype survive longer; reproduce and pass on alleles; reproductive isolation.

Maximum 5 marks

9a Reproductive isolation/would need to lose ability to interbreed; to produce fertile offspring.

9b Sympatric; geographical isolation has not occurred/both types of fly in same area.

Exam-style questions

1a Animals with lethal dwarfism/disadvantageous allele do not survive; do not reproduce/have fewer offspring; allele not passed on/reduces in frequency.

1b Genetic bottleneck/small number of condors left from original population; by chance, just one or two individuals carried this allele; idea that this is a high proportion of the condor gene pool.

2a q2 = 102/200 = 0.51

q = √0.51 = 0.71

p + q = 1, so p = 1 – q

p = 1 – 0.71 = 0.29

Therefore, frequency of M allele = 0.71 and frequency of N allele = 0.29.

2b German Baptists migrating into USA were a small sample of the German population/founder population; therefore allele frequencies may not have been typical of whole German population; marry within their community so there has been little gene flow.

Populations in ecosystems

1 1C, 2D, 3B, 4A

2 The maximum size of a population; that can be supported in a particular ecosystem.

3a Good food supply; no predators

3b Intraspecific; competing for food/limited food supply

3c When moose population grows, more food for wolves; wolf population increases, reducing moose population; less food for wolves so wolf population falls; fewer moose eaten so moose population increases; both moose and wolf populations increase and decrease around a certain population size/populations stay relatively stable.

Maximum 4 marks

3d The stable population size is the carrying capacity; when numbers increase above the carrying capacity, predation/lack of food/biotic factors bring about a reduction in population size.

4 Pioneer species colonise the area; seeds brought in by wind/dropped by animals/survive after fire; shrubs/woody plants start to grow; compete with pioneer species/herbaceous plants for light; eventually trees colonise, which outcompete lower-growing vegetation for light, etc.; climax community remains stable unless environment changes; over time, more species present/biodiversity.

Maximum 6 marks

5a Succession would occur; woody plants would grow and outcompete grassland plants; eventually climax community of forest would grow.

5b Sheep eat woody plants/prevent them growing; allows species to grow/preserves habitat for some species; idea that maintaining biodiversity is important, e.g. aesthetic value/may have valuable properties.

6 Ploughing breaks up roots, preventing woody species growing; animals graze field so taller/woody species cannot grow; crops grown and removed so other species cannot establish.

Maximum 2 marks

7a No predators; plenty of food

7b

i Around 1700–1800

ii Intraspecific competition; for food

8 60/x = 30/120 where x is population size

Therefore, 60 = 30x/120

30x = 60 × 120

x = (60 × 120)/30 = 240

2 marks for correct answer, 1 mark for correct working with arithmetic error

9 No migration; no births/deaths; marked individuals are not at increased risk of predation/not harmed; population mixes freely.

Maximum 2 marks

10 Marking does not harm mice/described; marking does not increase predation risk; should not get removed/wash off.

Maximum 2 marks

11 For allowing licensed hunting: money from hunting improves lives of local people; loss of a few male lions will not harm lion population in general; population has reached carrying capacity/lions might die anyway if hunting did not occur.

Against allowing licensed hunting: lions should be allowed to live naturally/it is wrong to interfere with populations; hunting for trophies is a behaviour we should discourage; when hunting was stopped, lion population grew.

Maximum 4 marks

Required practical

1a Kite graph/bar chart for species; line graph/bar chart for abiotic factors; distance along dunes on x axis; idea of all graphs/data on one piece of paper for comparison.

1b (original mass – final mass)/original mass × 100%

1c Elymus

1d Grows taller so outcompetes Elymus for light; branching roots allow it to gain water from larger area/stabilise sand so it can grow better/can withstand sand being blown over it.

1e Vegetation shows changes; as you go from front of dunes to back.

1f Use random numbers to generate x,y coordinates; place quadrat on ground at the place indicated by coordinates.

Exam-style questions

1a

i Very harsh conditions/no soil/nutrients.

ii When plants die they provide humus/nutrients; enables other plants to grow.

iii More food available for animals; provides habitats.

Section 8

The control of gene expression

Alteration of the sequence of bases in DNA can alter the structure of proteins

1 Any suitable examples, e.g. ultraviolet radiation; ionising radiation; chemicals such as benzene.

Maximum 2 marks

2

|Base sequence of DNA |Type of mutation |

|ATTGCCGAT |Inversion |

|ATTCCGAT |Deletion |

|GCGATATTC |Translocation |

|ATCTCGCGAT |Addition |

|ATACGCGAT |Substitution |

|ATTCGCGCGAT |Duplication (accept addition) |

3 Different amino acid coded for; gives altered protein/non-functional protein; amino acid may not change; TAG and TAT may be alternative codons for same amino acid/DNA code is degenerate.

Maximum 3 marks

4 Frame-shift; affects all codons from point of mutation; different amino acids coded for; non-functional/much changed protein.

Maximum 3 marks

Gene expression is controlled by a number of features

1 Can become any kind of specialised cell; very early embryo.

2 Can become most types of specialised cell; found in embryos.

3 Multipotent can form many kinds of specialised cells; unipotent can form only one kind of specialised cell.

4 Adult unipotent cell; induced to become pluripotent; by using transcription factors.

5a Smaller; able to divide.

5b They could be injected into damaged tissue where they would divide; become specialised cardiac muscle cells.

6a Using transcription factors.

6b These disorders affect brain cells so the right type of cell is needed; cannot test drugs on or investigate living people for ethical reasons; can obtain large number of (identical) cells for investigation; can compare these with normal brain cells.

Maximum 2 marks

7 Oestrogen binds to receptor in cells; forms oestrogen–receptor complex; complex acts as transcriptional factor; binds to promoter regions of genes; stimulates cell division.

Maximum 4 marks

8 Heritable changes in gene function; that do not involve changes in DNA base sequence.

9a Increased methylation of DNA inhibits transcription.

9b Decreased acetylation of histone proteins inhibits transcription.

10 Methyl groups removed from DNA over a lifetime; environmental factors influence DNA methylation.

11a Methylation increases with age; more genes ‘switched off’ as people age.

11b Have same DNA base sequence/alleles; same age and sex; differences can be due to environmental factors only.

12 C, A, B, E, D

2 marks for fully correct, 1 mark for single error, no marks for more than 2 errors

Gene expression and cancer

1a Regulate cell division by coding for growth factors; so cells only divide when necessary/for growth and repair.

1b Slow down cell division; cause cells with damaged DNA to be destroyed.

2 Mutated proto-oncogenes become oncogenes; code for mutated receptor proteins or excessive growth factors; triggers cell division too frequently; mutated tumour suppressor genes allow cells with damaged DNA to survive.

Maximum 3 marks

3 Benign tumours remain in one place; malignant tumours invade adjacent tissues; cells break off and travel round body; start secondary tumours/metastases elsewhere.

4 Inactivates tumour suppressor genes; allows cells with abnormal DNA to survive.

5a The more meat consumed per person per day, the greater the incidence of colon cancer in women/positively correlated; correlation is not perfect/data from graph, e.g. two countries with similar meat consumption but different rates of cancer.

5b There is a positive correlation between meat consumption and colon cancer; correlation does not prove cause and effect; do not know type of meat eaten/other foods eaten; only measures colon cancer in women; cannot be sure that the people who ate most meat were those who developed cancer.

Maximum 4 marks

6 Tamoxifen stops oestrogen binding to its receptors; therefore oestrogen–receptor complexes not formed; DNA transcription not stimulated.

Exam-style questions

1a Slow down cell division; cause cells with damaged DNA to be destroyed.

1b Stops cells with damaged DNA from dividing; methyl groups bind to DNA; inactivate TCF21.

Maximum 2 marks

1c This is likely to be before the tumour has spread/metastasised; therefore all cells can be removed.

1d Tumour suppressor genes when active slow cell division; cause cells with damaged DNA to die; cancer cells have damaged DNA; TCF21 could trigger cancer cell death.

Maximum 3 marks

Using genome projects

1 Genome is sequence of bases in DNA of an organism; proteome is the proteins coded for by the DNA.

2 From the sequence of bases we can work out the amino acids/proteins coded for; three bases code for one amino acid; gives primary sequence of proteins; most of the DNA codes for proteins.

Maximum 3 marks

3 Non-coding DNA present in higher organism; e.g. introns/regulatory genes/promoter sequences.

4a Large molecule/protein on outside of parasite; stimulates immune response.

4b Antigens of parasite stimulate antibody production in host; antibodies in host will not bind to new antigens; new antigens different shape; therefore parasite survives/reproduces.

4c Needs RH5 to bind to basigin; mutated gene for RH5 would produce non-functional/different shaped protein; parasite would not enter red blood cells; parasite would not survive and reproduce (passing on alleles).

4d Present in all malaria parasites; gene unlikely to mutate; therefore vaccine would be effective for long time/for all parasites.

Gene technologies

1

| Enzyme |Function |

|Restriction enzyme/endonuclease |Cuts DNA at a specific base sequence |

|Reverse transcriptase |Makes a single-stranded piece of complementary DNA from a messenger RNA template |

|(DNA) ligase |Joins two pieces of DNA together |

|DNA polymerase |Joins DNA nucleotides together to make a piece of DNA |

2 DNA has same structure in all living organisms; codons code for same amino acids.

3 Sequence of amino acids in insulin/primary structure known; use genetic code table to work out codons/triplets needed in DNA; enter information into ‘gene machine’ to synthesise DNA.

4 Use reverse transcriptase to make (single-stranded) DNA from mRNA; DNA polymerase to make complementary strand/make it double stranded.

5 Overhanging ends/staggered cut/single strand of DNA; bind to complementary sticky ends; allow new DNA to bind to vector/host DNA.

6

[pic]

7 Gene for identifiable/visible feature, e.g., antibiotic resistance, fluorescence; transferred at same time as target gene; allows recombinant cells to be identified.

8 Means of transferring new gene into host cell; e.g., virus/plasmid/gene gun (any two).

9 Gene inserted into a plasmid; using restriction enzyme/ligase; inserted into bacteria; bacteria reproduce making a copy of the plasmid/gene every time they divide.

Maximum 3 marks

10 New nucleotides; DNA polymerase enzyme; primers.

11 Short, single-stranded piece of DNA; complementary to start/end of piece of DNA to be copied; one primer needed for each end.

Maximum 2 marks

12a Stage B: hydrogen bonds break; DNA becomes single-stranded.

12b Stage C: primers attach; by complementary base pairing.

12c Stage A: new nucleotides attach by complementary base pairing; joined to make complementary DNA strand using DNA polymerase.

13 Inserting a new gene into a host cell; to treat a genetic condition.

14a Vector/to insert new gene/gene for ADA.

14b In some children the virus only modified a small proportion of cells; so immune system was only partly functional/immune system still non-functional.

14c

i New gene inserted into tumour suppressor gene/proto-oncogene; disrupted gene; caused cells to divide uncontrollably/cells with faulty DNA to divide.

ii Yes: Most of the children were cured/only a few developed leukaemia; SCID is such a serious condition that it is worth risking cancer; helps to increase medical knowledge.

No: Unethical to use a treatment that can kill some children; children unable to give informed consent.

Maximum 4 marks

15 Single-stranded piece of DNA; complementary to (part of) a gene.

16 So it can be located; indicates whether a gene is present; DNA is invisible on gel/membrane.

Maximum 2 marks

17a DNA is negatively charged; because of phosphate groups.

17b The band nearest the positive electrode/band that has travelled furthest.

[pic]

17c So that the probe can bind (by complementary base pairing).

18 The likelihood of developing breast cancer/it is not certain she will develop breast cancer; effect on her confidence/self esteem of removing breast tissue at such a young age; what effect reconstructive surgery may have on her; likelihood of successful treatment of breast cancer if she does develop it later.

19 Allows scientists to find the exact mutation that has occurred; may be able to give patient a drug that works specifically on the mutation/effects that the mutation causes.

20 Variable number tandem repeats; short sequences of DNA that are repeated several/many times; between genes/non-coding DNA.

Maximum 2 marks

21 The number of repeats on one chromosome was inherited from the mother; the other homologous chromosome was inherited from the father (therefore different number of VNTRs).

22 All the bands in the child’s DNA fingerprint that do not come from the mother; must come from the father; unless all the non-maternal bands match his DNA, he cannot be the father.

23 DNA sample cut into fragments using restriction enzyme; separated by gel electrophoresis; DNA transferred to a membrane; made single-stranded (using alkali); radioactive probe applied; washed to remove excess/unbound probe; use of photographic/X-ray film to highlight/visualise bands.

Maximum 6 marks

24 Carry out genetic fingerprinting on all the chimpanzees; the fewer bands that two chimps have in common the less related they are; these will be less genetically similar/more genetically varied.

25 Cut DNA using restriction enzyme; carry out gel electrophoresis; make DNA single-stranded; apply probe that is complementary to part of CF allele; wash off excess/unbound probe; if probe binds the woman carries the CF allele.

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