Download AQA Biology: Genetics, populations, evolution

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
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Hodder Education
1
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
Manx
:
Normal tail
2
:
1
(do not survive)
Offspring phenotypes
Ratio
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
3
:
1
Ratio
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
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7
Parent phenotypes
Roan, hornless
Roan, hornless
Parent genotypes
CRCWHh
CRCWHh
×
C RH C Rh C W H C W h
Gametes
CRH CRh CWH CWh
C RH
C Rh
C WH
C Wh
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
Roan,
hornless
Roan,
horned
White;
hornless
White,
horned
Ratio
3
1
6
2
3
1
8
Parent phenotypes
A Rhesus positive
A Rhesus negative
Parent genotypes
IAIODd
IBIOdd
×
IAD IAd IOD IOd
Gametes
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–
B Rh+
B Rh–
AB Rh+
AB Rh–
O Rh+
O Rh–
Ratio
1
1
1
1
1
1
1
1
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Hodder Education
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9
Parent phenotypes
Tortoiseshell female
Black male
Parent genotypes
XGXB
X BY
XG
Gametes
×
XB
XB
Y
XG
XB
XB
XGXB
X BX B
Y
XGY
X BY
Offspring genotypes
XGXB
Offspring phenotypes
Tortoiseshell
female
:
Black
female
:
Ginger male
:
Black
male
1
:
1
:
1
:
1
Ratio
10a
XBXB
X GY
XBY
No allele A present, so enzyme A not made/non-functional enzyme A.
Colourless precursor 2 not made; so no substrate for enzyme B.
10b
Genotype
Phenotype
AaBb
Purple
AAbb
White
aaBB
White
11
Parent phenotypes
Black
Black
Parent genotypes
BbEe
BbEe
BE
Gametes
Be
bE
be
×
BE
Be
BE
Be
bE
be
BE
BBEE
BBEe
BbEE
BbEe
Be
BBEe
BBee
BbEe
Bbee
bE
BbEE
BbEe
bbEE
bbEe
be
BbEe
Bbee
bbEe
bbee
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bE
be
Hodder Education
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Offspring genotypes
9 B_E_
3 bbE_
Offspring phenotypes
Black
Brown
:
3
:
Ratio
12a
9
:
3 B_ee
1 bbee
Golden
4
The phenotypes will be present in the expected 9:3:3:1 ratio.
12b
i
3
ii
The probability of getting these results by chance is less than 0.05/5%; therefore the
null hypothesis must be rejected.
12c
The genes for flower colour and for pollen shape are on the same chromosome show
autosomal linkage; therefore these alleles tend to be inherited together; recombinants
only occur when crossing over occurs between the two loci.
13a
13b
Phenotype
Number
Expected number
Tall, green seeds
80
90
Tall, yellow seeds
40
30
Short, green seeds
32
30
Short, yellow seeds
8
10
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
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1c
Parent phenotypes
Normal male
Normal female
Parent genotypes
XGY
XGXg
XG
Gametes
×
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
×
AB aB Ab ab
Gametes
AB aB Ab ab
AB
aB
Ab
ab
AB
AABB
AaBB
AABb
AaBb
aB
AaBB
aaBB
AaBb
aaBb
Ab
AABb
AaBb
AAbb
Aabb
ab
AaBb
aaBb
Aabb
aabb
Ratio of phenotypes is 9 red : 3 brown : 3 orange : 1 white.
© Pauline Lowrie 2016
Hodder Education
6
Populations
Homozygous recessives = q2 = 1/10 000 = 0.0001
1
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.
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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.
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Hodder Education
8
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 lowergrowing 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
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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.
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Hodder Education
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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
3
Base sequence of DNA
Type of mutation
ATTGCCGAT
Inversion
ATTCCGAT
Deletion
GCGATATTC
Translocation
ATCTCGCGAT
Addition
ATACGCGAT
Substitution
ATTCGCGCGAT
Duplication (accept addition)
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
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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
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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
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Hodder Education
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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 nonfunctional/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.
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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
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.
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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
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17a
DNA is negatively charged; because of phosphate groups.
17b
The band nearest the positive electrode/band that has travelled furthest.
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 singlestranded; 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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Hodder Education
17