Genetics, Evolution and Ecosystems Exam Questions

Question 1

Many industrial processes use immobilised enzymes.
Which of the options is not an advantage of using immobilised enzymes rather than free
enzymes?
A Enzymes can be reused.
B Enzymes remain active over a wider range of temperatures.
C Set-up costs are low.
D The product is not contaminated by enzymes.

Answer 1

C

Question 2

Nitrobacter is a bacterium that is involved in recycling nitrogen in an ecosystem.
Which nitrogen cycle reaction is carried out by Nitrobacter?
A 2NO2− + O2 → 2NO3−
B N2 + 8H+ + 8e− → 2NH3 + H2
C NH2CONH2 + H2O → 2NH3 + CO2
D NO3− → NO2− → N2O → N2

Answer 2

A

Question 3

Some students incubated plasmid DNA with a restriction enzyme.
After 24 h they used gel electrophoresis to analyse the products of the incubation.
Which option shows the correct procedure for gel electrophoresis?
A Load the sample onto agarose gel → apply voltage for a set time → photograph the gel
B Load the sample onto agarose gel → apply voltage for a set time → stain the gel →
photograph the gel
C Photograph the agarose gel → load the sample onto gel → apply voltage for a set time →
stain the gel
D Stain the agarose gel → apply voltage for a set time → load the sample onto agarose gel →
photograph the gel

Answer 3

B

Question 4

Evolutionary relationships can be determined by comparing certain biological molecules between
species.
Which option is commonly used to determine evolutionary relationships?
A The amino acid sequence of collagen
B The amino acid sequence of messenger RNA
C The base sequence of cytochrome c
D The base sequence of ribosomal RNA

Answer 4

D

Question 5

Competition is an important factor in determining population size.
Which statement about competition is not correct?
A Competition between two species can result in the extinction of the less well-adapted
species.
B Competition occurs between individuals of the same species.
C Predators only compete within their own species.
D Species with overlapping niches will compete with one another.

Answer 5

C

Question 6

When growing bacteria in culture, it is important that aseptic techniques are used.
(i) State why it is important that the technique used for culturing microorganisms be
aseptic. (1 mark)

Answer 6

• prevent contamination by unwanted microorganisms
• to prevent entry/growth of unwanted
  microorganisms

Question 7

The students prepared the culture by adding a suspension of bacteria to a flask
containing nutrient broth.
List two precautions that should be taken when preparing a bacterial culture in order to
ensure that the procedure is aseptic. (2 marks)

Answer 7

• use sterile/autoclaved flask/pipette/equipment
• stopper flask to prevent contamination
• disinfect/sterilise , surfaces
• nearby Bunsen flame to create upward air flow

Question 8

When counting the number of bacteria, the students performed serial dilutions on samples
removed from each small flask. In each serial dilution, the students removed 0.1 cm3 and
added it to 9.9 cm3 of water.
To estimate the total number of bacteria, the students used a light microscope to count the
number of bacterial cells in a 0.01 cm3 sample of the final serial dilution.
To estimate the number of viable bacteria, the students spread 0.1 cm3 of the final serial
dilution on an agar plate and counted the number of colonies that had grown after 24 h.
(i) The students shook each flask before they removed the samples for counting.
Suggest why the students shook the flasks. (1 mark)

Answer 8

• So the bacterial cells are evenly distributed

Question 9

It can be more difficult to count bacterial cells using a light microscope than it is to count
human cells.
Suggest one reason why bacterial cells are difficult to count using a light microscope.

Answer 9

• Bacterial cells are smaller than human cells.

Question 10

Some students investigated the effect of time on the growth of bacterial populations.
The students prepared a large flask of bacterial culture.
They divided this large culture into a number of smaller flasks each containing 50 cm3 of bacterial
culture.
They then incubated the smaller flasks at 20 °C for up to 48 h.
Every 4 h the students removed one of the flasks and counted the bacteria.
The students recorded the total number of bacteria and the number of viable bacteria in each
flask.
When counting the number of bacteria, the students performed serial dilutions on samples
removed from each small flask. In each serial dilution, the students removed 0.1 cm3 and
added it to 9.9 cm3 of water.
To estimate the total number of bacteria, the students used a light microscope to count the
number of bacterial cells in a 0.01 cm3 sample of the final serial dilution.
To estimate the number of viable bacteria, the students spread 0.1 cm3 of the final serial
dilution on an agar plate and counted the number of colonies that had grown after 24 h.
In one 0.01 cm3 sample the students counted 52 bacterial cells under the microscope.
Describe the calculation steps the students would then need to make to estimate the
total number of bacteria in the small flask. (3 marks)

Answer 10

• Calculate the number in 10 cm3
• multiply 52/number of bacteria in sample, by 1000
• Correct treatment of serial dilutions
  multiply by 100^n
  (where n is the number of serial
  dilutions)
• Calculate the total in 50 cm3
• multiply (answer to 1) by 5

52 x 1000 x 100^n x 5

Question 11

Explain, with reference to the graph, why the students used a logarithmic scale on the
y-axis. (log10 number of bacterial cells by time) (2 marks)

Answer 11

• differences in numbers would be too big to
  represent on paper
• two figures quoted in support (cannot do)

Question 12

Explain, with reference to the graph, the decrease in the population of viable bacteria
between 40 h and 48 h.
(initially population increases, then stabilises before decreasing) (4 marks)

Answer 12

• reproduction rate lower than death rate
• total count/dead bacteria is much higher than
  viable bacteria
• use of figures with units (cannot do)
• increased/high level of , waste products (eg. carbon dioxide)
• less oxygen/fewer (named) nutrients
• increased intraspecific competition
• dead cells/lack of space reduces surface
  area for access to nutrients/oxygen

Question 13

Microorganisms can be used to produce a variety of food products.
(a) Microorganisms have simple nutrient requirements, which helps to reduce production costs.
List two other advantages of using microorganisms in food production. (2 marks)

Answer 13

• no welfare/ethical issues
• can be genetically modified relatively easily
• rapid growth/production
• can be easily changed to
  meet demand
• non-seasonal/year-round production
• take up little space
• low costs because work at low temperatures

Question 14

i) Use the diagram to name the type of fermentation process used for mycoprotein
production. (Fermenter is going round in a loop like an assembly line) (1 mark)
ii) Suggest and explain why a cooling system is necessary. (2 marks)
iii) The air inlet provides the fungus with oxygen for respiration, and ammonia.
Suggest and explain why the fungus is provided with ammonia. (2 marks)

Answer 14

i) Name: continuous
Justification: there is an outlet for continuous collection of a product
ii) - temperature affects , rate of growth/enzyme activity
- fungal metabolic reactions generate heat
- to inhibit growth of pathogenic bacteria
iii) - a source of nitrogen
- for producing amino acids/polypeptides/proteins

Question 15

Body plan is important in multicellular organisms.
(a) Complete the following sentences about control of body plan using the most appropriate
terms.
Body plan is under genetic and ……………………………………….. control. Internal and
external……………………………………….. can influence the expression of genes that regulate
the cell cycle. Such genes can promote or inhibit programmed cell death, known as
……………………………………….. . During programmed cell death ………………………………………..
digest the cell contents and the products are removed by ……………………………………….. so
that they do not damage the surrounding tissues.

Answer 15

• Environmental
• Stimuli
• Apoptosis
• Enzymes
• Phagocytes/phagocytosis

Question 16

State the name of the type of gene responsible for controlling body plan in multicellular
organisms. (1 mark)

Answer 16

Homeobox genes

Question 17

Which of the following processes is important in determining the overall body plan of an organism?
A endocytosis
B exocytosis
C meiosis
D mitosis

Answer 17

D

Question 18

Genes are not expressed during cell division because chromosomes are more tightly wound
around histone proteins than during interphase.
Which of the following shows the level at which gene expression is being controlled when DNA is
more tightly wound during cell division?
A post-transcriptional
B post-translational
C transcriptional
D translational

Answer 18

C

Question 19

Which of the following substances is not required in DNA sequencing?
A DNA polymerase
B primers
C RNA nucleotides
D terminator bases

Answer 19

C

Question 20

DNA fragments can be separated using gel electrophoresis.
Which of the following explains how gel electrophoresis is able to separate DNA fragments?
A DNA carries a negative charge and large fragments are pulled more strongly than small
fragments towards the positive electrode.
B DNA carries a negative charge and small fragments are able to travel more quickly than large
fragments towards the positive electrode.
C DNA carries a positive charge and large fragments are pulled more strongly than small
fragments towards the negative electrode.
D DNA carries a positive charge and small fragments are able to travel more quickly than large
fragments towards the negative electrode.

Answer 20

B

Question 21

Gene sequencing has a number of uses.
Which of the following is not a use of gene sequencing?
A determining the amino acid sequence of a polypeptide
B the classification of newly-discovered organisms
C the polymerase chain reaction
D the selection of the correct vaccine in a disease outbreak

Answer 21

C

Question 22

Moving North from a large solitary tree in the school field, some students studied changes in
plant species. They laid a tape measure due North from the base of the tree trunk and dropped a
quadrat at 1 m intervals for 15 m.
Which of the following correctly describes the students’ sampling method?
A arbitrary
B random
C stratified
D systematic

Answer 22

D

Question 23

Which of the following best defines the term species richness?
A the distribution of species over an area
B the number of species in an area
C the relative abundance of each species in an area
D the relative number of individuals of a species in an area

Answer 23

B

Question 24

Biodiversity can be considered at different levels.
An area of woodland habitat has a high Simpson’s Index of Diversity.
Which of the following describes an area with a high Simpson’s Index of Diversity?
A the area has a high genetic biodiversity
B the area has a high habitat biodiversity
C the area has a high species biodiversity
D the area is high in all levels of biodiversity

Answer 24

C

Question 25

DNA profiling makes use of the fact that DNA varies between individuals.
When creating a DNA profile, DNA is first extracted from a sample of tissue.
Outline the subsequent steps involved in producing a DNA profile. (4 marks)

Answer 25

• amplify DNA fragment with PCR
• cut/digest the fragment with restriction enzymes
• separate using, gel electrophoresis
• transfer fragments to paper/a membrane
• add a radioactive/fluorescent probe
• use x-rays/UV light to view the position of DNA
  fragments

Question 26

DNA profiling can be used in cases of paternity and forensics.
State one other use of DNA profiling. (1 mark)

Answer 26

• assessing disease risk
• identifying/classifying a species

Question 27

DNA codes for proteins within the cell. Some regions of DNA are described as non-coding.
Explain why some regions of DNA can be described as ‘non-coding’. (2 marks)

Answer 27

• editing of primary mRNA (introns are removed)
• therefore codons are not present in mature mRNA
• therefore not translated
• these are regulatory sequences/genes

Question 28

Non-coding regions of DNA show more variation than coding regions. This makes
non-coding regions useful in DNA profiling.
Suggest why non-coding regions of DNA show more variation. (1 mark)

Answer 28

They are not selected against in evolution

Question 29

DNA sequencing has allowed scientists to create a strain of Escherichia coli bacteria with an
entirely artificial genome.
Complete the passage using the most appropriate words.
The creation of an organism with an artificial genome is known as …………………………………….
biology. The bacterium created has been classified in the genus …………………………………….
However, the classification of this organism is problematic because the basis of classification
is ……………………………………. .

Answer 29

Synthetic
Escherichia
Phylogeny/phylogenetics

Question 30

Scientists were able to estimate the increase in biomass in heather plants in one year.
Suggest how the increase in biomass over time in a plant such as heather could be
determined experimentally. (1 mark)

Answer 30

Measuring changes in dry mass over time

Question 31

Sunlight that can potentially be used in photosynthesis by green plants such as heather is
called photosynthetically active radiation (PAR).
Some of the solar radiation that falls on the leaves of plants is reflected. Some solar
radiation is of a wavelength that is not suitable for use in photosynthesis.
List one other reason why much of the PAR is not used by the plant in the production of
biomass. (1 mark)

Answer 31

• does not have chloroplasts/parts that photosynthesise
• the rate of photosynthesis is limited by another factor

Question 32

i) Suggest and explain why the percentage of biomass transferred between heather and
grouse is smaller than the percentage of biomass transferred between grouse and hen
harrier. (heather –> grouse –> hen harrier) (2 marks)

ii) Since 2004, the population of red grouse in the UK has been relatively stable and it is not
thought that the population has been affected by changes in climate.
Suggest an explanation for the decrease in hen harrier numbers since 2004 (1 mark)

Answer 32

i) - heather is less easily digested
- because of cellulose (cell walls)/lignin

ii)- deliberate killing to maintain grouse numbers - pollution
- pesticides
- disease
- loss of another food
source
- competition from new predator

Question 33

Heather moorland in the UK is managed in an attempt to conserve the habitat. One of the
procedures carried out as part of this management is regular burning of the moorland. Small
areas are burnt in the winter and new shoots begin to grow the following year. This helps
to maintain a variety of heights of heather plants, and prevents the growth of other larger
species of plant.

(i) State why the management of heather moorland is known as in situ conservation. (1 mark)

(ii) Apart from regular burning, suggest another procedure that could be carried out to
conserve the heather moorland habitat (1 mark)

Answer 33

i) - species are conserved in their natural habitat

ii) - controlled grazing
- monitoring of populations
- restricting human access
- remove invasive species

Question 34

Heather moorland in the UK is managed in an attempt to conserve the habitat. One of the
procedures carried out as part of this management is regular burning of the moorland. Small
areas are burnt in the winter and new shoots begin to grow the following year. This helps
to maintain a variety of heights of heather plants, and prevents the growth of other larger
species of plant.

In a newly-available area of land, the communities change over time. The process of change
is known as succession.

Outline the process of primary succession and explain why heather moorland is an example
of deflected succession. (6 marks)

Answer 34

• Pioneer community:
• begins with bare rock
• arrival as seeds or spores
• pioneer species have certain adaptations
  e.g. nitrogen-fixation
• Intermediate community:
• herb species, including grasses
• followed by shrubs and trees
• Climax community:
• dominance by a few tree species
• little change over time
• General principles:
• seral stages
• community and decomposition changes
  composition of soil
• increased organic nitrate or water content
• Heather moorland is
  deflected succession
  because a climax community is prevented from developing.
• As a result: plagioclimax, heather is a shrub

Question 35

The inheritance of some alleles depends on the sex of the individual.
In birds, sex is determined by a combination of Z and W chromosomes.
Male birds have two Z chromosomes and female birds have one Z chromosome and one W
chromosome.
(i) The chromosomes used to determine sex inheritance are given different letters in birds
and mammals.
Identify one other way in which the sex determination in birds is different from sex
determination in mammals. (1 mark)

Answer 35

• Female birds have two different sex chromosomes.

Question 36

A pigeon is a bird. The colour of pigeon feathers is determined by a single gene on the Z
chromosome.
The feather colour gene has three alleles.
ZA = red
ZB = blue
Zb = brown
ZA is dominant to ZB and Zb
ZB is dominant to Zb
The W chromosome contains no gene for feather colour.
A pigeon with the genotype ZA Zb was crossed with a pigeon with genotype ZBW.
Complete the answer lines below to show this genetic cross.
Parent genotypes: ZA Zb × ZBW
Parent phenotypes:
Gametes:
Offspring genotypes:
Offspring phenotypes:

Answer 36

• Parents’ phenotype: red male, blue female
• Gametes: ZA, Zb ZB, W
• Offspring genotypes: ZAZB, ZBZb, ZAW, ZbW
• Offspring phenotypes:
  red male, blue male, red female, brown female

Question 37

Name the type of speciation that occurs when two populations live in the same location. (1 mark)

Answer 37

Sympatric speciation

Question 38

Suggest how G. fortis with large beaks could become reproductively isolated from
G. fortis with small beaks despite living in the same location. (1 mark)

Answer 38

• individuals choose to mate only with other
  individuals with similar sized beaks

Question 39

Comparing anatomy between species such as beak size in finches can be used to
provide evidence to support the theory of evolution by natural selection.
Describe how DNA can be used to provide evidence to support the theory of evolution by
natural selection. (2 marks)

Answer 39

• DNA is found in all organisms
• some sequences are highly conserved
• comparison of DNA between species
• similar base sequence indicates recent common
  ancestor

Question 40

Alfred Russel Wallace is another important figure in the understanding of evolution.
Outline the way in which Wallace contributed to the acceptance of Darwin’s theory of natural
selection by the wider scientific community. (2 marks)

Answer 40

• similar ideas to Darwin
• arrived at his ideas independently
• presented/published a paper, together they had an increased weight of evidence

Question 45

Strawberry plants produce clones using runners. This is an example of natural cloning.
State one other method of natural cloning in plants. (1 mark)

Answer 45

• suckers
• stolons
• tubers
• rhizomes
• bulbs

Question 45

Tissue culture greatly increases the number of cloned plants that can be produced from a
single parent plant.
Outline how it is possible to produce many clones from a single original parent plant. (2 marks)

Answer 45

• take many explants from the original parent plant
• calluses subdivided
• meristems from plantlets can be subdivided

Question 45

Name the type of plant tissue from which natural clones are produced. (1 mark)

Answer 45

Meristematic

Question 46

Pineapples are plants that can be cloned by tissue culture.
Plant hormones are used during the tissue culture process. One of these plant hormones is
known as BAP.
The table below shows the effect of the concentration of BAP on the length of pineapple
shoots.
The valid investigation that generated the results shown above featured an independent
variable, a dependent variable and several controlled variables.
State the independent and dependent variables and suggest two appropriate controlled
variables.

Answer 46

• Independent variable:
  BAP concentration
• Dependent variable:
  shoot length
• Control variables:
  time
  light intensity
  lighting regime
  temperature
  water availability
  concentration of other plant hormones
  pH of growth medium

Question 47

Scientists self-pollinated some pea plants that were heterozygous for the gene controlling height.
They expected a 3:1 ratio of tall plants to short plants in the offspring.
1046 plants grew in the next generation. 798 were tall and 248 were short.
Which of the following, A to D, is a statistical test that could be used to determine if these numbers
are significantly different from a 3:1 ratio?
A chi-squared
B Spearman’s rank
C standard deviation
D Student’s t-test

Answer 47

A

Question 48

Polar bears, Ursus maritimus, and giant pandas, Ailuropoda melanoleuca, both belong to the
family Ursidae.
Which of the following, A to D, is not true about the classification of polar bears and giant pandas?
A They each belong to a different class.
B They each belong to a different species.
C They each belong to the same order, carnivora.
D They each belong to the same phylum, chordata.

Answer 48

A

Question 49

The nitrogen cycle involves a range of reactions and microorganisms.
Which of the following processes, A to D, usually occurs under anaerobic conditions?
A conversion of amino acids to ammonium compounds
B conversion of urea to ammonium compounds
C nitrification
D nitrogen fixation

Answer 49

D

Question 50

Which of the following bacteria, A to D, convert ammonium compounds to nitrites?
A Azotobacter
B Nitrobacter
C Nitrosomonas
D Rhizobium

Answer 50

C

Question 51

The body plan of multicellular organisms is under genetic control.
(a) Complete the passage below using the most appropriate words from the list.
analogous
archaea
development
DNA
domains
homeobox
homologous
homozygous
kingdoms
operon
phyla
plant
preserved
prokaryotes
regulator
ribosomes
transcription
translation

The development of body plan in eukaryotic organisms is controlled by
……….. genes. These genes code for proteins that are able to bind
to ……. and turn specific genes on and off and are known as ………… factors. These proteins contain a sequence of base pairs
that varies little between species within the animal, …………………. or
fungus ………………..

Answer 51

homeobox
DNA
transcription
plant
kingdoms

Question 52

Investigations into the activity of genes that control body plan frequently use fruit flies and
mice.
One reason fruit flies are used is that there are fewer public concerns about the ethics of
using flies.
(i) Suggest two other reasons why fruit flies are chosen for research into genes controlling
the development of body plan. (2 marks)

Answer 52

• low cost
• rapid reproduction rate
• fruit fly genetics/development is well understood
• simple genetics/body plan
• many mutations/structures observable with light microscopes

Question 53

Investigations into the activity of genes that control body plan frequently use fruit flies and
mice.
There are some public concerns about the ethics of using mice in these investigations.
Suggest two reasons why mice are chosen as a suitable species for investigation. (2 marks)

Answer 53

• low cost/rapid reproduction rate
• genetics/development are well understood
• more similar to humans
• effects are generalisable to more than one species
• more than one species is needed to demonstrate
  conservation of base sequence

Question 54

Tannase is an enzyme produced by some microorganisms. Tannase is useful in many industrial
applications including food production.
The tannase used in food production can be free in solution or immobilised.
(a) State one method by which tannase could be immobilised. (1 mark)

b) The results suggested that immobilised tannase was more stable over a range of pH
values than free tannase.
Explain why immobilised tannase is more active at pH8 than free tannase. (2 marks)

Answer 54

a) - in a matrix
- adsorption
- membrane separation
- cross-linking/covalent bonding

b) - immobilised enzymes are less easily denatured
- their tertiary structure is supported (by the support
material)
- part of enzyme was not fully exposed to pH 8

Question 55

Immobilised enzymes are often active over a greater pH and temperature range than free
enzymes.
Using immobilised enzymes can be cheaper than using free enzymes.

(i) Suggest two reasons why using immobilised enzymes in industrial processes could be
cheaper than using free enzymes. (2 marks)

(ii) State one disadvantage of using immobilised enzymes in industrial processes.

Answer 55

i) - product not contaminated with enzyme
- extraction of product/enzyme is not needed
- you can recycle the enzyme
- process can be run over wider temperature
range
- bioreactors can be run continuously for long
periods, so less emptying/cleaning is needed

ii) - higher initial/set-up costs
- fewer exposed active sites
- immobilization method might affect shape of the active site
- leakage

Question 56

Individuals within populations vary. Much of this variation is under genetic control.
(a) Two groups of scientists were studying genetic polymorphism in fruit flies.
They extracted DNA from two different species of fruit fly, A and B.
The first group of scientists studied 26 gene loci from species A. They calculated the genetic
polymorphism of species A to be 0.35.
The second group of scientists studied 32 gene loci from species B. They found that 13 of the
gene loci were polymorphic.
(i) Calculate the proportion of genetic polymorphic gene loci of species B.

(ii) Evaluate the conclusion
that species B shows greater genetic polymorphism than
species A (3 marks)

Answer 56

i) 13/32 = 0.41

ii) Supports because:
- species B has greater genetic polymorphism than species A

Might not support because:
- polymorphisms are similar
- there is no statistical test performed
- might not have sampled same loci
- no indication of the fruit flies sample size

Question 57

Since 1940, people have believed that the ability to roll the tongue is caused by a single
gene with two alleles.
R is dominant and allows tongue-rolling.
r is recessive and does not allow tongue-rolling.
The genotype of students who can roll their tongue could be either RR or Rr.
In the results shown in the table opposite
* the total number of students who could roll their tongue = 171
* the total number of students who could not roll their tongue = 77.
The Hardy–Weinberg principle allows us to estimate the proportion of each genotype.
Use the Hardy–Weinberg principle to estimate the proportion of heterozygous individuals
in the school survey in the table.
Use the equations:
p2 + 2pq + q2 = 1
p + q = 1

Answer 57

q^2 = 77/248 = 0.31
q = √ 0.31 = 0.557
p = 1 – 0.557 = 0.443
2pq = 2 x 0.443 x 0.557 = 0.494

Question 58

The Hardy–Weinberg principle might not give an accurate estimate of the proportion of
genotypes for the results of the student’s investigation.
The population of students varies from year to year and so cannot be said to be stable.
State two other reasons why it might be inappropriate to use the Hardy–Weinberg
principle to estimate allele frequencies for the results in the table. (2 marks)

Answer 58

• population is not large enough
• population is not randomly mating