Identify the genotype of the male wire-haired dog.

Using a Punnett square, determine how a smooth-haired puppy could be produced in the offspring.

WwKk

WK Wk wK and wk for male gamete genotypes and wk for female;

wwKk shown in Punnett square and identified as smooth;

Only 40 % of candidates answered this correctly. A surprisingly high number of candidates gave a genotype with only one allele for each gene, suggesting that they were not familiar with dihybrid crosses. In many cases it was hard to distinguish between upper and lower case letters -W/w and K/k. Candidates should be encouraged to make letters used for alleles as clear as possible.

There were some complaints from teachers that epistasis is not included in the program. It was not expected that candidates would have studied ratios derived from interaction between genes and instead it was intended that they work out the pattern of inheritance themselves from the information provided. About a third of candidates successfully did this. The Punnett squares drawn up by other candidates often revealed fundamental misunderstandings of dihybrid inheritance.

Discuss the role of genes and chromosomes in determining individual and shared character features of the members of a species.

Outline the process of speciation.

Describe, using one example, how homologous structures provide evidence for evolution.

Genes
a. mutation changes genes/causes genetic differences 

b. genes can have more than one allele/multiple alleles
OR
alleles are different forms/versions of a gene 

c. different alleles «of a gene» give different characters
OR
variation in alleles between individuals 

d. eye colour/other example of «alleles of» a gene affecting a character 

e. alleles may be dominant or recessive
OR
dominant alleles determine trait even if recessive allele is present 

f. both alleles influence the characteristic with codominance
OR
reference to polygenic inheritance 

g. all members of a species are genetically similar/have shared genes
OR
certain genes expressed in all members of a species 

h. reference to epigenetics/methylation/acetylation / not all genes are expressed «in an individual» 

i. genes are inherited from parents/passed on to offspring/passed from generation to generation

Chromosomes

j. same locus/same position of genes
OR
same sequence of genes/same genes on each chromosome «in a species» 

k. same number of chromosomes «in a species»/all humans have 46 chromosomes/differences in chromosome number between species 

l. some individuals have an extra chromosome/Down syndrome/other example of aneuploidy
OR
polyploidy divides a species/creates a new species 

m. X and Y/sex chromosomes determine the sex/gender of an individual 

n. meiosis/independent assortment/fertilization/sexual reproduction give new combinations «of chromosomes/genes»

a. speciation is the splitting of a species «into two species» 

b. reproductive isolation/lack of interbreeding 

c. isolation due to geography/«reproductive» behavior/«reproductive» timing 

d. polyploidy can cause isolation 

e. gene pools separated 

f. differences in/disruptive selection cause traits/gene pools to change/diverge 

g. gradualism / speciation/changes accumulating over long periods 

h. punctuated equilibrium / speciation/changes over a short time period

a. similar structure but different function «in homologous structures» 

b. pentadactyl limbs/limb with five digits/toes / other example 

c. similar bone structure/example of similarity of bones «in pentadactyl limbs» but different uses/functions 

d. two examples of use of pentadactyl limb by a vertebrate group 

e. suggests a common ancestor «and evolutionary divergence» 

f. process called adaptive radiation

State the type of inheritance shown.

Identify the recombinants.

The chi-squared value was calculated as shown. Deduce, with reasons, whether the observed ratio differed significantly from the expected Mendelian ratio.

«gene/autosomal» linkage ✔

Reject sex linkage

grey vestigial and black normal ✔

Accept Ggvv and ggVv or alternative acceptable upper/lower case genotypes.

a. yes/observed ratio did differ significantly «from the expected Mendelian ratio»
OR
expected ratio is 1:1:1:1 / 575 of each type / 25 % of each type ✔ Correct ratio not needed in first alternative of mpa

b. 3 degrees of freedom ✔

c. critical value is 7.815 «at the 5 % level / 11.345 «at the 1 % level» ✔ Accept mpc if candidates indicate the critical value of chi squared by circling it.

d. chi-squared value «of 1002.6» exceeds the critical value ✔

Allow other levels of significance as long as the critical value is correctly stated for the chosen level.

Outline the roles of helicase and ligase in DNA replication.

Explain how natural selection can lead to speciation.

Outline the features of ecosystems that make them sustainable.

helicase:

a. unwinds/uncoils the DNA «double helix» ✔

b. breaks hydrogen bonds «between bases» ✔

c. separates the «two» strands/unzips the DNA/creates replication fork ✔

ligase:

d. seals nicks/forms a continuous «sugar-phosphate» backbone/strand ✔

e. makes sugar-phosphate bonds/covalent bonds between adjacent nucleotides ✔

f. after «RNA» primers are removed/where an «RNA» primer was replaced by DNA ✔

g. «helps to» join Okazaki fragments ✔

a. variation is required for natural selection/evolution/variation in species/populations ✔

b. mutation/meiosis/sexual reproduction is a source of variation ✔

c. competition/more offspring than the environment can support ✔

d. adaptations make individuals suited to their environment/way of life ✔

e. survival of better adapted «individuals)/survival of fittest/converse ✔

f. inheritance of traits/passing on genes of better adapted «individuals»
OR
reproduction/more reproduction of better adapted/fittest «individuals» ✔

g. speciation is formation of a new species/splitting of a species/one population becoming a separate species ✔

h. reproductive isolation of separated populations ✔

i. geographic isolation «of populations can lead to speciation» ✔

j. temporal/behavioral isolation «of populations can lead to speciation» ✔

k. disruptive selection/differences in selection «between populations can lead to speciation» ✔

l. gradual divergence of populations due to natural selection/due to differences in environment ✔

m. changes in the gene pools «of separated populations»/separation of gene pools ✔

n. interbreeding becomes impossible/no fertile offspring «so speciation has happened» ✔

a. recycling of nutrients/elements/components/materials ✔

b. carbon/nitrogen/another example of recycled nutrient/element ✔

c. decomposers/saprotrophs break down organic matter/release «inorganic» nutrients ✔

d. energy supplied by the sun
OR
energy cannot be recycled «so ongoing supply is needed»
OR
energy is lost from ecosystems as heat ✔

e. energy flow along food chains/through food web/through trophic levels ✔

f. photosynthesis/autotrophs make foods/trap energy
OR
autotrophs supply the food that supports primary consumers ✔

g. oxygen «for aerobic respiration» released by autotrophs/photosynthesis/plants ✔

h. carbon dioxide «for photosynthesis» released by respiration ✔

i. populations limited by food supply/predator-prey/interactions/competition
OR
populations regulated by negative feedback
OR
fewer/less of each successive trophic level «along the food chain»/OWTTE ✔

j. supplies of water from rainfall/precipitation/rivers/water cycle ✔

This was generally well answered, with most candidates knowing at least something of the roles of these two enzymes. Most candidates knew that ligase connects Okazaki fragments but some claimed that it creates hydrogen bonds between nucleotides on template and the new strand. Many candidates did not distinguish between unwinding of DNA and separating the strands. Two details that should be more widely known are that helicase separates the two strands of a DNA molecule by encouraging the breakage of hydrogen bonds between bases and that ligase seals nicks by making sugar phosphate bonds.

Most candidates think they understand evolution by natural selection but many do not. Here the focus was on speciation - the splitting of a species into two or more species. Often answers described the evolution of one species over time, rather than speciation itself. An idea central to natural selection that was frequently missing from an answer is adaptation or fitness. Often traits were referred to as ‘favourable’ and therefore likely to lead to survival and reproduction but there is a circularity of argument there. Survival depends on traits fitting the environment, hence being an adaptation to it. The mostly common ideas seen in answers were differential survival and reproduction, due to differences in traits. A common fault was to confuse individuals and species and to refer to a whole species surviving and reproducing more successfully than another species.

There were some vague answers to this question but also some impressive ones that explained ecological processes including nutrient recycling, energy flow and regulation of population sizes.

List two causes of variation within a gene pool.

Describe how variation contributes to evolution by natural selection.

Outline what is required for speciation to occur.

a. sexual reproduction / random fertilization / meiosis

b. mutation

No mark for crossing over unqualified.
Reject natural selection/evolution as causes of variation.

a. (variation is) different phenotypes/differences between individuals in a population/species

b. struggle/competition for survival

c. some individuals have advantageous characteristics/are better adapted/have greater chance of survival/reproduction (than others)

d. favourable alleles/genetic variations passed on/inherited by offspring/next generation

Reject “pass on phenotypes”.

a. divided species/gene pool / part of species/gene pool becomes separated / species splits into separate populations

b. reproductive isolation / lack of interbreeding 

Mark point b refers to a lack of interbreeding between separated populations in a species, not the lack of interbreeding after speciation.

c. may be due temporal/behavioural/geographic isolation

d. different natural selection/different selective pressures

Draw a Punnett square to show all the possible genotypes of Queen Victoria’s children.

Deduce the genotype of Queen Victoria’s daughter Alice.

Draw the same chromosomes to show their structure at the same stage of meiosis if there had been one chiasma between two gene loci.

State the stage of meiosis where chiasmata formation may occur.

Explain gene linkage and its effects on inheritance.

Explain the mechanism that prevents polyspermy during fertilization.

a. parental alleles shown as X^H and X^h (female) and X^H and Y (male);
b. Punnett square with genotypes of offspring shown as X^HX^H and X^HY and X^HX^h and X^hY;

X^H X^h;

all four upper arms with one A and both chromosomes with one B and one b on the lower arms;

The chromatids can be shown as single lines rather than the wider versions in the question.

prophase I;

a. located on the same chromosome;
b. genes/gene loci close together (on the same chromosome);
c. do not follow (the law of) independent assortment;
d. more chance of recombination if genes are further apart;
e. inherited together unless crossing over/recombination occurs;
f. ratios of offspring in dihybrid crosses are different from expected/non-Mendelian
OR
more offspring with parental phenotype combinations than expected;

a. cortical reaction (after first sperm nucleus enters the egg);
b. vesicles/cortical granules release their contents/enzymes (from the egg/zygote);
c. zona pellucida/glycoprotein coat/outer coat hardened / fertilization membrane formed;
d. enzymes of sperm/acrosome cannot digest (hardened coat)
OR
glycoproteins/ZP3 (in zona pellucida) altered so sperm cannot bind;

Generally this was well answered. Common errors were to omit X and or Y chromosomes, show alleles on both X and Y chromosomes, or show the male X chromosome with a recessive h allele.

Any symbols were accepted for the alleles, as long as they were shown superscript to X and the genotype was heterozygous. 70 % of candidates answered correctly.

This was mostly well answered but some candidates showed one chromatid flopped over a non-sister chromatid, like crossed legs of a seated person, without any recombination having occurred. This showed a misunderstanding of the process of crossing over and chiasma formation. There were also a wide range of other answers showing confusion about events in meiosis.

Mostly Prophase I was given as the answer but all other phases of Meiosis I were given by some candidates and others failed to specify 1st or 2nd division of meiosis.

The was some confusion between autosomal and sex linkage. Candidates tended either to have a secure understanding of the mechanisms at work in autosomal gene linkage, or none. This question and (d) correlated closely with candidates’ overall score on the paper, so distinguished particularly effectively between stronger and weaker candidates.

Answers were mixed. Most knew that there is a mechanism that prevents more sperm fusing with the egg after the first one and some gave accurate and detailed accounts of it.

Describe the process of crossing over.

Explain the reason for linked genes not following the pattern of inheritance discovered by Mendel.

a. occurs during prophase I/during meiosis

b. homologous chromosomes form bivalents/pair up

c. breakage and rejoining of chromatids

d. exchange «of DNA/alleles» between non-sister chromatids/homologous chromosomes

[Max 2 Marks]

a. «linked genes are» on the same chromosome

b. Mendel's genes were on different chromosomes

c. linked genes are inherited together

    OR

    no independent assortment

d. «linked genes» only separated by crossing over

    OR

    fewer recombinants than with unlinked genes

Reject sex-linkage

Extensive areas of the rainforest in Cambodia are being cleared for large-scale rubber plantations. Distinguish between the sustainability of natural ecosystems such as rainforests and the sustainability of areas used for agriculture.

Describe the roles of the shoot apex in the growth of plants.

Research suggests that many living plant species are polyploid. Explain how polyploidy occurs and, using a named example, how polyploidy can lead to speciation.

a. sustainable communities/ecosystems allow continued survival of organisms/OWTTE ✔

b. natural ecosystems can be sustainable over long periods of time/OWTTE ✔

c. natural ecosystems/rainforest more sustainable than agricultural areas/plantations ✔

d. diverse community/high biodiversity/higher biodiversity in natural ecosystems/rainforest
OR
less/low biodiversity in agricultural areas/agricultural soils ✔

e. agricultural areas/monocultures more affected by pests/diseases ✔

f. nutrient recycling «efficient» in natural ecosystems/rainforest ✔

g. nutrients removed with crops/nutrients removed when crops are harvested
OR
less formation of humus/less organic matter in agricultural soils ✔

h. more water recycling/more rainfall/more transpiration in natural ecosystems/rainforest ✔

i. larger biomass/more carbon stored «in biomass» in natural ecosystems/rainforest ✔

j. shallower soils/less soil erosion/degraded soils/infertile soils in agricultural areas ✔

a. shoot apex is an «apical» meristem/has undifferentiated cells ✔

b. mitosis «in shoot apex» ✔

c. cell division/cytokinesis/cells produced «in shoot apex» ✔

d. cell elongation «in shoot apex» ✔

e. stem/shoot growth «due to the cell division and elongation in the shoot apex» ✔

f. produces auxin ✔

g. auxin stimulates growth/cell elongation ✔

h. growth towards light ✔

i. differentiation of cells «produced by the shoot apex» ✔

j. leaf initiation/leaf development begins/leaf «primordia» formation «at shoot apex» ✔

k. flowers produced «by shoot apex» ✔

a. polyploidy is having more than two sets of «homologous» chromosomes ✔

b. triploid has three sets/is 3n ✔

c. tetraploid has four sets/is 4n ✔

d. Allium/vizcacha rats/other named example» ✔

e. details of chromosome numbers in diploid and polyploid species in the example ✔

f. non-disjunction/failure of chromosome pairs to separate during meiosis ✔

g. diploid gamete «can lead to polyploidy» ✔

h. fusion of diploid and haploid gamete produces triploid cells ✔

i. DNA replication but no subsequent mitosis doubles the chromosome number/produces tetraploid «from diploid»
OR
fusion of two diploid gametes produces tetraploid/4n ✔

j. polyploid/tetraploid «crossed» with diploid/non-polyploid produces infertile offspring ✔

k. meiosis fails in triploids because «homologous» chromosomes cannot pair up ✔

l. polyploid individuals are reproductively isolated
OR
polyploidy causes instant/immediate speciation
OR
tetraploids can form a new species because they can cross with each other
OR
polyploids cannot cross/produce fertile offspring with diploids ✔

m. speciation by polyploidy is common in plants/commoner in plants than animals ✔

n. polyploid individuals tend to be larger ✔

State the alternative hypothesis for this study.

To calculate chi-squared, expected values must first be calculated. Assuming that there is no association between the two species, calculate the expected number of quadrats in which both species would be present, showing your working.

State the number of degrees of freedom for this test to determine the critical value of chi-squared.

When the data in the table were used to calculate chi-squared, the calculated value was 0.056. The critical value is 3.84. Explain the conclusion that can be drawn from the calculated and critical values for chi-squared.

there is a positive / negative association between the two species;
they tend to grow together / they tend to grow apart;

OWTTE

1. $\frac{70 \times 55}{150}$ ;
2. 25.7;

Award [1] for proper values chosen/equation or [1] for answer.

1 (df)
OR
(r-1) (c-1);

1. (when the calculated value is smaller than the critical value) there is no significant association between the two species / H₀/null hypothesis accepted;
2. it is random chance if both species are either present or absent in most quadrats;

There were many correct answers, but a number confused the alternate and null hypotheses.

Calculating the degrees of freedom proved problematic for many.

Distinguish between the structure of the chromosomes of prokaryotes and eukaryotes.

Outline the causes of sickle cell anemia.

Identify, with a reason, the sex of this individual.

State the haploid number for this nucleus.

A fly that is homozygous dominant for both body colour and wing size mates with a fly that is recessive for both characteristics. In the table, draw the arrangement of alleles for the offspring of this mating and for the homozygous recessive parent.

The offspring, which were all heterozygous for grey body and normal wings, were crossed with flies that were homozygous recessive for both genes. The table shows the percentages of offspring produced.

Explain these results, based on the knowledge that the genes for body colour and wing size are autosomally linked.

a. prokaryotes have circular DNA/chromosome but eukaryote chromosomes linear/OWTTE ✔
OR
eukaryotes have telomeres/centromeres whereas prokaryotes do not ✔

b. some prokaryotes have plasmids whereas eukaryotes do not ✔

c. eukaryotes have multiple chromosomes whereas prokaryotes «typically» have only one ✔

d. histones/nucleosomes/proteins associated with DNA in eukaryotes but not in prokaryotes/naked DNA in prokaryotes
OR
eukaryote DNA can coil/supercoil/condense «due to histones» but not prokaryote DNA ✔

a. genetic disease/caused by a gene
OR
inherited «from parents»
OR
caused by mutation «of a gene» ✔

b. base substitution
OR
GAG → GTG ✔

c. hemoglobin gene mutated / different allele/form/version of hemoglobin gene
OR
Hb^A → Hb^S ✔

d. leads to change in amino acid sequence «in hemoglobin»
OR
glutamic acid → valine ✔

e. only homozygotes have full disease/sickled cells / heterozygote has milder form
OR
hemoglobin crystallizes at low oxygen concentration ✔

f. «selected for/spreads in population» as it gives resistance to malaria ✔

male because «X and» Y chromosome present
OR
male because sex chromosomes/last two chromosomes/pair 21 are unpaired/different «from each other»/not homologous ✔

The answer must include “male” and the reason.

21

a. not a 1:1:1:1 ratio «because of linkage»
OR
not independent assortment
OR
grey normal and black vestigial types/parental combinations/double dominant and double recessive were commoner than 25 %/commoner than expected ✔

b. «linked genes» so were on the same chromosome ✔

c. grey body vestigial wing and black body normal wing are recombinants
OR
2 % plus 3 % of the offspring are recombinants ✔

d. recombinants due to crossing over/exchange of genes between «non-sister» chromatids
OR
2 % and 3 % of offspring were due to crossing over
OR
genes inherited together unless separated by crossing over ✔

e. crossing over between the two loci/between the two genes on the chromosomes ✔

f. few recombinants/not much crossing over because genes/gene loci close together ✔

Accept any of these points from an annotated diagram.

Answers to this question were variable. Some candidates wrote about cell structure rather than chromosome and many wrote about the location of chromosomes in the cell rather than their structure. Weaker candidates did not know the difference between prokaryotes and eukaryotes and some think that prokaryotes are plants and eukaryotes are animals. The term ‘naked’ was often used for DNA not enclosed in a nuclear membrane, when in this context it should be reserved for DNA that is not associated with histone proteins.

This question also elicited a wide range of answers. Some were accurate and detailed but some showed no knowledge of this genetic disease. There was lack of clarity in many answers, for example the disease was described as the mutation and sickling of cells was known but not linked to haemoglobin. The weakest answers tended to describe anemia in general, with iron deficiency sometimes given as the cause.

90 % of candidates correctly identified the individual as male, because the two sex chromosomes were different in size so must be an X and a Y. The other 10% mostly thought that there were two X chromosomes present, perhaps because pair 20 were X shaped.

Slightly more than 50 % of candidates got this right. A common wrong answer was 42 – diploid number. There were a variety of other answers, some of which must have been guesses from candidates who did not understand the term haploid.

This was generally well answered, with more than two thirds of candidates getting at least one of the genotypes correct. The best answers gave the alleles on lines that symbolised linked genes on a chromosome, but there was a potential pitfall here. If the alleles were arranged in a way that was impossible, given the nature of the cross, one mark was deducted. For example, each chromosome must have one of each gene, not two copies of one gene.

Most candidates found it very hard to explain the non-Mendelian ratio. A good start would have been to state the expected Mendelian ratio for unlinked genes in this type of cross and then compare this with the actual percentages. Answers tended to state which traits tended to be inherited together rather than explain the mechanism. An obvious answer was that that the two genes are located on the same chromosome but only a minority of candidates stated this and even fewer made the point that crossing over between the two gene loci results in the small percentage of recombinants. Many students referred to the 9:3:3:1 ratio, even though the cross performed in this case would not have given this ratio, even if there had not been gene linkage. Also many candidates claimed that more individuals showed dominant phenotypes than recessive, which was not supported by the percentages.

Cells go through a repeating cycle of events in growth regions such as plant root tips and animal embryos. Outline this cell cycle.

Draw a labelled diagram of the formation of a chiasma by crossing over.

Explain the control of gene expression in eukaryotes.

a. mitosis is the division of a nucleus to produce two genetically identical daughter nuclei 

b. consists of four phases: prophase, metaphase, anaphase, telophase 

c. cytokinesis occurs after mitosis 

d. interphase is the metabolically active phase between cell divisions  OWTTE

e. the interphase consists of the S phase, G1 and G2 

f. DNA replicates in the S phase 

g. cell growth
OR
preparation for mitosis
OR
duplication of organelles in G1 and G2

a. «crossing over/chiasmata shown between» homologous chromosomes  

b. centromere drawn and labelled  

c. single strand break «SSB»/DNA cut between homologous chromosomes 

d. non-sister chromatids labelled
OR
sister chromatids labelled 

e. chiasma between homologous chromosomes labelled «shown forming after SSB»

Homologous chromosomes must be labelled and correctly drawn.

It is likely that more than one diagram will need to be included to demonstrate the stages.

a. mRNA conveys genetic information from DNA to the ribosomes «where it guides polypeptide production» 

b. gene expression requires the production of specific mRNA «through transcription» 

c. most genes are turned off/not being transcribed at any one time/regulated
OR
some genes are only expressed at certain times 

d. some genes are only expressed in certain cells/tissues
OR
«cell» differentiation involves changes in gene expression 

e. transcription factors/proteins can increase/decrease transcription 

f. hormones/chemical environment of cell can affect gene expression 

g. example of cell environment
eg: auxin/insulin/cytoplasmic gradient in embryo

h. transcription factors/proteins may prevent or enhance the binding of RNA polymerase 

i. nucleosomes limit access of transcription factors to DNA/regulate gene expression/transcription
OR
activate or silence genes 

j. DNA methylation/acetylation appears to control gene expression «as epigenetic factor»
OR
methylated genes are silenced 

k. «some» DNA methylation patterns are inherited 

l. introns may contain positive or negative gene regulators
OR
gene expression can be regulated by post-transcriptional modification/splicing/mRNA processing

Autosomal genes are located in chromosomes that are not sex chromosomes. The inheritance of autosomal genes is affected by whether the genes are linked or unlinked. Explain the two types of inheritance, using the example of parents that are heterozygous for two genes A and B.

Outline how sperm are produced from diploid cells in the testis and how this production can be sustained over many decades of adult life.

Testis cells are eukaryotic cells. Identify the structures seen under the electron microscope in testis cells that are not present in prokaryotic cells.

a. unlinked genes are on different chromosomes / vice versa ✔

b. unlinked alleles migrate/segregate/are inherited independently (during meiosis) / vice versa ✔

c. (In unlinked inheritance) there is an equal chance for all 4 options to occur / AB, Ab, aB, ab / vice versa ✔

d. (dihybrid crosses involving) linked genes do not produce Mendelian ratios ✔

e. (excluding recombinants) there is a 1:1 chance of inheriting the different options/AB or ab ✔

f. in linked characteristics alleles might not migrate together if there is crossing over/ recombinants are formed ✔

g. crossing over occurs in prophase I of meiosis ✔

h. when the sister chromatids migrate in meiosis II the characteristics forming gametes are different/Ab, aB ✔

i. formation of recombinants causes changes in ratio/probability of inheritance/genetic variation ✔

j. correct named example of inheritance of linked/unlinked characteristics ✔

k. Punnett/paired diagrams of both unlinked and linked characteristics ✔

l. genes which are linked but are far apart on the chromosome can display independent assortment ✔

Mp a could be awarded from an annotated diagram.

Allow annotated diagram of inheritance / could be shown in a Punnett square.

Allow annotated diagram of linked inheritance for mp f.

For mp K, accept sex linked examples involving two genes.

If the student interprets the question as sex-linked and autosomal inheritance, look for WTTE marks from the scheme.

a. germinal epithelium divide endlessly (by mitosis giving rise to spermatagonia)

b. spermatogonia are diploid/2n ✔

c. spermatogonia divide by mitosis / provide a continuous supply throughout adult life ✔

d. (some) spermatogonia enlarge forming primary spermatocytes ✔

e. primary spermatocytes undergo the first division of meiosis/meiosis I ✔

f. secondary spermatocytes produced are haploid/n ✔

g. secondary spermatocytes undergo the second division of meiosis (to produce spermatids) ✔

h. spermatids develop tails
OR
spermatids differentiate into spermatozoa / spermatids associate with Sertoli cells ✔

Marks can be awarded to an annotated diagram.

Do not accept sperm or spermatozoa as equivalent to spermatagonia or spermatocytes.

a. nucleus/nuclear membrane ✔

b. membrane bound organelles ✔

c. mitochondria ✔

d. rough ER/smooth ER/golgi apparatus ✔

e. lysosomes / centrioles ✔

f. large/80S ribosomes / ribosomes attached to a membrane ✔

g. linear chromosomes / histones ✔

Question 5 was not a popular question choice. Few candidates performed well on this question overall. Many students wrote vaguely about dominant and recessive alleles or confused gene linkage with "sex-linkage" or even multiple alleles and blood groups. 

Although many candidates performed very well, many answers were vague and simply stated that sperm cells were made by meiosis without any further details. Commonly, the stages where meiosis occurs and where mitosis occurs were muddled as well as whether primary or secondary spermatocytes were diploid or haploid. The terms spermatozoa and spermatids were commonly used interchangeably.

Outline the criteria that should be used to assess whether a group of organisms is a species.

Describe the changes that occur in gene pools during speciation.

Discuss the process, including potential risks and benefits, of using bacteria to genetically modify plant crop species.

a. organisms can potentially interbreed;

b. to produce fertile offspring;

c. same sequence of genes (on chromosomes) / same types of chromosomes;

d. similar traits/phenotype/WTTE;

e. same chromosome number/karyotype;

a. gene pool is all genes/alleles in an (interbreeding) population;

b. gene pool splits/divides/separated during speciation;

c. due to reproductive isolation (of groups within a species);

d. temporal/behavioral/geographic isolation (can cause reproductive isolation);

e. divergence of gene pools;

f. allele frequencies change;

g. natural selection different (in the isolated groups so there is divergence);

h. different (random) mutations occur (in the isolated populations so there is divergence);

i. speciation has occurred when differences between populations prevent interbreeding;

Do not award both mpc and mpi for the same idea (reproductive isolation separating populations vs speciation due to interbreeding not being possible).

Process:
a. genetic modification by gene transfer between species;

b. gene/Bt gene/DNA segment transferred from bacterium to plant/crop;

c. gene/DNA codes for/responsible for desired protein/gene product;

d. bacteria have/produce plasmids / gene/DNA inserted into plasmid;

e. using restriction enzymes/endonucleases to cut DNA;

f. using DNA ligase to join DNA;

g. bacterium transfers (modified) plasmid to plant cell;

Benefits:
h. increase crop yields / more food produced / less land needed to grow food;

i. increase pest/disease resistance / use less pesticides/insecticides/fungicides;

j. improves crops to be more nutritious/increased vitamin content;

k. increased tolerance to saline soils/drought/high temperatures/low temperatures;

Risks:
l. GM organisms could spread to sites (where they will cause harm);

m. transferred gene could spread to other species / spread of herbicide resistance to weeds;

n. GM crops that produce pesticide could kill non-pest insects/monarch butterflies / insect pests could develop resistance to pesticides/insecticides/Bt toxin;

This was the most successfully answered part of Question 8. Many knew that members of a species can interbreed and produce fertile offspring. Surprisingly few mentioned that similarities in characteristics or phenotype are found in species.

Answers to this part of the question were mostly poor. Gene pools and speciation are the subject of sub-topic 10.3, but many candidates struggled to link these two concepts. Various misunderstandings were seen. As in previous exams, some candidates thought that speciation is evolutionary change over time in a species, rather than the splitting of a species into two or more separate species. Mutations were sometimes described as though they happen in response to a need for new traits, rather than them happening spontaneously and occasionally being selected for.

This part of the question was also poorly answered by many candidates. Only a simple account of the procedures used to genetically modify plants was expected and not the details of the use of Agrobacterium that are part of Option B. There was confusion among many candidates about how bacteria might be involved in genetic modification of plants. Two possibilities that were rewarded were the transfer of genes from bacteria or of plasmids derived from bacteria. Where marks were scored, they were mostly for risks and benefits, but many accounts of these were unconvincing. Some of the risks that were suggested were not evidence-based. As with vaccination, it is important that myths about the dangers of procedures such as genetic modification are not propagated.

Draw a labelled diagram to show the structure of a single nucleotide of RNA.

Describe how DNA profiling can be used to establish paternity.

Explain the reasons for variation in human height.

1. ribose drawn as a pentagon and labelled;
2. base linked correctly (to C₁) of ribose and labelled;
3. phosphate linked correctly (to C₅) of ribose and labelled;

1. DNA sample is collected from the child and its (potential) parents;
2. from saliva/mouth swab/blood/other body cells;
3. PCR used to amplify/produce more copies of the DNA;
4. short tandem repeats/genes consisting of a repeating sequence of bases repeats copied/used;
5. number of repeats varies between individuals;
6. unlikely that two individuals have same number of repeats for every gene included;
7. gel electrophoresis used to separate DNA fragments according to length/number of repeats;
8. gel electrophoresis generates a unique pattern of bands
9. DNA profile is the pattern of bands / diagram showing pattern of bands as in a DNA profile;
10. all bands in the child’s profile must be in one of the parents’ profiles / OWTTE;

1. environment affects height;
2. nutrition/malnutrition affects growth rate / other example of environmental factor affecting height;
3. genes/alleles affect height / height is partly heritable;
4. polygenic / many genes influence height;
5. continuous variation;
6. normal/bell-shaped distribution of height;
7. some alleles (of these genes) increase height and some reduce it;
8. many possible combinations of alleles of these genes;
9. specific gene mutations/alleles cause dwarfism/extreme height;
10. meiosis generates variation (in height);
11. mutations generate variation (in height);
12. males tend to be/are on average taller than females;
13. loss of height during aging;

Many candidates were able to draw a nucleotide and label the subunits correctly. Weaker candidates did not know what a nucleotide was, so often drew a diagram of either a DNA or RNA polynucleotide.

Answers to this question were very variable, with some very well-informed accounts including the use of genes containing of tandem repeats and details of the collection of DNA samples from both parents and the child. There were also many accounts with errors of understanding. A common misconception is that paternity is established by finding the male whose profile has most similarities to that of the child. This does not prove that a man is the father of the child and instead the father's profile must contain all bands in the child's profile that do not occur in that of the mother. 

This was a relatively high scoring question, with a mean mark of 2.4 (out of 7). Teachers expressed surprise in G2 forms that their students were being expected to make 7 valid points in their answer, but the wide-ranging mark scheme ensured that this was possible. There was some confusion between polygenic inheritance and multiple alleles. Weaker candidates tended to think that dominant alleles make us taller and recessive alleles cause shortness. Another fault in many answers was to focus on natural selection and evolution of height — if anything natural selection will reduce variation in human height rather than cause it.

The micrograph shows a plant cell of Lilium grandiflorum during meiosis.

Identify, giving reasons, the stage of meiosis shown by this cell.

Outline the law of independent assortment.

The genes for cystic fibrosis and blood group are not linked. Two parents are heterozygous for cystic fibrosis. One parent has blood group O and the other has blood group AB. Using a Punnett square, determine the probability that their child will have both cystic fibrosis and blood group A.

a. anaphase II

b. as four daughter cells are being formed
OR
the centromeres split / sister chromatids separate
OR
sister chromatids/ chromosomes are pulled «by the spindle microtubules» to opposite poles

a. two «or more» traits/genes are inherited independently of one another. Accept vice versa.

b. observed for traits/genes that are not linked/far apart on the chromosome 
Can be shown in annotated diagram.

c. «due to homologous» chromosomes aligning independently/randomly on equator during metaphase I/meiosis I 

d. during anaphase I homologues pulled to separate poles

Can be shown in annotated diagram.

a. correct parental genotypes shown / Ff ii, FfI^A I^B 

b. Punnett square with correct gametes shown 

c. correct probability:  $\frac{1}{8}$

                                   OR
                                   12.5%

Must use blood type symbols I^B I^A and i. No specific letters required to represent cystic fibrosis allele though dominant and recessive must be apparent.

Outline how reproductive isolation can occur in an animal population.

Describe the different cell types in the seminiferous tubules that are involved in the process of spermatogenesis.

Explain the roles of specific hormones in the menstrual cycle, including positive and negative feedback mechanisms.

a. can be sympatric or allopatric 

b. temporal isolation by members of difference populations reproducing at different times  OWTTE

c. behavioural isolation by difference in courtship behaviours  OWTTE

d. geographic isolation by a population being separated by river/mountain/barrier to contact  
An example of a geographic barrier is required.

e. polyploidy

a. spermatogonia «2n» are undifferentiated germ cells  OWTTE

b. spermatogonia mature and divide «by mitosis» into primary spermatocytes «2n» 

c. primary spermatocytes divide by meiosis I into secondary spermatocytes «1n» 

d. secondary spermatocytes divide by meiosis II into spermatids «1n» 

e. spermatids differentiate/mature into spermatozoa/sperm 

f. Sertoli/nurse cells provide nourishment/support to these developing cells 

g. Leydig/interstitial cells produce testosterone

a. anterior pituitary/hypophysis secretes FSH which stimulates ovary for follicles to develop 

b. follicles secrete estrogen 

c. estrogen stimulates more FSH receptors on follicle cells so respond more to FSH 

d. increased estrogen results in positive feedback on «anterior» pituitary 

e. estrogen stimulates LH secretion 

f. estrogen promotes development of endometrium/uterine lining 

g. LH levels increase and cause ovulation 

h. LH results in negative feedback on follicle cells/estrogen production 

i. LH causes follicle to develop into corpus luteum
OR
follicle cells produce more progesterone 

j. progesterone thickens the uterus lining 

k. high progesterone results in negative feedback on pituitary/prevents FSH/LH secretion 

l. progesterone levels drop and allow FSH secretion 

m. falling progesterone leads to menstruation/degradation of uterine lining

Award [5 max] if no reference to feedback is made.

Outline, using graphs, the effect of different factors that influence enzyme activity.

Describe the function of three named enzymes involved in DNA replication.

Explain how speciation occurs, including the different processes of isolation and selection.

1. enzymes have active sites that bind specific substrates;
2. act as catalysts to speed up reactions
   OR
   lower activation energy;
3. rate/activity increases with temperature;
4. up to an optimum temperature;
5. sharp decline in activity above (optimum temperature);
   (Graph has to be clearly asymmetrical for mpe)
   (Graph shown would earn mpc, mpd and mpe)
   ;
6. rate/activity declines at a pH above and below the optimum pH
   (Graph shown would earn mpf)
   ;
7. rate/activity increases sharply as substrate concentration goes up;
8. above a certain concentration, the rate reaches a maximum/plateau;
   (Graph shown would earn mpg and mph)
   ; 
9. competitive inhibitor present, as substrate concentration increases, enzyme requires higher concentration to achieve maximum rate / graph showing this with and without inhibitor;
10. non-competitive inhibitor present, as substrate concentration rises, enzyme activity is lower at all substrate concentrations / graph showing shape with and without inhibitor;

Award [3 max] if there are no graphs.

For each graph, axes must be correctly labelled, the shape must be correctly drawn.

Marks can be awarded to a correctly annotated graph.

1. Helicase separates/unwinds DNA from double helix;
2. (DNA) gyrase / topoisomerase releases tension/strain (caused by super coiling);
3. (DNA) primase builds/forms/adds an RNA primer;
4. DNA polymerase I is a (5’ $\to$ 3’) exonuclease/removes RNA primers/replacement of RNA by DNA;
5. DNA polymerase III synthesizes DNA (5’ $\to$ 3’) on leading/lagging strands/forms bonds between DNA nucleotides;
6. (DNA) ligase connects/seals nick between Okazaki fragments to make continuous DNA strand;

Only mark the first three answers in sequence.

DNA polymerase I required for mpd and DNA polymerase III required for mpe (i.e., the numbers are required).

1. Species is a group of organisms that interbreeds (normally in the wild) and produce fertile offspring;
2. within an interbreeding population there is variety / variation exists;
3. some adaptations favour survival to reproductive age /survival of the fittest / natural selection;
4. alleles for these adaptations become more frequent/are inherited in the population /change with time;
5. speciation is the formation of new species;
6. (speciation) occurs because populations have become reproductively isolated / no longer able to interbreed / exposed to different selection pressures;
7. behavioural isolation involves differences in courtship or mating behaviours;
8. temporal isolation involves differences in the timing of courtship or mating behaviours;
9. geographical isolation / allopatric refers to the physical barriers that exist that keep two populations from mating;
10. polyploidy can lead to reproductive isolation;
11. stabilizing selection is when the two extremes of a trait have lower reproductive fitness (OWTTE) / favours average phenotype;
12. directional selection is when one extreme of the trait has lower reproductive fitness (OWTTE);
13. disruptive/diversifying selection favours both extreme phenotypes / intermediate phenotype has lower fitness;

This question was the most popular choice of essay as well as the most successful.

Many earned full marks, but many others contained errors. Some of these errors included inverted axes, no abrupt curve drop for high temperatures, an asymmetric graph shown for pH, the optimum temperature or pH not mentioned or labelled.

This question about the enzymes involved in replication was generally well done. Some omitted the I and the III for polymerase and/or focused on base pairing instead of elongation of strands with DNA polymerase III. The details around the relative role of base pairing and covalent bond formation was sometimes lacking.

Many had a "good idea" of speciation and knew about the different types of isolation and selection. Very few discussed the role of polyploidy.

Draw labelled diagrams to show the structure of RNA nucleotides and how they are linked together to form a molecule of RNA.

Explain transcription.

Distinguish between continuous and discrete variation, using examples.

a. ribose drawn as pentagon and labelled sugar/ribose;
b. base drawn with correct link to (C₁ of) ribose and labelled base/nitrogenous base;
c. phosphate drawn with correct link to (C₅ of) ribose and labelled P/phosphate;
d. two (or more) ribonucleotides drawn with correct link (C₃ to C₅)

a. synthesis of RNA/mRNA / transcription of DNA to RNA;
b. RNA nucleotides linked together to form a strand/chain;
c. RNA strand assembled on DNA template/antisense strand / copy made of sense strand;
d. RNA polymerase carries out transcription/links RNA nucleotides;
e. uncoiling/separation of DNA strands;
f.  5’ end of nucleotides linked to 3’ end of (growing RNA) strand;
g. complementary base pairing (is the basis of copying the base sequence);
h. uracil instead of thymine in RNA;
i. starts at/RNA polymerase binds to a promoter;
j. regulated by transcription factors/DNA binding proteins/nucleosomes;

Annotated diagrams can be used.

Many candidates were able to draw the structure of an RNA nucleotide and link it correctly by a 5’ to 3’ bond to another nucleotide. The commonest error was to show two strands of nucleotides linked by base pairing, indicating confusion between RNA and DNA.

Answers were mixed. Strong candidates had no difficulty in describing transcription in detail but weaker ones tended to get confused with replication and/or translation. Helicase was often stated as the enzyme that uncoils and splits the double helix, rather than RNA polymerase. Marks were not awarded merely for stating that transcription is 5’ to 3’ unless it was clear that the candidate understood that the 5’ terminal of a free nucleotide is linked to the 3’ terminal of the chain of nucleotides already linked up. Candidates are expected to show understanding in their answers, rather than just state memorized phrases.

Differences between continuous and discrete variation were not well known and the average performance for this question was the lowest for any part of Section B. A general fault was to describe the two types of variation separately and in consequence forget to include both sides of a distinction. For example, some candidates stated that the environment can cause continuous variation, but most did not then also state that discrete variation is generally unaffected by environment. A mark was awarded for examples of the two types of variation, but in some cases an inappropriate example was chosen, such as hair colour for discrete variation or eye colour for continuous variation. The latter example was best avoided entirely as there are aspects of both continuous and discrete variation in the pigmentation of the iris.