Download 1. Cellular control Booklet [A2]

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A2 BIOLOGY
Cellular control
Learning objectives:
 State that genes code for polypeptides, including enzymes;
 Explain the meaning of the term genetic code;
 Describe, with the aid of diagrams, the way in which a nucleotide sequence codes for the
amino acid sequence in a polypeptide;
 Describe, with the aid of diagrams, how the sequence of nucleotides within a gene is used to
construct a polypeptide, including the roles of messenger RNA, transfer RNA and ribosomes;
 State that mutations cause changes to the sequence of nucleotides in DNA molecules;
 Explain how mutations can have beneficial, neutral or harmful effects on the way a protein
functions;
 State that cyclic AMP activates proteins by altering their three-dimensional structure;
 Explain genetic control of protein production in a prokaryote using the lac operon;
 Explain that the genes that control development of body plans are similar in plants, animals
and fungi, with reference to homeobox sequences;
 Outline how apoptosis (programmed cell death) can act as a mechanism to change body
plans;
Key definitions:
Compile a glossary by writing your own definitions for the following key terms related to the
learning objectives above.
Key term
gene
polypeptide
genome
locus
protein
translation
codon
Definition
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Key term
anticodon
chromosome mutation
DNA mutation
point mutation
(substitution)
insertion/deletion
mutation
frameshift
allele
β-galactosidase
lactose permease
lac operon
structural genes
operator region
promoter region
repressor protein
homeobox genes
apoptosis
Definition
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From DNA to proteins
Transcription is the process by which the code contained in the DNA molecule is transcribed
(rewritten) into a mRNA molecule. Transcription is under the control of the cell’s metabolic
processes which must activate a gene before this process can begin. The enzyme that directly
controls the process is RNA polymerase, which makes a strand of mRNA using the single strand of
DNA (the template strand) as a template (hence the term). The enzyme transcribes only a gene
length of DNA at a time and therefore recognises start and stop signals (codes) at the beginning and
end of the gene. Only RNA polymerase is involved in mRNA synthesis; it causes the unwinding of the
DNA as well. It is common to find several RNA polymerase enzymes molecules on the same gene at
any one time, allowing a high rate of mRNA synthesis to occur.
A copy of the genetic information for making a protein is made in the form of messenger RNA
(mRNA). Many mRNA copies may be made from a single gene on the DNA molecule. Once the
mRNA is complete and has been released from the chromosome, it travels to the edge of the
nucleus where it gains access to the cytoplasm through a tiny hole called a nuclear pore. In
prokaryotic cells (bacteria) there is no nucleus, and the chromosomes are in direct contact with the
cytoplasm. This means that the next stage (translation) can begin immediately, with the mRNA still
being synthesised by enzymes on the DNA molecule.
Translation
The diagram on the next page shows the translation phase of protein synthesis. A single mRNA
molecule can be ‘serviced’ by many ribosomes at the same time.
Ribosomes are made up of a complex of ribosomal RNA (rRNA) and proteins. They exist as two
separate sub-units until they are attracted to a binding site on the mRNA molecule, when they join
together. Ribosomes have binding sites that attract transfer RNA (tRNA) molecules loaded with
amino acids. The tRNA molecules are about 80 nucleotides in length and are made under the
direction of genes in the chromosomes. There is a different tRNA molecule for each of the different
possible anticodons and, because of the degeneracy of the genetic code, there may be up to six
different tRNAs carrying the same amino acid.
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The anticodon at the base of each tRNA must make a perfect complementary match with the codon
on the mRNA before the amino acid is released. Once released, the amino acid is added to the
growing polypeptide chain by enzymes.
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How the polypeptide is assembled [summary]
A molecule of mRNA binds to a ribosome. Two codons (six bases) are attached to the small
subunit of the ribosome and exposed to the large subunit. The first exposed mRNA codon is
always AUG. using ATP energy and an enzyme, a tRNA with methionine and the anticodon
UAC forms hydrogen bonds with this codon.
A second tRNA, bearing a different amino acid, binds to the second exposed codon with its
complimentary anticodon.
A peptide bond forms between the two adjacent amino acids. An enzyme, present in the
small ribosomal subunit, catalyses the reaction.
The ribosome now moves along the mRNA, reading the next codon. A third tRNA brings
another amino acid, and a peptide bond forms between it and the dipeptide. The first tRNA
leaves and is able to collect and bring another of its amino acids.
The polypeptide chain grows until a stop codon is reached. There are no corresponding
tRNAs for these three codons, UAA, UAC or UGA, so the polypeptide chain is now complete.
Some proteins have to be activated by a chemical, cyclic AMP (cyclic adenosine monophosphate or
cAMP) that, like ATP, is a nucleotide derivative. It activates proteins by changing their 3D shape so
that their shape is a better fit to their complimentary molecules.
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Mutations
Mutations add, delete, or rearrange genetic material. Not all mutations are inherited. Mutations
can happen spontaneously due to DNA replication errors, or they can be induced by mutagens. Only
those mutations taking place in cells that produce gametes will be inherited. If they occur in a body
cell after the organism has begun to develop beyond the zygote stage, then they may give rise to
chimaeras (mixture of gene types in a single organism).
Gene mutations
Gene mutations are small, localised changes in the structure of a DNA strand. These mutations may
involve change in a single nucleotide (often called point mutations), or they may involve changes to
a triplet (e.g. deletion or triplet repeat). If one amino acid in a protein is wrong, the biological
function of the entire protein can be disrupted.
Not all mutations may result in altered proteins; because of the degeneracy of the genetic code, a
substitution of the 3rd base in a codon may code for the same amino acid. These alterations in the
DNA are at the nucleotide level where individual codons are affected. Alteration of the precise
nucleotide sequence of a coded gene in turn alters the mRNA transcribed from the mutated DNA
and may affect the polypeptide chain that is normally creates.
In some cases, mutations trigger the onset of cancer, through the disruption of the normal controls
regulating cell division. It is not correct to assume that all mutations are harmful. There are many
documented cases where mutations conferring a survival advantage have arisen in a population.
Such beneficial mutations are most common among viruses and bacteria, but occur in multicellular
organisms also e.g. the development of pesticide resistance in insects.
Sometimes, a mutation may be neutral and have no immediate effect. If there is no selective
pressure against it, a mutation may be carried in the population and be of benefit (or harm) at some
future time.
Harmful mutations
There are many well-documented examples of mutations that cause harmful effects. Examples are
the mutations giving rise to cystic fibrosis (CF) and sickle cell disease. The sickle cell mutation
involves a change to only one base in the DNA sequence, whereas the CF mutation involves the loss
of a single triplet (three nucleotides). The malformed proteins that result from these mutations
cannot carry out their normal biological functions. Albinism is caused by a mutation in the gene that
produces an enzyme in the metabolic pathway to produce melanin. It occurs in a large number of
animals. Albinos are uncommon in the wild because they tend to be more vulnerable to predation
and damaging UV radiation.
Cystic fibrosis mutation
Cystic fibrosis is an inherited disorder caused by a mutation of the CF gene. It is one of the most
common lethal autosomal recessive conditions affecting Caucasians, with an incidence in the UK of
1 in 2500 births and a carrier frequency of 4%. The CF gene’s protein product, CTFR, is a membranebased protein with a function in regulating the transport of chloride across the membrane. A faulty
gene in turn codes for a faulty CFTR. More than 500 mutations of the CF gene have been described,
giving rise to disease symptoms of varying severity. The CF gene is located on chromosome 7. The
δF508 mutation of the CF gene describes a deletion of the 508 th triplet, which in turn causes the loss
of a single amino acid from the gene’s protein product, the cystic fibrosis transmembrane
conductance regulator (CFTR).
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The CFTR protein consists of 1480 amino acids. This protein normally regulates the chloride
channels in cell membranes, but the mutant form fails to achieve this.
One mutation is particularly common and accounts for more than 70% of all defective CF genes.
This mutation, called δF508, leads to the absence of CFTR from its proper position in the
membrane. Its absence results in defective chloride transport and leads to a net increase in water
absorption by the cell. This accounts for the symptoms of CF, where mucus-secreting glands,
particularly in the lungs and pancreas, become fibrous and produce abnormally thick mucus. The
widespread presence of CFTR throughout the body also explains why CF is a multisystem condition
affecting many organs.
Another CF mutation, R117H, which is also relatively common, produces a partially functional CFTR
protein.
Sickle cell mutation
Formerly called sickle cell anaemia, sickle cell disease is an inherited disorder caused by a gene
mutation which codes for a faulty β chain haemoglobin protein. This in turn causes the red blood
cells to deform causing a whole range of medical problems.
Each red blood cell contains about 270 million haemoglobin molecules. In their normal state, the
red blood cells have a flattened disc shape. Normal haemoglobin produces normal red blood cells.
Haemoglobin molecules are made up of 2 α-chains and 2 β-chains linked together. The gene coding
for the β-chain of haemoglobin protein is located on chromosome 11 and consists of 438 bases. The
438 nucleotides produce a protein made of 146 amino acids.
The sickle cell mutation involves the substitution of one base for another (T is substituted by A) in
the HBB gene, causing a single amino acid to be altered. This new amino acid has different
properties and makes the haemoglobin behave in a different manner.
The mutated form of haemoglobin is less soluble and precipitates when deprived of oxygen. This
deforms the red blood cells to give them their sickle shape. The effects of the sickle cells on human
metabolism are considerable: sickle cells are rapidly removed from the circulation leading to
jaundice and anaemia; a cascade of other ailments include heart defects, brain damage, kidney
defects, skin lesions and enlargement of the spleen.
Beneficial mutations
Tolerance to high cholesterol levels in humans: in the small village of Limone, in Italy, about 40
villagers have extraordinarily high levels of blood cholesterol, with no apparent harmful effects on
their coronary arteries. The village has a population of 980 inhabitants and was, until recently,
largely isolated from the rest of the world. The villagers possess a mutation that alters the protein
produced by just one amino acid. This improved protein is ten times more effective at mopping up
excess cholesterol. No matter how much excess cholesterol is taken in by eating, it can always be
disposed of. All carriers of the mutation are related and descended from one couple who arrived in
Limone in 1636. Generally, the people of Limone live longer and show a high resistance to heart
disease.
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The lac operon
The operon mechanism was proposed by Jacob and Monod to account for the regulation of gene
activity in response to the needs of the cell. Their work was carried out with the bacterium
Escherichia coli and the model is not applicable to eukaryotic cells where the genes are not found as
operons. An operon consists of a group of closely linked genes that act together and code for the
enzymes that control a particular metabolic pathway. These may be for the metabolism of an
energy source e.g. lactose or the synthesis of a molecule such as an amino acid.
The structural genes contain the information for the production of the enzymes themselves and
they are transcribed as a single transcription unit. These structural genes are controlled by a
promoter, which initiates the formation of the mRNA, and a region of the DNA in front of the
structural genes called the operator. A gene outside the operon, called the regulator gene,
produces a repressor molecule that can bind to the operator, and block the transcription of the
structural genes. It is the repressor that switches the structural genes on or off and controls the
metabolic pathway.
Two mechanisms operate in the operon model: gene induction and gene repression. Gene
induction occurs when genes are switched on by an inducer binding to the repressor molecule and
deactivating it. In the lac operon model based on E. coli, lactose acts as an inducer, binding to the
repressor and permitting transcription of the structural genes for the utilisation of lactose (an
infrequently encountered substrate). Gene repression occurs when genes that are normally on e.g.
genes for synthesis of an amino acid) are switched off by activation of the repressor.
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Lactose is not a common energy source for E. coli and the genes for the metabolism of lactose by
the cell are normally switched off. With lactose absent, the repressor molecule binds tightly to the
operator. This prevents RNA polymerase from transcribing the adjacent structural genes and the
enzymes for lactose metabolism are not produced.
When lactose is available, some of it is converted into the inducer allolactose. Allolactose binds to
the repressor molecule, altering its shape and preventing it from binding to the operator. The
structural genes can then be transcribed, and the enzymes for the metabolism of lactose are
produced.
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Apoptosis
Apoptosis is programmed cell death that occurs in multicellular organisms. Cells should undergo
about 50 mitotic divisions (the Hayflick constant) and then undergoes a series of biochemical events
that leads to an orderly and tidy cell death. This is contrast to cell necrosis, an untidy and damaging
cell death that occurs after trauma and releases hydrolytic enzymes.
Enzymes break down the cell cytoskeleton.
The cytoplasm becomes dense, with organelles tightly packed.
The cell surface membrane changes and small bits called blebs form.
Chromatin condenses and the nuclear envelope breaks. DNA breaks into fragments.
The cell breaks into vesicles that are taken up by phagocytosis. The cellular debris is
disposed of and does not damage any other cells or tissues.
The whole process occurs very quickly.
The process is controlled by a diverse range of cell signals, some of which come from inside the cells
and some from outside. The signals include cytokines made by cells of the immune system,
hormones, growth factors and nitric oxide. Nitric oxide can induce apoptosis by making the inner
mitochondrial membrane more permeable to hydrogen ions and dissipating the proton gradient.
Proteins are released into the cytosol. These proteins bind to apoptosis inhibitor proteins and allow
the process to take place.
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1. Describe how the information coded on genes is used to synthesis polypeptides
and how these polypeptides control the physical development of an organism.
In your answer, you should consider both the synthesis of polypeptides and their
roles.
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_____________________________________________________________ [8]
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2. Fig. 5.1 is a circular representation of the genetic code.
(a) Fig. 5.2 shows a sequence of bases coding for a sequence of amino acids.
The name of the third amino acid in the sequence has been filled in.
Identify the remaining amino acids in the sequence.
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1 ____________________________________________________________
2 ____________________________________________________________
3 lysine
4 ____________________________________________________________
5 _________________________________________________________ [2]
(b) State the name of the stage of protein synthesis represented in Fig. 5.2 and
name the organelle in the cell where this takes place.
_____________________________________________________________
___________________________________________________________ [2]
(c) Identify the type of nucleic acid that holds the sequence of bases shown in
Fig. 5.2.
___________________________________________________________ [2]
(d) Using the information in Fig. 5.1 list the three triplet codons that would
cause termination of a polypeptide chain (stop codons) and explain why
these codons have this effect.
_____________________________________________________________
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_____________________________________________________________
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___________________________________________________________ [2]
(e) What name would be given to a mutation that resulted in a change of the
codon UUU to UUC?
___________________________________________________________ [1]
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3. Fig. 1.1 shows a metabolic pathway involving the amino acid, phenylalanine.
One of the products of this pathway is melanin, the pigment that gives a brown
colour to hair, skin and the iris of the eyes. This metabolic pathway also
produces thyroid hormones.
(a) Use Fig. 1.1 to name:
the enzyme that catalyses the last step in melanin production
___________________________________________________________ [1]
the genetic disorder resulting from the absence of the enzyme at the start
of the metabolic pathway for melanin production
___________________________________________________________ [1]
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(b) Phenylalanine and tyrosine are both amino acids.
Explain why phenylalanine and tyrosine are classified as amino acids.
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___________________________________________________________ [2]
(c) One effect of thyroid hormones is to increase the activity of mitochondria
within cells. Suggest how the metabolism of a person with the condition
congenital hypothyroidism might differ from that of a person who does not
have this condition.
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___________________________________________________________ [3]
4. Homeobox genes show astonishing similarity across widely different species of
animal, from fruit flies, which are insects, to mice and humans, which are
mammals. The sequences of these genes have remained relatively unchanged
throughout evolutionary history and the same genes control embryonic
development in flies and mammals.
(a) State what is meant by a homeobox gene.
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___________________________________________________________ [2]
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(b) Homeobox genes show ‘astonishing similarity across widely different
species of animal’. Explain why there has been very little change by
mutation in these genes.
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___________________________________________________________ [2]
(c) Frogs reproduce by laying eggs in water. Each egg develops into a tadpole,
which has external gills to extract oxygen from the water, and a tail to help
it swim. The tadpole gradually changes into an adult frog as it grows. During
this time its gills and tail disappear.
List two cellular processes that must occur during the development of a
tadpole into a frog.
_____________________________________________________________
___________________________________________________________ [2]
(d) Name another kingdom of organisms, other than animals, that have similar
homeotic genes.
___________________________________________________________ [1]
5. Growth and development in organisms is controlled by a number of
mechanisms that operate at the cellular level. The control elements involved in
these mechanisms include hormones, the second messenger molecule cyclic
AMP and regulatory genes.
 In eukaryotes the most important regulatory genes contain homeobox
sequences and are called homeotic genes.
 The regulatory genes of the lac operon in prokaryotes are studied to help
us understand how regulatory genes and their products interact to
switch structural genes on and off.
(a) Use your understanding of the biochemical identity and interactions of
these control elements to complete Table 5.1 (on the next page) by putting
a tick () or a cross (x) in each box.
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Some of the boxes have been completed for you.
(b) RNA polymerase and DNA polymerase are both enzymes. RNA polymerase
is involved in the action of some control elements, whereas DNA
polymerase is not.
Describe and explain the difference between the function of these two
enzymes.
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___________________________________________________________ [4]
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(c) Another mechanism that can act to change the body plan of an organism
during its development is programmed cell death.
Fill in the gaps of the following passage describing this process and the
importance of its regulation.
Programmed cell death is known as ____________________. Firstly the fine
network of protein filaments and microtubules known as the
____________________, which gives structure to the cell, is broken down
and digested by ____________________.
The plasma (cell surface) membrane then changes, forming small bulges
called ‘blebs’. The cell breaks into membrane-bound fragments that are
removed by the process of ____________________ so that harmful
substances are not released into surrounding tissues.
Programmed cell death is a controlled process. However, mutation in a gene
called p53 can prevent programmed cell death. When this occurs, the rate
at which somatic cells are produced by the process of
____________________ becomes greater than the rate at which cells die,
resulting in the formation of a mass of cells known as a
____________________.
[6]
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Meiosis and variation
Learning objectives:
 Describe, with the aid of diagrams and photographs, the behaviour of chromosomes during
meiosis, and the associated behaviour of the nuclear envelope, cell membrane and
centrioles (names of the main stages are expected, but not the subdivisions of prophase);
 Explain the terms allele, locus, phenotype, genotype, dominant, codominant and recessive;
 Explain the terms linkage and crossing-over;
 Explain how meiosis and fertilisation can lead to variation through the independent
assortment of alleles;
 Use genetic diagrams to solve problems involving sex linkage and codominance;
 Describe the interactions between loci (epistasis) (production of genetic diagrams is not
required);
 Predict phenotypic ratios in problems involving epistasis;
 Use the chi-squared (Χ2) test to test the significance of the difference between observed and
expected results (the formula for the chi-squared test will be provided);
 Describe the differences between continuous and discontinuous variation;
 Explain the basis of continuous and discontinuous variation by reference to the number of
genes which influence the variation;
 Explain that both genotype and environment contribute to phenotypic variation (no
calculations of heritability will be expected);
 Explain why variation is essential in selection;
 Use the Hardy-Weinberg principle to calculate allele frequencies in populations;
 Explain, with examples, how environmental factors can act as stabilising or evolutionary
forces of natural selection;
 Explain how genetic drift can cause large changes in small populations;
 Explain the role of isolating mechanisms in the evolution of new species, with reference to
ecological (geographic), seasonal (temporal) and reproductive mechanisms;
 Explain the significance of the various concepts of the species, with reference to the
biological species concept and the phylogenetic (cladistic/evolutionary) species concept;
 Compare and contrast natural selection and artificial selection;
 Describe how artificial selection has been used to produce the modern dairy cow and to
produce bread wheat (Triticum aestivum);
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Key definitions:
Compile a glossary by writing your own definitions for the following key terms related to the
learning objectives above.
Key term
meiosis
haploid
homologous pairs
bivalent
chiasmata
crossing-over
maternal chromosomes
paternal chromosomes
homozygous
heterozygous
dominant
recessive
codominant
epistasis
chi-squared test
Definition
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Key term
polygenes
polygenic
Hardy-Weinberg
equation
population genetics
gene pool
environmental
resistance
selection pressure
stabilising selection
directional selection
evolutionary force
isolating mechanisms
genetic drift
biological species
concept
polyploidy
Definition
A2 BIOLOGY
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Meiosis
Meiosis is a special type of cell division concerned with producing sex cells (gametes) for the
purpose of sexual reproduction. It involves a single chromosomal duplication followed by two
successive nuclear divisions, and results in a halving of the diploid chromosome number. Meiosis
occurs in the sex organs of plants and animals. If genetic mistakes (gene and chromosome
mutations) occur here, they will be passed on to the offspring (they will be inherited).
At the onset of meiosis, DNA strands thicken into chromosomes. Homologous, or like,
chromosomes begin to approach each other.
Homologous chromosomes pair to form bivalents. The centrioles divide and move to
opposite poles of the cell.
The bivalents duplicate to form tetrads, or four-chromatid groups. The nuclear membrane
disintegrates. Crossing over (recombination) occurs.
In metaphase I, the tetrads attach to spindle fibres at their centromeres,
and line up at mid cell.
In early anaphase I, the tetrads separate, and the paired chromatids
move along the spindle to their respective centrioles.
In late anaphase I, the chromatids have almost reached the
spindle poles. The cell membrane begins to constrict.
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In telophase I, nuclear membranes enclose the separated chromatids.
The cell membrane completes its constriction.
The first meiotic division ends. There are now two cells, each with the
same number of chromatids as the parent cell.
Prophase II begins. In the second meiotic division, homologous
chromatids do not duplicate but merely separate.
In metaphase II, the chromatids line up at mid-cell. The centrioles and
asters are at the poles. A spindle has formed.
In anaphase II, the now-separated chromatids approach their respective
poles. The cell membrane begins to constrict.
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Telophase II has been completed. There are now four cells, each with
half the number of chromosomes of the parent cell.
Meiosis as seen under a microscope:
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Alleles
Sexually reproducing organisms in nearly all cases have paired sets of chromosomes, one set coming
from each parent. The equivalent chromosomes that form a pair are termed homologues. They
contain equivalent sets of genes on them but there is the potential for different versions of a gene
to exist in a population and these are termed alleles.
Homologous chromosomes: in sexually reproducing organisms, most cells have a
homologous pair of chromosomes (one coming from each parent). Each homologue carries
an identical assortment of genes, but the version (allele) of the gene from each parent may
differ.
Two different versions of a gene create a condition known as heterozygous. Only the
dominant allele will be expressed.
When both chromosomes have identical copies of the dominant allele, the organism is said
to be homozygous dominant for that gene.
When both chromosomes have identical copies of the recessive allele, the organism is said
to be homozygous recessive for that gene.
Genes occupying the same locus or position on a chromosome code for the same trait.
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Genetic diagrams
The following problems involve Mendelian crosses through to the F2 generation. The alleles
involved are associated with various phenotypic traits in domestic breeds.
1. The Himalayan colour-pointed, long-haired cat is a breed developed by crossing a pedigree
(true breeding), uniform-coloured, long-haired Persian with a pedigree colour-pointed
(darker face, ears, paws and tail) short-haired Siamese.
Persian _______________
Siamese ______________
Himalayan _____________
The genes controlling hair colouring and length are on separate chromosomes: uniform
colour U, colour pointed u, short hair S, long hair s.
(a) Using the symbols above, indicate the genotype of each breed below its photograph.
(b) State the genotype of the F1 (Siamese x Persian): _____________________________
(c) State the phenotype of the F1 _____________________________________________
(d) Use the Punnett square to show the outcome of the cross between the F1 (the F2):
(e) State the ratio of the F2 that would be Himalayan:
___________
(f) State whether the Himalayan would be true
breeding:
___________
(g) State the ratio of the F2 that would be colour-point, short-haired cats: __________
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(h) Explain how two F2 cats of the same phenotype could have different genotypes:
_______________________________________________________________________
_______________________________________________________________________
(i) Explain how you could determine which of the F2 colour-point, short hairs were true
breeding for these characters:
_______________________________________________________________________
_______________________________________________________________________
2. In rabbits, spotted coat S is dominant to solid colour s, while for coat colour: black B is
dominant to brown b. a brown spotted rabbit is mated with a solid black one and all the
offspring are black spotted (the genes are not linked).
(a) State the genotypes: male parent __________ female parent __________
offspring __________
(b) Use the Punnett square to show the outcome of a cross between the F1 (the F2):
(c) Use ratios, state the phenotypes of the F2 generation: ____________________________
________________________________________________________________________
(d) State the name given to this type of cross: ______________________________________
3. In guinea pigs, rough coat R is dominant over smooth coat r and black coat B is dominant
over white coat b. the genes for coat texture and colour are not linked. In a cross of a
homozygous rough black animal with a homozygous smooth white:
(a) State the genotype of the F1: ________________________________________________
(b) State the phenotype of the F1: _______________________________________________
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(c) Use the Punnett square to show the outcome of a cross between the F1 (the F2):
(d) Use ratios, state the phenotypes of the F2 generation: ____________________________
________________________________________________________________________
(e) Use the Punnett square to show the outcome of a cross between the offspring of a back
cross of the F1 to the rough, black parent:
(f) Use ratios, state the phenotypes of the F2 generation: ____________________________
________________________________________________________________________
(g) Use the Punnett square to show the outcome of a cross between the offspring of a test
cross of the F1 to the smooth, white parent:
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(h) Use ratios, state the phenotypes of the F2 generation: ____________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
(i) A rough black guinea pig was crossed with a rough white one produced the following
offspring: 28 rough black, 31 rough white, 11 smooth black, and 10 smooth white.
Determine the genotypes of the parents:
________________________________________________________________________
________________________________________________________________________
4. Chickens with shortened wings and legs are called creepers. When creepers are mated to
normal birds, they produce creepers and normal birds with equal frequency. When creepers
are mated to creepers they produce two creepers to one normal. Crosses between normal
birds produce only normal progeny. Explain these results.
___________________________________________________________________________
___________________________________________________________________________
5. Black wool of sheep is due to a recessive allele (b), and white wool to its dominant allele (B).
A white ram is crossed to a white ewe. Both animals carry the allele for black (b). They
produce a white ram lamb, which is then back crossed to the female parent. Determine the
probability of the back cross offspring being black.
___________________________________________________________________________
6. Mallard ducks have their plumage colour controlled by a gene with three alleles: MR
restricted mallard pattern, M mallard pattern, and m dusky mallard pattern. The dominance
hierarchy is: MR > M > m (i.e. MR is more dominant than M, which is more dominant than
m). Determine the genotypic and phenotypic ratios expected in the F1 of the following
crosses:
(a) MRMR x MRM:
genotypes: ________________________________________________
phenotypes: _______________________________________________
(b) MRMR x MRm:
genotypes: ________________________________________________
phenotypes: _______________________________________________
(c) MRM x MRm:
genotypes: ________________________________________________
phenotypes: _______________________________________________
SACKVILLE SCIENCE DEPARTMENT
(d) MRm x Mm:
A2 BIOLOGY
genotypes: ________________________________________________
phenotypes: _______________________________________________
(e) Mm x mm:
genotypes: ________________________________________________
phenotypes: _______________________________________________
7. A dominant gene (W) produces wire-haired texture in dogs; its recessive allele (w) produces
smooth hair. A group of heterozygous wire-haired individuals are crossed and their F1
progeny are then test-crossed. Determine the expected genotypic and phenotypic ratios
among the test cross progeny:
___________________________________________________________________________
___________________________________________________________________________
Using Chi-Squared in genetics
The chi-squared test, X2, is frequently used for testing the outcome of dihybrid crosses against
expected (predicted) Mendelian ratio, and it is appropriate for use in this way. When using the chisquared test for this purpose, the null hypothesis predicts the ratio of offspring of different
phenotypes according to the expected Mendelian ratio for the cross, assuming independent
assortment of alleles (no linkage). Significant departures from the predicted Mendelian ratio
indicate linkage of the alleles in question. Raw counts should be used and a large sample size is
required for the test to be valid.
SACKVILLE SCIENCE DEPARTMENT
A2 BIOLOGY
1. Four different eye pigments in the fruit fly, Drosophila melanogaster, are made
from the amino acid tryptophan. A simplified metabolic pathway of pign=ment
production is shown in Fig. 2.1. Three different gene loci control the pathway.
Each locus has two alleles. These alleles are V or v, C or c and B or b, as shown
in Fig. 2.1.
(a) Using the information in Fig. 2.1, deduce the phenotypes of flies with the
following genotypes:
genotype
phenotype
VvCcBb
___________________________________________
vvCCBB
___________________________________________
VvccBB
________________________________________ [3]
State the term that is applied to this type of gene interaction.
__________________________________________________________ [1]
SACKVILLE SCIENCE DEPARTMENT
A2 BIOLOGY
Explain how the products coded for by the genes interact to give the
different pigments.
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
___________________________________________________________ [3]
(b) A mutation in another gene at another locus in Drosophila gives rise to
white-eyed flies. The red eye allele of this gene (R) is known to be dominant
to the white eye allele (r).
A student crossed a red-eyed fly with a white-eyed fly, expecting to get an
F1 generation of red-eyed flies. In fact, the results were as shown in Table
2.1.
The student first suggested that the reason for there being red-eyed and
white-eyed flies in the offspring was that the red-eyed parent was
heterozygous.
Explain why this cannot be the correct explanation for the results shown in
Table 2.1.
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
__________________________________________________________ [2]
SACKVILLE SCIENCE DEPARTMENT
A2 BIOLOGY
(c) In Drosophila, the males are the heterogametic sex, possessing two
different sex chromosomes, X and Y.
Draw a genetic diagram to show how the results shown in Table 2.1 could
have been produced.
Parental genotypes ____________________ ____________________
Gametes
____________________ ____________________
F1 genotypes
____________________ ____________________
[3]
(d) The chi-square (X2) test can be used to analyse the results in Table 2.1.
The expected ratio of red-eyed females to white-eyed males is 1 : 1.
Use Table 2.2 to calculate a value for chi-squared (X2).
SACKVILLE SCIENCE DEPARTMENT
A2 BIOLOGY
Use your calculated value of X2 and the table of probabilities shown in Table
2.3 to test the significance of the difference between the observed and
expected results.
State your conclusion in the space below:
Conclusion ____________________________________________________
_____________________________________________________________
_____________________________________________________________
__________________________________________________________ [4]
2. The fruit fly, Drosophila melanogaster, the zebra fish, Danio rerio, and the
mouse, Mus musculus, have all been used by scientists to find out more about
how genes control development in all animals, including humans. They are
described as ‘model organisms’.
(a) Suggest why information gained from studying such model organisms can
be applied to humans.
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
__________________________________________________________ [2]
(b) Suggest two characteristics that researchers should look for when choosing
an organism for research into how genes control development.
1. ___________________________________________________________
2. ________________________________________________________ [2]
SACKVILLE SCIENCE DEPARTMENT
A2 BIOLOGY
(c) Fig. 3.1 and Fig. 3.2, below, show the heads of two Drosophila fruit flies.
Fig. 3.1 shows a normal wild type fly.
Fig. 3.2 shows a mutant fly.
Name the type of microscope used to take the two pictures.
__________________________________________________________ [2]
SACKVILLE SCIENCE DEPARTMENT
A2 BIOLOGY
State one significant difference between the two heads.
_____________________________________________________________
__________________________________________________________ [1]
Name the type of gene which, if mutated, gives rise to dramatic changes in
body plan.
__________________________________________________________ [1]
3. Albinism is a genetic disorder in which a person lacks melanin pigment in their
skin, hair and the iris of their eyes. A person with this disorder is called an
albino. The genotype of an albino has two copies of a recessive allele of the
gene for an enzyme involved in melanin production.
(a) State the term used to describe a genotype that has two copies of the same
allele at a particular gene locus.
__________________________________________________________ [1]
(b) Explain what is meant by the following terms:
genotype _____________________________________________________
_____________________________________________________________
_____________________________________________________________
allele ________________________________________________________
_____________________________________________________________
__________________________________________________________ [4]
4. Domestic chickens have been bred for many years to increase the number of
eggs laid by the females. It is useful to be able to identify the young female
chicks on the day after they hatch as only the females need to be kept for
laying eggs.
SACKVILLE SCIENCE DEPARTMENT
A2 BIOLOGY
Unlike mammals, where the sex chromosomes are known as X and Y, in
chickens the sex chromosomes are known as Z and W.
 Male chickens have two Z chromosomes (ZZ).
 Female chickens have one Z chromosome and one W chromosome (ZW).
(a) Some genes for feather colour and pattern in chickens are carried on the Z
chromosome but not on the W chromosome. One such example is the gene
for striped feathers (barring).
State the name given to this type of inheritance.
__________________________________________________________ [1]
(b) Inheritance of the barring pattern can be used to identify female chicks
when they are one day old.
The phenotypes associated with the two alleles of the barring gene are
shown in Table 1.1.
State the adult phenotypes and sex of the following individuals:
ZBZb
_____________________________________________________
ZBW
_____________________________________________________
ZbW
__________________________________________________ [3]
SACKVILLE SCIENCE DEPARTMENT
A2 BIOLOGY
(c) A cross was carried out between a barred female and a non-barred male.
Complete the genetic diagram to show the parental genotypes, their
gametes and the F1 genotypes. State the phenotypes of the offspring as
day-old chicks.
Parent phenotypes
Barred female
Non-barred male
Parent genotypes
_____________
_______________
Gametes
_____________
_______________
F1 genotypes
_____________
_______________
F1 day-old chick
phenotypes
_____________
_______________
male ________________________________________________________
_____________________________________________________________
female ______________________________________________________
_____________________________________________________________
[5]
(d) The autosomal gene I/I shows epistasis over all other genes affecting
feather colour in chickens.
Individuals carrying the dominant allele I have white feathers.
Chickens that are not white have the genotype ii.
State the precise term used to describe the genotype ii.
__________________________________________________________ [1]
Predict the colour(s) of the offspring of a cross between a male homozygous
barred chicken and a white female chicken with the genotype II.
__________________________________________________________ [1]
SACKVILLE SCIENCE DEPARTMENT
A2 BIOLOGY
5. The following boxes show the names of different stages that occur during
meiosis.
(a) State the stage(s) in which the following events occur:
independent assortment
____________________________
formation of the spindle apparatus
____________________________
separation of sister chromatids
____________________________
formation of nuclear membranes
____________________________
chromosomes pulled to opposite poles ____________________________
[5]
(b) Meiosis is used in many organisms for the production of gametes.
Explain why meiosis needs to have twice as many stages as mitosis.
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
___________________________________________________________ [2]
(c) Meiosis is a source of genetic variation. Mutation is another source of
variation.
What feature of the DNA molecule is changed as a result of mutation?
__________________________________________________________ [1]
SACKVILLE SCIENCE DEPARTMENT
A2 BIOLOGY
Discuss the possible effects that mutation can have on the structure and
function of a protein.
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
___________________________________________________________ [3]
6. Describe the differences between prophase 1 of meiosis and prophase 2 of
meiosis.
________________________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
_____________________________________________________________ [3]