TOPIC 7 + TOPIC 4 EXAM QUESTIONS Flashcards
.In cats, males are XY and females are XX. A gene on the X chromosome controls fur colour
in cats. The allele G codes for ginger fur and the allele B codes for black fur. These alleles
are codominant. Heterozygous females have ginger and black patches of fur and their
phenotype is described as tortoiseshell.
(a) Explain what is meant by codominant alleles.
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(1)
(b) Male cats with a tortoiseshell phenotype do not usually occur. Explain why.
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(1)
(c) A tortoiseshell female was crossed with a black male. Use a genetic diagram to
show all the possible genotypes and the ratio of phenotypes expected in the
offspring of this cross.
Use X
G
to indicate the allele G on an X chromosome.
Use X
B
to indicate the allele B on an X chromosome.
Genotypes of offspring ………………………………………………………………………
Phenotypes of offspring ……………………………………………………………………..
Ratio of phenotypes …………………………………………………………………………..
(d) Polydactyly in cats is an inherited condition in which cats have extra toes. The allele
for polydactyly is dominant.
(i) In a population, 19% of cats had extra toes. Use the Hardy-Weinberg equation
to calculate the frequency of the recessive allele for this gene in this
population.
Show your working.
Answer = ……………………….
(2)
(ii) Some cat breeders select for polydactyly. Describe how this would affect the
frequencies of the homozygous genotypes for this gene in their breeding
populations over time.
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(
M1.(a) Both alleles are expressed / shown (in the phenotype).
Accept: both alleles contribute (to the phenotype)
Neutral: both alleles are dominant
1
(b) Only possess one allele / Y chromosome does not carry allele / gene / can’t
be heterozygous.
Accept: only possess one gene (for condition)
Neutral: only 1 X chromosome (unqualified)
1
(c) 1. XGX
B
, XBX
B
, XGY, XBY;
Accept: equivalent genotypes where the Y chromosome is
shown as a dash e.g. XG
-, or is omitted e.g. XG
Reject: GB, BB, GY, BY as this contravenes the rubric
2. Tortoiseshell female, black female, ginger male, black male;
3. (Ratio) 1:1:1:1
2 and 3. Award one mark for following phenotypes
tortoiseshell, black, (black) ginger in any order with ratio of
1:2:1 in any order.
Allow one mark for answers in which mark points 1, 2 and 3
are not awarded but show parents with correct genotypes i.e.
X
GX
B
and XBY or gametes as XG
, XB
and XB
, Y
3. Neutral: percentages and fractions
3. Accept: equivalent ratios e.g. for 1:1:1:1 allow 0.25 : 0.25 :
0.25 : 0.25
3
(d) (i) Correct answer of 0.9 = 2 marks;
Incorrect answer but shows q2
= 0.81 = one mark.
Note: 0.9% = one mark
2
(ii) Homozygous dominant increases and homozygous recessive
decreases.
The scientists wanted to know on which chromosome the gene with alleles R and r
was located. From the flies with genotype RR, they obtained cells that were in
mitosis and added a labelled DNA probe specific for allele R. They then looked at
the cells under an optical microscope.
Explain why they used cells that were in mitosis.
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(Cells in mitosis) chromosomes visible;
2. (So) can see which chromosome DNA probe attached to.
) Which statistical test could the scientist use to determine whether his observed
results were significantly different from the expected results?
Give the reason for your choice of statistical test.
- Chi squared test;
- Categorical data.
.In birds, males are XX and females are XY.
(a) Use this information to explain why recessive, sex-linked characteristics are more
common in female birds than in male birds.
(Recessive) allele is always expressed in females / females have one
(recessive) allele / males need two recessive alleles / males need to be
homozygous recessive / males could have dominant and recessive alleles /
be heterozygous / carriers;
Accept: Y chromosome does not carry a dominant allele.
Other answers must be in context of allele not chromosome
or gene.
Species of plant Number counted in
1 m^2
Common heather 2
Red fescue 14
Vetch 2
White clover 8
(a) What is the species richness of this sample?
(1)
(b) Calculate the index of diversity of this sample. Show your working.
Use the following formula to calculate the index of diversity.
d =
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where N is the total number of organisms of all species
and n is the total number of organisms of each species
Index of diversity = ……………………………..
(2)
(c) Suggest how this student would obtain data to give a more precise value for the
index of diversity of this habitat.
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2.68(6).
If answer incorrect:
Σn(n-1) = 242 = 1 mark
N(N-1) = 650 = 1 mark
2
(c) 1. Take more samples and find mean;
2. Method for randomised samples described.
Allow larger area = 1 mark
Species richness and an index of diversity can be used to measure biodiversity within a
community.
(a) What is the difference between these two measures of biodiversity?
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Species richness measures only number of (different) species / does not measure
number of individuals.
Ecologists investigated the size of an insect population on a small island. They used a
mark-release-recapture method. To mark the insects they used a fluorescent powder. This
powder glows bright red when exposed to ultraviolet (UV) light.
(a) The ecologists captured insects from a number of sites on the island. Suggest how
they decided where to take their samples.
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(2)
(b) Give two assumptions made when using the mark-release-recapture method.
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1 ………………………………………………………………………………………………………
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2 ………………………………………………………………………………………………………
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(2)
(c) Suggest the advantage of using the fluorescent powder in this experiment.
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Draw grid over (map of) area;
2. Select squares / coordinates at random.
2
(b) 1. No emigration / immigration;
2. No losses to predation;
3. Marking does not affect survival;
4. Birth rate and death rate equal;
5. (In this case) all belong to one population.
2 max
(c) 1. Only glows brightly with UV, so doesn’t make insects more visible;
2. So doesn’t affect / increase predation;
OR
1. Glows brightly with UV marking visible;
2. So makes it easy to pick out labelled insects.
What two measurements are needed to calculate an index of diversity?
1 ………………………………………………………………………………………………………
2 ………………………………………………………………………………………………………
Number of (individuals of) each species;
Accept: ‘population’ for ‘number
structure of protein
- Polymer of amino acids;
- Joined by peptide bonds;
- Formed by condensation;
- Primary structure is order of amino acids;
- Secondary structure is folding of polypeptide chain due to hydrogen
bonding;
Accept alpha helix / pleated sheet - Tertiary structure is 3-D folding due to hydrogen bonding and ionic /
disulfide bonds; - Quaternary structure is two or more polypeptide chains.
transcription steps
- Helicase;
- Breaks hydrogen bonds;
- Only one DNA strand acts as template;
- RNA nucleotides attracted to exposed bases;
- (Attraction) according to base pairing rule;
- RNA polymerase joins (RNA) nucleotides together;
- Pre-mRNA spliced to remove introns.
how are proteins digested in a human gut
- Hydrolysis of peptide bonds;
- Endopeptidases break polypeptides into smaller peptide chains;
- Exopeptidases remove terminal amino acids;
- Dipeptidases hydrolyse / break down dipeptides into amino acids.
