7.1- Genestic, inheritance

Question 1

Q1.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.

Answer 1

Both alleles are expressed / shown (in the phenotype).
Accept: both alleles contribute (to the phenotype)
Neutral: both alleles are dominant

Question 2

Q1.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.

Male cats with a tortoiseshell phenotype do not usually occur. Explain why (1)

Answer 2

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)

Question 3

(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.

Answer 3

(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

Question 4

(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. (1)

Answer 4

Homozygous dominant increases and homozygous recessive

decreases.

Question 5

Q2.Some populations of flies are becoming resistant to insecticides intended to kill them.
Scientists developed a method for finding out whether a fly was carrying a recessive
allele, r, that gives resistance to an insecticide. The dominant allele, R, of this gene does
not give resistance.
The scientists:
• crossed flies with genotype RR with flies with genotype rr
• obtained DNA samples from the parents and offspring
• used the same restriction endonuclease enzymes on each sample, to obtain DNA fragments.
(a) Explain why the scientists used the same restriction endonuclease enzymes on
each DNA sample. (2)

Answer 5

1. Cut (DNA) at same (base) sequence / (recognition) sequence;
   Accept: cut DNA at same place
2. (So) get (fragments with gene) R / required gene.
   Accept: ‘allele’ for ‘gene’ / same gene

Question 6

The scientists added two different primers to each sample of DNA fragments for the
polymerase chain reaction (PCR).
• Primer A3 only binds to a 195 base-pair fragment from allele r.
• Primer A4 only binds to a 135 base-pair fragment from allele R.
The scientists separated the DNA fragments produced by the PCR on a gel where
shorter fragments move further in a given time.
Their results are shown in Figure 1.
Figure 1
(b) Explain why primer A3 and primer A4 only bind to specific DNA fragments.

Answer 6

1. Each has / they have a specific base sequence;
2. That is complementary (to allele r or R).
   Accept description of ‘complementary’

Question 7

(c) Use all the information given to explain the results in Figure 1.

Answer 7

1. Fragments L from parent rr, because all longer fragments / 195
   base pair fragments;
   Ignore: references to fragments that move further / less,
   require identification of longer / shorter or 195 / 135
   Accept: (homozygous) recessive
2. Fragments N from parent RR, because all shorter fragments / 135 base
   pair fragments;
   1 and 2 Accept: A3 for 195 and A4 for 135
3. Accept: (homozygous) dominant
4. (M from) offspring heterozygous / Rr / have both 195 and 135 base pair
   fragments.
   Accept: have both bands / strips
   Reject: primer longer / shorter

Question 8

(d) 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. (2)

Answer 8

1. (Cells in mitosis) chromosomes visible;

2. (So) can see which chromosome DNA probe attached to

Question 9

(e) Another group of scientists thought that pesticide resistance in some flies was
related to increased activity of an enzyme called P450 monooxygenase (PM).
This enzyme breaks down insecticides.
The scientists obtained large numbers of resistant and non-resistant flies. They then
set up the following experiments.
• Non-resistant flies exposed to insecticide.
• Resistant flies exposed to insecticide.
• Resistant flies treated with an inhibitor of PM and then exposed to insecticide.
They then determined the percentage of flies that were dead at different times after
being exposed to insecticide.
Figure 2 shows their results.

(i) Explain why the scientists carried out the control experiment with the
non-resistant flies(2)

Answer 9

(i) 1. For comparison with resistant flies / other (two) experiments
/ groups;
Ignore: compare results / data / no other factors

2. To see death rate (in non-resistant) / to see effect of insecticide in
   non-resistant / normal flies.
   Accept: ‘pesticide’ as ‘insecticide’
   Accept to see that insecticide worked / to see effect of
   enzyme

Question 10

The scientists concluded that the resistance of the flies to the insecticide is
partly due to increased activity of PM but other factors are also involved.
Explain how these data support this conclusion. (4)

Answer 10

(ii) (PM must be involved because)
1. Few resistant flies die (without inhibitor);
2. More inhibited flies die than resistant flies;
3. (PM) inhibited flies die faster (than resistant flies);
(Other factors must be involved because)
4. Some resistant flies die;
5. But (with inhibitor) still have greater resistance / die slower than
non-resistant flies.
Accept: (with inhibitor) die slower than non-resistant flies

Question 11

Q3.(a) In fruit flies, the genes for body colour and wing length are linked. Explain what this
means

Answer 11

(Genes / loci) on same chromosome.

Question 12

A scientist investigated linkage between the genes for body colour and wing length.
He carried out crosses between fruit flies with grey bodies and long wings and fruit
flies with black bodies and short wings.
Figure 1 shows his crosses and the results.
• G represents the dominant allele for grey body and g represents the recessive
allele for black body.
• N represents the dominant allele for long wings and n represents the
recessive allele for short wings.
Figure 1

Phenotype of parents grey body,
long wings ×
black
body,
short
wings
Genotype of parents GGNN ggnn
Genotype of offspring GgNn

Phenotype of offspring all grey body, long wings
These offspring were crossed with flies homozygous for black body and short wings.
The scientist’s results are shown in Figure 2.
Figure 2
GgNn crossed with ggnn
Grey body,
long wings
Black body,
short wings
Grey body,
short wings
Black body,
long wings
Number of
offspring 975 963 186 194

(b) Use your knowledge of gene linkage to explain these results. (4)

Answer 12

1. GN and gn linked;
2. GgNn individual produces mainly GN and gn gametes;
3. Crossing over produces some / few Gn and gN gametes;
4. So few(er) Ggnn and ggNn individuals

Question 13

(c) If these genes were not linked, what ratio of phenotypes would the scientist have
expected to obtain in the offspring?

Answer 13

(Grey long:grey short:black long:black short) =1:1:1:1

Question 14

(d) 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.

Answer 14

1. Chi squared test;

2. Categorical data.

Question 15

Q4.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.

Answer 15

(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.

Question 16

(b) In chickens, a gene on the X chromosome controls the rate of feather production.
The allele for slow feather production, F, is dominant to the allele for rapid feather
production, f. The following figure shows the results produced from crosses carried
out by a farmer.

(i) Explain one piece of evidence from the figure which shows that the allele for
rapid feather production is recessive (2)

Answer 16

1. 1, (2) and 5;
   Accept: for 1 mark that 1 and 2 have slow (feather
   production) but produce one offspring with rapid (feather
   production).
   Neutral: any reference to 3 being offspring of 1.
2. 1 must possess / pass on the recessive allele / 1 must be a carrier
   / heterozygous / if slow (feather production) is recessive all
   offspring of (1 and 2) would be slow (feather production) / if rapid
   (feather production) was dominant 1 would have rapid (feather
   production);
   Reject: both parents must be carriers / possess the
   recessive allele.
   Reject: one of the parents (i.e. not specified) must be a
   carrier / heterozygous.

Question 17

0 6 . 1 In genetic crosses, the observed phenotypic ratios obtained in the offspring are often box
not the same as the expected ratios.
Suggest two reasons why.
[2 marks]

Answer 17

1. Small sample size;
2. Fusion/fertilisation of gametes is random;
3. Linked Genes;
4. Epistasis;
5. Lethal genotypes

Approximately 72% of students gained at least one mark for this question. The most
common correct responses related to linkage and the random fusion of gametes. A
significant number of students mentioned small sample size. Epistasis was infrequently
suggested. The most common incorrect response was ‘mutation’, followed by
environmental factors and epigenetics.

Question 18

In tomato plants, the genes for height and for the type of leaf are on the same
homologous pair of chromosomes. The allele T, for a tall plant, is dominant to the
allele t, for a dwarf plant. The allele M, for normal leaves, is dominant to the allele m,
for mottled leaves.
A biologist carried out crosses between parent plants heterozygous for both genes
and examined the offspring produced. The position of the two alleles for both genes
was the same in each parent plant as shown in Figure 7. The phenotypes and
number of offspring produced are shown in Table 3.
Figure 7
Table 3
Phenotype of offspring Number of offspring
Tall plants and normal leaves 1860
Tall plants and mottled leaves 68
Dwarf plants and normal leaves 57
Dwarf plants and mottled leaves 580
0 6 . 2 What would be the genotype of the offspring with dwarf plants and mottled leaves?
[1 mark]

Answer 18

1. ttmm;

Question 19

In tomato plants, the genes for height and for the type of leaf are on the same
homologous pair of chromosomes. The allele T, for a tall plant, is dominant to the
allele t, for a dwarf plant. The allele M, for normal leaves, is dominant to the allele m,
for mottled leaves.
A biologist carried out crosses between parent plants heterozygous for both genes
and examined the offspring produced. The position of the two alleles for both genes
was the same in each parent plant as shown in Figure 7. The phenotypes and
number of offspring produced are shown in Table 3.
Figure 7
Table 3
Phenotype of offspring Number of offspring
Tall plants and normal leaves 1860
Tall plants and mottled leaves 68
Dwarf plants and normal leaves 57
Dwarf plants and mottled leaves 580
0 6 . 3 Use the information provided to explain the results in Table 3. box
[3 marks]
[1 mark]

Answer 19

1. Genes are linked;
2. Produces few(er) tall, mottled and dwarf, normal
   offspring;
3. Crossing over (has occurred)

1. Accept ‘Alleles are
linked’ but reject if
context suggests
alleles of the ‘same
gene’
2. Accept produces
few Tm and tM
gametes
2. Accept ‘fewer
recombinants’ 

06.3 This question tested a new topic on the specification and proved challenging for many
students. Despite the stem of the question and Figure 7 providing evidence of linkage, this
was not mentioned by many students in their explanations of the results of the genetic
cross. This resulted in 45% of students scoring zero. Students who did refer to linked
genes did not always mention crossing over or the low number of phenotypes produced
because of this. Consequently, only 31% of students obtained at least two of the three
marks available. The best answers explained that if the linked genes were inherited
together, the offspring would be in a 3:1 ratio, but that crossing over produced small
number of Tm and tM gametes that resulted in the low numbers of recombinant tall and
mottled, and dwarf and normal plants. Many students who scored zero suggested that the
ratio should be 9:3:3:1 and explained why the plants with the dominant phenotypes were
most frequent. A few students suggested that mutations had caused the phenotypes which
occurred in low numbers. Some students suggested that crossing over occurs in mitosis.

Question 20

6 . 4 Complete Table 4 to show the expected ratio of phenotypes if the same cross had
been carried out but the genes for height of plant and for the type of leaf were on
different homologous pairs of chromosomes.
[2 marks]

Answer 20

One mark for each correct column;;
Phenotype of
offspring
Ratio of
offspring
Tall (plant and) normal
(leaves)
9
Tall (plant and) mottled
(leaves)
3
Dwarf (plant and)
normal (leaves)
3
Dwarf (plant and)
mottled (leaves)
1

06.4 Almost 50% of students obtained both marks for this question. Students gaining a single
mark (30%) usually correctly completed the phenotypes of the offspring in Table 4.
Incorrect responses for the ratio of offspring in Table 4 varied considerably, however a ratio
of 1:1:1:1 was seen relatively frequently

Question 21

0 Read the following passage. box
Complete achromatopsia is a form of complete colour blindness. It is caused
by having only rods and no functional cone cells. People with complete
achromatopsia have difficulty in seeing detail. Complete achromatopsia is
caused by an autosomal recessive allele and is usually very rare in
populations with only one in 40 000 being affected. However on the Pacific
island of Pingelap ten percent of the population are affected.
One form of red-green colour blindness is caused by a sex-linked recessive
allele which affects more men than women. People with this red-green
colour blindness are unable to distinguish between red and green, and also
between other colours. They have green-sensitive cones but the
photoreceptive pigment they contain does not function.
Scientists investigated the use of gene therapy to correct red-green colour
blindness in monkeys. They injected viruses containing the gene for the
green-sensitive pigment directly into the eyes of the monkeys. Although the
monkeys maintained two years of colour vision, there is debate on whether
this form of gene therapy is worthwhile. No clinical trials of this procedure
have been carried out on humans. Current research into the treatment of
red-green colour blindness involves the use of induced pluripotent stem cells
(iPS cells). The use of iPS cells could have advantages over the use of gene
therapy.
Use the information in the passage and your own knowledge to answer the
following questions.
5
10
15
20
1 0 . 1 People with complete achromatopsia have difficulty in seeing detail (lines 2–3).
Explain why.
[3 marks

Answer 21

1. No (functional) cones
   OR
   Only rods;
2. Cones are connected to a single neurone
   OR
   Several rods connected to a single neurone;
3. (Cones) Separate (sets of) impulses to brain
   OR
   (Rods) Single (set of) impulse/s to brain;

Question 22

Identify label A-C on the chromosome in figure 1

Answer 22

Sister chromatids
Centromere
Chromosome

Question 23

Define the term Genotype

Answer 23

The complete set of genes in an organism