Download Biology Answers Cr8 moderation

Part 4 – Cr 8: Survival of Change
Cutoffs:
A > 23
B > 16
C > 12
C- = 10 to 12
t < 10
Question 1
Consider the diagram showing the reproductive cycle of a fern (type of plant):
a) Identify the processes in the diagram:
(1.5 marks)
a. Process A: Fertilisation
b. Process B: Mitosis
c. Process C: Meiosis
b) Referring to the processes and stages shown in the diagram, explain the benefits of
sexual reproduction.
(4 marks)
The meiotic division, process C, creates genetic diversity as it allows crossing over, the
exchange of genetic material between homologous chromosomes (1) and independent
assortment, the random lining up of homologous pairs for daughter cells to be produced
(1).
Further genetic diversity occurs in process A, fertilization , when two haploid cells from
the two individuals come together and fuse creating a new diploid zygote containing
genetic material from each parent (1). Process B, mitosis, then occurs so that the zygote
goes from a single cell to a multicellular fern (1).
COMMENTS:
This question was poorly attempted. Many students did not list processes in part (a), and
instead had a massive information dump of random concepts.
Part (b) was not well responded too, and many students compared and contrasted asexual
reproduction with sexual reproduction, which is not what the question wants them to do.
Question 2
a) Label the chart to show the genotypes for each individual.
½ mark taken off for each incorrect genotype; no marks for X linked notation
(2 marks)
b) What is the likely mode of inheritance for Falconi’s anaemia? Justify your response.
(2 marks)
The likely mode of inheritance is autosomal recessive. (½)
As Ann and Michael are unaffected, but their daughter Carla is affected means it is a
recessive allele (1). It cannot be X-linked, as otherwise Sam would have inherited it from
Arlene. (½)
Other reasons also accepted.
c) List four people from the chart who are carriers of Falconi’s anemia
(2 marks)
Definitely Ff: George, Sam, Ann, Michael, Daniel, Alan (½ mark per correct response)
Could be either Ff or FF (so may not be carriers): Sandra, Tina or Christopher (0)
COMMENTS:
Attempted reasonably well by students. Part (a) needed to show the difference between F
and f more clearly. Part (b) many could say it was recessive, but not indicate using examples
from the chart – needed to do this to gain full marks.
However, many did not understand part (c) to be able to indicate who were carriers of the
disease and listed Arlene, Tom, Wilma and Carla who are affected, and not carriers.
Question 3
A female horse and a male donkey can breed to create offspring known as a mule.


Diploid cells of a horse contain 64 chromosomes
Diploid cells of a donkey contain 62 chromosomes
Use this information to explain why a mule would be infertile.
-
(4 marks)
Meiotic divisions in a horse would yield 32 chromosomes in the gametes (haploid
cells),( ½ ) whereas in a donkey would yield 31 chromosomes in the gametes (haploid
cells). ( ½ )
-
Therefore, when a horse and a donkey mate, the resulting zygote (the mule) would
contain 63 chromosomes. (1)
-
For sexual reproduction in a mule to create a mule, the gametes would need to
contain 31.5 chromosomes, to create a baby mule with 63 chromosomes which can
clearly not occur (1)
-
Meiosis could not create haploid cells with 31.5 chromosomes therefore no sexual
reproduction and no offspring meaning mules are infertile (1)
OR
-
Horses and donkeys are different species, their chromosomes may not be the same
and therefore different chromosomes may contain different genes (alleles) between
each corresponding chromosome (1)
-
Having different numbers of chromosomes means that there will not be matched
pairs or homologous pairs in meiosis (1), this prevents the production of viable
haploid cells therefore mules would be sterile (1)
Question 4
a) i) An example of incomplete dominance:
(2 marks)
Scenario: (3) Snapdragon plants
Explanation: When the two homozygous traits, that is a yellow and a blue phenotype are
present in the P1 generation (1) but are not seen in the resulting cross F1 generation, but a
combination phenotype (trait somewhere between the parents), in this case green (1), is
seen this is incomplete dominance. OR Neither trait is recessive to the other, and while both
are expressed – they are not completely expressed forming a phenotype that is a
combination of both (1)
ii) An example of co-dominance:
(2 marks)
Scenario: (1) Bloodtypes OR (2) Roan cattle
Explanation: When both alleles IA and IB when inherited together are seen in the phenotype
as they are expressed together/equally , giving the blood type AB (1). This is co-dominance
where neither A or B phenotype is recessive to the other (1).
OR
When a cross between two pure breeding individuals such as the white (homozygous) cow
and red (homozygous) bull are mated producing offspring that are heterozygous –the Roan
cow - which expresses both phenotypes (1) having both red and white hairs displaying
codominance where neither trait is recessive to the other, but seen individually in the
heterozygous form (1).
iii) An example of multiple alleles:
(2 marks)
Scenario: (1) Blood types
Explanation: There are three alleles, IA, IB or i , and not two alleles that can be expressed to
determine blood type (1) so that instead of 2 possible phenotypes being produced there are
four possible phenotypes –
-
Type A blood can be an expression of the genotypes IAIA, IAi
Type B blood can be an expression of the genotypes IBIB, IBi
Type AB blood is expressed by the genotype IAIB
Type O blood is expressed by the genotype ii
(1)
COMMENTS:
Most students identified the correct scenario, however, did not explain in relation to that
scenario. Some students made up their own scenario to explain and many just used the
information straight from the information sheet. If the latter was the case, a maximum of 1
mark could be achieved. If the scenario was not used in the explanation, maximum marks
could not be obtained.
Question 5
a) Using information from the pedigree, analyse and describe the likely mode of
transmission for retinosa pigementosa.
(3 marks)
The mode of inheritance of retinosa pigmentosa is likely to be X-linked ( ½ ) dominant. ( ½ )
Individual I-1 (the father), passes the trait to all daughters and no sons have the trait (1), as
all daughters must receive an affected X from their father (males do not receive an X from
the father, therefore unaffected). (1).
Responses might also include other rationales such as:
This can also be seen in generation IV where all affected fathers pass on the trait to all
daughters (1).
Generation III show male and females are equally affected, meaning that an affected X has
to come from the mother in generation II and as the fathers are unaffected, for daughters to
express the trait, it must be X linked dominant (1).
b) If individual IV-6 was to have children with a partner who was affected, what is the
likelihood that their children would be affected? Show all working and explain your
response.
(4 marks)
XR
Y
XR
XR XR
XRY
Xr
XR Xr
XrY
(1)
Individual IV-6 has the genotype XR Xr and a partner who was affected would have the
genotype XRY.
In the above cross, the possible outcomes for any offspring resulting from this match
would be:
-
25% chance of the offspring being of each of the following genotypes: XR XR , XR Xr,
XRY or XrY (1)
-
All daughters would be affected (1)
-
50% chance of an affected son; 50% chance of an unaffected son. (1)
COMMENTS:
Part (a) most students indicated it was X-linked dominant. Examples from the chart must be
specifically given to obtain full marks.
Part (b) was reasonably well attempted – I gave marks if the working out followed an
alternate mode of inheritance as given in part (a) – even if that was incorrect.
Only 1 mark could be given if student said “75% of children would be affected” without
breaking it down as indicated in the solution.