DBQ on p. 153 – February 15, 2013
1) Explain why there are no male carriers of the allele for hemophilia in this pedigree
chart.
Hemophilia is a sex-linked condition, which means that the gene related to hemophilia
is not present on the Y chromosome. Whether or not the allele is present on the X
chromosome determines if the male will have hemophilia.
2) Explain why there are no females with hemophilia in this pedigree chart.
There are no females with hemophilia in this pedigree chart because the gene for
hemophilia is recessive, so if they have at least one allele for non-hemophilia then they
will not have the disease. Therefore, the chances of females getting hemophilia is
lower than for males.
3) Explain the evidence from the pedigree chart for hemophilia being a sex-linked
condition.
The evidence from the pedigree chart for hemophilia being a sex-link condition is that
there are no carrier males. Because there are seven carrier females in total, it can be
assumed that the lack of carrier males is not a coincidence. Furthermore, none of the
females were hemophiliac. This difference in phenotype between males and females
suggests that there is a sex linkage.
4) Deduce the possible genotypes and phenotypes of the daughters of Nicholas II of
Russia and Alexandria. Show your working and indicate the probabilities.

Heterozygous – non-hemophilia
o Chances: 1/2

Two alleles for non-hemophilia (homozygous dominant)
o Chances: 1/2
5) Deduce the possible genotypes and phenotypes of the second and third sons of
Alfonso XIII of Spain and Victoria Eugenia. Show your working and indicate the
probabilities.
 Allele for hemophilia
o Chances: ½
 Allele for non-hemophilia
o Chance: ½
DBQ on p. 147 – February 13, 2013
1) Compare the typica and annulata forms of Adalia 2-punata.
The typica form of Adalia 2-punata has a black dot on each of its wings, whereas the
annulata form has a connected pattern from its head to the bottom of its wings. Half of
the head of the typica is yellow, but the head of the annulata form has less yellow
(around ¼ of its head is yellow).
2) The differences between the two forms are due to a single gene. If male and female
typica are mated together, all the offspring are typica. Similarly, the offspring
produced when annulata forms are mated are all annulata. Explain the conclusions
that can be drawn.
The alleles for the typica and annulata form are co-dominant. For a ladybird to be
typica, it would have to have two alleles that coded for that form. A typica would be
homogeneous, so the offspring of two typica would also be homogeneous and
therefore a typica. The same applies for annulata.
3) When typica is mated with annulata, the F1 hybrid offspring are not identical to
either parent. Examples of these F1 hybrid offspring are shown in Figure 10a.
Distinguish between the F1 hybrid offspring and the typica and annulata parents.
The typica have two black dots on each of their wings, and the annulata have what
looks like an upside down T with a line through the middle on their wings. On the
other hand, the F1 hybrid offspring having markings that are somewhat similar to
those of the parents, but not quite the same. For example, there might be an identical
mark on each of the wings, but the shape of the mark wouldn’t be a circle.
4) If F1 hybrid offspring are mated with each other, the offspring include both typica
and annulata forms, and also offspring with the same wing case markings as the F1
hybrid offspring.
a) Use a genetic diagram to explain this pattern of inheritance.
b) Predict the expected ratio of phenotypes.
1 typica: 2 mixed features: 1 annulata
DBQ on p. 145 – February 8, 2013
1) Calculate the ratio between grey and albino offspring, showing your working.
Second generation  1:0
Third generation  11:4
2) Deduce the colour of coat that is due to a recessive allele, with two reasons for your
answer.
The color of coat that is due to a recessive allele is white fur. The first reason is
because none of the hybrid mice had white fur even though one of their parents had
white fur – this shows that allele for grey fur (which is what all the hybrid mice had)
dominated over the allele for white fur. Furthermore, even in the third generation, the
number of mice with grey fur was greater than the number of mice with white fur.
3) Choose suitable symbols for the alleles for grey and albino coat and list the possible
genotypes of mice using your symbols, together with the phenotype for each
genotype.
T: grey coat – black eyes
t: albino coat – red eyes
4) Using the headings shown to the right, explain how the observed ratio of grey and
albino mice was produced.
The parental phenotypes were grey fur with black eyes and white fur with red eyes,
and the genotypes were TT, Tt/tT, and tt. The alleles in the gametes were the alleles
for grey fur with black eyes and white fur with red eyes. The hybrid phenotype was
grey fur with black eyes, and the genotypes was tT/Tt. The alleles in their gametes
were the same as the alleles in their parents’ gametes. The genotypes of the offspring
of the hybrid mice were grey fur with black eyes and white fur with red eyes, and the
genotypes were tt, TT, Tt/tT.
5) Suggest how one gene can determine whether the mice had grey fur and black eyes
or white fur and red eyes.
One gene could have a dominant allele that is responsible for the grey fur with black
eyes and a recessive allele that is responsible for the white fur and red eyes.
DBQ on p. 138 – February 6, 2013
1)
a) In figure 9, distinguish between:
i) chromosome 5 and chromosome 6
Chromosome 5
ii) chromosome 17 and chromosome 18
Chromosome 18 has a thicker yellow tip than chromosome 17.
Chromosome 18 also has a white band at the other end whereas chromosome 17
doesn’t.
iii) the X and Y chromosome
The X chromosome is larger than the Y chromosome.
2)
a) State the gender of the subject of the human karyotype in Figure 9.
Female.
b) State whether the karyotype shows any abnormalities.
DBQ on p. 137 – February 6, 2013
1) Outline the relationship between maternal age and the incidence of chromosomal
abnormalities in live births.
There is an exponential relationship between maternal age and the incidence of
chromosomal abnormalities in live births; as maternal age increases, incidence
increase at an accelerating rate.
2)
a) For mothers 40 years of age, determine the probability that they will give
birth to a child with trisomy 21.
The probability is approximately 1%.
b) Using the data in Figure 5, calculate the probability that a mother of 40
years of age will give birth to a child with a chromosomal abnormality other than
trisomy 21.
The probability is approximately 0.8%.
3) Only a small number of possible chromosomal abnormalities are ever found among
live births, and trisomy 21 is much the commonest. Suggest reasons for these trends.
If there is an abnormality in chromosomes 1 to 12, spontaneous abortion always
occurs. Furthermore, the mother can undergo prenatal tests to identify whether or not
the embryo has chromosomal abnormalities, and if so, choose to abort it.
4) Discuss the risks parents face when choosing to postpone having children.
When parents choose to postpone having children, they increase the risks of their
children having chromosomal abnormalities. As the mother gets older than 40 years,
the chances of her baby having an abnormality increases rapidly.
DBQ on p. 136 – February 4, 2013
1) Compare the DNA content of the bog mosses.
The S. aongstroemii, S. balticum, S. fibriatum, S. teres, S. tundrae, and S. warnstorfii all
have 19 chromosomes and masses ranging from 0.42 to 0.48 DNA/pg.
2) Suggest a reason for six of the species of bog moss on the Svalbard islands all
having the same number of chromosomes in their nuclei.
A reason why the six species of bog moss on the Svalbard islands all had the same
number of chromosomes could be because they all live in the same environment.
3) S. arcticum and S. olafii probably arose as new species when meiosis failed to occur
in one their ancestors.
a) Deduce the chromosome number of nuclei in their leaf cells. Give two
reasons for your answer.
The chromosomes number of nuclei in their leaf cells is 38 because if meiosis
failed to occur then there would be twice as many chromosomes and also because the
mass of the DNA for those two species is approximately double the masses of the
other species (with 19 chromosomes).
b) Suggest a disadvantage to S. arcticum and S. olafii of having more DNA
than other bog mosses.
A disadvantage could be that there would be increased chances of a mutation
occurring because there are more possibilities for error.
4) It is unusual for plants and animals to have an odd number of chromosomes in their
nuclei. Explain how mosses can have odd numbers of chromosomes in their leaf cells.
Mosses can have odd numbers of chromosomes in their leaf cells because they have a
haploid number of chromosomes.
DBQ on p. 134 – February 4, 2013
1) Outline five similarities between the life cycle of a moss and of a human.
Both involve bitosis, meiosis, and fertilization. Both also have eggs, sperm, and
zygotes.
2) Distinguish between the life cycles of a moss and a human by giving five
differences.
The human life cycle has a male and female but moss doesn’t. Human cycles are two
cycles (female and male) that meet together but moss is just one cycle. Meiosis
happens twice in the human cycle but only once in the moss cycle, and mitosis
happens four times in the moss cycle but only twice in the human cycle. Moss plants
have a haploid number of chromosomes whereas humans have a diploid number.
DBQ on p. 145 – January 30, 2013
1) There are many different chromosome numbers in the table, but some numbers are
missing, for example, 5, 7, 11, 13. Explain why none of the species has 13
chromosomes.
None of the species have 13 chromosomes because in the body cells of most
eukaryotes there are two chromosomes of each type.
2) Discuss, using the data in the table, the hypothesis that the more complex an
organism is, the more chromosomes it has.
The data shown in the table seems to support this hypothesis up to a certain point
because in general, larger mammals have more chromosomes than insects. It’s
assumed that larger mammals are more complex than insects, so this makes sense.
However, there are exceptions because organisms that are thought to be the most
intelligent (i.e. modern humans and chimpanzees) have fewer chromosomes than
those that aren’t considered to be as intelligent (i.e. domestic sheep and goats).
3) Explain why the size of the genome of a species cannot be deduced from the
number of chromosomes.
The number of chromosomes doesn’t reveal the size of the whole of the genetic
information of an organism.
4) Suggest, using the data in Table 3, a change in chromosome structure that may have
occurred during human evolution.
Humans may have lost some of their chromosomes.