6 Answers to end-of-chapter questions

1 C [1]
2 D [1]
3 D [1]
4 D [1]
5 C [1]
6 C [1]
7 D [1]
8 D [1]
9 A [1]
10 D [1]
Structured questions

11 a V – late anaphase
W – prophase
X – metaphase All 5 stages correct [3]
Y – late anaphase/early telophase 3–4 stages [2]
Z – interphase 1–2 stages [1]

cell wall

nuclear membrane

chromosome

cytoplasm

Well drawn representation [1]

Well labelled [1]
Drawing showing prophase in mitosis

Biology for CAPE Original material © Cambridge University Press 2011 1

 cell wall

cytoplasm

chromosome

Well drawn representation [1]

Well labelled [1]
Drawing showing metaphase in mitosis

c i II
• Cell plate can be seen in the middle
• 4 nuclei are forming
IV
• Cell plate formation in middle of cell so there are Identify both cells [2]
2 cells still undergoing cell division. Reason [1]

ii I – anaphase I: separation of homologous chromosomes /

chromosomes have moved to the poles / no cell plate in the
middle of the cell as yet
II – anaphase II: cell plate seen / 2 cells; sister chromatids
separating to form 4 nuclei
III – metaphase I: alignment of chromosomes at the
equator; some chromosomes are facing one pole, others
face the opposite pole; no cell plate dividing mother cell as
yet
IV – metaphase II: cell plate in middle to indicate 2 cells 3–4 stages correct [2]
formed already; alignment of chromosomes at the 1–2 stages correct [1]
equator Any 3 reasons [3]
[max 5]

12 a • Cut off 5 mm of the ends of 2 garlic roots

• Put tips into watch glass containing 10 drops of acetic orcein and 1drop of 1mol
dm–3 HCl
• Warm the glass by moving through flame
• Use a dissecting needle to lift each root tip onto a slide
• Cut off 2 mm at the end of tip (discard rest) Logical sequence [3]
• Put 2 drops of acetic orcein onto each root tip 3–4 points [2]
• Place coverslip over the top 1–2 points [1]

Biology for CAPE Original material © Cambridge University Press 2011 2

b i

Student Total number of cells in each stage of mitosis

interphase prophase metaphase anaphase telophase
Aneela 14 13 8 8 7
Faeryal 17 10 9 10 4
Alex 15 12 7 9 7
Damian 15 8 9 9 9
Leon 13 11 9 9 8
class total 74 54 42 45 35
class 14.8 10.8 8.4 9.0 7.0
average
% number of 29.6 21.6 16.8 18 14 3 points correct [1]
cells Max [4]

ii

35
30
25
% of cells

20
15
10
5
0
inter. pro. met. ana. telo.
stage of mitosis

Bar graph showing the percentage of cells in allium root tip at

various stages of mitosis
Axes labelled correctly [2]
Bars drawn accurately [1]
Title (descriptive) [1]
[max 4]

c Difficulty in deciding which of the cells were ending telophase

and now entering interphase. [1]

d • Cells spend different lengths of time in each stage of

mitosis
• Cells spend the longest time in interphase
• The order of time spent in each stage from longest to
shortest is
interphase → prophase → anaphase → metaphase →
telophase Any 1 point [1]

e
Phase Number of Number of
chromosomes chromatids
prophase 16 32
metaphase 16 32
anaphase 32 32 4 points correct [1]
telophase 32 32 [max 2]

Biology for CAPE Original material © Cambridge University Press 2011 3

Essay questions

13 a The process of meiosis

• Produces genetically different cells
• With haploid number of chromosomes
• 2 divisions
• Meiosis I leads to separation of homologous chromosomes / reduction
division
• Meiosis II leads to separation of sister chromatids
Prophase I
• The chromosomes shorten and thicken by coiling / nuclear envelope
disintegrates / nucleolus disappears / formation of spindle fibres / centrioles
begin to move towards the poles
• Homologous chromosomes pair up; each pair is called a bivalent
• During the coiling and shortening process, non-sister chromatids frequently
break
• Broken ends rejoin more or less immediately
• A join between different chromatids is a chiasma, and joining event is
known as a crossing over
• This leads to an exchange of equal and corresponding parts of non-sister
chromatids / source of variation
Metaphase 1
• The spindle fibres attach to the centromere of pair of homologous
chromosomes
• Homologous chromosomes align themselves at equator
Anaphase 1
• Spindle fibres pull the homologous chromosomes apart, to opposite
poles
• The individual chromatids in a chromosome remain attached to each
other by their centromeres
Telophase 1
• Two nuclei formed containing a single set of chromosomes that
are made of two chromatids / one set of homologous
chromosomes
• Nuclear membrane reforms
• Cytokinesis – two cells formed with a single set of
chromosomes
• Centrioles divide and new spindles start to form
Meiosis II
• Starts immediately / separation of sister chromatids
• In metaphase II, the spindles are at right angles to those in
metaphase I, aligned at equator
• In anaphase II, sister chromatids separate
• In telophase II, nuclear membrane reforms / spindle
disintegrates / nucleolus reforms / spindle fibres
disintegrate / decondensation Correct diagrams [2]
• Cytokinesis – 4 genetically different cells with Correction description of meiosis I [3]
haploid number of chromosomes Correct description of meiosis [2]
[Max 7]
Information can be obtained from diagrams
[Should have appropriate diagrams – they are requested in the
question.]

Biology for CAPE Original material © Cambridge University Press 2011 4

 b • Produces gametes which contain haploid number of chromosomes
• This is important because in sexual reproduction fertilisation of
gametes occurs
• This allows for restoring chromosome number / maintenance of a
constant chromosome number / meiosis allows the chromosome
number to remain constant in each generation
• If there were no reduction division in meiosis, fertilisation would
lead to doubling of the chromosome number in each generation
• As a result, there would be no physical room for the chromosomes
• There would be conflict with too many chromosomes
• There would also be problems with sterility and mating with
individuals with a different chromosome number
• A major point of meiosis and sex is to introduce genetic
variation by independent assortment and crossing over in
meiosis; random fusion of genetically different gametes
• Which allows species to adapt to their environment and 1 well-described point [1]
so to evolve [max 6]

c • in the pollen sacs / anther [1]

• in the ovule / ovary [1]

14 a • Chromosomes align at equator [1]

• Attach to spindle fibre at centromere [1]
• Spindle fibres shorten [1]
• Centromeres split [1]
• Chromatids separate and are pulled towards the pole [1]
[with the aid of diagrams – none supplied?]
b • Apical meristem: root and shoot apex
• Lateral meristem: in the cambium usually associated with
older parts of the plant (e.g. cork cambium and vascular
cambium)
• Intercalary meristem: found between regions of permanent
tissue (e.g. nodes of many monocotyledons such as bases of
grass leaves)
• It may occur where damage has occurred during the formation
of callus tissue 2 points equals [1]
• It also occurs in the pollen and embryo sacs [max 2]

c • Formation of genetically identical cells / clone

• Asexual reproduction – production of offspring from a single
parent using mitosis. This can be advantageous in an unchanging
environment
• Growth – the number of cells within an organism increases by
mitosis. This is the basis of growth in multicellular organisms
• Cell replacement – replacement of cells and tissues
• Regeneration – some animals are able to regenerate whole parts of
the body (e.g. legs in crustaceans and arms in starfish)
• Genetic stability – mitosis produces two nuclei which have 1 point [1]
the same number of chromosomes as the parent cell [max 4]

d • May disrupt the formation of spindles

• If spindles fibres were not formed, chromatids would not separate
• Cells would not receive the newly copied DNA

Biology for CAPE Original material © Cambridge University Press 2011 5

 • Cell division would be interrupted
• Therefore 2 identical cells / clones would not be formed 1 point [1]
• No tumour / undifferentiated mass of cells formed [max 4]

15 a • A pair of chromosomes containing the same linear gene

sequences, each derived from one parent [2]
• Homologous chromosomes may have different alleles / have
different sequence of bases / nucleotides as each has come from
different parent. [1]

b • In early prophase I, replication occurred in interphase hence sister

chromatids are identical and are now condensing [1]
• In metaphase I, homologous chromosomes have lined up in late prophase and
crossing over of equal and corresponding parts of non-sister chromatids has
occurred [1]
• Hence in metaphase I, sister chromatids are no longer identical. [1]

c i Variations of genomes / genotypes between members of

species or groups of a species living in different habitat [2]

ii Chiasma formation
• Allows for mixing of genes on same chromosome
• This is a join between different chromatids / non-sister chromatids
• Leads to crossing over / joining of non-sister chromatids
• Leads to exchange between bivalents which involves the breakage and
rejoining of equal and corresponding parts of non-sister chromatids of
bivalents
• Leading to new combinations of alleles / mixing of maternal and
paternal alleles / linkage groups broken
• Homologous chromosomes would be genetically different from each Any point [1]
other / different from parent cell [max 3]

Independent assortment
• Allows for different combinations of chromosomes
• In metaphase I, bivalents line up at equator
• There are 2n possible orientation positions at the equator
(n = haploid number) Any point [1]
• Leads to many different possible gametes [max 4]

Information can be obtained from annotated diagrams

Biology for CAPE Original material © Cambridge University Press 2011 6