Download 1 - Biology Mad

1.
(a)
When a cell divides, the genetic material can divide by mitosis, by meiosis or by neither
of these processes. Complete the table with a tick to show the process by which you
would expect the genetic material to divide in each of these examples.
mitosis
meiosis

The division of plasmids in bacterial reproduction.
The stage in the formation of male gametes in a plant in which
haploid daughter cells are formed from a haploid parent cell.

Cell division which takes place in the growth of a human testis
between birth and five years of age.

The stage in the lifecycle of a protoctistan in which a large
number of genetically different spores are produced.
neither

(2)
(b)
The diagram shows a cell during the first division of meiosis.
Complete the diagram below to show the appearance of the chromosomes in each of the
four daughter cells formed at the end of the second division of meiosis.
Four haploid cells, each with two chromosomes;
Correct combinations of chromosomes;
(2)
(c)
In an insect, 16 chromatids were visible in a cell at the start of the first division of
meiosis. How many chromosomes would there be in a normal body cell from this insect?
8 chromosomes/4 pairs;.
(1)
(Total 5 marks)
2.
The diagram shows stages of meiosis in a human testis. Each circle represents a cell.
46
Meiosis I
Meiosis II
(a)
In each empty circle, write the number of chromosomes that would be found in the cell.
Meiosis I
23
23
meiosis II
23
(b)
23
23
23;
(1)
Describe two ways in which meiosis contributes to genetic variation.
Random assortment of chromosomes/correct description;
crossing over/chiasma formation/correct description;
by reducing chromosome number it makes possible random fertilisation;
(2)
(c)
Explain the importance of genetic variation in the process of evolution.
Causes variation in phenotype/characteristics of organisms/some better
adapted/have more favourable characteristics;
correct reference to (natural) selection of better adapted organisms;
selection of different phenotypes in different environments;
eventually leads to species change/change in gene pool/change in gene
frequencies;
(2)
(Total 5 marks)
3.
A queen honey bee can lay both fertilised and unfertilised eggs. Fertilised eggs develop into
diploid females and unfertilised eggs develop into haploid males. The diagram shows the
formation of gametes in female bees and in male bees.
Cell of adult
female bee
Cell of adult
male bee
n
2n
Cell division
Female n
gametes
n
n
n
Male
gametes
Giving a reason for your answer in each case, name the type of cell division in the bee that
produces:
(i)
female gametes;
Meiosis as number of chromosomes halved;
(1)
(ii)
male gametes.
Mitosis as number of chromosomes stays the same;
(1)
(b)
The table shows some features which contribute to variation in the offspring of bees.
Complete the table with a tick if the feature may contribute or a cross if it does not.
Feature
Female offspring
Male offspring
Crossing over


Independent segregation of
chromosomes


Random fusion of gametes

X
(2)
(c)
Body colour in bees is determined by a single gene. The allele B for yellow body is
dominant to the allele b for black body. Explain why, in the offspring of a mating
between a pure-breeding black female and a yellow male, all the males will be black.
Offspring will contain a single b allele only;
(1)
(Total 5 marks)
4.
The diagram below shows the length of cobs in two pure breeding varieties of maize plant and
the F1 and F2 generations derived from a cross between them.
Parental Generation
Pure breeding
variety A
50
F1 generation
30
Pure breeding
variety B
40
Percentage 20
of cobs
10
30
Percentage
of cobs
20
10
0
0
2
(a)
6
10
14 18 22 26
Cob length/cm
30
34
38
42
F2 generation
20
Percentage
10
of cobs
2
6
10
14 18 22 26
Cob length/cm
30
34
38
0
42
2
6
10
14 18 22 26
Cob length/cm
30
34
38
42
What information does the term pure breeding give about genotype?
Homozygous (for alleles concerned);
(1)
(b)
Give the modal class of the F1 generation.
24 – 26cm;
(1)
(c)
(i)
Explain what is meant by polygenic inheritance.
Single characteristic controlled by many genes/more than one gene;
(1)
(ii)
Explain the evidence from the diagram below which suggests that inheritance of
cob length is polygenic.
Many lengths/wide range of lengths/phenotypes;
in F2 generation;
(d)
(2)
What is the evidence that differences in cob length in the parental generation are partly
due to:
(i)
genetic difference;
Two different groups/produce intermediate offspring;
(1)
(ii)
environmental differences?
Each parent shows variation although genotype the same;
(1)
In maize a single gene with two alleles controls the type of carbohydrate stored in the cells of
the plant. Starchy varieties of maize have starch grains which stain blue-black with iodine
solution; waxy varieties have starch grains which stain red. The allele for starch, W, is dominant
to that for waxy, w.
(e)
Explain what is meant by:
(i)
a gene;
A piece of DNA which codes for a single protein/polypeptide;
(1)
(ii)
an allele.
A particular form of a gene
(1)
(f)
Pollen from a single maize plant was dusted on a microscope slide and stained with iodine
solution. The results are shown in the table.
Pollen grains stained blue-black
with iodine solution
Pollen grains stained red with
iodine solution
58
64
What is the genotype of:
(i)
the pollen grains stained red with iodine solution;
w;
(1)
(ii)
the parent plant from which these pollen grains were taken?
Ww;
(1)
(g)
In a field of maize the frequency of allele W was 0.7 and the frequency of allele w was
0.3.
(i)
If the maize plants were randomly fertilised, what frequencies of these two alleles
would be expected in the next generation?
The same/0.7 and 0.3;
(1)
(ii)
Use the Hardy Weinberg equation to calculate the percentage of heterozygous
plants in the field of maize.
Hardy-Weinberg equation given correctly as p2 + 2pq + q2 (= 1);
understands p = 0.7 and q = 0.3;
percentage of heterozygotes = 42
(3)
(Total 15 marks)
5.
In cats, one of the genes for coat colour is present only on the X chromosome. This gene has two
alleles. The allele for ginger fur, XB, is dominant to that for black fur, Xb
(a)
All the cells in the body of a female mammal carry two X chromosomes. During an early
stage of development one of these becomes inactive and is not expressed. Therefore
female mammals have patches of cells with one X chromosome expressed and patches of
cells with the other X chromosome expressed. Tortoiseshell cats have coats with patches
of ginger and patches of black fur.
(i)
What is the genotype of a tortoiseshell cat?
XBXb /XBXb
(1)
(ii)
Explain why there are no male tortoiseshell cats.
Need two X chromosomes / males have only one X / male has a Y / X and Y
chromosome / no allele on Y;
(1)
(b)
A cat breeder who wished to produce tortoiseshell cats crossed a black female cat with a
ginger male. Complete the genetic diagram and predict the percentage of tortoiseshell
kittens expected from this cross.
Parental phenotypes:
Parental genotypes
gamete genotypes
offspring genotypes
black female
XbXb
Xb
ginger male
XBY
XB
Y;
XbY / correct from
given gametes
if sex-linked;
percentage of tortoiseshell kittens = 50% / correct from offspring genotype
given;
XbXB
(3)
(Total 5 marks)
6.
The formation of sperm cells involves mitosis and meiosis. The graphs show the amount of
DNA in the nuclei of cells taken from different parts of a mammalian testis.
20
20
15
15
Number of
nuclei
Number of
nuclei
10
10
5
5
0
0
2
2
6
4
Amount of DNA/arbitrary units
6
4
Amount of DNA/arbitrary units
Graph 2
Graph 1
(a)
(i)
Explain the evidence which shows that Graph 1 represents cells undergoing
mitosis.
DNA will be halved / less / DNA doubles / increases / there are 2 groups of
nuclei;
(1)
(ii)
Explain how the events of meiosis result in three groups of nuclei in Graph 2.
Group with most DNA is before division;
in first division amount of DNA halved;
second division halves this again;
three different amounts of DNA / quoted figures;
(3)
(b)
Explain one way in which meiosis can produce daughter cells each with different DNA.
Random segregation / independent assortment;
allows different combinations DNA / chromosomes;
OR
crossing over / chiasmata;
allows mixing of / swapping / recombination of DNA / mixing alleles / genes
/ exchange of parts of chromatids / chromosomes;
(2)
The flow chart shows one method of finding the sequence of bases in a small piece of DNA. A
large number of identical pieces of DNA are produced and treated in the same way.
One strand labelled with a
radioactive phosphorus atom, 32 P
Piece of
DNA
The strands are separated and the one
32
that has been labelled with P is analysed
The labelled strands are given one of four different chemical treatments
1
2
One guanine (G)
on each strand is
destroyed.
One guanine or
one adenine (A)
is destroyed.
3
One thymine (T)
or one cytosine (C)
is destroyed.
4
One cytosine
is destroyed.
Only enough chemical is added to destroy one of the bases concerned. Destroying a base results
in the strand breaking into two separate pieces. In treatment 1, for example, the strand of DNA
shown below
G
G
will give one of two different radioactively labelled pieces.
Either
or
G
(c)
(i)
A nucleotide is made up of three components joined by condensation reactions.
Which component would be labelled with 32P?
Phosphate / phosphoric acid / P ;
(ii)
(1)
Name the type of chemical bond which is broken when the two strands in a DNA
molecule are separated.
Hydrogen (bond);
(1)
The samples of treated DNA are analysed by electrophoresis. The results in the diagram show
the position of the radioactive fragments.
1
One G
destroyed
2
One G or one
A destroyed
3
One T or one
C destroyed
4
One G
destroyed
1
2
3
4
5
6
Direction of
movement
8
9
10
11
12
13
14
(d)
How many times in the strand of DNA being investigated does:
(i)
(ii)
the base cytosine occur;
5;
(1)
the base thymine occur?
3;
(1)
(e)
Which of the radioactive fragments labelled 1 to 14 in the diagram only contains one
base?
Explain how you arrived at your answer.
Fragment number
Fragment 14 / 13;
Explanation single base represents smallest / lightest piece of DNA;
smallest piece moves furthest / fastest;
(3)
(Total 13 marks)
7.
(a)
Explain the meaning of the term gene pool.
(All) the alleles/genes in a population/ definition;
(1)
In two-spot ladybirds, the colour of the wing-cases is genetically determined. The allele for red
wing-cases, R, is dominant to that for black wing-cases, r.
(b)
In a ladybird with black wing-cases, how many copies of the r allele would you expect to
find in a nucleus taken from a muscle cell in the insect's leg? Explain your answer.
Two. All somatic/ body cells are diploid/have two copies of each
chromosome/ must be homozygous for phenotype to be expressed;
(1)
(c)
In a population of two-spot ladybirds found on a patch of nettles, 168 had red wing-cases
and 32 had black wing-cases. Showing your working in each case, calculate the frequency
of:
(i)
the r allele;
2 marks
1 mark
0.4 / 40%
incorrect figure obtained from 0.16
Answer: ............................................
(2)
(ii)
the R allele.
60% / 0.6 but accept answer derived correctly from (c)(i);
Answer: ............................................
(1)
(Total 5 marks)
8.
(a)
The inheritance of the ability to produce hydrogen cyanide is controlled by two genes which are
located on different chromosomes. The dominant allele of one gene, G, controls the production of
enzyme G which converts a precursor to linamarin. The dominant allele of the other gene, E,
controls the production of enzyme E which converts linamarin to hydrogen cyanide. This is
summarised in the diagram.
Precursor
Allele G
Enzyme G
Allele A
Enzyme E
Linamarin
Hydrogen cyanide
(i)
Explain why plants homozygous for the allele g will not produce hydrogen cyanide when
their tissues are damaged.
Do not produce enzyme G/produce non-functional enzyme G;
No linamarin formed;
(2)
(ii)
Plants with genotypes GgEe and GgEE were crossed. Use a genetic diagram to predict the
percentage of plants produced from this cross which would release hydrogen cyanide when
their tissues were damaged.
(Parental genotypes:
GgEe
GgEE)
GE Ge gE ge
GE gE;
GE
Ge
gE
ge
GE
GGEE
GGEe
GgEE
GgEe
gE
GgEE
GgEe
ggEE
ggEe
Gametes:
75% producing cyanide;
Set out in logical genetic diagram / with gamete genotypes labelled;
(4)
Recently a strain of genetically engineered clover has been developed which has a high concentration of
proteins rich in sulphur-containing amino-acids. A piece of DNA was prepared which contained the three
different genes. This was inserted into a clover plant.
Gene 1 obtained from sunflower seeds. This gene codes for
a protein rich in sulphur-containing amino acids.
Gene 2 ensures that the protein rich in sulphur-containing
amino acids is produced in leaf cells.
Gene 3 prevents this protein being digested in the rumen
of sheep.
(b)
The copy of Gene 1 used in this experiment was obtained from the mRNA of the sunflower seeds.
(i)
Explain how enzymes could be used to obtain the gene from the mRNA.
Reverse transcriptase;
makes single strand of DNA/cDNA from RNA;
Double strand then formed;
DNA polymerase;
(ii)
(3)
Explain why it would be an advantage to obtain the gene from the mRNA rather than from
the DNA of the sunflower seeds.
If from DNA all genes are present in cell;
mRNA from activated genes only/codes for one protein;
DNA has introns/non-coding /junk DNA;
Have been edited out in mRNA;
(2)