ANSWERS 2 (57 Marks) - Cerebralenhancementzone

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ANSWERS 2 (57 Marks)

IB Standard and Higher level Biology Dulwich College Shanghai

Topic 4 (SL) and 10 (HL): Genetics


4.1

Chromosomes, genes, alleles and mutations (SL)

4.1.1

State that eukaryote chromosomes are made of DNA and proteins.

4.1.2

Define gene, allele and gen
ome.

4.1.3

Define gene mutation.

4.1.4

Explain the consequences of a base substitution mutation in relation to the processes of
transcription and translation, using the example of sickle
-
cell anaemia.


4.2

Meiosis (SL)

4.2.1

State that meiosis is a reductive division of a diplo
id nucleus to form a haploid nuclei.

4.2.2

Define homologous chromosomes.

4.2.3

Outline the process of meiosis, including pairing of homologous chromosomes and
crossing over, followed by two divisions, which results in four haploid cells.

4.2.4

Explain that non
-
disjunction
can lead to changes in chromosomes number, illustrated by
references to Down syndrome (trisomy 21).

4.2.5

State that, in karyotyping, chromosomes are arranged in pairs according to their size
and structure.

4.2.6

State that karyotyping is performed using cells collect
ed by chorionic villus sampling or
amniocentesis, for pre
-
natal diagnosis of chromosome abnormalities.

4.2.7

Analyse a human karyotype to determine gender and whether non
-
disjunction has
occurred.


10.1

Meiosis (HL)

10.1.1

Describe the behaviour of the chromosomes in the ph
ases of meiosis.

10.1.2

Describe the behaviour of the chromosomes in the phases of meiosis.

10.1.3

Explain how meiosis results in an effectively infinite genetic variety in gametes through
crossing over in prophase 1 and random orientation in metaphase 1.

10.1.4

State Mendel’s

law of independent assortment.

10.1.5

Explain the relationship between Mendel’s law of independent assortment and meiosis


4.3

Theoretical Genetics (SL)

4.3.1

Define genotype, phenotype, dominant allele, recessive allele, codominant alleles, locus,
homologous, heterozygou
s, carrier and test cross.

4.3.2

Determine the genotypes and phenotypes of the offspring of a monohybrid cross using a
Punnett grid.

4.3.3

State that some genes have more than two alleles (multiple alleles).

4.3.4

Describe ABO blood groups as an example of codominance and m
ultiple alleles.

4.3.5

Explain how the sex chromosomes control gender by referring to the inheritance of X
and Y chromosomes in humans.

4.3.6

State that some genes are present on the X
-
chromosome and absent from the shorter Y
chromosome in humans.

4.3.7

Define sex linkage.

4.3.8

Describe the inheritance of colour blindness and haemophilia as examples of sex linkage.

4.3.9

State that a human female can be homozygous or heterozygous with respect to sex
-
linked genes.

4.3.10

Explain that female carriers are heterozygous for X
-
linked recessive alle
les.

4.3.11

Predict the genotypic and phenotypic ratios of offspring of monohybrid crosses
involving any of the above patterns of inheritance.

4.3.12

Deduce the geneotypes and phenotypes of individuals in pedigree charts.


10.2

Dihybrid Crosses and Gene Linkage (HL)

10.2.1

Calculat
e and predict the genotypic and phenotypic ratio of offspring of dihybrid
crosses involving unlinked autosomal genes

10.2.2

Distinguish between autosomes and sex chromosomes

10.2.3

Explain how crossing over between non
-
sister chromatids of a homologous pair in
prophase
I can result in an exchange of alleles.

10.2.4

Define linkage group

10.2.5

Explain an example of a cross between two linked genes. Alleles are usually shown side
by side in dihybrid crosses, for example TtBb. In representing crosses involving linkage,
it is more commo
n to show them as vertical pairs for example:




10.2.6

Identify which of the offspring are recombinants in a dihybrid cross involving linked
genes.



10.3

Polygenic Inheritance

10.3.1

Define polygenic inheritance

10.3.2

Explain that polygenic inheritance can contribute to contin
uous variation using two
examples, one of which must be human skin colour



4.4

Genetic Engineering and Biotechnology (SL)

4.4.1

Outline the use of polymerase chain reaction (PCR) to copy and amplify minute quantities
of DNA.

4.4.2

State that, in gel electrophoresis, frag
ments of DNA move in an electric field and are
separated according to their size.

4.4.3

State that gel electrophoresis of DNA is used in DNA profiling.

4.4.4

Describe the application of DNA profiling to determine paternity and also in forensic
investigations.

4.4.5

Analyse

DNA profiles to draw conclusions about paternity or forensic investigations.

4.4.6

Outline three outcomes of the sequencing of the complete human genome.

4.4.7

State that, when genes are transferred between species, the amino acid sequence of
polypeptides translate
d from them is unchanged because the genetic code is universal.

4.4.8

Outline a basic technique used for gene transfer involving plasmids, a host cell
(bacterium, yeast or other cell), restriction enzymes (endonucleases) and DNA ligase.

4.4.9

State two examples of t
he current uses of genetically modified crops or animals.

4.4.10

Discuss the potential benefits and possible harmful effects of one example of genetic
modification.

4.4.11

Define clone.

4.4.12

Outline a technique for cloning using differentiated animal cells.

4.4.13

Discuss the
ethical issues of therapeutic cloning in humans.


Paper 1

Multiple Choice (10 Marks)

1.

Two genes A and B are linked together as shown below.



If the genes are far enough apart such that crossing over between the alleles
occurs occasionally, which stat
ement is true of the gametes?

A.

All of the gametes will be Ab and aB.

B.

There will be 25% Ab, 25% aB, 25% ab and 25% AB.

C.

There will be approximately equal numbers of Ab and ab gametes.

D.

The number of Ab gametes will be greater than the number of ab
gametes.


2.

A polygenic character is controlled by two genes each with two alleles. How many
different possible genotypes are there for this character?

A.

2

B.

4

C.

9

D.

16


3.

A cross is performed between two organisms with the genotypes AaBb and aabb.


What genotypes in the offspring are the result of recombination?

A.

Aabb, AaBb

B.

AaBb, aabb

C.

aabb, Aabb

D.

Aabb, aaBb


4.

The diagram below shows a cell during meiosis.



How many chromosomes would each daughter cell have at the end of meiosis?

A.

1

B.

2

C.

4

D.

8


5.

The pedigree below shows which members o
f a family were Rhesus positive

and Rhesus
negative. The allele for Rhesus positive blood (Rh
+
) is dominant over the allele for Rhesus
negative blood (R
-
).



Which are possible genotypes of the ind
ividuals numbered I, II and III?


I

II

III

A.

Rh
+

Rh
+

Rh
+

Rh
+

Rh
+

Rh


B.

Rh
+

Rh
+

Rh
+

Rh


Rh
+

Rh
+

C.

Rh
+

Rh
+

Rh
+

Rh


Rh
+

Rh


D.

Rh
+

Rh


Rh
+

Rh


Rh
+

Rh
+


6.

The diagram below shows the results of DNA profiling using gel electrophoresis.



What conclusi
on can be drawn about the DNA in bands I and II?

A.

The DNA in the two bands has the same base sequence.

B.

The DNA in the two bands consists of fragments of the same length.

C.

The DNA in the two bands has the same ratio of bases.

D.

The DNA in the two ba
nds came from the same source.




7.

The diagram below shows chromosomes during prophase I of meiosis. How many
chromosomes and chiasmata are visible?



Number of chromosomes

Number of chiasmata

A.

2

2

B.

4

2

C.

2

4

D.

4

4



8.

In peas the allele for

round seed (
R
) is dominant over the allele for wrinkled seed
(
r
). The allele for yellow seed (
Y
) is dominant over the allele for green seed (
y
).


If two pea plants with the genotypes
YyRr

and
Yyrr

are crossed together, what
ratio of phenotypes is expected

in the offspring?

A.

9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green

B.

3 round yellow : 3 round green : 1 wrinkled yellow : 1 wrinkled green

C.

3 round yellow : 1 round green : 3 wrinkled yellow : 1 wrinkled green

D.

1 round yellow
: 1 round green : 1 wrinkled yellow : 1 wrinkled green



9.

What is a difference between autosomes and sex chromosomes?

A.

Autosomes are not found in gametes but sex chromosomes are.

B.

Sex chromosomes are found in animal cells and autosomes are found in p
lant
cells.

C.

Autosomes are diploid and sex chromosomes are haploid.

D.

Sex chromosomes determine gender and autosomes do not.


10.

The diagram below shows a cell in meiosis. What can be deduced from this diagram?


[Source: J W Saunders, (1968),
Animal M
orphogenesis
, MacMillan, page 7]


Stage of meiosis shown

Haploid number of

chromosomes in this cell

A.

Metaphase I

6

B.

Prophase I

3

C.

Prophase I

6

D.

Metaphase I

3





Paper 2

Section A

Data Analysis (17 marks)

1.

Rats were bred for several generat
ions to prefer alcohol (ethanol) consumption.
When tested, it was discovered that the brains of these rats possessed lower
quantities of the chemical neuropeptide Y (NPY).


To test the hypothesis that lower quantities of NPY leads to a preference for
alcoh
ol, rats were genetically engineered to be NPY deficient (genotype NPY

/

), or
to produce an excess of NPY (NPY
-
EX). In separate experiments, the two groups
were compared to normal rats (in terms of their alcohol preference) possessing the
genotype NPY +/
+. The groups were offered solutions of increasing alcohol
concentration. The quantity of each solution consumed per day was measured.

Figure 1

Figure 2


[Source: adapted from Thiele et al, Nature, (1998), 396 pages 366

369]


(a)

Calculate the difference
in consumption of the 6% alcohol solution between
the

(i)

NPY

/


and NPY +/+ rats (figure 1);






(1)


2.8 (±0.5) g kg

1

day

1


(ii)

NPY
-
EX and NPY+/+ rats (figure 2).






(1)




2.0 (± 0.5) g kg

1

day

1


(b)

Compare the alcohol consumption of t
he NPY

/


rats with the NPY
-
EX rats.


(3)


NPY


/


consumes more alcohol

(1)

than NPY
-
EX (at all concentrations);

consumption increases at a (relatively) constant rate (above 6%)



wit
h concentration for NPY
-
EX, but
levels off at higher concentrat
ions for
NPY


/


(1)
;



as
alcohol concentration increases
both NPY


/


and NPY
-
EX rats consume
more;

(1)



NPY


/


consumes more al
cohol than NPY + / + and NPY
-
EX
consumes less
than NPY + / +;
(Accept use of word control.)



biggest difference from N
PY + / + for NPY
-
E
X is
at 20%, but for NPY


/


it
is at 10%;


(c)

Identify the relationship between NPY levels and alcohol consumption.



(1)



NPY levels are inversely r
elated to alcohol consumption



the lower the NPY levels,
the more alcohol consump
tion
(Accept the converse.)



An experiment was carried out to test the hypothesis that an increase in preference for
alcohol might be related to a decrease in sensitivity to its effects.


Rats were injected with a sample of alcohol and then assessed for t
he length of
time it took for them to regain the righting reflex. (The righting reflex refers to
the ability of the rat to return to its feet after being placed on its back.)

Figure 3


[Source: adapted from Thiele
et al, Nature
, (1998),
396

pages 366

369]


(d)

Deduce the relationship between NPY levels and the time required to regain the
righting reflex.











(3)



NPY


/


takes least time to regain the reflex;

NPY
-
EX takes most time to regain the reflex;

NPY
-
EX is the most sensitive to effects of
alcohol /

NPY


/


is the least sensitive to effects of alcohol;

the higher the NPY levels the more time taken to regain the reflex;

NPY


/


/ under
-
expression has more of an effect than NPY
-
EX / over
-
expression;


An additional experiment was carried ou
t to determine whether differences in
sensitivity to the effects of alcohol might be related to differences in the rats’
ability to remove alcohol from their blood. Rats were injected with alcohol and
blood samples were taken one hour and three hours later

to determine alcohol
levels. The results are shown below.

Figure 4






(e)

Evaluate the hypothesis that differences in sensitivity to the effects of alcohol
might be related to differences in the ability of the rats to remove alcohol from
their blood.










(2)


there is no difference / small difference

in blood levels between
groups at 1 hour and 3 hours /
decrease from 1 to 3
hours is the same for both;



therefore, the hypothesis
does not appear to be justified;


(f)

Using all the data, outline th
e relationship between preference for alcohol and
sensitivity to the effects of alcohol.








(2)


NPY
-
EX does not prefer alcohol and is se
nsitive to effects of alcohol


NPY


/


prefers alcohol and is not s
ensitive to effects of alcohol;


therefore, a
lcohol preference i
s inversely related
to sensi
tivity to effects of
alcohol


the less sensitive the rats are to alcohol, the more they consume it;


(g)

(i)

Define the term
homozygous
.







(1)


having
two identical alleles

(of a gene)


(ii)

State the ph
enotype of a rat with the genotype NPY +/+.





(1)


normal

preference for alcohol /

moderate

preference for alcohol /


standard

phenotype for alcohol preference /

normal levels of NPY


Do not accept "normal rats" only.


(iii)

Using a Punnett grid, pr
edict the fraction of offspring that would have the
genotype NPY +/


if two rats were crossed, one homozygous for the NPY+
allele and one homozygous for the NPY


allele.






(2)

Punnett square correctly drawn;

Can be two by two or one by one, accept +
/


without

NPY, square or diamond shape acceptable but clear cells expected,

male or female symbols not necessary. 100% / all
;

Paper 2

Section A

Short Structured (13 Marks)

1.

The diagram below shows the pedigree of a family with red green colour
-
blindne
ss,
a sex
-
linked condition.




(a)

Define the term
sex
-
linkage
.








(1)


a gene / trait / allele carried
on a sex chromosome

/ X and Y / X / Y;


(b)

Deduce, with a reason, whether the allele producing the condition is dominant or
recessive.











(2)


recessive
;

(1)


evidence from the pedigree;
eg

2nd generation

2 and 3 do not have the
condition but have one child who does
;

(1)


(c)

(i)

Determine all the possible genotypes of the individual (2nd generation

1)
using appropriate symbols.








(1)




X
a

Y (where a = condition);



(ii)

Determine all the possible genotypes of the individual (3rd generation

4)
using appropriate symbols.









(1)

X
A
X
a

or X
A
X
A

where A = normal, a = condition (
must have both
);


(
If upper case letter and lower cas
e letter are reversed then the ECF rule

applies
.)


2.

In
Zea mays
, the allele for coloured seed (C) is dominant over the allele for
colourless seed (c). The allele for starchy endosperm (W) is dominant over the allele
for waxy endosperm (w). Pure breeding
plants with coloured seeds and starchy
endosperm were crossed with pure breeding plants with colourless seeds and waxy
endosperm.

(a)

State the genotype and the phenotype of the F
1

individuals produced as a result of
this cross.










(2)


C c W w;

(
1)

all are coloured starchy;

(1)


(b)

The F
1

plants were crossed with plants that had the genotype c c w w. Calculate the
expected ratio of phenotypes in the F
2

generation, assuming that there is
independent assortment. Use the space below to show your wor
king.



(3)

gametes are C W, C w, c W, c w and c w;

(1)

F
2

genotypes are CcWw, Ccww, ccWw and ccww;

(1)

1 coloured starchy: 1 coloured waxy: 1 colourle
ss starchy: 1 colourless waxy;
(1)

Phenotypes must be unambiguously indicated, but not necessarily on th
e line.



The observed percentages of phenotypes in the F
2

generation are shown below.

coloured starchy

37%

colourless starchy

14%

coloured waxy

16%

colourless waxy

33%


The
observed

results differ significantly from the results
expected

on the basis of
in
dependent assortment.



(c)

State the name of a statistical test that could be used to show that the observed
and the expected results are significantly different.





(1)


chi
-
squared test


(d)

Explain the reasons for the observed results of the cross d
iffering significantly
from the expected results.









(2)

(autosomal)
linkage

(
reject sex linkage
) / genes are on the same chromosome /

genes do not assort independently;

coloured starchy and colourless waxy are parentals / coloured waxy and

colou
rless starchy are the recombinants;

recombinants produced by crossing over
;


Section B

Extended Response (17 Marks)

1.

Describe how sexual reproduction promotes genetic variation within a species.



(4)

random orientation

of bivalents / pairs of chromosom
es;

maternal and paternal chromosome could go to either pole;

2
n

combinations;

eg

over 8 million in humans;

crossing over
;

exchange of material be
tween homologous chromosomes /
non
-
sister chromatids;

segregation of alleles in meiosis;

combinations of allel
es are broken up
;

fertilization brings toget
her genes /
alleles from two different parents;

fertilization generates new combinations of genes

/ a
lleles;

random fertilization /
many possible combinations of male and female gamete;

eg

over 64 million million

in humans (ignoring crossing over);






2.

Outline the differences between the behaviour of the chromosomes in mitosis and
meiosis.












(5)

two

divisions in meiosis, only
one

in mitosis;

meiosis results in
haploid

cells, mitosis in
diploid

cells
;

crossing over

only occurs in
meiosis
;

no S phase precedes meiosis II;

chromosome behaviour in meiosis II and mitosis is similar / chromosome

behaviour in meiosis I and mitosis is different;

chiasmata only form during meiosis
;

homologous chromosomes

move
to the equator in
pairs only in meiosis
;

Do not accept number of cells produced
-

it is a result not a behaviour.


3.

Explain the use of
two

named enzymes in biotechnology.





(8)


Examples and application
:

pectinase;

obtained from citrus fruits / toma
toes / apples;

used in fruit juice production;

breaks down pectin allowing cells to separate;

assisting in juice formation;

juice formed is clear;

high juice yield using enzyme;




2nd enzyme:

name
;

(1) e.g restriction enzymes/ lactase/ amylase, proteases,

lipases

source;

(1)

use
;

(1) genetic engineering/ for lactose intolerance/ biological washing powders

mode of action; (1)

cutting DNA fragments/ converts lactose into galactose and
glucose/ digests food stains e.g. starch to maltose, proteins to amino aci
ds, lipids to
fatty acids and glycerol

advantage of using enzyme
;

(1) allows recombination of DNA/ allows lactose
intolerant people to use milk products without consequences/ low temperatures for
washing

details of enzyme use;

8 max


Possible second exampl
e could be meat tenderising


papain from papaya fruit /
bromelain from the pineapple plant, biological washing powders


amylases /
proteases / lipases, glucose biosensors


glucose oxidase / peroxidase, cheese
making


rennin, high fructose syrups


gluc
ose isomerase, breadmaking


fungal
amylases / fungal proteases, DNA profiling


DNA ligase, etc.


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