Download Markscheme

Chapter 3 - Genetics
1.
[185 marks]
The feather colour of a certain breed of chicken is controlled by codominant alleles. A cross between a
[1 mark]
homozygous black-feathered chicken and a homozygous white-feathered chicken produces all speckled chickens. What
phenotypic ratios would be expected from a cross between two speckled chickens?
A. All speckled
B. 1 black feathers : 1 white feathers
C. Speckled, black feathers and white feathers in equal numbers
D. 1 black feathers : 2 speckled feathers : 1 white feathers
Markscheme
D
2.
The presence of freckles is a characteristic controlled by a dominant gene. Two parents who are heterozygous for [1 mark]
the characteristic have three children, all of whom have freckles. Which statement is true if they have a fourth child?
A. There is a 100 % chance that their next child will have freckles.
B. There is a 75 % chance that their next child will have freckles.
C. There is a 50 % chance that their next child will have freckles.
D. The next child will have no freckles as the ratio is 3 with freckles to 1 without freckles.
Markscheme
B
3.
Which process can be used to amplify small fragments of DNA?
A. Gel electrophoresis
B. Polymerase chain reaction
C. DNA profiling
D. Electron microscopy
[1 mark]
Markscheme
B
4.
If a man with blood group O and a woman with blood group AB have children, which blood group(s) could the
children have?
A. Group O only
B. Groups A and B only
C. Group AB only
D. Groups O, A, B and AB
Markscheme
B
[1 mark]
5.
Which individuals are colour blind in this Punnett grid?
[1 mark]
A. X B Y
B. X B X B
C. X b Y
D. X B X b
Markscheme
C
6.
What was an aim of genetic modification of organisms?
A. To provide stem cells from embryos for medical use
B. To make crop plants resistant to herbicides
C. To provide sperm cells for in vitro fertilization (IVF)
D. To produce genetically identical sheep
[1 mark]
Markscheme
B
7.
What causes the presence of three chromosomes 21 in Down syndrome?
A. Crossing over
B. Allele change
C. Non-disjunction
D. Gene mutation
[1 mark]
Markscheme
C
8.
What is a characteristic of the human Y chromosome?
[1 mark]
A. It is made of DNA and histones covered by phospholipids.
B. It contains some genes that are not present on the X chromosome.
C. It is the largest chromosome in the human karyotype.
D. It has a condensed length of approximately 100 µm.
Markscheme
B
9.
What is a definition of a clone?
A. A group of cells derived from a single parent cell
B. Differentiated cells that retain the capacity to divide
C. A fetus developed specifically for medical use
D. A group of cells that have lost the ability to differentiate
[1 mark]
Markscheme
A
10. A colour blind man and a woman carrier for colour blindness have a son. What is the probability that their son will [1 mark]
be colour blind?
A. 25 %
B. 50 %
C. 75 %
D. 100 %
Markscheme
B
11.
Laboratory analysis of DNA from a 40 000 year old woolly mammoth used the polymerase chain reaction (PCR).
What role did the PCR have in the analysis?
[1 mark]
A. DNA denaturation
B. DNA comparison
C. DNA separation
D. DNA amplification
Markscheme
D
12.
What is the difference between dominant, recessive and codominant alleles?
Markscheme
C
[1 mark]
13.
The following shows a pedigree chart.
[1 mark]
What type of inheritance is shown in this pedigree chart?
A. X-linked recessive
B. Y-linked dominant
C. X-linked dominant
D. Y-linked recessive
Markscheme
A
14.
Which of the following involves meiosis?
[1 mark]
A. Tissue repair
B. Production of gametes
C. Asexual reproduction
D. Growth
Markscheme
B
15.
In guinea pigs black coat colour is dominant to white. In a test cross between a black and a white guinea pig both [1 mark]
black and white offspring were produced. What percentage of the offspring would be expected to be white?
A. 75 %
B. 50 %
C. 33.3 %
D. 25 %
Markscheme
B
16. What term describes the failure of sister chromatids to separate during anaphase II?
A. Sex linkage
B. Karyotyping
C. Non-disjunction
D. Semi-conservative replication
[1 mark]
Markscheme
C
17.
The image shows a human karyotype.
[1 mark]
According to the image, what conditions can be determined?
A. Non-disjunction has occurred and the individual is male.
B. Non-disjunction has occurred and the individual is female.
C. The individual is female and has Down syndrome.
D. The individual is male and has Down syndrome.
Markscheme
A
18.
The diagram shows results of electrophoresis of DNA from a crime scene.
Which suspect could be implicated as the criminal, according to the gel of DNA shown?
A. Suspect 1
B. Suspect 2
C. Suspect 3
D. Suspect 4
[1 mark]
Markscheme
D
19.
What is a gene mutation?
[1 mark]
A. Failure of chromosome pairs to separate properly during cell division
B. Changes to genes caused by natural selection
C. Changes to the nucleotide sequence of the genetic material
D. Changes in karyotypes
Markscheme
C
20.
Which is a feature of sex-linked genes in humans?
[1 mark]
A. Males can only be heterozygous for the gene.
B. Females can only be homozygous for the gene.
C. Males can be either heterozygous or homozygous for the gene.
D. Females can be either heterozygous or homozygous for the gene.
Markscheme
D
21.
What is amplified using the polymerase chain reaction (PCR)?
[1 mark]
A. Large amounts of RNA
B. Small amounts of DNA
C. Small amounts of protein
D. Large amounts of polymers
Markscheme
B
22.
What is a plasmid?
A. Chloroplast DNA
B. Mitochondrial DNA
C. Small circle of DNA that can transfer genes to or from a prokaryote
D. The bacterial chromosome
Markscheme
C
[1 mark]
23.
If there are 16 chromosomes in a cell that is about to divide, what will be the number of chromosomes in a
daughter cell after division by mitosis or meiosis?
[1 mark]
Markscheme
B
24.
Which of his grandparents must be a carrier if none of them had the disease?
[1 mark]
A. Maternal grandmother (his mother’s mother)
B. Maternal grandfather (his mother’s father)
C. Paternal grandmother (his father’s mother)
D. Paternal grandfather (his father’s father)
Markscheme
A
25.
What makes gene transfer between species possible?
[1 mark]
A. All species use the same genetic code.
B. All species have the same genetic material.
C. All species produce the same polypeptides.
D. All species transcribe genes using plasmids.
Markscheme
A
26.
What stage of meiosis is shown in the micrograph?
A. Prophase I
B. Metaphase II
C. Anaphase II
D. Telophase I
[1 mark]
Markscheme
C
27. What is the name given to a heritable factor which controls a specific characteristic?
[1 mark]
A. Allele
B. Chromosome
C. Gene
D. Mutation
Markscheme
C
28.
What would be the expected result if a woman carrier for colour blindness and a colour blind man had many
children?
[1 mark]
A. All offspring will be colour blind.
B. All male offspring will be colour blind and all females normal.
C. All males will be normal and all females will be colour blind.
D. All females will be carriers of colour blindness or colour blind
Markscheme
D
29.
In peas, tall is dominant to dwarf. In a cross between a dwarf plant and a heterozygous tall plant what percentage [1 mark]
of the offspring will be dwarf?
A. 0 %
B. 25 %
C. 50 %
D. 100 %
Markscheme
C
30.
What is the difference between the alleles of a gene?
[1 mark]
A. Their position on the chromosome
B. Their amino acid sequence
C. Their pentose sugars
D. Their base sequence
Markscheme
D
31.
Which of the following statements about homologous chromosomes is correct?
A. Each gene is at the same locus on both chromosomes.
B. They are two identical copies of a parent chromosome which are attached to one another at the centromere.
C. They always produce identical phenotypes.
D. They are chromosomes that have identical genes and alleles.
[1 mark]
Markscheme
A
32.
The pedigree chart below shows the blood types of three members of a family.
[1 mark]
Which could be the blood types of individuals 1 and 2?
Markscheme
D
33. Which of the following statements about homologous chromosomes is correct?
A. Each gene is at the same locus on both chromosomes.
B. They are two identical copies of a parent chromosome which are attached to one another at the centromere.
C. They always produce identical phenotypes.
D. They are chromosomes that have identical genes and alleles.
Markscheme
A
[1 mark]
34.
In the following diagram, which pair represents homologous chromosomes?
[1 mark]
A. 1 and 2
B. 3 and 4
C. 2 and 5
D. 4 and 6
Markscheme
D
35.
Rhesus factor is an antigen present on the surface of red blood cells of Rhesus positive individuals. Rhesus
[1 mark]
positive (Rh+ ) is dominant to Rhesus negative (Rh– ). A mother with Rhesus negative blood gives birth to a baby with
Rhesus positive blood and there are concerns that subsequent pregnancies will trigger an immune response.
What are the genotypes of the mother and her first baby?
Markscheme
B
36.
What is meiosis?
[1 mark]
A. Division of a diploid nucleus to form diploid nuclei
B. Reduction division of a haploid nucleus to form diploid nuclei
C. Reduction division of a diploid nucleus to form haploid nuclei
D. Division of a haploid nucleus to form haploid nuclei
Markscheme
C
37.
What causes genetic variety in the formation of gametes during meiosis?
A. Crossing over in prophase I and random orientation of homologous chromosomes in metaphase I
B. Crossing over in metaphase I and random orientation of homologous chromosomes in metaphase II
C. Linkage of genes in prophase I and crossing over in metaphase I
D. Linkage of genes in metaphase I and random orientation of homologous chromosomes in metaphase II
[1 mark]
Markscheme
A
38.
What type of inheritance is shown in this pedigree chart?
[1 mark]
A. X-linked dominant
B. Y-linked dominant
C. X-linked recessive
D. Y-linked recessive
Markscheme
C
39.
In humans a V-shaped hair line is dominant to a straight hair line. A woman with a V-shaped hair line and a man [1 mark]
with a straight hair line have children. The woman has a mother with a straight hair line. What is the proportion of children
who are likely to have a V-shaped hair line?
A. Half of the children
B. A quarter of the children
C. All of the children
D. None of the children
Markscheme
A
40a.
State the name of the process used to produce the pattern of bands seen in the image.
[1 mark]
Markscheme
gel electrophoresis/DNA profiling
40b.
Determine, with a reason, which male is the father of the child.
[1 mark]
Markscheme
male 1 because all child’s bands / alleles match either mother or male 1 / (approximately) half of bands match male 1
[1]
Do not accept reference to genes.
41a.
Define linked genes.
[1 mark]
Markscheme
Genes located on the same chromosomes.
In cats, the allele for curled ears (C) is dominant over the allele for normal ears (c). The allele for black colour
[3 marks]
(B) is dominant over the allele for grey colour (b). A cross occurs between two cats that are both heterozygous for these
unlinked traits.
41b.
Using a Punnett grid, predict the ratio of phenotypes of offspring in the next generation.
Markscheme
a. Punnett grid correctly drawn / correct gametes;
b. phenotype ratio 9:3:3:1;
c. correct identification of expected phenotypes (9) black cats with curled ears, (3) grey cats with curled ears, (3) black
cats with normal ears, (1) gray cat with normal ears;
In a strain of soybeans, high oil content (H) in seeds is dominant to low oil content (h) and four seeds in a pod (F) [1 mark]
is dominant to two seeds in a pod (f). A farmer crosses two soybean plants, both with high oil content and four seeds in a
pod. The offspring have a phenotypic ratio of 9 : 3 : 3 : 1.
42a.
Identify the genotypes of the soybean plants with high oil content and four seeds in a pod that were used in the cross.
Markscheme
HhFf HhFf
/ (both) HhFf;
In a strain of soybeans, high oil content (H) in seeds is dominant to low oil content (h) and four seeds in a pod (F)[2 marks]
is dominant to two seeds in a pod (f). A farmer crosses two soybean plants, both with high oil content and four seeds in a
pod. The offspring have a phenotypic ratio of 9 : 3 : 3 : 1.
42b.
Determine the genotypes of the gametes and offspring using a Punnett grid.
Markscheme
all gametes shown correctly on Punnett grid;
all offspring genotypes correct;
In a strain of soybeans, high oil content (H) in seeds is dominant to low oil content (h) and four seeds in a pod (F) [2 marks]
is dominant to two seeds in a pod (f). A farmer crosses two soybean plants, both with high oil content and four seeds in a
pod. The offspring have a phenotypic ratio of 9 : 3 : 3 : 1.
42c.
Identify the phenotypes of each part of the phenotypic ratio.
Markscheme
Award [1] for any two correct phenotypes.
42d.
Deduce the reason for the person developing as a female.
[1 mark]
Markscheme
no Y chromosome.
42e.
Determine, with a reason, whether this karyotype shows that non-disjunction has occurred.
[1 mark]
Markscheme
yes as there is only one X chromosome/chromosome missing/only 45 chromosomes
43a.
State three processes occurring in a cell during interphase of the cell cycle but not in mitosis.
1. ....................................................................
2. ....................................................................
3. ....................................................................
Markscheme
a. growth (of cells);
b. transcription/protein synthesis/translation;
c. DNA replication / genetic material copied;
d. production of organelles/mitochondria/chloroplasts;
e. named normal activity of cell (eg active transport, movement, secretion);
NB Do not accept G1, S, G2 unless linked to correct process.
[3 marks]
43b.
Explain how sexual reproduction can allow evolution to occur.
[3 marks]
Markscheme
a. sexual reproduction promotes variation in species;
b. independent assortment of genes / random orientation of chromosomes in metaphase/meiosis;
c. crossing-over provides new combinations of alleles;
d. production of great variety of gametes (by meiosis) / different combinations of chromosomes in gametes;
e. (random) combination of gametes from both parents (in fertilization);
f. (genetic) variation allows natural selection which leads to evolution;
44a.
(i) State the technique used to collect cells for pre-natal testing.
[3 marks]
(ii) State the method used to arrange the chromosomes in a karyotype.
(iii) State at what stage in the cell cycle the cells would be when this photograph was taken.
Markscheme
(i) amniocentesis/sampling amniotic liquid/fluid (via needle)/chorionic villus sampling
(ii) chromosomes are grouped by pairs according to size and structure/band pattern/location of centromeres
(iii) metaphase/late prophase of mitosis
Albinism is inherited as a recessive trait; the alleles of the gene involved are A and a. An individual with albinism [3 marks]
produces little or no pigment in the eyes, skin and hair. In a family, one sister has albinism while the parents and other
sister have
normal pigmentation.
44b.
(i) Determine, using a Punnett grid to show your reasoning, the possible genotypes of the sister with normal pigmentation.
(ii) Deduce the probability that the next child of this couple will have albinism.
Markscheme
(i) Punnett grid shows the gametes (A and a) on one axis and the gametes (A and a) on the other axis and genotypes
(AA, Aa, Aa, and aa) of offspring;
AA/homozygous dominant and Aa/heterozygous (show normal
pigmentation); Both needed
Do not award marks to any answer suggesting sex linkage.
(ii) 1/4 /25 %/0.25 probability of albinism / 1 in 4 chance
45a.
Analyse this karyotype.
[2 marks]
Markscheme
Male has (one X and) one Y chromosome / X chromosome is bigger than Y chromosome;
non-disjunction leads to three copies of chromosome 13/trisomy 13.
45b.
Outline the inheritance of hemophilia in humans.
[2 marks]
Markscheme
sex-linked/on X chromosome;
recessive allele / Xh;
more common in males than females;
heterozygous females are carriers / only females can be carriers;
45c.
Using an example, describe polygenic inheritance.
[3 marks]
Markscheme
more than one gene contribute to/control same characteristic;
as number of genes increase so does possible number of phenotypes;
leads to continuous variation;
specific example; (eg human skin color (due to differing amounts of melanin))
Award [2 max] for general points with no example.
46a.
Determine the genotype of person 1.
[1 mark]
Markscheme
X HY
46b.
Deduce the genotype of the mother of person 2.
[1 mark]
Markscheme
X HX h
Apply ECF if upper case and lower case forms of another letter are used to correctly denote hemophilia in female
genotype.
46c.
If person 3 has a son, and the father is a hemophiliac male, predict the son’s phenotype.
[1 mark]
Markscheme
normal (male) / not affected / no hemophilia
Do not accept X HY by itself, since question asks for phenotype.
46d.
Suggest how sheep could be genetically modified to help the treatment of hemophilia in humans.
[1 mark]
Markscheme
genetically modify sheep to produce (blood) clotting factors (e.g. factor IX) in milk
47a. The correlation shown in the data above can be explained by natural selection. Outline how the process of
natural selection can lead to evolution.
[3 marks]
Markscheme
offspring compete/environment cannot support all offspring;
(genetic) variation in the offspring;
natural selection /survival of better adapted/fittest organisms;
reproduction passes characteristics to other generations;
allele frequencies change;
malaria causes selection pressure (in Africa/worldwide);
different hemoglobin/sickle-cell genotypes exist / normal hemoglobin and sicklecell alleles exist;
natural selection/resistance to malaria of sickle-cell heterozygotes/allele;
survivors pass on sickle-cell allele to offspring; (do not accept sickle-cell anemia)
frequency of sickle-cell allele highest in areas of high malaria incidence;
47b.
Explain how a base substitution mutation, such as GAG to GTG, can lead to a disease like sickle-cell anemia.
[2 marks]
Markscheme
change in the codon (of the mRNA);
tRNA with a different anticodon attaches;
(if codon changed) wrong/different amino acid is joined to peptide/glutamic acid replaced by valine;
distorted hemoglobin molecule alters red blood cell shape/reduces ability to carry oxygen;
47c.
Using a Punnett grid, determine the possible genotypes and phenotypes of a cross between a man and a woman [2 marks]
who are both carriers of the sickle-cell allele. Use the symbol HbS for the sickle-cell allele and HbA for the normal allele.
Phenotypes:
Markscheme
(genotypes shown in a Punnett grid eg)
(phenotypes)
(HbA HbA ) normal and (HbA HbS) normal carrier/intermediate/sickle-cell trait and (HbS HbS) sickle-cell
anemia/diseased / (HbA HbA and HbA HbS) normal /symptomless and (HbS HbS) sickle-cell anemia/diseased;
To award the mark all phenotypes must be mentioned.
48a.
State what would be used as the control in this experiment.
[1 mark]
Markscheme
maize not modified/transformed with Bt (genes) / maize that did not have Bt gene added / not genetically modified /
untreated maize
48b.
Outline the effects of the three species of stem borer on Bt maize type A.
[2 marks]
Markscheme
there was a decrease in damage by all three types of stem borers compared to control;
there was almost no change in damage by Eldana compared to control;
there was almost no damage/little effect (to Bt maize type A) by Sesamia (and Eldana);
Busseola caused the most damage (to Bt maize type A);
48c.
Evaluate the efficiency of the types of Bt maize studied, in controlling the three species of stem borers.
[2 marks]
Markscheme
very efficient at controlling Sesamia;
type B is the most effective against the three stem borers collectively;
no type of Bt maize controlled Busseola well / vice versa i.e. Busseola not well controlled by any types of Bt maize;
all types of Bt maize decreased Sesamia damage (significantly) / Bt maize type E not damaged by Sesamia / vice
versa;
Bt maize types C/H/I had more damage caused by Busseola (than was caused in the control) / vice versa;
all types of Bt maize decreased Eldana damage (to some extent) / type B was not damaged by Eldana / vice versa;
Eldana damage low in control / less effect;
cannot determine efficiency since data is about leaf damage and stem borers may feed (preferentially) on other
structures/stems/roots;
48d.
Describe the change in mean mass for the female rats during the 90-day experiment.
Markscheme
mass increases in all three groups;
increase is more rapid in beginning and tapers off later in the study;
mass seems to be levelling off in rats fed Bt and non-Bt maize / rate of increase in mass is slowing down;
rats fed rat food always have higher mass/greater mass increase than those fed either type of maize;
[2 marks]
48e.
Evaluate the use of Bt maize as a food source compared to the other diets tested.
[3 marks]
Markscheme
all three foods result in the same pattern of growth/mass gain / highest rate of growth at start of study / tapering off
later in the study;
Bt maize causes same amount of growth as non-Bt maize / appears to be as good a food source as non-Bt maize /
there is no significant difference between Bt and non-Bt maize (in terms of mass gain);
corn (both types) appears to cause less growth/mass gain than rat food / vice versa;
genetic modification does not affect growth/mass gain;
no evidence to support risk of Bt maize to growth/mass gain;
study does not investigate other possible risks of Bt maize to rats;
sample size is small / only 12 rats (in each group) so this may not be enough to give trends;
only female rats tested, no males;
49a.
Explain why carriers of sex-linked (X-linked) genes must be heterozygous.
[2 marks]
Markscheme
carrier has (one copy of) a recessive allele;
must also have a dominant allele to prevent having the condition/disease;
or
cannot be homozygous dominant or they would not carry the recessive allele;
cannot be homozygous recessive or they would have the condition/disease;
49b.
Label the diagram below which shows a basic gene transfer.
[2 marks]
I. .............................................................
II. .............................................................
III. .............................................................
IV. .............................................................
Markscheme
Award [1] for every two correct answers.
I. bacterial cell/bacterium/prokaryote;
II. plasmid;
III. inserted/engineered/cloned/desired DNA/gene / DNA/gene from donor cell;
IV. genetically modified/transformed/GM/recombinant organism/cell/ bacterium/host cell containing recombinant
plasmid;
49c.
State two general types of enzymes used in gene transfer.
[1 mark]
Markscheme
restriction enzymes / endonucleases;
ligases;
reverse transcriptase;
Award [1] for two correct answers.
50a.
Define the term allele as used in genetics.
[1 mark]
Markscheme
allele: one specific form of a gene (occupying the same gene locus as other alleles of the same gene)
50b. List the possible genotypes for blood group B.
[1 mark]
Markscheme
IB IB and IB i
51a.
Predict the genotypic and phenotypic ratios of the possible offspring of a male hemophiliac and a female carrier[3 marks]
using suitable symbols for the alleles in a Punnett grid.
Genotypic ratio:
Phenotypic
ratio:
Markscheme
correctly constructed Punnett square with correct gamete genotypes;
genotypic ratio: 1 X HX h : 1X hX h : 1 X HY : 1 X hY; (can be inferred from cells of Punnett square)
phenotypic ratio: 1 female hemophiliac : 1 female carrier/non-hemophiliac :
1 male hemophiliac : 1 male normal/non-hemophiliac / 50 % hemophiliac :
50 % non-hemophiliac;
Allow ECF. Award [2 max] if notation used does not indicate sex linkage, i.e. if cross is Hh×hh.
51b.
Hemophilia is a disorder where the ability to control blood clotting or coagulation is impaired. Describe the
process of blood clotting.
[2 marks]
Markscheme
release of clotting factors from platelets/damaged cells;
conversion of prothrombin to thrombin;
thrombin catalyses the conversion of fibrinogen into fibrin;
(insoluble) fibrin (net) captures blood cells;
52a.
State the stage of mitosis typified by image II.
[1 mark]
Markscheme
anaphase
52b.
List two processes that involve mitosis.
[2 marks]
Markscheme
a. growth (through increasing cell number);
b. embryonic development;
c. tissue production/repair;
d. (asexual) reproduction;
52c.
State the process that results in tumour (cancer) formation or development.
[1 mark]
Markscheme
uncontrolled mitosis/cell division
52d.
Explain, using one example, how non-disjunction in meiosis can lead to changes in chromosome number.
[2 marks]
Markscheme
a. pair of homologous chromosomes moves in same direction/does not separate during anaphase I / chromatids move
in same direction/do not separate during anaphase II;
b. leaving a cell with an (some) extra chromosome(s)/missing chromosome(s);
c. an example; (e.g. Down syndrome where there is an extra chromosome number 21);
53a.
Describe karyotyping and one application of its use.
Markscheme
cells undergoing mitosis are used for karyotyping;
process of mitosis is stopped at (mitotic) metaphase;
chromosomes (cut from photographs) are arranged in pairs of similar structure/ homologous chromosomes;
allows abnormalities in the chromosome number/appearance to be seen;
any valid example (e.g. in Down syndrome / gender of fetus);
detected by identifying unique feature (e.g. trisomy 21 / one extra chromosome / 47 chromosomes);
Award [3 max] for an example with no description of karyotyping.
[4 marks]
53b.
Describe a technique used for gene transfer.
[5 marks]
Markscheme
restriction enzymes/endonucleases cut a small fragment of DNA from an organism;
same restriction enzymes used to cut DNA of plasmid / e.g. E. coli;
sticky ends are the same in both cases;
fragment of DNA is inserted into the plasmid;
spliced together by ligase;
to make recombinant DNA/plasmids;
recombinants can be inserted into host cell and cloned;
53c.
Describe a technique used for gene transfer.
[5 marks]
Markscheme
restriction enzymes/endonucleases cut a small fragment of DNA from an organism;
same restriction enzymes used to cut DNA of plasmid / e.g. E. coli;
sticky ends are the same in both cases;
fragment of DNA is inserted into the plasmid;
spliced together by ligase;
to make recombinant DNA/plasmids;
recombinants can be inserted into host cell and cloned;
53d.
Using a named example, discuss the benefits and harmful effects of genetic modification.
[9 marks]
Markscheme
genetic modification is when the DNA/genotype of an organism is artificially changed;
genetic modification alters some characteristic/phenotype of the organism;
named example with modification (e.g. salt tolerance in tomato plants);
benefits: [5 max]
allows crops to be grown where they would not grow naturally;
provides more food;
economic benefits;
expands world’s productive farmland;
reduces the need to clear rainforests to grow crops;
lowers cost of production;
less pesticides/fertilizers/chemicals needed so better for environment;
Award marks for any valid benefit consistent with a named example.
harmful effects: [5 max]
may be released into natural environment;
may affect food chains / unintended effects on other organisms;
may affect consumers e.g. allergies/health risks;
unfair to smaller farmers who cannot compete;
long-term effects are unknown; risk of cross-pollination;
risk of long-term contamination of soil;
Award marks for any harmful effect consistent with the named example.
53e.
Using a named example, discuss the benefits and harmful effects of genetic modification.
[9 marks]
Markscheme
genetic modification is when the DNA/genotype of an organism is artificially changed;
genetic modification alters some characteristic/phenotype of the organism;
named example with modification (e.g. salt tolerance in tomato plants);
benefits: [5 max]
allows crops to be grown where they would not grow naturally;
provides more food;
economic benefits;
expands world’s productive farmland;
reduces the need to clear rainforests to grow crops;
lowers cost of production;
less pesticides/fertilizers/chemicals needed so better for environment;
Award marks for any valid benefit consistent with a named example.
harmful effects: [5 max]
may be released into natural environment;
may affect food chains / unintended effects on other organisms;
may affect consumers e.g. allergies/health risks;
unfair to smaller farmers who cannot compete;
long-term effects are unknown; risk of cross-pollination;
risk of long-term contamination of soil;
Award marks for any harmful effect consistent with the named example.
54a.
Explain, using a named example, how polygenic inheritance gives rise to continuous variation.
[2 marks]
Markscheme
human skin colour can vary from pale to very dark / amount of melanin varies;
skin colour/melanin controlled by (alleles from) at least three/several genes;
no alleles are dominant / alleles are co-dominant / incomplete dominance;
many different possible combinations of alleles;
skin colour controlled by cumulative effect/combination of genes/alleles;
Award the above marking points for any other valid example.
54b.
Describe the inheritance of colour blindness in humans.
[3 marks]
Markscheme
sex linked condition;
carried on an X chromosome / absent from Y chromosome;
if present in male causes colour blindness;
(allele is) recessive so heterozygous females are not colour blind;
homozygous females are colour blind;
Do not allow carried on sex chromosome.
55a. State the property of stem cells that makes them useful in medical treatment.
[1 mark]
Markscheme
has the ability to differentiate (into specialized tissue)
55b. Explain how multicellular organisms develop specialized tissues.
[2 marks]
Markscheme
only some genes are expressed in each cell type/tissue;
tissues therefore develop differently/become differentiated;
example of differentiated cell and the function of tissues;
55c. Outline some of the outcomes of the sequencing of the human genome.
[3 marks]
Markscheme
knowledge of location of human genes / position of human genes on chromosomes;
knowledge of number of genes/interaction of genes / understanding the mechanism of mutations;
evolutionary relationships between humans and other animals;
discovery of proteins / understanding protein function / detection of genetic disease;
leads to the development of medical treatment/enhanced research techniques;
knowledge of the base sequence of genes/study of variation within genome;
56a.
Outline a possible cause of Down syndrome.
[4 marks]
Markscheme
non-disjunction (can cause Down syndrome);
occurs when pair of homologous chromosomes fails to separate during meiosis;
one gamete/daughter cell receives two chromosomes / diagram showing this;
occurs in anaphase I/II of meiosis;
fertilization involving this gamete leads to change in chromosome number/47 chromosomes;
most common form of Down is trisomy 21/extra chromosome 21;
increase risk of Down syndrome with increased age of mother;
56b.
Outline the processes involved in oogenesis within the human ovary.
Markscheme
oogenesis is process by which female gametes/eggs are produced;
begins during fetal development; oogonia/large number of cells formed by mitosis;
oogonia/cells enlarge/undergo cell growth/become primary oocytes;
begin first meiotic division but stop in Prophase I;
until puberty;
(at puberty) some follicles develop each month in response to FSH;
(primary oocyte) completes first meiotic division;
forms two cells of different/unequal sizes / unequal distribution of cytoplasm;
(creating a) polar body;
polar body eventually degenerates;
larger cell/secondary oocyte proceeds to meiosis II;
stops at prophase II;
meiosis II completed if cell is fertilized;
ovum and second polar body formed;
[8 marks]
56c. Discuss the ethical issues surrounding IVF.
[6 marks]
Markscheme
To award full marks, discussion must contain both pro and con considerations.
pros/positive considerations: [3 max]
chance for infertile couples to have children;
decision to have children is clearly a conscious one due to difficulty of becoming pregnant;
genetic screening of embryos could decrease suffering from genetic diseases;
spare embryos can safely be stored for future pregnancies/used for stem cell research;
cons/negative considerations: [3 max]
IVF is expensive and might not be equally accessible;
success rate is low therefore it is stressful for the couple;
it is not natural/cultural/religious objections;
could lead to eugenics/gender choice;
could lead to (unwanted) multiple pregnancies with associated risks;
production and storage of unused embryos / associated legal issues / extra embryos may be used for (stem cell)
research;
inherited forms of infertility might be passed on to children;
Accept any other reasonable answers.
57a.
Describe how natural selection leads to evolution.
[6 marks]
Markscheme
populations produce more offspring than can survive;
individuals show variation;
limited resources;
create a struggle for survival/competition;
survival of the fittest / some are better suited to the environment and survive;
variation/characteristic must be heritable;
best fitted individuals survive to reproduce;
advantageous variation/characteristic/allele passed on;
over time advantageous variation/characteristic/allele increases in the population;
57b. Explain the consequences of altering a DNA base in the genome of an organism.
Markscheme
altering a base (in DNA) is a (point) mutation;
only has an effect if base is in a gene;
when mRNA is produced by transcription one mRNA base is different;
one codon in mRNA is different;
one amino acid is different in the polypeptide;
polypeptide produced by translation of mRNA;
some base changes do not change the amino acid coded for;
structure of polypeptide /protein may be altered;
usually the polypeptide/protein does not function as well;
example given:
disease: sickle cell anemia;
mutation: GAG to GTG;
consequence in translation: glutamic acid to valine;
consequence for protein: hemoglobin altered so sickle cell formed;
consequence for individual: less oxygen can be carried;
© International Baccalaureate Organization 2017
International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®
[8 marks]
Printed for Highland High School