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UNIT 4
BIOLOGY NOTES
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CHAPTER 7 – GREEN SHEETS
Mendel – Heredity
1 History of
Watson and Crick, Franklin and Chargraff – Structure of DNA
Heredity
Somatic cells, Gamete, Genes, Locus, Alleles, Diploid,Haploid, Chromatin,
Definitions of
Chromatids, Histones, Centromere, Homologous, Autosomes, Sex
Chromosomes
chromosomes(X,Y) Karyotype,
2
*
3
4
5
6
7
8
9
Structure in
Eukaryotes
Karyotype
Structure in
Prokaryotes
Cell division
Overview
Monohybrid
crosses
Monohybrid cross,
Punnet square,
Probabilities
Phenotype
TEST CROSS
CODOMINANCE
MULTIPLE
ALLELES
LETHAL
ALLELES
POLYGENES
DIHYBRID
CROSS
INHERITANCE
PATTERNS
PEDIGREES
Picture of chromosome pairs normally and when dividing, with parts labelled.
Cut out the chromosomes (doubled) and finding their homologous pair.
Structure of chromosomes and genetic material in prokaryotes (bacteria)
Asexual Versus Sexual reproduction
Overview of Mitosis and meiosis.
Mendels study of pea plants and words used -Purebreeding, Parent, F1,F2
Dominant, Recessive, Genotype, Phenotype,Homozygous, Heterozygous
An example of a monohybrid cross used by Mendel to show that characteristics
where inherited from each parent and weren’t a blend of the parents
(codominance not discovered then) but were discrete as dominant and recessive.
The F1 generation carried the hidded recessive which revealed itself again in F2
Phenotype = Genotype + Environment
Test Cross – To find out if an organism displaying a DOMINANT phenotype is
HOMOZYGOUS or HETEROZYGOUS
Or Incomplete dominance result in a THIRD phenotype for the Heterozygote.
When there are more than two alleles for a gene
eg blood type has 3 alleles A,B and O
This is when a particular combination of alleles is lethal (kills) the organisms so
the ratio of offspring will be different to the expected because some die.
This is when there are many genes for one characteristic which shows a
CONTINUOUS change eg HEIGHT in humans and skin COLOUR.
This the study of two characteristics at a time. In the text book it was height and
flower colour of pea plants. You need a large punnet square.
If the characteristic is studied on the AUTOSOMAL chromosomes of the
X –LINKED (Sex linked)
This studies inheritance over generations to establish if a characteristic is
Autosomal, X-linked , Dominant or Recessive.
Working through two Autosomal conditions to find the genotype of the people
in the various generations. Recessive can skip generations.
Working through two X-linked conditions to find the genotype of the people in
the various generations. Recessive Mother-son, Dominant Father-Daughter.
Practice working through pedigrees
How to see if a characteristic is carried on the Autosomal chromosomes or the X
chromosomes.
PEDIGREE
AUTOSOMAL
10 PEDIGREE
X-LINKED
11
12 RECIPROCAL
CROSS
Genotype symbols
2 Diploid Daughter cells)
REPRODUCTION. MITOSIS (Somatic-Normal
MEIOSIS (Gametes-Eggs/Sperm 4 Haploid Daughter cells)
Semi-conservative Replication, DNA helicase, DNA polymerase,
DNA
REPLICATION
I.P.M.A.T Interphase, Prophase, Metaphase, Anaphase, Telophase. Cytokinesis
STAGES
1 Division- Chromosomes condense Chromotids, Chromatin, Centrioles/Spindle
MITOSIS
2 Divisions- IPMAT 1 and 2. Homologous pairs cross over at Prophase,
MEIOSIS
Homologous pairs line up along the equator at Metaphase 1
MENDELS LAWS Law of Segregation, Independent Assortment, Linked genes ,
Recombination/Crossover
Dihybrid cross results if genes on separate chromosomes and if linked.
LINKED GENES
CHAPTER 8
DNA
1
Text book questions. ***Must be done.
Transcription(premRNA) -Splicing (introns,exons)- Translation (polypeptide)
A summary of the processes and structures used in
Transcription (RNA polymerase, Codons, mRNA, Splicing Exon/Intron,
Translation (Ribosome, tRNA, Anticodon, Amino Acid,
Lactose metabolism in bacteria
Switches on/off, Regulatory Gene, Regulatory protein, Promoter, Operator,
GENE –(Point)Substitution, Inversion, Insertion/Deletion (Frameshift)
CHROMOSOME –Deletions, Inversions, TRANSLOCATION, Duplications
(NON-DISJUNCTION) -Chr 21 Downs Syndrome
Klinefelters XXY, Turners XO, Polyploidy(Plants)
Mutation rate, Mutagens, Xrays, U.V light,
2
3
4
TRANSCRIPTION
TRANSLATION
PROTEIN
SYNTHESIS
BIOZONE
GENE
REGULATION
MUTATIONS
Somatic/Germline
ANEUPLOIDY
CAUSES
YELLOW SHEETS
CHAPTER 9
BIOTECHNOLOGY
BLUE SHEETS
Prokaryotes (Plasmids)/ Eukaryotes, Overview of Tools and Techniques
1
Cutting DNA at specific recognitions sites
2 RESTRICTION ENZYMES
Separating DNA pieces based on their size and charge
GEL ELECTROPHORESIS
(Polymerase chain reaction)-making multiple copies of DNA pieces.
PCR
Glue. Rejoining DNA)
3 DNA ligase
Foreign DNA inserted Eg A gene inserted into a plasmid
RECOMBINANT DNA
Foreign DNA inserted into a bacteria.
TRANSFORMATION
Finding the order of the nucleotides (A,T,C,G)
4 DNA SEQUENCING
using STR’s Short tandem Repeats eg ATCCG ATCCG ATCCG
DNA PROFILING
– searching DNA for a complementary piece
5 GENE PROBES
6 GENETIC TESTING
Restriction Fragment Length Polymorphism (RFLP)
Restriction sites lost/gained so can use it to identify organisms.
Transformations with Eukaryotes
7 TRANSGENIC ORGANISM
Embryo splitting, Nuclear transfer
CLONING
Undifferentiated cells which can turn into all cell types
STEM CELLS
CHAPTER 10
EVIDENCE OF CHANGE
PINK SHEETS
1 FOSSILS and DATING
Divergent evolution (Homologous features)
COMPARITIVE ANATOMY
Convergent evolution (Analogous features)
Vestigial Features, Embryology
Closely related species live close together. If separated it may have been
BIOGEOGRAPHY
due to ancestors on Gondwana or Pangea before they split.
MOLECULAR COMPARISON
Nuclear DNA (Sequencing and Hybridisation)
Mitochondrial DNA, Chloroplasts and Proteins.
2 FOSSILS AND DATING
FOSSIL FORMATION
DATING – Comparitive techniques,
Absolute Techniques- Radio-isotopes C14 , K40
A worksheet summary of evolution.
Video - EVOLUTION
CHAPTER 11
CHANGING POPULATIONS
PINK SHEETS
1 DARWINS THEORY
NATURAL SELECTION
Adaptation, Variation, Selective pressures,Evolution,
Gene pool-Mutation, Natural selection, Gene flow, Genetic drift
Change- Species, Allopatric Speciation
2 ALLELE FREQUENCY
Factors Preventing Reproduction –Pre and Post zygotic
1
2
3
1
2
3
Game-Selecting for a lack of aggression causes allele change
Wolves to Dogs
Homologous features Primates, monkeys apes, Chromosome
HUMAN EVOLUTION
Hominins, BIPEDALISM, LARGER BRAIN
HOMININ PHYLOGENY
Spread from Africa, Nomadic-settled, Cultural change
HOMININS
HUMAN INTERVENTION IN EVOLUTIONARY PROCESS
SELECTIVE BREEDING
Issues on the use of techniques which could alter human
TRANSGENIC ORGANISMS
Evolution.
4CLONING,SCREENING
STEM CELLS, GENE THERAPY
CH.
1 7
2
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3
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4
7
5
6
8
8
7
8
8
8
9
9
10 9
11 9
12
12 10
13 10
14 11
15 12
MATERIAL
STRUCTURE OF CHROMOSOMES
Prokaryotes, Eukaryotes, Genes, Locus, alleles, Genotype/Phenotype,
Phenotype= genotype + environment
Homozygous/Heterozygous, homologous, Autosomes, Sex chromosomes,
KARYOTYPE, Chromatids, Polyploid, Apoptosis.
CELL DIVISION
DNA. Replication (Semi-conservative, DNA helicase, DNA polymerase,
DNA ligase, MEIOSIS -I.P.M.A.T x2 (Haploid) ,MITOSIS (Diploid)
Crossover,
CROSSES
Codominant, Homozygous/Heterozygous, Genotype/Phenotype)
Test Cross for genotype(or Back cross), Dihybrid cross, Linked genes
Multiple alleles (Blood type) , Lethal alleles,
Polygenes (continuous variation – Height ,colour)
Monohybrid cross Mendels crosses.(Dominant, Recessive)
X linked cross (Reciprocal)
Dominant/recessive, Autosomal/Sex linked
PEDIGREES
Dominant/recessive, Autosomal/Sex linked
PROTEIN SYNTHESIS
TRANSCRIPTION- Nucleus,RNA polymerase,Codons, mRNA Exon/Intron
TRANSLATION – Ribosomes(rRNA and Proteins),tRNA, Anticodons,
Reverse Transcriptase – RNA to DNA
DNA Structure and Discovery Watson/Crick, Rosalind Franklin,
Chargraff’s bases, Chromatin, Histones,
GENE REGULATION – Gene switched on/off, Promoter, Operator,
regulatory genes ,
MUTATIONS – Chromosome (Crossover during meiosis,
Deletions,Inversions, TRANSLOCATION, Duplication ANEUPLOIDY
(Non-Disjuction-Trisomy, Klinfelters, Turners) Polyploidy
Gene- Substitution, Inversion, Insertion/deletion, Frameshift, Mutagens,
Mutation rates, Somatic or Germ line
EXPERT
Matthew
Dino
Nadia
Casey
Kandace
Kayla
Shirelle
Natalia
GENETIC ENGINEERING- Tools – Restriction enzymes(sticky/blunt ends),
DNA ligase, Recombinant DNA , Vectors- plasmids, viruses, Transgenic
organisms, Transformations, Antibiotic resistance markers,
Jessica
P.C.R ,Gel electrophoresis, DNA sequencing, Human Genome project,
DNA profiling(fingerprinting)-STR’s and VNTR’s ,
Genetic Testing- Restriction length polymorphisms(RFLP), Gene Probes,
Artificial evolution– Selective breeding,Cloning(Embryo splitting, nuclear transfer)
Stem cells, Transformation, Screening, Gene therapy
FOSSILS- sedimentary rock, trace,
Dating- absolute (radioisotopes), Comparative (index fossils)
Pangea-Gondwana Biogeography, Wallaces line
EVOLUTION – Adaptation, Recent common ancestors, ,
Divergent- homologous features, Convergent-analogous features, Phylogenetic tree
(cladogram) COMPARISONS- Structural(Embryology, homologous features)
Biochemical(DNA Nuclear/ Mitochondrial, Molecular clock, proteins-cytochrome
C, haemoglobin, Hybridisation) Biogeography,
Extinction,
NATURAL SELECTION–Variation, Selection pressure, Gene Pool, Wild-type
allele, Polymorphism, Mutation, Genetic drift, Bottleneck effect, Founder effect,
Darwin, Lamarck, Wallace, - Speciation, Isolating mechanisms (pre-reproductive
and post reproductive) Allopatric speciation, Heterozygote fitness (sickle cell)
Hominids to Hominins – Primates, Apes and monkeys, Fossil structure,
Bipedalism, opposable thumb, Australopithecus(afarensis,africanus), Paranthropus,
Homo habilis, Homo erectus, Homo neanderthal, Homo florensis, Homo sapiens,
Cultural evolution, tools, fire, painting, rituals, language,
Kireeti
Alula
Rachel
Robert
Blake
David
Done
CHAPTER 7
STRUCTURE OF CHROMOSOMES
PART 1
Prokaryotes, Eukaryotes, Genes, Locus, alleles, Genotype/Phenotype, Centromere, homologous,
Chromatids, Diploid, Haploid, Polyploid, Autosomes, Sex chromosomes, KARYOTYPE,
Questions
1. Describe the relationship between DNA, Genes and chromosomes.
2. What is the difference between a gene and an allele?
3. What is the difference between the terms genotype and phenotype?
Do two organisms with the same genotype necessarily have the same phenotype. Give an example to
support your answer.
Do two individuals with the same phenotype necessarily have the same genotype? Give an example to
support your answer.
4. How many chromosomes do humans have in a) normal body cells (somatic cells) b) sex cells (gametes).
5. The chromosomes sets of normal human males and females are similar but different. Explain.
6. What differences in chromosome structure are used to sort chromosomes for a karyotype?
7. What features visible under a microscope can be used to distinguish a human X chromosome from a Y
chromosome?
8. Referring to Recessive conditions what is meant by a carrier?
9. What is Apoptosis and how is it different to necrosis?
APOPTOSIS
Apoptosis is controlled cell death. It can happen during development when the body is being shaped (cells
between fingers undergo apoptosis to separate fingers), when the cell is infected or not functioning
correctly or as a control for rapidly dividing cells like blood or skin. It is initiated by chemicals either from
in or out-side the cell and results in enzymes dismantling the cell whose parts are consumed by phagocytes.
KARYOTYPE
The analysis involves comparing chromosomes for their length, the placement of centromeres (areas where
the two chromatids are joined), and the location and sizes of
_______ ________ ________
1
2
3
________ ________
4
5
________ ________ ________ ________
6
7
8
9
________ ________ ________
10
11
12
________ ________ ________
13
14
15
________ ________
19
20
________ ________ ________
16
17
18
________ ________
21
22
________
XX/XY
How many chromosomes? What sex? Are there any chromosome mutations – non disjunction? What is this
condition called?
Question 1
From the photograph it is reasonable to conclude
that
A. the fetus is a female.
B. there are 44 autosomes.
C. the DNA sequence is identical along each of the
pair of chromosomes labelled 13.
D. the centromeres of the chromosomes in Group A
are near the ends of the chromosomes.
2. In rabbits, a single autosomal gene determining coat colour has four different alleles. At this locus the
maximum number of different alleles a rabbit can have isA one
B two
C three
D four
3. (Q122 Q7 2000)
The number of different genotypes possible with respect to this gene for coat colour is
A four B six
C eight
D ten
4. (15.
From a karyotype it is
A
impossible to identify the sex of the organism.
B
normal to have 44 autosomes in humans.
C
possible to say the DNA sequence along a homologous pair is identical.
D
normal to view the centromeres in the middle of the chromosomes.
5. (Question 1)
The domestic cat, Felis catus, has a diploid number of 38.
a. How many chromosomes would a gamete of a cat contain?
6. (Question 22)
In a group of organisms, individuals genetically identical at a particular single gene locus show a variety of
phenotypes for the trait. It is reasonable to conclude that the variation in phenotypes for this trait is the result of
A. codominance.
B. polygenic inheritance.
C. environmental influences.
D. multiple alleles at the locus.
7. ( Question 22)
Homologous chromosomes contain the same
A. DNA sequences. B. number of guanine and adenine nucleotides. C. alleles.
8. Alleles
A are made of genes.
C code for specific proteins
B are usually expressed individually
D are made of proteins
9. (Question 9)
For humans, the term nuclear genome refers to all the genes in
A. an autosome.
B. an X chromosome.
C. a Y chromosome.
D. a set of autosomes plus the sex chromosomes.
D. genes.
10. (Question 1)
In eukaryotic organisms genes are
A. composed of DNA.
C. composed of DNA and protein.
B. alternative forms of an allele.
D. the same length as a chromosome.
11. (Question 2)
In prokaryotic organisms
A. translation occurs at the ribosome.
C. chromosomes are usually linear.
B. transcription occurs in the nucleus.
D. DNA is only found in plasmids.
12. (Question 14)
This process is called
A. mitosis.
B. necrosis.
C. apoptosis.
D. binary fission.
13. (Question 15)
The process shown in the diagram above
A. is initiated only within the cell itself.
B. involves the total destruction of structures Y.
C. is under the control of a single enzyme.
D. involves the destruction of parts by phagocytes such as X.
14. (Q1, 2001) The diploid number of the platypus is 52. The number of chromosomes present in a somatic
cell of a platypus is
A 13
B 26
C 52
D 104
15.
(Q8, 2000)
A goldfish has 94 chromosomes in its somatic cells the diploid number for the gold fish is
A 47
B 94
C 188 D 376
16.
The diagram shows
A One chromosome
C An homologous pair of chromosomes
B Two chromosomes
D One chromatid
The open circle in the middle is a symbol for the
A Locus
C Chromatid
B Centromere
D Histone
17. Question 2
Bay scallops (Argopecten irradians) have three shell colours: orange, yellow and black. It is known that the
colour is under the control of one gene locus with three alleles.
a. What is a gene locus?
b. If you compared the alleles of this locus at the molecular level, what would be different?
CHAPTER 7
CELL DIVISION
PART 2
DNA. Replication (Semi-conservative, DNA helicase, DNA polymerase, DNA ligase,
MEIOSIS -I.P.M.A.T x2 , Daughter cells – Haploid MITOSIS X1
Daughter cells - Diploid
Linked genes , Crossover,
Questions
1. Describe the DNA replication process.
2. When does replication occur in the life cycle of the cell?
3. Compare DNA replication and PCR by listing the similarities and differences between the two processes.
4. Why is meiosis significant to sexually reproducing organisms?
5. Compare Meiosis and Mitosis – number of cell divisions, chromosome number in daughter cells (n,2n)
Stages when crossover occurs, Non-disjunction can happen. Note- when chromosomes are doubled the
whole thing is called a chromosome and the parts are called chromatids.
6. How are the terms ‘ segregation of alleles” and ‘independent assortment of chromosomes’ related?
7. During meiosis, crossing over and recombination occur between homologous chromosomes. Describe the outcome
of recombination.
8. What are linked genes?
9. Describe what may happen to linked genes during the ‘crossing over’ process. Why is crossing over
important?
10. Which is most likely to be affected by crossing over- closely or distantly linked genes?
MEIOSIS
I___________
P__________
M_________
A_________
T_________
C_________
I__________
P_________
M________
A________
T_________
C_________
1. (Q110 (Q8,2000)
A goldfish has 94 chromosomes in its somatic cells. The diploid number for gold fish is
A 47
B 94
C 188
D 376
2
(Question 4)
The cell could come from
A. skin.
B. liver.
C. testes.
D. bone marrow.
3. During DNA replication
A. messenger RNA (mRNA) is produced.
B. reverse transcriptase enzymes play an important role.
C. bonds between phosphate and sugar molecules break.
D. each of the DNA strands acts as a template strand.
4.
(Question 3)
In specialised cells in the ovary and testis, cells divide by the process of meiosis to produce gametes. A cell with a
diploid number of 4 underwent meiosis. The following images illustrate different stages throughout the total process
of meiosis in this cell
a. Using the letters under each cell (A–F),
put the cells in order commencing with the
earliest stage of meiosis shown.
b. During meiosis, crossing over and
recombination occur between homologous
chromosomes. Describe the outcome of
recombination.
c. During meiosis the nucleus undergoes two
divisions.
i. Which of the cells, E or F, represents
anaphase 1? Explain
The list 1–4 below describes events and outcomes of the replication of DNA within a eukaryotic cell.
1. Complementary nucleotides bind to each of the two strands.
2. Sugar phosphate bonds form between the nucleotides.
3. The newly formed DNA molecules are semi-conserved.
4. Unwinding of the DNA molecule forms two single strands.
5.
(Question 12)
The correct order of these events during DNA replication, with the earliest event first, is
A. 1, 2, 3, 4
B. 1, 4, 3, 2
C. 4, 2, 1, 3
D. 4, 1, 2, 3
6.
(Question 1)
In some species of locusts the female has a ‘diploid chromosome number’ of 16.
a. What is meant by the diploid chromosome number?
b. In this species a male locust has a diploid number of 15.
What is the chromosome number of each of the two daughter cells produced during a mitotic division in a male?
The difference in chromosome number between the sexes is the result of a difference in the number of sex
chromosomes (X chromosomes in this case). Female locusts have two X chromosomes (XX) and male locusts have
only one X chromosome and no other sex chromosome (XO).
c. At the end of a meiotic division, how many chromosomes would you expect in the gametes of
i. a female?
ii. a male?
7. (Question 5)
In bees, females are diploid and males are haploid.
This means that male bees
A. produce gametes with half the haploid number of chromosomes.
B. produce gametes by meiosis.
C. produce gametes by mitosis.
D. do not produce gametes.
8. During meiosis
A Genetically identical daughter cells are produced.
B Homologous chromosomes line up along the equator
C DNA is replicated when the chromosomes are
visible.
D One diploid cell makes four haploid cells.
9. During mitosis
A Homologous chromosomes line up along the equator
B Haploid cells are produced
C Non-disjunction produces greater variation.
D Crossing over occurs.
10. Which of the following
statements is correct?
A Cell 1 could be either mitosis or
meiosis
B Cell 11 could be either mitosis
or meiosis
C Cell 11 could be either mitosis
or meiosis.
D The order of cells in mitosis
would be II,I, III.
I
II
III
11. Question 14
DNA was incubated with radioactive nucleotides. After one cycle of replication the distribution of radioactive
and non-radioactive nucleotides in the DNA would be
12. Question 1
The Zenkey is a hybrid animal produced from a cross between a species of zebra with a diploid number of 44
and a donkey with a diploid number of 62.
a. What is the diploid number of the Zenkey?
b. By what process are gametes formed?
c. Starting with the cell shown below with a pair of homologous chromosomes, draw what happens to the
cell and its chromosomes during the process by which gametes are produced.
d. Zenkeys are unable to produce offspring. Using your knowledge of gamete formation, suggest why the
Zenkey is sterile.
CHAPTER 7
CROSSES
PART 3
Codominant, Homozygous/Heterozygous, Genotype/Phenotype) Test Cross for genotype(or Back cross),
X linked cross (Reciprocal)
Dihybrid cross,
Multiple alleles (Blood type) ,
Lethal alleles,
Polygenes (continuous variation – Height ,colour)
Questions
1. Distinguish between the terms complete dominance and partial dominance.
2. What is the major difference between monohybrid and dihybrid cross problems?
3. Does the genotype of one offspring influence the genotype of future offpring?
4. Coat colour in cattle is co-dominant. Cattle can have 3 phenotypes –red, white or roan. Using ‘CR’ as the
symbol for allele for red hair and ‘CW’ for white hair. Calculate the following probabilities if a roan cow
“CRCW’ is crossed with a roan ‘CRCW’ bull.
a) the cattle produce a heterozygous calf,
b) the cattle produce a white calf.
c) the cattle produce a red calf.
d) the cattle produce two roan calves.
e) the cattle produce three red calves.
5. Think carefully in natural populations are individuals with a dominant phenotype always more common
than individuals with a recessive phenotype? (remember Natural selection!)
6. Consider two genes with two alternative alleles. Both are on different chromosomes. A gene with the
alleles B (abnormally high cholesterol and b (normal cholesterol) is on chromosome 19. Another gene
controls Rhesus blood type (D- Rhesus positive and d- Rhesus negative ) Its locus is on chromosome 1.
A couple are heterozygous for the two genes. What is the expected phenotypic ratio amongst their offspring
given they produce a huge number of children.
In another couple the father is heterozygous for both characteristic and the mother has low cholesterol and is
rhesus negative. Calculate the expected phenotypic ratios.
7. Genetic variation can be monogenic or polygenic. The resulting variation can be discontinuous or
continuous. Explain the terms using examples.
1.
(Question 6)
In the Australian human population, when collecting data about the frequency of different phenotypes at the ABO
blood group locus, it is possible to group all members of the population into four phenotypic classes.
This is an example of
A. hybridisation.
B. continuous variation.
C. polygenic inheritance.
D. discontinuous variation.
2.
(Question 7)
A fisherman was surprised to catch a fish which had no scales (nude). To investigate the origin of this phenotype
the nude fish was mated several times to fish with scales and the result of each cross was recorded. In the crosses
of nude with scaled, a third phenotype appeared, which was later called linear. The linear phenotype has only
a single line of scales down one side of the body. The outcomes of these crosses are shown in the table.
From the data it can be concluded that
A. there is incomplete dominance between the nude and scaled phenotype.
B. the environment is the reason for the loss of scales in the nude Þ sh.
C. all of the linear fish are homozygous.
D. the nude fish are heterozygous.
3. (Question 8)
In guinea pigs two genes have the following alleles.
Gene 1: hair length Gene 2: hair type
S : long hair W : straight hair
s : short hair w : wavy hair
A breeder carried out the cross SSWW × ssww and obtained a number of SsWw offspring.
The breeder then carried out many test crosses involving these offspring to find out if the two genes were on the same
chromosome. If the genes were on the same chromosome, close to each other, you could reasonably expect the
offspring from the test crosses to include
A. more with long, wavy hair than short, wavy hair.
B. more with short, straight hair than short, wavy hair.
C. approximately equal numbers of long, straight hair and long, wavy hair.
D. approximately equal numbers of long, straight hair and short, wavy hair.
One of the human blood groups is the MN group. There are two alleles, LM and LN, at this gene locus which determine
the presence of an antigen, M or N, on the surface of the red blood cells. The heterozygote L MLN has a different
phenotype from each of the homozygotes.
4.
(Question 13)
If an individual of blood type M and one of blood type MN have children, the number of different phenotypes possible
in their offspring is
A. 1
B. 2
C. 3
D. 4
5. (Q 119 Q2 ,2000)
A cross that could confirm that a prize-winning dog was a carrier for albinism is
A AA x AA
B aa x aa
C Aa x Aa
D Aa x
aa
6. (Q 127 Q6,2001)
A trait such as egg size in chickens is polygenic. This means that
A one gene locus with many alleles is involved.
B the population can be sorted into one or two phenotypic classes
C the distribution of phenotypes is continuous.
D one gene locus with incomplete dominance is involved
7.
(Q142 (Q2, 2003)
Purebreeding guinea pigs with rough textured black coats were crossed to purebreeding guinea pigs with
smooth textured white coats. The F1 were all rough textured with black coats. The F1 were allowed to
interbreed to produce an F2. The numbers of each phenotype produced in the F2 are shown in the table
F2 offspring
Number
Rough, black
95
Rough, white
32
Smooth, black
27
Smooth,white
11
Total
165
a
There are two genes involved in these crosses. What information provided in these results confirms that
there are two genes each with two alleles?
b Use allelic symbols R and r for the texture locus, and B and b for the colour locus.
i) What are the genotypes of the purebreeding parental guinea pigs?
Indicate in your answer which genotype matches which phenotype.
ii What is the genotype of the F1 guinea pigs?
iii Give one genotype for an F2 smooth, black guinea pig
8.
(Question 7 2005)
Variation in some traits is due to the action of many genes.
These traits are said to be
A. polymorphic. B. polypeptic.
C. polygenic.
D. polyploid.
The ABO blood group system is under the control of a single gene with the alleles
IA : presence of protein A on red blood cells IB : presence of protein B on red blood cells
i : neither protein A or B on red blood cells
9.
(Question 9)
With respect to this gene
A. an individual could have one of eight different genotypes.
B. an individual could have one of six different phenotypes.
C. a child could have protein A even if both parents lacked protein A on their red blood cells.
D. a child with neither protein A nor B could have a mother with protein A and a father with protein B.
10. A particular species of plant has the following genes and alleles.
Leaf shape
L : normal shape
l : wrinkled shape
Number of seed cases P : one seed case p : three seed cases
Question 10
The genotypes of the two parents would be
A. ppll x PPLL
B. Ppll x PPLL
C. PpLl x PpLl
D. PPLl x PpLl
The hair colour of Australian shepherd dogs is under genetic control. The colour of the hair found inside the ears, on
the legs and under the tails, is under the control of a gene that has the following alleles.
AW white
aS sable colour
ac copper colour
White is dominant to both sable and copper colour. Sable colour is dominant to copper colour.
11.
(Question 2)
If two dogs with the genotypes AW ac and aS ac are mated, the resulting offspring could have
A. 4 genotypes and 4 phenotypes.
B. 4 genotypes and 3 phenotypes.
C. 3 genotypes and 4 phenotypes.
D. 3 genotypes and 3 phenotypes.
12.
(Question 12)
In the Australian sheep blowfly, the length of the wing is a polygenic trait.
This means that
A. there is a small number of clearly defined phenotypes.
B. wing length is controlled by one gene with many alleles.
C. the wing length phenotype shows continuous variation.
D. the phenotype is controlled by many genes on the same chromosome.
13.
(Question 4)
With respect to the ABO blood group locus it is possible to produce children of four different phenotypes if the
parents are
A. type B x type B.
B. type A x type B.
C. type O x type AB.
D. type AB x type AB.
Two genes in watermelons with their alternate alleles are
Gene 1 S : spots s : solid colour
Gene 2 B : bitter fruit b : sweet fruit
The two genes assort independently. Two plants, both heterozygous at each gene locus, were crossed and 1600 seeds
were collected.
14.
(Question 3)
When plants were grown from these seeds, it would be reasonable to expect that about
A. 100 of the plants produced spotted-coloured, bitter fruit.
B. 300 of the plants produced solid-coloured, bitter fruit.
C. 300 of the plants produced solid-coloured, sweet fruit.
D. 1600 of the plants produced spotted-coloured, sweet fruit.
Milk production in cows is under genetic control. The daily volume of milk produced by each cow in a large herd of
cows was recorded and the results graphed. The results were as follows.
15.
(Question 8)
A trait showing this kind of distribution is
A. not found in humans.
B. called a discontinuous trait.
C. under the control of many genes.
D. not influenced by environmental factors.
In lentils the seed coat pattern is determined by a gene with three alleles. The phenotypes are marbled, spotted and
clear. Four crosses were repeated many times. The crosses and the outcomes of these crosses are shown in the table
below.
16.
Question 3
The number of different genotypes possible at the locus for seed coat pattern is
A. 3 B. 4 C. 5 D. 6
17.
Question 4
From the data it is possible to conclude that
A. spotted is recessive to clear.
B. all of the clear offspring are heterozygous.
C. two thirds of the marbled offspring in cross 3 are heterozygous.
D. the marbled parents in cross 1 have the same genotype as the marbled parents in cross 3.
18.
Coat colour in mice is under the control of a single gene with two alleles. Many crosses between yellow coated mice
and mice with grey coats gave the following results. The mice with grey coats were known to be homozygous.
Parental cross yellow x grey First generation 50% yellow : 50% grey
Many crosses were carried out between the first generation yellow mice.
d. What genotypic ratio and phenotypic ratio would we expect to see in the offspring of the cross between
the first generation yellow mice? Make sure you indicate the allelic symbols you are using for this gene
locus.
Scientists performed this cross many times and the result they observed was always a ratio of 2 yellow to
1 grey mouse.
e. How can this result be explained?
Question 3
In a particular insect species, sex is determined by a single gene. The male insect has the genotype Mm and the
female insect is mm. This gene is linked to another gene that determines the body colour of the insect.
a. What is meant by linked genes?
The linked gene determining body colour in these insects has two phenotypes, black body colour and bronze
body colour. Black body colour is the dominant phenotype.
When a heterozygous black male was crossed with a bronze female the following offspring were produced.
b. What are the expected ratio’s for this cross? Which traits are linked?
CHAPTER 7
MONOHYBRID/SEXLINKED
PART 4
Monohybrid cross Mendels crosses.(Dominant, Recessive)
X linked cross (Reciprocal)
Dominant/recessive, Autosomal/Sex linked
Questions
1. A gene’s locus is on the X chromosome. Does this affect its inheritance patterns?
2. A man and woman are heterozygous for a gene. The gene has two possible alleles and there are two
phenotypes. Use ‘D’ to represent the dominant allele and ‘d’ for the recessive allele.
Calculate the following probabilities
a) the two individuals produce a heterozygous dominant offspring.
b) the two individuals produce a homozygous dominant offspring.
c) the two individuals produce a homozygous recessive offspring.
d) the two individuals produces two homozygous recessive offspring.
e) the probability the child is heterozygous given the two individuals produce a child with the dominant
phenotype.
f) the expected ratio for dominant to recessive phenotypes.
3. Show a cross involving a homozygous recessive man and a woman heterozygous for the gene. What are
the possible genotypes and phenotypes of their offspring and in what proportions.?
4. Explain why sex-linked recessive conditions, like colour blindness, are more often expressed in men than
woman
5. A normal woman with a colour-blind father has children with a man of normal vision.
a) What is the probability that a child is colour blind?
b) What is the probability that a son is colour blind?
c) What is the probability that a daughter is colour blind?
6. Geneticists, when experimenting with fruit flies, pea plants, mice etc like large numbers of offspring
before making conclusions. Relate this to differences between expected results, observed results, sample size
and probability.
7. Describe the drawing of pedigrees. What symbols are used and how are generations numbered?
8. What patterns do you look for in a pedigree to distinguish between autosomal recessive and autosomal
dominant inheritance?
9. What patterns do you look for in a pedigree to distinguish between x-linked recessive and X-linked
dominant inheritance?
PEDIGREES - Is the trait Dominant D,
ex-linked dominant XD,
RULES –
Recessive r
sex-linked recessive Xr ?
Does it skip generations?
If YES – RECESSIVE – Autosomal or Sex – linked ?
You can only prove that it is autosomal, you just suspect it is sex-linked. If a mother has the
recessive trait then her sons must get it to be sex- linked – otherwise its autosomal.
x
Xr Xr
XY
Xr Y
If NO – It may be DOMINANT (although can still be RECESSIVE)
DOMINANT
- If the offspring have a trait that is dominant at least one parent must also have it.
- If two parents have it but not all their offspring then the trait can’t be recessive it must be dominant.
1.
SEX-LINKED? – If the father has it his daughter must have it for a sex-linked dominant trait.
XD Y X XX
↓
XDX
(Q5, 2002)
Colour blindness is inherited as an X-linked recessive condition. It is reasonable to claim that
A colour - blind male must have a colour blind mother.
B colour - blind male must have a colour-blind grandfather.
C colour - blind female must have a colour-blind father.
B colour – blind female must have a colour – blind grandmother.
2. (Q117 Q7 2001)
Numbers of offspring of each colour
Cross
brown
Green
Cross A
Brown x green
12
11
Cross B
Brown x brown
7
1
Cross C
Brown x brown
18
0
Cross D
Green x green
0
21
It is possible to conclude from these results that the brown phenotype is dominant becauseA approximately equal numbers of brown and green offspring are produced in cross A
B one green offspring is produced in cross B
C only brown offspring are produced in Cross C
D only green offspring are produced in Cross D.
3. (SA Q4, 2001)
In tomatoes the shape of the fruit is an inherited trait. In a cross between plants which are pure breeding for
spherical fruit, and other which are pure breeding for oval fruit, all offspring have spherical fruit.
From this information you can conclude thatA a cross between tomatoes, each heterozygous, for spherical fruit, could produce offspring with two
phenotypes.
B oval-shaped fruit is the dominant trait.
C any cross between tomatoes with spherical and oval fruit will produce offspring with one phenotype.
D all tomatoes with spherical fruit are homozygous.
(1 mark)
4. (Q76, Q 1,2004). An autosomal gene controls the shapes of tabby stripes in a cat’s fur. It has the
alternative alleles
T : vertical strips of colour (called mackerel tabby)
t : swirly stripes of colour (called blotched tabby)
Several crosses were carried out between mackerel and blotched cats. All of the mackerel tabby cats were
heterozygous. In total, the cats produces 100 kittens. Among these kittens, it would be reasonable to expectA approximately 50 mackerel tabby females
(1 mark)
B approximately 25 blotched tabby females
C more mackerel tabby females than mackerel tabby males
D more blotched tabby males than blotched tabby females.
5. (Question 1)
In humans the presence of a dimple in the chin is dominant.
The first child of a couple, each with a dimple, does not have a dimple but their second child does have a dimple.
a. Use alleles D and d to show the genotype corresponding to the phenotype of the parents and the child without a
dimple.
Genotype of parents
genotype of child without a dimple
6. A plant has two phenotypes, herbicide resistant and herbicide sensitive. A farmer wanted to establish the pattern
of inheritance for this trait and performed the following crosses. Assume this trait is from one gene with two alleles.
b. i. Which one of the crosses, on its own, allows
you to conclude which is the dominant phenotype?
ii. Explain your choice in b.i.
c. Use appropriate allelic symbols to show the
genotypes of the parents and offspring of cross 3.
7.
(2005)
The gene responsible for the autosomal recessive condition PKU controls the production of an enzyme that
converts the amino acid phenylalanine to tyrosine. The gene has the alleles
P : produced enzyme
p : no enzyme produced
In a person lacking the enzyme, phenylalanine accumulates in toxic levels throughout the body and the disease
develops. Development of the disease is prevented by placing a baby, who lacks the enzyme, on a special diet a few
days after birth. This diet is low in phenylalanine and is maintained for as long as possible, preferably for life.
Question 3
With respect to PKU,
A. the special diet for babies with the disease would also contain reduced levels of tyrosine.
B. each person in a population would have one of three possible genotypes at the PKU gene locus.
C. a PKU individual treated from soon after birth would not be able to pass allele p onto their offspring.
D. because the disease can be treated, one would expect the number of babies born with PKU to decline over time.
8. (Q141 Q1, 2003)
The back of the leopard frog can be either patterned or non- patterned.
Several patterned frogs were allowed to breed and they produced 75 patterned offspring and 25 nonpatterned offspring.
a i Which of the phenotypes, patterned or non-patterned, is dominant?
ii Explain your answer to i. (BE CAREFUL!)
b Using your own allelic notation, show the genotypes with their respective phenotype for the parents and
offspring of the cross between the patterned frogs described above.
Draw a punnet square using these alleles and give the expected genotype and phenotype ratios.
Crosses between different patterned and non-patterned frogs were performed. Not all crosses produced the
same outcome. The results are shown in the table below. For both cross A and cross B there were large
numbers of offspring produced.
Parents
Offspring
Patterned
x
non-patterned
All patterned
Cross A
Patterned
x
non-patterned
½ patterned ; ½ non patterned
Cross B
c The parents in crosses A and B have the same phenotypes. Explain why the outcome of the crosses A
and B are different .
CHAPTER 7
PEDIGREES
PART 5
1. (Question 2)
Ptosis is a weakness in the muscles of the eyelid.
a. Using the information below, construct a pedigree to fit the description of the inheritance of ptosis in the family.
Bill had ptosis. Bill married Jill who did not have ptosis and they had two children, a boy Ben who was unaffected and
a girl Daisy who had ptosis. Daisy married Bob who did not have ptosis and they had three daughters, two girls with
ptosis and one girl without ptosis.
b. The pedigree below shows another family in which ptosis was found.
i. Based on the information in this pedigree, what
is the most likely mode of inheritance of ptosis?
ii. Provide two pieces of evidence from the
pedigree that support this conclusion.
2 (Q157 Q3, 1999) Human pedigrees
A survey of male and female Biology students found that some individuals have one whorl of hair at the
crown of the head whereas others have two whorls. This was investigated further in two families, Family A
and Family B. Data collected from these families were recorded in two pedigrees shown in Figure 2.40.
I
I
1
2
1
I
2
II
2
I
1
1
1
3
4
5
6
1
1
1
1
2
3
4
5
III
PEDIGREE A
- one whorl
PEDIGREE B
- two whorls
a i Which of the pedigrees, A or B will fit more than one pattern of inheritance for one whorl?
ii Explain how the pedigree chosen in a i fits more than one pattern of inheritance.
b i Consider the pedigree not chosen in part a i. Explain what feature of this pedigree allows you to
decide that the inheritance of a single whorl is dominant
ii Describe one feature of the pedigree considered in part b i which indicates that the inheritance of
one whorl is autosomal dominant and not X-linked dominant.
3. (Question 1)
With respect to the PKU gene locus, you could reasonably conclude that individual
A. I – 1 is homozygous.
B. I – 2 is homozygous. C. II – 5 is heterozygous. D. III – 1 is heterozygous.
4. (Q124 Q3, 2000)
The pedigree consistent with X-linked recessive inheritance of the trait represented by the shaded symbols is
I
I
I
I
1
1
2
1
I
II
I
3
2
I
2
3
4
1
2
3
4
D
I
I
I
I
1
1
2
1
I
I
3
(Question 2)
3
4
3
2
I
2
II
4
III
1
2
1
II
2
5.
4
III
B
C
I
3
2
I
1
II
4
III
A
2
1
4
III
1
2
3
4
Coat colour in cocker spaniel dogs varies. Four of these colours are black, liver, red and lemon. These four
colours result from the interaction of two particular autosomal genes.
The pedigree below (Figure 9) shows the inheritance of coat colour in a group of cocker spaniels.
a. I–1 and I–2 are heterozygous at both the R and B locus. What evidence from the pedigree supports this conclusion?
b. i. What is the specific genotype of II–4?
ii. What is the specific genotype of III–4?
c. Explain how many different phenotypes could be expected in the offspring of a mating between individuals
II–4 and III–4. Show all working.
6. (Question 3)
Haemophilia is an X-linked recessive condition. The following pedigree shows a portion of a family in which
some members have haemophilia. Those on the pedigree with haemophilia are shaded.
a. Use appropriate allele symbols from XH, Xh and Y to indicate the genotype of each of the following
individuals.
I1
and I2
b. The couple II1 and II2 have a son. What is the probability that the child has haemophilia?
c. The couple II5 and II6 have a son. What is the probability that the child has haemophilia?
The following pedigree is of a family in which one member (shaded) has an autosomal recessive condition. The alleles
of the gene locus involved are G and g.
d. Give a possible genotype for each of the four members of the family.
I1
I2
II1
II2
CHAPTER 8
PROTEIN SYNTHESIS
PART 6
TRANSCRIPTION- Nucleus, RNA polymerase, Codons, mRNA (exon/Intron)
TRANSLATION – Ribosomes(rRNA and Proteins),tRNA, Anticodons,
Questions
1. How do prokaryote and Eukaryote genes differ?
2. What are triplets, codons and anticodons?
3. In terms of nucleotides and triplets, what is a gene?
4. What are the major differences between RNA and DNA?
5. How many different codons are possible in the genetic code?
6. Are all codons needed? How many amino acids are there?
7. Do all codons code for amino acids?
8. List the major steps in protein synthesis including the names of each stage, where each stage occurs , the
products formed and any enzymes required.
9. Introns and exons are parts of the eukaryotic gene sequences. Which parts seem to have the most
importance?
10 What does Reverse transcriptase do?
TRANSCRIPTION / TRANSLATION
Use the photocopied Biozone sheet for the overview
Table of mRNA codons pg 253
What is the difference between RNA and DNA codes and structures?
Convert the following DNA code to an Amino Acid chain
D.N.A
TAC TTT TGC CTA ATC
_____
A.A
CODONS - need 3 nucleotides bases to code for 1 amino acid – Why 3?
1 base
codes for 4 Amino Acids because there are four different bases A,C,T,G
2 bases
code for 4 x 4 = 16 different combinations so 16 A.A’s
3 bases
code for ________________________________________
Figure 3 shows portion of a cell engaged in protein synthesis. The various parts of the cell are not drawn to scale.
1. (Question 10)
It is reasonable to conclude that
A. process N is translation.
B. structure P is made of t-RNA.
C. compound Q is messenger RNA.
D. structure R is the site of protein synthesis
Haemoglobin consists of four polypeptide chains. In normal haemoglobin, two of these chains are beta chains, each
comprising 146 amino acids. Variations exist in the amino acid composition of these chains and this results in
different kinds of haemoglobins. One of these variants is called haemoglobin S.
The first seven amino acids in the beta chains of these two haemoglobins, and the amino acid at position 143 are given
below. The amino acids at each of the remaining positions are the same for each haemoglobin.
2. (Question 13)
It is reasonable to
conclude that during
transcription
of
normal haemoglobin,
the mRNA codon
sequence could be
A. GTT for amino
acid 1.
B. ACT for amino
acid 4.
C. CTT for amino
acid 3.
D. CAC for amino
acid 2.
3.
(Question 2)
The following diagram outlines events associated with the production of a polypeptide chain in a eukaryotic cell.
a. What is the name of the process at step 1?
b. i. Name the product of step 1
ii. Outline what occurs at step 2.
c. Name the event that occurs at structure 3.
d. i. Name the structure at 4.
ii. Outline the function of the structure you named in
d.i.
4. The first six amino acids of the P34H protein are
Met
Pro
Lys
Tryp
Val
cys
A mRNA sequence that corresponds to this amino acid sequence is
AUG
CCA
AAG
UGG
GUA
UGC
Write dowm the double stranded DNA sequence corresponding to this section of the gene.
Explain if this is the only base sequence possible for this section of this gene.
5. Question 15
Translation of the genetic code occurs in the cytosol of a cell. The following diagram is one representation of
translation.
In the model presented
A. M represents a ribosome.
B. N represents messenger RNA.
C. O represents transfer RNA.
D. P represents an amino acid.
CHAPTER 8
DNA STRUCTURE/ GENE REGULATION
PART 7
DNA Structure and Discovery Watson/Crick, Rosalind Franklin, Chargraff’s bases, Chromatin, Histones,
GENE REGULATION – Gene switched on/off, Promoter, Operator, regulatory genes ,
Questions
1. What is the basic building block of DNA?
2. What are the 3 major components of a DNA nucleotide?
3. Two of the above components are constant. Which?
4. The third component varies. Name the four variations.
5. What are ‘Complementary bases’?
6. What type of chemical bond holds the base pairs together?
7. How can the above bonds be broken?
8. Do all genes make proteins which will be used outside the nucleus?
1. (Q128 Q13 2001)
The base composition of the DNA of a particular bacterium was analysed. It was found that 18 per cent of
the bases were adenine. You could reasonably conclude that
A the DNA comprised 18 percent adenine-thymine pairs.
B 32 percent of the bases were cytosine
C the DNA comprised 32 percent of cytosine-guanine pairs.
D 36 percent of the bases were thymine
2. (Question 8)
In DNA, the number of
A. phosphate groups equals the number of nitrogen bases.
B. adenine nucleotides equals the number of cytosine nucleotides.
C. phosphate groups equals twice the number of sugar molecules.
D. guanine nucleotides equals the number of uracil nucleotides.
The following nucleotide sequence forms part of the template strand of a gene coding for protein X.
TGGATGAC
DNA template strand
*
3. (Question 9)
The complementary base found at the fourth nucleotide (marked *) in a sequence transcribed from this sequence
would be
A. C B. G
C. T
D. U
4. (Question 10)
In a double-stranded molecule formed from this DNA template strand (shown above) the number of deoxyribose sugar
units you would expect to find is
A. 4
B. 8
C. 16
D. 32
5.
(Question 11)
Reverse transcriptase catalyses the production of
A. DNA from an mRNA template.
B. DNA from a protein template.
C. mRNA from a DNA template.
D. tRNA from a DNA template.
6.
(Question 20)
One section of a polypeptide has the amino acid sequence Ala . Cys . Lys . Ile . Asn
The codons for these amino acids areAla - GCA GCC GCG GCU
Cys - UGC UGU
Lys - AAA AAG
Ile - AUU AUC AUA
Asn - AAC AAU
The sequence of DNA coding for this section of the polypeptide could be
A. CGTACGTTTTATTTG
B. CGTTCGTTTTATTTG
C. CGTACTTTTTACTTG
D. CGAACATTCTATTTT
7. (Question 7)
Organisms can regulate the expression of their genes in a number of ways.
a. Suggest why an organism regulates the expression of its genes.
One example in bacteria is the regulation of the expression of a gene which produces an enzyme (enzyme X)
involved in the metabolism of the amino acid tryptophan. Enzyme X is only produced when tryptophan is in high
concentration. This gene regulation involves several genes. Two of the genes include a gene for the production of
enzyme X and an operator gene. If a protein, called a repressor protein, binds to the operator gene, transcription
of the gene for enzyme X is stopped. If no repressor protein is bound to the operator, transcription of the gene
for enzyme X occurs. A summary of this regulation is shown in Figure 1.
b. The gene coding for enzyme X is not transcribed when the repressor protein binds to the operator gene.
What enzyme is prevented from functioning during this binding?
When tryptophan binds to the repressor protein, the repressor protein can no longer bind to the operator gene.
c. When tryptophan binds to the repressor protein what will happen to the production of enzyme X?
d. Based on Figure 2, suggest how tryptophan prevents repressor protein function.
CHAPTER 8
MUTATIONS
PART 8
MUTATIONS – Chromosome (Crossover during meiosis, Deletions,Inversions, TRANSLOCATION,
Duplication ANEUPLOIDY (Non-Disjuction-Trisomy, Klinfelters, Turners) Polyploidy
Gene- Substitution, Inversion, Insertion/deletion, Frameshift, Mutagens, Mutation rates, Somatic or Germ
line
Questions
1. Gene mutations are alterations to the base sequences in genes. Describe the three major types of base
mutations – Substitution, Insertion, Deletion and there effects.
2. Explain why mutations are usually harmful. Which of the 3 above types is the least harmful. Why?
3 Distinguish between germ line and somatic mutations.
4 What sort of mutation can be beneficial to an organisms long term survival?
5. What is Translocation
6. Can genes on different chromosomes become linked? (Think)
7. What is non-disjunction.
8. Make a list of examples of a condition that arises as a result of non-disjunction. Explain what must have
happened during meiosis and at what stage it happens.
9. What is polyploidy. How does it help sterile hybrids?
10. List some causes of mutation.
1. (Question 8)
The origin of new alleles at a gene locus is the result of
A. migration.
B. mutation.
C. recombination.
D. independent assortment.
2.
(Question 13)
Achondroplasia is an autosomal dominant trait in humans that results in a form of dwarfism. In some cases a child
with achondroplasia is born to parents who have normal height.
The most likely reason for the appearance of the child with achondroplasia is that
A. the parents are carriers and the child has inherited the mutant allele from each parent.
B. a mutation has occurred in a gamete of either the mother or the father.
C. a mutation has occurred in a somatic cell of one of the parents.
D. a mutation has occurred in the tissues of the child.
3
(Question 6)
Cabbage (Brassica oleracea) and radish (Raphanus sativus) both have a diploid number of 18. However they
do not naturally hybridise with each other.
a. How many chromosomes would be expected in the gametes of the cabbage?
In the laboratory, the two species can be forced to mate and produce offspring. The offspring are sterile.
b. i. What would be the diploid number of the hybrid?
ii. Explain why the hybrid of the cabbage and radish is sterile.
An occasional spontaneous event produces a doubling of each chromosome set in the hybrid. The new plants are able
to grow and produce fertile offspring.
c. What term is used to describe cells with more than two sets of chromosomes?
d. Explain, with reference to the events of meiosis, why the new plants are fertile.
4
(3.) Genetic ‘accidents’ can occur to chromosomes. One type of accident is when one chromosome becomes
permanently attached to another. This kind of arrangement is called a translocation.
A scientist observed a translocation involving chromosomes 9 and 18 in somatic cells of a male cat. Figure 8
shows chromosomes 9 and 18 in a normal male cat and their arrangement in the cat carrying the translocation.
Note that the centromere of the translocated number 18 chromosome has been lost.
In some cases translocations lead to abnormalities, but that was not the case here. The cat had a normal phenotype.
b. i. What is meant by the phenotype of an organism?
ii. The cat with the translocation had only 37 chromosomes in each somatic cell. Explain why it still had a normal
phenotype.
The cat’s reproductive tissue also contained the translocation, and investigation showed that he produced four
different kinds of sperm with respect to chromosomes 9 and 18. Two of these four types are shown in Table 2.
c. Complete Table 2 by naming the chromosome
make-up of sperm types 3 and 4 respectively.
One type of sperm produced by the male carrying the
translocation does not survive.
d. i. Explain which type this is most likely to be.
ii. The male cat with the translocation is mated to a
normal female. What is the chance that he will father
a kitten with the same kind of translocation involving
chromosomes 9 and 18, and a normal phenotype?
5. Streptococcus pneumonia is a bacterium which causes pneumonia in humans and may show resistance to
antibiotics.
a. How would antibiotic resistance have first occurred in the Streptococcus pneumonia population?
6. In some Australian insects, new species have arisen through changes that occurred to chromosomes in an ancestral
species. Such changes may involve the joining together of chromosomes, the loss of whole or parts of chromosomes,
and rearrangement of the genetic material within chromosomes.
One ancestral species has the following haploid set of chromosomes.
As the changes in chromosomes accumulate, a number of different species can
result from a single ancestral species.
Three species that have evolved from the ancestral species shown above have
the haploid sets of chromosomes shown below.
6. (Question 25)
The most likely order of evolution of these species is
A. ancestral species, species Z, species G, species M.
B. ancestral species, species G, species M, species Z.
C. ancestral species, species M, species G, species Z.
D. ancestral species, species G, species Z, species M.
Below is the DNA sequence from the beginning of a gene coding for an enzyme involved in photosynthesis. The
upper line of bases (in bold) represents the template strand.
a. Write the mRNA sequence that would be transcribed from this DNA sequence.
7. (b. i. )The 6th base on the template strand of the sequence above is substituted by C.
What type of mutation is this?
ii. Explain the effect this mutation will have on the amino acid sequence of the protein produced
c. The 11th base pair of the sequence is deleted. Explain the effect that this mutation will have on the amino acid
sequence of the protein produced.
CHAPTER 9
GENETIC ENGINEERING
PART 9
Tools – Restriction enzymes(sticky/blunt ends), Recognition site.
DNA ligase, Recombinant DNA , Vectors- plasmids, viruses, Transgenic organisms, Transformations, Antibiotic
resistance markers, PCR
Questions
What is the aim of genetic engineering?
What are restriction enzymes and how are they used?
Given that geneticists can cut and rejoin pieces of DNA, how do they transport DNA from source to target cells?
How do they detect if the gene has been inserted into the new organism?
Explain the steps of the PCR (what does it stand for?) and give an example of its use.
How is it similar to DNA replication in cell division?
1. (Question 11)
Incubation of this length of DNA in a tube containing
A. Spe I would result in three pieces of DNA.
B. Hin dIII would result in two pieces of DNA.
C. Spe I and Eco RI would result in five pieces of DNA.
D. Bgl II and Hin dIII would result in four pieces of DNA.
2. (Question 4)
Genes can be transferred from one species to another in different ways. One method is to use plasmids, circular pieces
of DNA found in some bacteria.
In this method, a plasmid is cut and a piece of foreign DNA inserted. The foreign piece of DNA usually contains more
than one gene. The process is shown in Figure 5 below.
Many copies of the new plasmid are then incubated with bacteria.
a. What is the name given to a plasmid that is used to transfer DNA from one organism to another?
b. What is used to cut the DNA of a plasmid?
c. What is used to join the inserted piece of DNA to the plasmid?
One of the foreign genes inserted into the plasmid, codes for resistance to a particular antibiotic.
d. Explain why it is important to include a gene for antibiotic resistance in the plasmid produced.
Bacteria containing plasmids, that are constructed in the way outlined in Figure 5, can be used in a variety of settings
with plants and animals. For example, some plants are resistant to attack by insects. The plants produce a protein that
poisons the larval stage of some insects that feed on them. The production of the protein is under genetic control.
A particular species of crop plant was genetically engineered to contain this gene. Such plants are referred to as GM
(genetically modified) plants.
e. Explain why a farmer might choose to grow a crop that was genetically engineered to be resistant to insects, rather
than spray the crop with insecticide.
Some plants are resistant to particular herbicides, chemicals that are used to kill plants. This trait is also under genetic
control. The gene that confers herbicide resistance has also been incorporated into some GM crop plants. This enables
a farmer to spray his GM crop with a herbicide that will not harm the GM crop but does kill weed plants growing
within the crop.
f. Suggest one advantage for a farmer to be able to spray his crops with a herbicide.
Two farmers have properties next door to each other. They grow the same cereal crop.
• Farmer X wishes to grow GM crops that are resistant to herbicide.
• Farmer Y wishes to continue to grow non-GM crops.
Farmer Y was concerned, and suggested to farmer X that pollen from the GM crop could fertilise the non-GM plants.
g. Explain why farmer Y might be concerned about the possibility of his crop being fertilised by pollen from farmer
X’s crop.
The farmers agreed to carry out field trials to establish whether leaving a gap between crops reduced the likelihood of
cross-pollination. A number of trials were planted so that the results of one trial did not interfere in any way with the
results of another. The percentage of seeds produced at various positions as a result of cross-pollination was measured
for each trial. The outline of these trials and the results gathered are shown in the following table.
h. From the data, what conclusions can be drawn
about cross-pollination and the gap between
crops?
3.
(Question 7)
Sickle cell anaemia is a serious inherited blood condition. It leads to tiredness and kidney or heart failure and without
treatment children usually die before the age of 10. Sickle cell anaemia is due to a change in the gene which codes for
beta haemoglobin. There are two alleles for the beta haemoglobin gene; HbA coding for normal beta haemoglobin and
HbS coding for the changed haemoglobin. An individual with two copies of the HbS allele will develop symptoms of
sickle cell anaemia.
It is now possible to genetically test people to see if they carry the HbS allele. This test uses PCR, the restriction
enzyme MstII and gel electrophoresis.
MstII is a restriction enzyme that recognises the 7-base sequence in DNA,
C C T G A G GG G A C T C C
and cuts it between the C and the T to produce
a. What term is used to describe the ends of the fragments produced by MstII?
4
(Question 4)
a. Describe the appearance of a bacterial plasmid.
A bacterial plasmid was modified in the laboratory so that it contained a gene for an enzyme which provided
resistance to the antibiotic tetracycline. Bacterial cells, which in their natural environment were sensitive to the
antibiotic tetracycline, were mixed with the modified plasmid. The bacterial cells were treated so that they could take
up the plasmid.
b. What is the name of the process in which a bacterial cell takes up a plasmid and expresses the genes of
the plasmid?
The outcome of an experiment is shown below.
With respect to the growth of bacteria the results of plates A and C are shown. On plate A there is a continuous
growth of bacteria over the surface of the agar. On plate C the colonies are distinguishable from each other.
c. i. What result would you expect on plate B with respect to the growth of the bacteria?
ii. Explain your answer to c.i.
d. Explain why there is a difference in the way the bacteria have grown on plates A and C.
Question 16
The following diagram indicates the cutting sites of three different restriction enzymes on a particular bacterial
plasmid.
If the plasmid was incubated with the restriction enzyme
Eco R1, the number of pieces of DNA obtained would
be
A. two.
B. three.
C. four.
D. seven.
CHAPTER 9
GENETIC ENGINEERING
PART 10 and 11
Gel electrophoresis, P.C.R,
DNA sequencing, Human Genome project,
DNA profiling(fingerprinting)-STR’s and VNTR’s ,
Genetic Testing- Restriction length polymorphisms (RFLP), Gene Probes,
Artificial evolution – Selective breeding, Cloning(Embryo splitting, nuclear transfer) Stem cells, Transformation,
Genetic Screening, Gene therapy
Questions
What is the purpose of Gel electrophoresis. How is this accomplished?
What do the letters PCR stand for?
What is the purpose of P.C.R. How does this work including enzymes required and the role of primers.
What does it mean to sequence DNA?
What was the Human genome project and why is this information important.
How could knowing the sequence of a worm’s genome help understand human development?
What do the initials STR and VNTR’s stand for? Why are these regions studied when comparing individuals?
What does the phrase Restriction length polymorphisms mean? How can it help detect for some genes of a disease?
What is a gene probe? How are they used? Mention autoradiography (radioactive) in your answer.
What is the difference between Natural Selection and Artificial Selection.
What is a clone? How is embryo splitting and nuclear transfer different?
What are Stem cells and why are they useful?
What is trying to be accomplished in gene therapy?
1. Q158 2002
There were three suspect in an assault case. A forensic scientist found blood other than the victim’s at the site. DNA
was extracted from five blood samples: the victim, the site and the three suspects. PCR was used on the extracted
DNA.
a) A DNA polymerase enzyme is involved in the PCR process. Explain the role of the polymerase enzyme in PCR.
One of the regions used in the forensic analysis was a short tandem repeat (STR) sequence of 4 bases, called
HUMTH01. This sequence, located on chromosome 11, has many alleles which differ from each other by the number
of times the sequence AATG is repeated. It was this region of chromosome11 which was amplified using PCR.
The amplified samples were loaded onto a gel and electrophoresis was performed to separate the fragments of DNA.
b) Name the 2 properties of the DNA fragments which allow them to be separated from each other during the gel
electrophoresis process.
c) Why is there only one band in lane 2 but two bands in lanes 3,4,5 and 6?
d) How many different alleles at the 1 HUMTHO 1 locus are represented on the gel in individuals 2,3,4,5,6?
e) Which piece of DNA, A or B, has the greater number of the 4 base repeat sequence?
f) Which of the suspects appears to have committed the assault ? Explain.
2. Question 11
Amplification of DNA in the polymerase chain reaction requires
A. nucleotides of uracil. B. DNA polymerase.
C. amino acids.
D. ribose sugar.
3. Molecular studies have shown that the sickle cell allele differs only by one base pair from the normal allele. This
base change occurs in a 7-base sequence that is recognised by the restriction enzyme MstII. This is the only MstII site
found within the region of the gene that is used in the genetic test.
The PCR products are digested using MstII. The resulting fragments undergo gel electrophoresis.
b. How is the action of the MstII enzyme affected by the HbS mutation?
c. Mark on the picture of the gel below the
banding patterns you could expect to see for
someone who has each of the following
genotypes.
i. HbA/HbS
ii. HbS/HbS
Question 5
CC. for Carbon Copy is the name of the Þ rst cloned kitten born in 2001. The nucleus of a cat.s egg cell was removed.
It was replaced by a nucleus from a somatic cell of a donor female cat. Once development commenced the egg cell
was transferred into a surrogate female.
a. What is meant by the term cloning?
The diploid number of a cat is 38.
b. i. How many chromosomes would have been in the nucleus that was removed from the egg cell?
ii. Is CC male or female? Explain.
To determine if CC was in fact a true clone, studies were made of specific variable regions in the DNA of the donor,
CC and surrogate. The results are shown in the table.
c. For each region of DNA there are two values, for example, 164/164. Suggest a reason for this.
d. In the case of DNA variable region 4 in the donor DNA, why are the pairs of values different?
e. From the data it was concluded that CC was a true clone. Explain the evidence in the table that supports this claim.
6. Selective breeding has been used to improve the milk yield of cattle herds in Australia.
e. Identify a key difference between selective breeding and random mating in a herd of cattle.
f. What is the impact of selective breeding on genetic variability in a herd of cattle?
The quality and yield of milk in cattle has been improved by artificial insemination in which semen from a
selected bull is used.
g. Explain how the use of artificial insemination may intervene in the evolutionary process.
7. Question 18
A restriction enzyme cutting site may be present or absent in a particular 200 kb region of human chromosome 1.
CHAPTER 10
FOSSILS
PART 12
Fossilisation process, Sedimentary rock, Trace fossils,
Dating- absolute (radioisotopes), Comparative (Index fossils) Transistional fossils
Biogeography Pangea/Gondwana, Wallaces line
Questions
Is it always true to say that fossils are the remains of dead organisms?
Under what environmental conditions are fossils formed? What conditions reduce the chances of good fossils
forming?
What information can fossils provide about the lifestyle of once living organisms?
Why is it difficult to identify different species using only fossil evidence?
Describe the difference between absolute dating (radioactive isotope dating) and relative dating (stratigraphy).
Different radioactive isotopes have different half-lives. What does this mean?
Different radioactive mentods are used to date fossils of different ages. List some commonly used isotopes and state
their use. Is it always the fossil that is dated?
What are “Transistional fossils” like Archaeopteryx and Tiktaalik and the recent Ida.
Use the following information to answer Question 12.
Radioisotopes may be used to establish the age of rocks and fossils. Potassium–40 is an isotope which decays
to argon–40. The half-life of potassium–40 is 1.3 million years.
Figure 4 shows the proportion of potassium–40 (%) remaining in rock over a period of millions of years.
1. Question 12
A rock was found to have 25 per cent potassium–40 remaining.
The approximate age of the rock is
A. 0.65 million years.
B. 1.3 million years.
C. 2.6 million years.
D. 3.9 million years.
2. Question 18
During a volcanic eruption molten material called magma comes out of the volcano, cools and solidifies on the surface
of the earth forming basalt. Volcanic ash, also from the eruption, is deposited near the basalt and may contain well
preserved fossils. The surrounding basalt can be useful to date fossils in the strata formed by the ash because
A. eruption dates of volcanoes are known from historical data.
B. organic remains are baked and preserved in the basalt.
C. radioactive elements within the basalt can be accurately dated.
D. the basalt may contain an index fossil.
3. Question 19
There is little fossil evidence of the earliest forms of life because the organisms
A. decayed quickly in the oxygen-rich atmosphere.
B. did not have hard parts which would fossilise easily.
C. evolved so quickly that they left few remains.
D. lived in water and were not preserved.
4. Question 21
Fossils of soft-bodied organisms are relatively rare because they
A. were never common in the environments in which they lived.
B. lived in environments where sedimentation did not occur.
C. are generally small in size.
D. readily decompose.
5. Trilobites (Figure 6) are an extinct group of marine arthropods. They are a very well-studied group due to the
abundance of fossils. Trilobites had a tough exoskeleton and bodies and legs divided into segments. They were
distributed worldwide and occupied a range of habitats. They existed for almost 300 million years before becoming
extinct around 250 million years ago.
a. i. What structural feature improves the chances of a trilobite becoming a fossil?
ii. Suggest one other reason why trilobite fossils are abundant
6. (Question 9)
Naracoorte Caves are located in southeast South Australia. These limestone caves contain the greatest number, most
diverse and best preserved fossils of the Pleistocene Epoch (1.8 million years to 10 000 years ago) in Australia. For
more than 300 000 years sediment and animals fell into one particular cave through an opening in its ceiling, forming
an enormous cone-shaped pile. Animals that fell in through the hole were unable to escape and died. The pile of
sediment and bones eventually grew up to the ceiling and blocked the hole about 15 000 years ago.
The diagram above represents the cone-shaped pile within the cave.
a. In which layer (1 to 5) would you expect to find the oldest group of fossils?
b. Scientists sometimes use an index fossil to date a rock layer. Describe two features of a fossil that would make it
useful as an index fossil.
c. Outline one other method that the scientists could use to date the fossils in the oldest layer within the cave.
d. This cave has been a good environment for the formation and preservation of fossils. Give one reason why fossils
have been well preserved in this cave.
To date 118 species of vertebrate animals have been found. One of these is the now extinct Thylacoleo carnifex,
referred to as a marsupial lion because of the cat-like nature of its skull and its carnivorous habit. The scientists have
only found hard parts of these extinct animals.
e. Describe a particular hard part that would be useful to the scientists to determine that Thylacoleo carnifex was
carnivorous.
After studying the fossilised hard parts of the Thylacoleo carnifex, the scientists have decided that the animal when
alive weighed around 120 kilograms and had a very muscular body. Drawings have been made to indicate the animal’s
appearance.
f. How do the scientists reach a conclusion about the animal’s appearance when only the hard parts of the extinct
animal are available?
Question 25
The diagrams below represent sedimentary rock strata from two different palaeontological sites.
Based on these diagrams it would be reasonable to conclude that fossils in
A. strata B are the same age as fossils in strata E.
B. strata A are younger than fossils in strata E.
C. strata D are younger than fossils in strata F.
D. strata I are older than fossils in strata G.
CHAPTER 10
EVOLUTION
PART 13
Adaptation, Recent common ancestors,
Divergent evolution- Homologous features, Convergent-Analogous features, Parallel, Phylogenetic tree (cladogram)
Comparisons- Structural, (Homologous features, Embryology) Biochemical(DNA, Nuclear/ mitochondrial,
Hybridisation,Molecular clock, proteins-cytochrome C, haemoglobin), Biogeography.
Extinction.
Questions
What is a ‘recent common ancestor’?
What is a phylogenetic tree?
What is evolution? Is it a process or a result? Explain
Apart from fossils what other evidence supports evolution.
Distinguish between convergent, parallel and divergent evolution.
Can an individual organism become extinct?
1. (Question 24)
The scales of reptiles and the feathers of birds are considered to be homologous structures because they
A. have arisen as a result of similar selection pressures.
B. have a common evolutionary origin.
C. are both forms of skin covering.
D. serve a similar function.
2. (Question 22)
The most recent common ancestor of Homo and Kenyanthropus is represented on the diagram at
A. Z
B. Y
C. X
D. W
3.
(Question 13)
Populations of bacteria can evolve rapidly in response to changes in the environment.
One factor which contributes to this is that bacteria
A. have a single chromosome.
B. have a short generation time.
C. are single-celled organisms.
D. have a low rate of mutation.
4. (Question 15)
From this diagram it would be reasonable to conclude that
A. white-lipped snakes and black snakes lack a common ancestor.
B. tiger snakes share more characteristics with brown snakes than with swamp snakes.
C. death adders are more closely related to golden-crowned snakes than they are to tiger snakes.
D. broad-headed snakes share a more recent common ancestor with white-lipped snakes than they do with
swamp snakes.
5. (Question 17)
The shingleback lizard is closely related to the bluetongue lizard. The bluetongue has a long pointed tail and
smooth scales. The shingleback has a short stumpy tail and enlarged rough scales.
The evolution of these characteristics in the shingleback is an example of which type of evolution?
A. cultural
B. parallel
C. Divergent
D. convergent
A comparison was made between human, rabbit, mouse and chimpanzee of the
• DNA coding sequence of the globin gene
• DNA sequence in the introns of the globin gene
• Amino acid sequence of the globin polypeptide.
6. (Question 18)
It is possible to conclude from this data that
A. a human is more closely related to a mouse than to a rabbit.
B. the variation between chimpanzees and humans occurs in a region of the globin gene which would code
for amino acids.
C. the variation in the intron sequence between human and mouse would account for some of the differences
in the amino acid sequence.
D. the comparison between rabbit and human indicates that the differences in their DNA did not always make
a difference to the amino acid produced.
7.
(Question 19)
The common evolutionary ancestry of many organisms is reflected by a geographic distribution consistent with
the former supercontinent Gondwanaland.
An example of this is the distribution of
A. parastacid crayfish in South America, New Zealand, Australia and New Guinea.
B. bears in North and South America, Europe and Asia.
C. flying foxes in Australia, Asia, Africa and Europe.
D. mockingbirds in South and North America.
8. (Question 15)
Evidence for evolution includes data from comparative anatomy and embryology. When comparing two species such
evidence would be obtained from
A. chromosome numbers.
C. shared habitats.
B. mitochondrial DNA.
D. the presence of gill slits during development
9.
(Question 19)
The linking of the present-day distributions of organisms with past movements of continental plates is referred to as
A. continuous variation.
B. biogeography.
C. genetic drift.
D. biodiversity.
.
Trilobites are thought to be closely related to three other groups of fossil arthropods; helmetids, tegopeltids and
naraoids. The tegopeltids and helmetids are the two most closely related groups. These two groups are more closely
related to trilobites than they are to naraoids. The diagram below illustrates the evolutionary relationships between
these four groups.
b. Write the names of these four groups in the boxes at the top of the diagram below so that the evolutionary
relationships between them are consistent with the information provided.
c. Suggest one reason why all species of trilobites became extinct.
10. (Question 16)
Convergent evolution may produce similar structures in two different species.
This may lead to
A. analogy. B. homology. C. divergence. D. embryology.
11. (Question 17
The shark (a fish) and the dolphin (a mammal) are an example of
A. convergent evolution. B. allopatric speciation. C. divergent evolution.
D. species radiation.
12. (Question 25
Comparisons of the amino acid sequences of the α-globin polypeptide have been made between humans and a number
of other vertebrates. The number of differences is shown in the table below.
Based on the information provided, the correct placement of each animal on the figure to show the evolutionary
relationship is
A. V = cow, W = kangaroo, X = newt, Y = carp, Z = shark
B. V = shark, W = carp, X = newt, Y = kangaroo, Z = cow
C. V = carp, W = shark, X = kangaroo , Y = newt, Z = cow
D. V = kangaroo, W = cow, X = newt, Y = shark, Z = carp
Thylacinus cyanocephalus (Tasmanian tiger) was the largest living marsupial carnivore in Australia at the time
of European settlement. The thylacine is believed to have become extinct on 7 September 1936 when the last
captive thylacine died in the Hobart Zoo. There are thylacine fossils found in Tasmania and mainland Australia, but
when Europeans arrived in Australia living thylacines were only found in Tasmania.
d. Suggest why thylacines were not found in mainland Australia at the time of European settlement.
Since 1936 there have been many reported sightings of thylacines in Tasmania and along the southern coast of
Victoria.
e. Explain why scientists still believe thylacines are extinct.
The dingo is a eutherian mammal and the thylacine is a marsupial mammal. Scientists regard these two carnivores as
an example of convergent evolution.
f. Explain why scientists would regard the thylacine and the dingo as an example of convergent evolution.
SECTION
CHAPTER 11
NATURAL SELECTION
PART 14
Variation, Selection pressure, Gene Pool, Wild-type allele, Polymorphism, Mutation, Genetic drift, Bottleneck effect,
Founder effect,
Darwin, Lamarck, Wallace, - Speciation, Isolating mechanisms (pre-reproductive and post reproductive) Allopatric
speciation,
Questions
What is the modern version of evolution by Natural selection? (refer to answer checkpoints pg158 3-35)
What is meant by variation when looking at members of a population? These can be listed under 3 main headings.
What is the most important source of variation –recombination during meiosis or mutation? Justify your opinion
explaining the significance of each.
Distinguish between genotypic and phenotypic variation.
How are Natural selection and Evolution related?
If Natural Selection is to occur, what must exist in a population?
Is selection at the phenotypic or genotypic level? Explain what sorts of individuals are most likely to survive in a
particular environment.
If an individual is to pass on its alleles to the next generation how long must it live?
What is meant by the terms gene pool and allele frequency?
In terms of the gene pool and allele frequency , what is the difference between gene flow and genetic drift?
When will the two processes have the greatest affect on allele frequency?
What is a species? (What is the most important test to find members of the same species)
What is allopatric speciation? What are some geographic barriers.
Name some mechanisms which could prevent interbreeding between closely related species.
1. Question 24
The following statements (not in correct order) summarise the steps in natural selection.
1. Some individuals are better suited to a particular environment.
2. Over time there is an increase in particular characteristics in the population.
3. There is variation within a population, some of which is genetic.
4. Individuals better suited to the environment are more successful at survival and reproduction.
The order of statements which best describe natural selection are
A. 1, 3, 2, 4
B. 3, 1, 4, 2
C. 2, 3, 1, 4
D. 1, 2, 4, 3
2. Question 17
Two types of bird were originally thought to be different species. However, recently a group of biologists has agreed
that they are the same species. The biologists must have found out that the two types of bird
A. look alike enough to be thought one species.
B. are separated by a geographical barrier.
C. successfully interbreed in nature.
D. live in the same habitat.
3. Question 14
Many frog species inhabiting tropical rainforests have evolved green skin colour.
It would be reasonable to conclude that the main selection pressure responsible for the evolution of green skin
colour is
A. predation. B. climate. C. reproduction.
D. infection by pathogens.
4. Question 16
Genetic drift is
A. most evident in large populations.
B. the gene flow between populations.
C. random changes in the gene pool of a population.
D. the gradual change in phenotypic frequency resulting from natural selection.
5. Question 20
For a species living in an unchanging environment
A. there are no selection pressures.
B. the selection pressures remain constant.
C. the only selection pressure is genetic drift.
D. all individuals are equally suited to the selection pressures.
6. Question 25
The variation in allele frequencies between several isolated populations can be due to genetic drift.
Genetic drift is likely to be observed when
A. there is gene flow.
B. the mutation rate is high.
C. there are strong selective pressures.
D. a population is reduced to a few individuals.
7.
Question 7
Warfarin is a poison used to control rat populations. Figure 13 shows changes in the proportion of rats resistant
to warfarin in a particular population over a period of about 4 years. High levels of warfarin were used on this
population during Year 2 but poisoning stopped at the end of this period. Rats are reproductively mature at an
age of three months and can breed about every three weeks.
a. Explain the process which led to the
increase in the percentage of resistant
rats during Year 2.
b. Using the data in Figure 13, explain
what can be concluded about the
selective advantage to a rat of being
warfarin-resistant compared to being
nonresistant in an environment
without warfarin.
8.
Question 8
Figure 14 shows the natural distribution of a mammal, the red-necked wallaby, Macropus rufogriseus.
a. Give two reasons why populations of this species in Tasmania have not
evolved into a separate species despite being geographically isolated by
the waters of Bass Strait.
b. Another mammalian species common in Tasmania is the Eastern Quoll, Dasyurus viverrinus. This species,
which is about the size of a domestic cat, was widely distributed in south-eastern mainland Australia until
about 50 years ago. It is now believed to be extinct in Victoria and possibly over the rest of its former
range in mainland Australia.
Give two possible reasons for the extinction of this species in mainland Australia.
Figure 4 shows the results of a breeding experiment with the vinegar fly Drosophila melanogaster. In each of the first
25 generations the smallest flies were selected to produce the next generation. After 25 generations the selection was
reversed. From generation 25–35 the largest flies were chosen to breed the next generation.
9.
(Question 21)
With respect to the genes controlling body size, the results of the experiment suggest that
A. after 25 generations there was no genetic variation.
B. for the first 25 generations there was no genetic variation.
C. selection between generations 25 and 35 had a significant effect on average body size.
D. if selection for small body size continued after generation 25, average body size would continue to fall.
10.
(Question 16)
Geographical isolation is important in assisting the process of
A. adaptation. B. gene flow. C. speciation. D. fossilisation.
11.
(Question 17)
Snakes and legless lizards evolved separately from ancestors with legs.
The lack of legs in these reptiles is an example of
A. analogy. B. divergence. C. founder effect. D. polymorphism.
A single gene locus with two alleles determines the colour of snapdragon flowers. A flower may have one of 3
phenotypes.
phenotype genotype
red
CRCR
pink
CRCW
white
CWCW
A student investigated the frequency of each of these phenotypes in a population of 100 snapdragon plants. The
frequency of each phenotype and genotype is shown in the table.
12.
(Question 23)
From this data it can be concluded that
A. the number of CW alleles in this population is 20.
B. the number of CR alleles in this population is 120.
C. the proportion of CR alleles in this population is 0.4.
D. the total allele pool for this locus in this population is 100.
In a species of marine snail the colour of the shell is controlled by a single gene with 2 alleles, G and g.
Allele frequencies for this locus were determined in six populations of snails. These populations were located close to
each other, but not able to interbreed. The frequencies of the G allele are shown in the table below.
13,
Question 24
The frequency of the g allele in population 5 is
A. 0.15 B. 0.3 C. 0.35 D. 0.7
14. b. The incidence of antibiotic resistant Streptococcus pneumonia has increased in the last 15 years.
Approximately 40% of infections by this bacterium are resistant to commonly used antibiotics.
i. Explain how the increase in bacteria resistant to antibiotics has occurred.
ii. What is the selective agent associated with the increase in antibiotic resistance in Streptococcus pneumonia?
15. Question 10
All the alleles in a population are referred to as the
A. phenotypic family.
B. proteome.
C. gene pool.
D. genotype.
16. Question 16
The founder effect and bottleneck are examples of
A. gene flow.
B. speciation.
C. genetic drift.
D. selection pressures.
17. Question 14
The frequencies of the phenotypes of the MN blood group were measured in a European population. Of 100
individuals, 40 were blood type M, 20 were blood type MN and 40 were blood type N.
From this data it is possible to conclude that
A. the frequency of the LN allele is 0.3.
B. the frequency of the LM allele is 0.5.
C. there are 60 LM alleles in this population.
D. there is a total pool of 100 alleles at this locus for this population.
16. (Question 15)
Natural selection acts upon an organism’s
A. habitat.
B. Genotype
C. phenotype.
D. environment.
17. (Question 23)
In some autosomal recessive conditions in humans, the homozygous recessive genotype results in death before
reproductive age. Despite this, the allele for the recessive trait is maintained in the population. Maintenance of this
allele in the population is most likely the result of
A. mutation.
B. migration between populations.
C. the heterozygote being biologically fitter than either of the homozygous genotypes.
D. individuals with the homozygous dominant genotype leaving more offspring in each generation.
18. Rock wallabies, Petrogale lateralis pearsonii, on Pearson Island off the coast of South Australia have had no
genetic contact with mainland rock wallabies since they were isolated by rising sea levels at the end of the last glacial
period, around 10 000 years ago. Scientists have taken blood samples from the wallabies and compared the
distribution of unique DNA sequences called microsatellites, which are scattered across the wallabies’ chromosomes.
These microsatellites give a measure of the population’s genetic diversity, or lack of it. In this case the microsatellite
data showed that the Pearson Island population has low genetic diversity.
The scientists concluded that the Pearson Island population of rock wallabies has been through a genetic bottleneck. A
genetic bottleneck is an example of genetic drift.
a. Explain how a genetic bottleneck may lead to a decrease in genetic diversity.
Despite the Pearson Island rock wallabies’ lack of genetic diversity, the population size has been maintained over
many generations. In fact, the wallabies appear to be thriving.
b. Suggest one reason for the wallabies’ success despite the lack of genetic diversity within the population.
c. The population of rock wallabies on Pearson Island is most closely related to small populations of rock wallabies in
southern Western Australia. Some scientists argue that some individuals from the southern Western Australian
populations should be released onto Pearson Island.
Give one reason for this suggestion.
18. Question 8
The Isthmus of Panama is a narrow strip of land that joins North and South America.
The land bridge formed approximately 3 million years ago.
Snapping shrimps, genus Alpheus, can be found on either side of the land bridge. The two groups are
phenotypically similar. However when the males and females from either side of the land bridge were brought
together they snapped aggressively at each other and would not mate. They are now considered to be two
different species.
a. Why is the inability to mate sufficient evidence to call the two groups different species?
b. What type of speciation has occurred in the snapping shrimp?
c. Explain how the differences between the shrimp on either side of the land bridge could have arisen.
Mesosaurus was a giant reptile that lived about 270 million years ago. The average Mesosaurus measured about
one metre in length, had webbed feet, a long tail and numerous sharp teeth. Fossils from Mesosaurus have been
found in only two places; the eastern side of South America and the west side of South Africa.
19. Question 23
Mesosaurus was most likely to be
A. a land animal that ate plants.
B. an aquatic animal that ate plants.
C. an aquatic animal that ate small fish.
D. a land animal that ate small animals
20. Question 7
The graph below shows changes in allele frequencies at a single locus with two alleles B and b in two very
large populations. The phenotype resulting from allele B is the dominant phenotype. In one population there
is selection against the homozygous recessive phenotype, and in the other population there is selection against
the homozygous dominant and heterozygous phenotypes.
a. In which population, X or Y, is selection ocurring against the homozygous recessive phenotype?Explain.
b. Describe what has occurred in population Y by the 60th generation.
c. Given the results in population Y, why has a similar pattern of events not occurred in population X?
CHAPTER 12
HOMINIDS AND HOMININS
PART 15
Primates, Apes and monkeys, Characteristics- fossil structure, Bipedalism, opposable thumb, Large brain,
Australopithecus(afarensis,africanus), Paranthropus, Homo habilis, Homo erectus, Homo neanderthal, Homo florensis,
Homo sapiens,
Cultural evolution, tools, fire, painting, rituals, language,
Questions
List the major characteristics that distinguish mammals from other animals.
List the major characteristics that distinguish primates from other animals.
Why are opposable thumbs important to primates?
Why do netballers and footballers require good stereoscopic vision?
What features distinguish hominins from other primates.
Describe the location changes of the foramen magnum.
What changes in jaw and tooth structure have occurred during human evolution? Describe how these changes may be
linked to an increase in brain size.
Skull shape and size have changed during human evolution. Describe their changes and their significance.
What has been the limiting factor to a further increase in skull size?
What selective pressures might have led to an increased brain size?
Is it true to say that man has evolved from chimps and gorillas. If not , rephrase the statement.
Biologists talk of ‘physical evolution’ and ‘cultural evolution’. Many would argue that the latter is now more
important. What is the definition of culture?
1. Question 23
The skull of Kenyanthropus was discovered in 1999 in sediments between 3.2 and 3.5 million years old. The
skull is unusual in that it shows the combination of a small braincase, flat face and small teeth.
The following characteristics would be seen in the skull of Homo.
A. a small braincase and small teeth
B. a small braincase and a flat face
C. a flat face and small teeth
D. a flat face and large teeth
2. Question 6
The table below shows the number of nucleotide differences between a region of mitochondrial DNA in humans,
chimpanzees and a Neanderthal.
a. Based on the data in the table, which individual is most closely related to the Neanderthal?
b. The differences between the mitochondrial DNA recorded are the result of base substitutions. There are 77
nucleotide differences between Human 1 and Chimpanzee 1.
Explain why 77 nucleotide differences is a minimum number of base substitutions.
c. The Neanderthal DNA was extracted from a fossil approximately 25 000 years old.
i. What other type of information obtained from the fossil could be used to assist in determining the evolutionary
relationship of Neanderthals with humans and chimpanzees?
ii. What method would be used to estimate the absolute age of the Neanderthal fossil?
iii. What method could be used to determine the relative age of the Neanderthal fossil?
3. a. i. Which of the two skulls is the more ancient? ii. List one characteristic of this skull in support of your choice.
Although extensive searches have occurred, fossils classified in the genus Australopithecus and other early hominid
genera have been found only in Africa. Assume that these fossils in fact only exist in Africa.
b. What is a possible explanation for these fossils being limited to Africa alone?
4. Question 18
Homo neanderthalensis lived in Europe and Western Asia from approximately 250 000 to 28 000 years ago. In
comparison to modern day humans the Homo neanderthalensis possessed relatively
A. smaller brains.
B. smaller noses.
C. higher foreheads.
D. more prominent brow ridges.
5. Question 19
Evidence of hominid cultural evolution can be found in the fossil record.
This evidence would include
A. position of the attachment of the spine to the head.
B. length of arm bones in comparison to leg bones.
C. number of teeth present in the skull.
D. presence of stone tools.
6. Question 21 Hominids are believed to have evolved in Africa because
A. the oldest hominid fossils have been found in Africa.
B. the most hominid fossils have been found in Africa.
C. monkey fossils were found in Africa.
D. Africa is the oldest continent.
7. Question 22 Consider the following diagrams of skulls.
The skull most likely to be that of a chimpanzee is
A. W
B. X
C. Y
D. Z
Question 23
There is evidence that Homo sapiens and Homo neanderthalensis coexisted in Europe more than 30 000 years
ago. Both of these species left signs of cultural evolution from this period. An example of evidence which would show
that cultural evolution was occurring in these groups at this time is
A. animal remains close to a Homo skeleton .
B.. male and female skeletons in the same area.
C. drawings and carvings on rocks
D. Homo sapiens and Homo neanderthalensis skeletons in the same area.
Question 22
Consider the following diagrams of Hominid skulls.
The correct sequence of evolution, from oldest to
youngest, of the Hominid species shown is
A. 2, 3, 1
B. 3, 1, 2
C. 1, 2, 3
D. 2, 1, 3
CHAPTER 7
STRUCTURE OF CHROMOSOMES
PART 1
1. B 2. B Each rabbit only has a pair of homologous chromosomes so it can only have 2 alleles.
3. D 1, 1 1,2 1,3 1,4 2,2 2,3 2,4 3,3 3,4 4,4 10 combinations
4. B
5. 19
6.C Phenotype= Genotype+Environment 7.D 8.C 9.D 10.A 11.A 12.C 13.D 14.C 15. B 16.aA bB
17 a. The gene locus is the position of the gene along the chromosome. b. The sequence of
bases/nucleotides would be different.
CHAPTER 7
CELL DIVISION
1. B 2.C 3. D 4. C, E, D, B, F, A
PART 2
4 b Any two of
• increases variation in the gametes or offspring
• the recombining of maternal and paternal alleles
• exchange of alleles or genetic material between homologous chromosomes or between non-sister chromatids resulting in new
combinations of alleles.
Some students incorrectly stated that the homologous chromosomes exchanged genes. Homologous chromosomes have
the same genes; they may have different forms of a gene. Therefore, they may exchange different forms of a gene. The
different forms of a gene are called alleles.
Ci E Cii In anaphase one chromosomes are double stranded. The chromosomes shown moving to the poles in E are double
stranded.
5. D
6. a The diploid chromosome number is the number of chromosomes found as homologous pairs in a cell (organism) or the
number of chromosomes in a cell (organism) that have two copies of each specific chromosome.
Responses that were not awarded marks included ‘number of chromosomes in a somatic cell’ and ‘2n = 16’.
b. Each daughter cell would contain 15 chromosomes.
C i. Each gamete of a female will contain eight chromosomes.
ii. Half of the male gametes will contain eight chromosomes and the other half will have seven chromosomes.
Students needed to have both answers to be awarded full marks. Some incorrect answers included seven and a half
chromosomes and 15 chromosomes.
7. a. 53 b. Meiosis
c. Students were expected to be able to accurately show the following:
• separation of homologues in meiosis I (at the end of the first divisions, showing one chromosome in each cell as doublestranded)
• separation of chromatids in meiosis II (at the end of the second division, showing chromosomes in four cells (two of each type)
as single-stranded)
• the process of division involving cytokinesis, the spindle, etc.
d. Chromosomes are not homologous and therefore will not pair during meiosis.
CHAPTER 7
CROSSES
PART 3
1 D 2 A 3 D (A or b could be true but you can’t tell because you don’t know which alleles are linked on
the same chromosome, C isn’t true because there will be more of one than the other) 4 B 5. D 6. C 7 A
8. The presence of four phenotypes in the F2 generation indicates that two genes are involved. One of the genes determines
coat colour. One allele for coat colour codes for black coats, the alternative allele codes for white coats. The gene for
coat texture has two alleles; one coding for rough texture, the other allele codes for smooth texture.
OR there are four phenotypes in the F2 generation with a ratio of 9:3:3:1 indicating the parents are heterozygous for two
gene loci each with two alleles.
bi RR BB rough, black rr bb smooth, white bii Bb Rr
biii BBrr or Bbrr
9. D (1,1 1,2 1,3 2,2 2,3 3,3 6 genotypes A, AB, B, O 4 Phenotypes) 10. C 9:3:3:1 11. B (AWac x aSac 4 genotypes
AWaS, AWac, acaS, acac White, White, Sable, copper 3 phenotypes . ) 12. C 13. B (could be AO x BO)
14. B 9:3:3:1= 16 into 1600 so 100 for 1. 15. C 16. D (1,1 1,2, 1,3 2,2 2,3 3,3 6 genotypes) 17.C
18. Y= Yellow y= grey Genotype ratio 1 YY: 2Yy : 1 yy Phenotype ratio: 3 Yellow: 1 grey. They ratio is not this because the
YY is lethal and these offspring die.
CHAPTER 7
MONOHYBRID CROSSES X LINKED
PART 4
1. C 2. B 3 A 4 B 5 DdxDd , dd 6.Cross 2 because two resistant plants are crossed and yet 12 offspring who are sensitive
are produced which indicates this trait was carried hidden in the parents (recessive). R=resistant r = sensitive
Rr x rr = ½ Rr : ½ rr 7. B 8. A Patterned is dominat because patterned parent frogs produced non patterned offspring indicating
this characteristic was carried by the parents (recessive). b) P=patterned p= non-patterned PpxPp 1PP:2Pp:1pp
3 patterned : 1 non patterned is the expected outcome.
CHAPTER 7
PEDIGREES
PART 5
1. Autosomal dominant. Dominant because two parents produce a child with a different phenotype to each of them which means
they must have been carrying the normal allele. Autosomal because if it was X-linked a father who had the dominant allele gives
it to his daughters so any girls should have psotosis.
2. ai Pedigree A ii) It can be autosomal dominant or recessive as long as the parent with the dominant phenotype is heterozygous.
It can’t be X-linked because there is a mix each of the sexes with and without the trait.
bi Pedigree B is dominant because two parents with the trait have had offspring without it indicating that the normal trait was
carried (recessive). ii. It is not X linked because a father with the trait gives the X to his daughters so all daughters would be
affected.
3. A 4 A
5. Rr x Rr because II 1 is liver coloured which requires rr and Bb x Bb because II 3 is red with
requires bb. They each have one R and B because of their own colouring. 5b i) Rr Bb ii) rrBb a cross would result in all the
phenotypes being possible 6a) I1 XHXh I2 XhY b) 100% c) 50% d) I1- Gg
I2- Gg II1 gg II2 GG or Gg
CHAPTER 8
PROTEIN SYNTHESIS
PART 5
1 C 2 D 3a) Transcription b) PremRNA c) the pre-mRNA has the introns removed and the exons are joined back together to
form the final mRNA d) Translation e) i t RNA ii) The role is to bring in the correct Amino Acid to join the polypeptide chain
based on the codon on the mRNA fitting the anticodon on the tRNA.
ATG CCA CCG TGG GTA TGC
TAC GGT GGC ACC CAT ACG
CHAPTER 8
5B
DNA STRUCTURE/ GENE REGULATION
PART 6
1. B 2. A 3. D 4.C 5.A 6 A 7 a) Organisms regulate the expression of their genes so that the correct proteins are present in the
cell for it to carry out its role in the organism. They are also only produced when needed so that energy and resources are not
wasted on producing proteins which can’t be utilized. B) RNA polymerase c) The production of enzyme X will commence as the
gene is free to be transcribed because the RNA polymerase is no longer blocked. D) The tryptophan binds to the repressor protein
and changes its shape so that it can no longer bind to the DNA.
CHAPTER 8
MUTATIONS
PART 7
1. B 2. B 3a)9 b)i. 18 ii. The chromosomes aren’t homologous so they don’t pair up during meiosis (prophase 1) c) polyploid
d) The new plants now have a pair for each chromosome so they canpair up at meiosis and form gametes with one copy of every
chromosome. 4. B. The phenotype are they characteristics expressed by the organism. ii. It had two copies of every chromosome
it’s just that two of the copies were stuck to each other. This didn’t alter their functioning.
c) Sperm 3- 9/18 by itself Sperm 4 – A 9 by itself 5. d) The 9 by itself doesn’t have any information for the 18. e) ¼ chance to
get the same as the father . The would have to be the one with the 9/18 by itself.
5. A mutation would have occurred which changed the phenotype so that it was no longer susceptible to the antibiotic 6. D
7a. AUG AAA UUC UCG AAU AGC bi. A point mutation called a substitution. Ii. The DNA code changes from TTT to TTC so
originally the mRNA was AAA not it is AAg. The original amino acid was Asn and its Lys. This change of an amino acid may
have no affect on the poly pepetide produced or it may make the protein more or less efficient. c) All of the code from then on
will be read out of order so it will affect eery amino acid from then on. This is called a ‘frameshift’ mutation. AGC TTA TCG
will now be ACT TAT and only CG left, so the codes have been totally changed and reduced to 2 codons rather than three.
CHAPTER 9
GENETIC ENGINEERING
PART 9
1. C 2. a) A recombinant plasmid (A vector). B) Restriction enzyme C) DNA ligase D) To identify the bacteria which have
taken up the plasmids you can include an antibiotic resistance gene and then grow the bacteria on a plate containing that antibiotic.
Only the bacteria with the plasmid will grow. E. There are many reasons why a farmer might want to do this – normally crop
spraying is done by air and the insecticide can be carried by the wind and kill insects over a wide area. The insects may be
beneficial in the food chain feeding birds etc. More insecticide in the environment can encourage selection of resistant insects. It
costs more in time and insecticide to spray. F. He can kill the weeds easily even when the crop is growing. He doesn’t have to
introduce heavy machinery amongst the crop to kill the weeds or hire costly labourers. G. Consumers in many countries are
concerned about using crops that have been genetically modified and some countries are banning the import of
GM foods. If the farmer is selling to those markets he would be unable to if his crops where fertilised by the GM crops h. When
there is no gap between the crops there is a 10% contamination rate on the edge of the crop and contamination even within the
crop. Moving the crops 5 metres or even 7 metres apart, lowers the contamination rate but it doesn’t totally eradicate it. So the
farmer could still not claim his crops where GM free at 7 metres apart and once one plant grew with the transgene it could
pollinate others the next year and increase in the crop.
3.a) Sticky ends 4a a small, circular chromosome. b) Transformation c) i. No growth ii. The bacteria are sensitive to the
antibiotic tetracycline and therefore do not grow. D) Plate A: The bacteria are able to grow as there is no tetracycline present. 
Plate C: Only those bacteria which take up the plasmid can grow in the presence of tetracycline. (It was important to only some of
the bacteria take up the plasmid) 5. C
CHAPTER 9
GENETIC ENGINEERING PART 2
PART 10 AND 11
1a) The polymerase enzyme catalyses the production of a new strand of DNA or is involved in making multiple copies of DNA or
amplification of DNA AND
DNA polymerase replicates the DNA by extending from the primer or by complementary base pairing or by using the original
DNA as a template. (there were 2 marks awarded for this question so 2 separate remarks needed to be made)
b) A DNA fragment will move according to its charge and molecular weight (size) or DNA is negatively charged and moves to
the positive pole; smaller DNA fragments move further or faster than larger fragments.
c) There is only one band in lane 2 because individual 2 is homozygous, the others on the gel are heterozygous or the two
fragments of DNA are the same size or the number of repeats in the two fragments is the same.
d) There are 5 different alleles at the HUMTHO1 locus represented on the gel.
e) DNA piece A has the greater number of the 4 base repeat sequence. The greater the molecular weight of the sample the smaller
distance the sample will move from the loading well.
f) The bands on the gel for suspect 5 match the sample of blood found on the victim, which was not the victim’s blood (lane 3).
2. B 3. b) The enzyme does not cut the DNA, or the enzyme no longer recognises the sequence
C)
5. a) Making an exact genetic copy of an organism. b) 19 ii) female . The donor cat was female so the genetic material would be
female c) There are two homologous chromosomes for each region studied. This part was not well answered. Many students
incorrectly thought the two numbers were representative of the two strands of DNA. D) The values are different because the
regions have slightly different lengths 154/160 One is 154 nucleotides long and the other 160. Most students attempted this
question, however, many students again related the numbers incorrectly to the two strands of DNA or made reference to the
occurrence of mutations having occurred, which was also incorrect. E) All of CC DNA which was studied was the same length
as the donors DNA but did not match any of the regions in the surrogates DNA. This implies that it was a clone of the donor
rather than and offspring of the surrogate. 6. Selective breeding means that humans control which bull will mate with which
cow. In random mating the animals themselves choose. F. Selective breeding means that the traits found in the herd will be those
considered desirable by the breeder. This will reduce the variability in the herd to these characteristics.
g. The animals may reach the stage where they are unable to reproduce themselves. They may lose the behaviour involved in
mating or some other factor of the mating process.
7. A
CHAPTER 10
FOSSILS
PART 12
Q1. C 2.C 3. B 4 D 5.a) They had a tough exoskeleton. b) Trilobite fossils may be abundant for any one of the following
reasons: • were widely distributed geographically • able to survive in a wide range of habitats • existed for 300 million years
• lived in an environment where fossilisation can occur readily.
Q6 A) Layer one b) Any two of the following answers were acceptable:• an index fossil must be distinctive and easily
recognisable • an index fossil should be abundant • an index fossil lived a short range through time • an index fossil lived in a
wide geographic distribution.
Many students defined an index fossil rather than describing the features of a fossil that would make it useful as an
index fossil. C) One of the following was an acceptable method:
• radioactive isotopes (for example, potassium/argon, argon/argon, lead/lead, uranium/lead)
• stratigraphy, which involves a study of the relative ages of the surrounding rock layers.
Carbon dating was not acceptable, as it only dates to within 30 000 – 50 000 years.
d) The remains have been undisturbed from both scavenging animals and the weather. Students were not awarded a mark if they
stated the cave protected the fossil and gave no further information as to what it was offering protection from.
e) The teeth or jaw would be used to determine that the animal was carnivorous. F) The scientists would compare the hard parts of
the extinct animal to living species. They would look for points of muscle attachment to animals living today and for indications
on the fossil bones where ligaments might have been attached to gain insight into musculature, or compare the sizes and types of
bones of the fossil to animals living today to help establish the weight of the fossil. 7. C
CHAPTER 10
1. B 2. B 3 B
4.D
EVOLUTION
5.C
6.D 7. A
8. D
9. B
PART 13
b)Trilobites, tegepeltids, helmetids, naraoids. C) The
environment altered and there were no variations within the gene pools which could cope with it.
A global climate change must have occurred to make all species of trilobites extinct. Some students mentioned a climate change
without any indication of the global nature of the change and therefore were not awarded a mark.
10. A 11. A 12. B 13.d) They had become extinct on the mainland due to not adapting to a new selection pressure (maybe the
arrival of the aboriginals? e) Sightings are not concrete enough evidence so show that they still exist. You would need some
physical evidence – an actual animal. f) Mammals and marsupials diverged a long time ago into a all the different forms of today.
These two have similar features because they are both carnivores and similar features (large teeth, strong jaws etc) would have
been separately selected for to arrive at a similar function.
CHAPTER 11
NATURAL SELECTION
PART 14
1. B 2.C 3.A 4.C 5.B 6.D
7. a)The process that led to the increase in the percentage of resistant rats included warfarin resistant rats existing in
the population before the use of warfarin and when warfarin is used non-resistant rats are killed and warfarin resistant
rats survive to reproduce, and pass on the allele for resistance or warfarin resistance is inherited and so is passed on to
next generation and over several generations the proportion of warfarin resistant rats increases.
Students who could clearly express their ideas in a logical way were more likely to be awarded full marks. Students
must be encouraged to formulate answers to questions that require a detailed account of a concept/s.
b) From the graph it can be seen that the percentage of resistant rats decreases when the use of warfarin is
discontinued (years 3 and 4) therefore resistant rats are at a disadvantage in a non warfarin environment or resistant
rats are less fit in a non warfarin environment or non-resistant rats are at a selective advantage in a non warfarin
environment.The question asked students to use the data in figure 13. Therefore, students were expected to explain
how the data was used in arriving at their conclusion, for example more successful answers specifically mentioned
that the number of rats decreased in years 3 and 4.
8a) The populations of red-necked wallabies in Tasmania have not been isolated long enough from the populations of
red-necked wallabies on the mainland for sufficient genetic differences to accumulate and the populations of the rednecked wallabies occupy similar habitats so similar selection pressures or a specific example of a selection pressure.
Students could score 1 mark if they gave a specific example of a similar selection pressure but were not awarded 2
marks if they gave two examples of similar selection pressures.
b)The Eastern Quoll may be extinct on the mainland of Australia because a disease may have spread through the
mainland populations and killed all quolls or a predator may have been introduced on the mainland which killed all of
the quolls or the quoll habitat may have been destroyed when humans cleared much of the mainland for farming.
One-word answers such as ‘hunting’ or ‘predators’ are unlikely to be awarded marks. The space provided for the
answers indicated the detail needed in the response.
9 A 10.C 11. A 12.B 13. B 14. B. When antibiotics are used to treat an infection if there are any resistant
bacteria present and the immune system does not destroy these they will multiply. The more times the same
antibiotics are used the more resistant bacteria are selected for and so their numbers and percent of the bacterial
population increases. C) The Antibiotics which are used to treat the infections. 15. C 16. C 17. B
16. C
17. C 18. A genetic bottleneck results in a severe reduction in population size and as a consequence allele
frequencies may change by chance, which will decrease genetic diversity.b) There were many acceptable answers to
this question. Some of these included: • the wallabies are well adapted to the environment • there have been no
changes in the selection pressures acting on the wallabies • the wallabies are protected from new predators as they are
on an island C). The scientists want to increase the genetic diversity of the wallabies, giving the population a better
chance of survival if the environment changes.
18 (again!) a) Inability to mate is not enough evidence; it is the
inability to produce viable offspring which is important. (Stupid question!) b) Allopatric speciation Incorrect answers
included divergent evolution and founder effect. C) There were different selection pressures in the two environments,
which allowed differences in allele frequencies to develop. •Over time there is an accumulation of genetic differences
which changes a trait, such as mating behaviour. This question was poorly answered and was generally not recognised
as a ‘natural selection’ question. Students needed to provide both of the above points for two marks
20a)
Population X has selection against the homozygous recessive phenotype, as the frequency of the b allele is decreasing
over time. b) Any of: • the allele b has become fixed • the frequency of allele b = 1 • only one allele in population
Y. Some students misread the graphs for parts a. and b.Students should be encouraged to take greater care when
reading the axes and applying this knowledge to the question asked. c) In population X there is selection against bb,
but there are b alleles still present in the heterozygotes that survive and will pass on these alleles to the next
generation. This question was not well done. Many students did not realise that the b allele would be present in
heterozygotes and would therefore persist in the population.
CHAPTER 12
HUMAN EVOLUTION PART 15
1. C 2. a) Human 1 b) Mutations are reversible and some substitutions may reverse an earlier change making 77
nucleotides the minimum number of base substitutions. C) Skeletal structure/morphology may be used to assist in
determining the evolutionary relationship of Neanderthals with humans and chimpanzees. ii) Carbon dating can be
used because the fossil is approximately 25 000 years old. iii)To determine the relative age of the fossil, an index
fossil may be used. Stratigraphy which states that the oldest stratum is at the bottom and younger layers lie above can
also be used 3 a) Skull one b) Skull one has a more rounded jaw line or large jaw relative to overall skull size, or skull
one has more prominent eyebrow ridges or skull one has a smaller brain case. Students need to be reminded that when
a question asks for one characteristic, they should only give one characteristic. If they include more and one is an
incorrect response they may be penalised. C) The genera may have become extinct before the migration out of Africa
could have occurred OR there were no selection pressures placed on the genera forcing them to move, so they
remained in Africa until extinct. 4. D 5. D 6. A 7. B 8. C 9. A