Past Papers

Question 1

1. From which stage of the cell cycle do cells begin mitosis and meiosis?

A. G0
B. G1
C. G2
D. S

Answer 1

D. S

Question 2

2. In humans, with 2n = 46, how many chromatids are present in each cell at Metaphase II of meiosis

A. 23
B. 46
C. 92
D. 184

Answer 2

-> Metaphase 1 -> Before fist division
B. 46

• Double normal number of chromosomes & chromatids.
  » Normal number = 23 chromosomes, 46 chromatids
  So during meiosis 1 = 46 chromosomes, 92 chromatids.

Metaphase 2 -> this number will have halved -> so 23 chromosomes, 46 chromatids

Then after metaphase 2, following meiosis 2 (end of meiosis) each cell has 11.5 chromosomes & 23 chromatids (haploid no.) so when inherit one of each gametes from parents = diploid no. of 23 chromosomes & 46 chromatids again.

Question 3

3. Which process makes the largest contribution to the genetic variability brought about through meiosis?

A. crossing-over
B. DNA mutation
C. independent assortment D. non-disjunction

Answer 3

C. independent assortment ?

Question 4

4. The AB blood type in humans is an example of

A. co-dominance
B. incomplete dominance
C. over-dominance
D. pre-dominance

Answer 4

A. co-dominance

Question 5

5. There are more than 10 million SNPs in the human genome. Current genome wide association studies (GWAS) typically involve genotyping all individuals under study with 300,000 to 500,000 of these SNPs. Using more SNPs would provide greater coverage. What is the rationale for genotyping only a subset of the known SNPs?

A. Not all SNPs are functional B. Not all SNPs are in genes
C. The existence of haplotype blocks means that some SNPs act as tags for others
D. The technology currently available for genotyping cannot accomodate more SNPs

Answer 5

C. The existence of haplotype blocks means that some SNPs act as tags for others

The haploblock structure identified by HapMap project also allowed the focus on the subset of SNPs that would describe most of the variation

• Genome Wide Association Studies (GWAS)
 Population – level approach to disease gene mapping.
 Identification of SNPs:
- Cases: C Allele -> 62%
T Allele -> 38%
- Controls: C Allele -> 49%
T Allele -> 51%
- Odds ratio for C allele:
 (Odds of disease with C) / (Odds of disease with T)
= [A/B] / [C/D] = [62/49] / [38/51]
= 1.7
- Allele odds ratio ; >1.0 -> Allele gives Higher risk of disease
< 1.0 -> Allele is protective
 Process:
 Coloured dots represent single SNP on DNA chip
 Odds ratio (OR) for each variant calculated
 Probability (P) OR is significantly higher in cases compared to controls i
illustrated -> Manhattan Plot.
 Manhattan plot
Illustrates the probability -> Odds ratio of an allele is significantly higher in cases compared to controls.
Variants of high significant associations with trait visible as skyscrapers.
 Haplotype Blocks:
Groups of closely linked haplotypes (SNPS on the same chromosome) which are inherited together.
 Each haplotype block -> defined by small no. tag SNPs
 Linkage Disequilibrium:
Co-inheritance of SNPs in a haplotype block

• Role of GWAS in Human Disease Genetics:
- Majority of disease-associated variants have small incr. risk of disease.
- >90% variants present in non-coding DNA
»Difficult to identify causal variant.
- Generate new biological hypotheses about causes of disease
Eg. Inflammatory Bowel Disease & Autophagy.
- Studies of rare SNPs & other types of DNA seq. variation - eg. Copy Number Variants (CNV) – necessary to further understand complex traits.

Question 6

6. Non-familial (sporadic) Down syndrome is predominantly the result of

A. non-disjunction occurring during the first meiotic division in the father
B. non-disjunction occurring during the first meiotic division in the mother
C. non-disjunction occurring during the second meiotic division in the mother
D. a Robertsonian translocation involving chromosomes 14 and 21

Answer 6

B. non-disjunction occurring during the first meiotic division in the mother

Trisomy 21 is caused by a meiotic nondisjunction event.
Gamete produced with an extra copy of chromosome 21 (24 chromosomes). When combined with a typical gamete from the other parent, the child now has 47 chromosomes, with three copies of chromosome 21.
Trisomy 21 is the cause of approximately 95% of observed Down syndrome, with 88% coming from nondisjunction in the maternal gamete and 8% coming from nondisjunction in the paternal gamete. Mitotic nondisjunction after conception would lead to mosaicism, and is discussed later.
Down syndrome, a trisomy of chromosome 21, is the most common anomaly of chromosome number in humans.[2] The majority of cases results from nondisjunction during maternal meiosis I.

• Maternal Age & Trisomy
 Incr. trisomy
 Responsible for incr. miscarriage -> older women.
 95% -> trisomy 21
 Maternal non-disjunction -> meiosis 1
 Human oocytes
 Paused -> late meiotic prophase I (diplotene)
»_space; Alongside paired, replicated chromosomes
 Begins before birth
»_space; Maintained -> decades
-> Until egg matures -> menstrual cycle
 Loss of cohesion -> Prophase I
 Aneuploidy -> older women
» Premature loss -> cohesion
-» 2 univalents -> segregate independently
> Aneuploid (n +1) gametes

-> Robertsian Translocation can be due to inheritance or other reasons, however most commonly familial.
Translocation Down syndrome is often referred to as familial Down syndrome. It is the cause of about 4.5% of the observed Down syndromes.[4] It does not show the maternal age effect, and is just as likely to have come from fathers as mothers.

Mosaic Down syndrome -> some cells are normal & some have trisomy 21, an arrangement called a mosaic. This can occur in one of two ways:
A nondisjunction event during an early cell division leads to a fraction of the cells with trisomy 21;
An anaphase lag of a chromosome 21 in a Down syndrome embryo leads to a fraction of euploid cells (2n cellsThis is the cause of 1–2% of the observed Down syndromes.

	Down’s Syndrome:
 Charcteristics:
-	Characteristic facial features
-	Short stature
-	Learning disabilities
-	Higher risk -> Heart defects
-	Alzheimers & some cancers -> 1/1000 

 Causes:
	 Trisomy (2n +1):
-	Trisomy 21
	Robertsian Translocation:
-	Chromosome 21 & 14
	Genetic mosaicism:
-	Mix of normal & trisomy 21 cells 
>>Non-disjunction -> early embryonic mitotic divisions

Question 7

7. Which of the following processes contributes to the maintenance of genetic variation in populations?

A. Founder effect
B. (Negative) frequency dependent selection
C. Inbreeding
D. Purifying selection

Answer 7

? Purifying keeps alleles relatively stable -> but could minimise / make alleles closer to mean .
Negative frequency doesn’t maintain alleles but does ensure it keeps changing in variation in continuous cycle.

Founder effect -> Population based on small group of individuals -> little genetic variation

i) Founder effect:
Small no. of individuals -> start new population

> > Frequency dependent selection
Allele selected against -> high frequency
Allele selected for -> low frequency
Eg. Scale-eating fish
Left & right mouthed predators
Left -> attack back right flank
Right -> attack back left flank
-»If high no. right mouthed
– Prey expecting to be attacked -> left flank
– Less likely to predict attack -> right flank
– Left mouthed have advantage -> incr. no. of left-mouthed individuals.
-»Low no. right mouthed
–High no left mouthed -> expect attack on right flank
– Incr. success of right mouthed -> pop. Incr.

Purifying = stabilisaing selection -> Maintains pop alleles

Question 8

8. For a particular trait, the narrow-sense heritability can be expressed in terms of the

A. additive genetic variance and the phenotypic variance B. additive genetic variance and the environmental variance
C. gene-environment interaction and the genetic variance
D. genetic variance and the phenotypic variance

Answer 8

A. additive genetic variance and the phenotypic variance

The narrow-sense heritability is the ratio of additive genetic variance to the total phenotypic variance. h 2 = V A /V P

Question 9

9. X chromosome inactivation

A. always affects the paternal X chromosome
B. begins shortly after birth
C. can never occur in males D. is a type of dosage compensation

Answer 9

D. is a type of dosage compensation

• Dosage compensation -> Animals:
- X chromosome activation -> mammals:
 Females -> 2 x X chromosomes
»Double the gene dosage -> X-linked genes compared -> males
 1 of X chromosomes in each female cell -> inactivated
->Dosage compensation.
»Becomes highly condensed
->No expression -> genes
-> Cytologically visible -> interphase as -> Barr body.
 Inactivation -> starts early -> development
»Random -> effects either X chromosome
»Persists -> all subsequent mitotic cell divisions
Therefore female mammals -> mosaics
»_space;Example -> epigenetic control -> gene expression.

• X chromosome inactivation can occur -> males -> more than one X chromosome

• Individuals -> multiple X chromsomes
Eg. XX females, XXY males, XXX females
»Each cell -> expresses only 1 X chromosome
Other X chromosomes -> Barr bodies

Question 10

10. Hamilton’s rule explains

A. only haplodiploid forms of eusociality
B. the role of eusociality in the maintenance of haplodiploidy
C. the origin of eusociality
D. the origin of haplodiploidy

Answer 10

B. the role of eusociality in the maintenance of haplodiploidy

-> Eusociality evolved as a result of haplodiploidy.
–> Haplodiploidy causes females to be closely related to siblings compared to offspring.
This causes development of eusociality -> more likely to raise siblings than offspring.
» Prevents / reduces breeding of any other females with males so maintains haplodiploidy of pop.

• Hamilton’s Rule -> Haplodiploidy of Hymenoptera:
 Hymenoptera -> more closely related
-> sister -> than to -> own offspring
 Genes more likely -> passed to next generation -> female rears sister rather than having own offspring.
 Relatedness may be lower -> queen -> mates -> multiple males
Ancestral form mated -> single male
->Explains evolution of system
 Close relatedness -> not precondition/determining factor
Ecology -> shapes evolution of species.
 Many hymenoptera -> not eusocial
Solitary bees, sawflies, parasitoid wasps etc.
Eg. Termites -> XY sex determination -> based on close relatedness of individuals.

Question 11

11. The molecular clock can be based on molecular differences in

A. genes
B. genomes
C. proteins
D. all of the above

Answer 11

D. all of the above

3. Molecular genetics:
   Can track evolutionary relationships by comparison of genetic sequences/proteins.
   - The Molecular Clock:
    Proteins from pair of species compared
    Differences in DNA code correlated with incr. distant evolutionary relationship from fossil evidence.
    Genetic differences accumulate at constant rate:
   Non-coding sequences of DNA not subject to selectional pressures mutate at a constant rate.
   What is it’s significance? What does it illustrate?
    Evolutionary tree of cytochrome c gene correlates with morphological trees.
    Rapid appearance of major bird species/groups after extinction events.
    Helps illustrate relationships – Eg. Falkland wolf -> only native mammal to Falklands. Found to split with fox species on mainland during last glacial maximum due to analysis of genetic sequences.

Question 12

12. Sympatric speciation occurs:
    1) rarely in animals;
    2) on islands;
    3) easily in plants;
    4) when species change at the same time.

A. 1 and 4
B. 1, 2 and 3
C. 1 and 3
D. 1, 2, 3 and 4

Answer 12

??

• > Can occur on islands but not exclusively -> can occur anywhere as long as same place.
• > Both species do not need to change behaviour, only necessary for one to change.

• Types of Speciation?:
1. Allopatric speciation:
 (Allos = other; patra =homeland) -> Classic, widespread.
 Geographical isolation of existing population
Eg. Antelope Squirrels -> Grand Canyon. -> Birds on both sides show no such effects/speciation.

2. Sympatric speciation:
    (Sym = same ; patra = homeland) -> Theoretically debated -> animals -> few examples
   -> Commonly found in plants
    No geographical isolation of groups.
   Eg. Change in behaviour -> nocturnalism of one group
   -> natural selecton -> use of different resources
   Eg. Shortcleuch Waters –(see also allypolyploid evolution below)
3. Paraphatic speciation:
    Somewhat but not complete geographical isolation -> reproductive barriers when restricted gene flow.

Question 13

13. Which of the following would be an example of evidence that humans are still subject to natural selection?

A. Changes in height in a given population over time
B. Changes in the frequencies of alleles for an adaptive character over time C. Different characters in different populations over time
D. Increased fitness for an adaptive character

Answer 13

D. Increased fitness for an adaptive character

• > Msintenance of higher frequencies of certain advantageous characteristics of a population which incease fitness / changes or survival enabling reproduction
• > Simple changes could be sue to any reason and not have selectional advantage.

Question 14

14. Which of the following processes are possible:

1) evolution by natural selection;
2) evolution without natural selection;
3) selection by adaptive evolution;
4) evolution by non-adaptive selection?

A. 1 only
B. 1 and 2
C. 1, 2 and 3
D. 1, 3 and 4

Answer 14

??

Evolution without natural selection possible as unadvantageous evolution may occur & organisms then just die out / evolution which has no effect on fitness of individual occurs.

Natural selection only acts on the population’s heritable traits: selecting for beneficial alleles and, thus, increasing their frequency in the population, while selecting against deleterious alleles and, thereby, decreasing their frequency. This process is known as adaptive evolution.

Random (non adaptive) process that contributes to genetic diversity - occurs simultaneously with mutations and gene flow in natural selection.

Question 15

15. All animal eyes are based on molecules called

A. opsins
B. opsitins
C. opticins
D. optins

Answer 15

A. opsins

Protein opsins -> molecular basis for all sight
First functioned as clocks more than light sensors -> combined with melatonin which dies upon contact with light -> enabling detection of daybreak by absence of melatonin.
 Continually generate melatonin for next day of use but chromophores could just change shape continuously.

Question 16

16. Human HOX11A and mouse Hox11A are examples of

A. cytoskeletal genes
B. orthologous genes
C. paralogous genes D. ribosomal genes

Answer 16

B. orthologous genes

• Hox11 paralogue mutants
   All sacral vertebrates transformed -> lumbar vertebrae

Cytoskeletal genes regulate
brain size
 Paralogous genes:
 Duplicated genes within single chromosome
 Orthologous genes:
 Same gene present -> diff. organisms

Question 17

17. What are some of the major advantages of planarians, as model organisms?

A. External development, tetrapod, and asexual reproduction
B. High regenerative capacity, adult stem cells, sexual and asexual reproduction
C. Is a mammal and has strong genetics
D. Is a vertebrate and has sexual and asexual reproduction

Answer 17

B. High regenerative capacity, adult stem cells, sexual and asexual reproduction

 •	Planaria:
-	Bilateral symmetry 
-	Triploblastic
Ectoderm, mesoderm & endoderm
-	RNAi knockdown technology
-	Adult stem cells
-	Sexual & Asexual reproduction
-	High regenerative capacity

Question 18

18. Transplantation of cells or tissues from one part of the embryo to another is an example of

A. descriptive embryology
B. epigenetics
C. experimental embryology
D. morphogenesis

Answer 18

C. experimental embryology

• Experimental embryology:
- Investigate how cells acquire fate / how they develop
- How cells know what they should develop into:
 Asymmetric inheritance of cytoplasmic determinants
 Cell-cell communication
 Induction via communication with other cells
»_space; Influence by signals -> surrounding cells
»_space;Fate also influenced by what parental cells were, location in embryo & signals
from env. (neighbouring cells)

Descriptive Embryology:
- Observation of the natural development of an embryo
 Internal & external observations
- Understand mechanisms of growth -> vary with stages & species
- Describe different stages of growth by naming them
- Use colours to describe types of tissues observing
 Ectoderm -> blue -> (outside layer of tissues)
 Mesoderm -> red
 Endoderm -> yellow -> (inner linings / linings of organs

epigenetics is the study of heritable phenotype changes that do not involve alterations in the DNA sequence.

Morphogenesis is the biological process that causes an organism to develop its shape. It is one of three fundamental aspects of developmental biology along with the control of cell growth and cellular differentiation, unified in evolutionary developmental biology (evo-devo).

Question 19

19. The Spemann’s organiser

A. is present in the dorsal mesoderm - induces the nervous system
B. is present in the dorsal vegetal pole - induces Nieuwkoop centre
C. is present in the ventral mesoderm - induces the blood
D. is present in the ventral vegetal pole - induces ventral mesoderm

Answer 19

?

Question 20

20. Autotetraploidy occurs by

A. genome duplication from failure in meiosis within a single species
B. genome duplication from failure in mitosis within a single species
C. hybridisation of two different species, followed by failure in meiosis
D. hybridisation of two different species, followed by failure in mitosis

Answer 20

?

Question 21

21. The graph above shows
    i) the relationship between infant mortality and birthweight, and
    ii) the frequency distribution of birth weights in the population with a mean weight of 7 pounds. What type of natural selection is most likely to be acting on birth weight?

A. Balancing selection B. Directional selection C. Disruptive selection D. Stabilising selection

[Birth weight goes upwards in bell-curve shape ; but is quite narrow.
Infant mortality decreases as birth weight increases. Once bell curve reaches it’s peak, infaznt mortality practically at zero.
Infant mortality slightly increases (very slow rate) following this point (during which birth rate on declining section of bell-curve)

Answer 21

Without looking at graph would say stabilisaing as too underweight or too overweight disadvantage so weight would tend around mean.
Birth weight in bell-curve.
Infant mortality incr. much more slowly than it decr. hence possible cause of directional selection over time?

Question 22

22. The human MN blood group has two alleles (M and N) at a single genetic locus. In a sample of 500 individuals from the UK, the number of individuals of each genotype was as follows:
    MM = 151,
    MN = 248,
    NN = 101.
    What is the observed frequency of the N allele?

A. 0.40 B. 0.45 C. 0.5 D. 0.55

Answer 22

?

Question 23

In squash plants the fruit can be white, yellow or green. A plant with white fruit was crossed to a plant with green fruit.
In the progeny 48 plants produced white fruits, 12 produced yellow fruits and 4 produced green fruits. These results can be explained by

A. dominant epistasis
B. genomic imprinting
C. incomplete dominance
D. recessive epistasis

Answer 23

C. incomplete dominance

• Dominant Epistasis:
 12:3:1 Phenotypic Ratio
Eg. Parental Phenotypes: White x Cinnamon
Genotypes: WW BB x ww bb
 100% White offspring (Ww Bb)

    Offspring F1 Phenotypes:  White x White
                         Genotypes:    Ww Bb x Ww Bb
                                          9 White              -> W_B_
                                          3 White              -> W_bb
                                          3 Agouti             -> wwB_
                                          1 Cinnamon       -> wwbb
                           Phenotypes: 12:3:1 
                                           12 White : 3 Agouti : 1 Cinnamon 
                                             Instead of normal 9:3:3:1 ratio. 

• Genomic imprinting:
 Affects limited no. genes (100 -> mouse)
 Many imprinted genes -> involved -> foetal growth
Paternally expressed genes -> promote growth
Maternally expressed genes -> suppress growth
***–»> Kinship / Parental conflict theory:
Conflict between sexual / reproductive interests -> maternal &
paternal genes in foetus.
»_space; Mother -> equally related to all offspring
>Wants to divide resources equally
»_space; Father -> likely related to subset of foetuses
>Wants to incr. survival chances of his offspring -> promoting
their growth.

Incomplete dominance is a form of intermediate inheritance in which one allele for a specific trait is not completely expressed over its paired allele. This results in a third phenotype in which the expressed physical trait is a combination of the phenotypes of both alleles.

• Recessive epistasis:
 9:4:3 phenotypic ratio
Eg. Parental Phenotypes: Cinnamon x Albino
Genotypes: bb CC x BB cc
 100% Agouti offspring (WT)

    Offspring F1 Phenotypes:  Agouti x Agouti
                         Genotypes:    Bb Cc x Bb Cc
                                          9 Agouti              -> B_C_
                                          3 Albino              -> B_cc
                                          3 Cinnamon        -> bbC_
                                          1 Albino               -> bbcc
                           Phenotypes: 9:4:3 
                                           9 Agouti : 4 Albino : 3 Cinnamon 
                                             Instead of normal 9:3:3:1 ratio.

Question 24

24. A dominant allele called creeper gives short, stunted legs in chickens. Two creeper birds were crossed and two-thirds of the progeny were creeper and onethird were normal. This result can be explained by

A. a lethal allele
B. co-dominance
C. dominant epistasis
D. incomplete penetrance

Answer 24

A. a lethal allele

•	Co-dominance:
	Heterozygotes illustrate phenotype -> both alleles
	Self tolerance
AA / AO / BB / BO 
	Universal recipients
AB
	Universal donors 
OO

•	Lethal alleles:
	Cause skewed phenotypic ratios 
	Dominant allele 
	Phenotypic ratio 2:1 instead of 3:1 -> heterozygotes – (AYA & AYA)
AYAY genotype -> not produced -> death 
 3 combinations produced:  AYA, AAY, AA
 2 phenotypes -> 2 x (AYA) & (AA)
>>3 produced -> 2:1
	Pleiotropy

• Dominant Epistasis:
 12:3:1 Phenotypic Ratio
Eg. Parental Phenotypes: White x Cinnamon
Genotypes: WW BB x ww bb
 100% White offspring (Ww Bb)

    Offspring F1 Phenotypes:  White x White
                         Genotypes:    Ww Bb x Ww Bb
                                          9 White              -> W_B_
                                          3 White              -> W_bb
                                          3 Agouti             -> wwB_
                                          1 Cinnamon       -> wwbb
                           Phenotypes: 12:3:1 
                                           12 White : 3 Agouti : 1 Cinnamon 
                                             Instead of normal 9:3:3:1 ratio. 

       Incomplete penetrance 
        Eg. Breast cancer susceptibility (BRCA genes)
       Variable expressivity 
        Eg. Agouti viable yellow
               --> Insertion -> transposable element -> promoter region of agouti gene
       Incomplete penetrance & variable expressivity

Question 25

25. Examine the RNA genetic code below. In a species of Paramecium the UGA stop codon codes for tryptophan; in some human genes the UGA stop codon codes for an amino acid – selenocysteine – that does not appear on the table below. Which of the following are true? A. Stop codons can be reassigned to code for amino acids B. The genetic code is not universal C. There are more than 20 naturally-occurring amino acids D. All of the above

Answer 25

A. Stop codons can be reassigned to code for amino acids - > Genetic code is universal - > There are 20 amino acids which can code for a protein, each of which is comprised fo a codon of nucleotide bases. - > Stop codons can be mutated & become a nucleotide sequence / codon which codes for an amino acid

Question 26

26. The human genome is around 3,000,000,000 base pairs in size. Many scientists think that only around 15% of our genome has a genetic function, with 85% of it being ‘junk’. On average, two humans differ from each other for about 0.1% of their base pairs. Assuming those differences are randomly distributed in the genome, if we compared two people, how many base pairs would we expect to differ in sequences that have a genetic function? A. 300,000 B. 450,000 C. 3,000,000 D. 4,500,000

Answer 26

B. 450,000 85% junk therefore 100-85 = 15% functional. 15% of 3,000,000,000 = 450,000,000 functional nucleotides. 0.1% difference between individuals -> 0.1% of 450,000,000 = 450,000 base pairs.

Question 27

Diagram showing a mammalian blastocyst embryo 28. If one wanted to establish cell lines of the extraembryonic lineages, which region would you want to isolate and culture? A. Region labelled A (Cells forming a line/layer in wall of uterus) B. Region labelled B (Cells forming circular barrier surrounding clump of cells) C. Region labelled C (Cells clustered together within a circle of other cells forming mem around them) D. Region labelled D (Cytoplasms / fluid filled cavity inside circular mem. of cells surrounding cell cluster) 29. Which region will give rise to the mammalian embryo proper? A. Region labelled A B. Region labelled B C. Region labelled C D Region labelled D

Answer 27

B. Region labelled B (Cells forming circular barrier surrounding clump of cells)?? Then ; C. Region labelled C (Cells clustered together within a circle of other cells forming mem around them) ?? Extraembryonic: Extraembryonic mesoderm, derived from the posterior of the primitive streak,is contrib- utes to the amnion, the mesoderm lining of the visceral yolk sac and to the allantois Outside the embryonic body; for example, the extraembryonic membranes involved with the embryo's protection and nutrition that are discarded at birth.

Question 28

30. Here are some possible explanations of the figure: 1. The selection pressure caused by malaria in some regions is weaker than in others 2. The sickle-cell mutation has not arisen in some areas. 3. Migration of individuals with the sickle-cell allele is not evenly distributed. 4. The regions with malaria but the lowest frequency of sickle-cell are unpopulated deserts. Which of these explanations may be true? A. 1 and 2 B. 1 and 3 C. 1, 2 and 3 D. All of them [ Map shown with shading on regions of map correlating to frequencies of sickle-cell allele. Darkest shade = highest frequency & lightest shade the lowest. - > Darkest shade around one particular coastal section of west Africa ; radiates out from this location with shade becoming lighter as distance increases. - > Dotted area illustrates distribution of malaria - present sporadically -> almost in all areas with prescence of sickle-cell allele & also in some regions where lowest freq. / no freq. of allele. ]

Answer 28

A. 1 and 2 ?? -> Cannot be sure sickle-cell hasn't arisen at all in some areas as shades on key are given in scale; where lightest shade indicates 0-2.5% frequency of allele So 2 likely untrue. -> Graph does not illustrate migration of individuals with sickle-cell allele; just frequencies. So 3 untrue. -> Regions with lowest freq. of allele may not have people living there & that reason; but then how did they measure presence of malaria in first place if these areas were completely unpopulated? So 4 untrue. -> Unsure and no option for just 1 so 1 & 2 must be possibly true. ??

Question 29

31. A European Homo sapiens living 40,000 years ago had as much Neanderthal DNA as you have DNA from one of your great-great-grandparents. What percentage is that? A. 12.5% B. 8.75% C. 6.25% D. 3.00%

Answer 29

C. 6.25% 50% parent -> 50% of which they obtained from one of their parents (your grandparent) -> 50% from their parent (your ggp) & 50% from gggp. --> 0.5 ^4 = 0.0625 = 6.25%

Question 30

33. These skeletal preparations show the phenotype of knocking out the paralogous Hox10 and Hox11 genes in the mouse forelimb and hindlimb. What conclusions can be made from these data? A. That Hox11 paralogous genes are required for the development of femur in the hindlimb B. That Hox11 paralogous genes are required for the development of the radius in the forelimb C. Answers A and B are both correct D. Neither answer A nor answer B is correct In controls, radius & femur are long, normal length. In mutated versions, they are barely even present, practically end of bone as soon at it starts st the joint of the limb. - > Removal of Hox11 gene -> arm - > Removal of Hox 10 gene -> leg

Answer 30

B. That Hox11 paralogous genes are required for the development of the radius in the forelimb - > Exp, knockout genes -> therefore mutations caused because of absence of these genes. - > Hox11 paralogous genes removed from radius & radius mutated after removal therefore must be required for dev of forelimb. - > Hox1o paralogous genes removed from femur & femur mutated after removal therefore must be required for dev of forelimb.

Question 31

34. What is the most likely mode of inheritance for the genetic trait indicated by filled symbols in pedigree above? A. Autosomal dominant B. Autosomal recessive C. X-linked dominant D. X-linked recessive - > Not present at all in F1 in either of families despite fact one of each of pairs of origin parents homozyogous for trait. - > No obvious pattern

Answer 31

B. Autosomal recessive - Autosomal dominant:  Phenotype present -> every generation  Equally affects both genders  Homozygous mutant -> sometimes lethal. - Autosomal recessive:  A genetic condition that appears only in individuals who have received two copies of an autosomal (non-sex chromosome) gene, one copy from each parent.  The parents are heterozygous carriers who have only one copy of the gene  Gene is recessive to its normal counterpart gene. - X-linked domiant:  Affected male parents >> All daughters affected >> No sons affected / carriers >> XdXd x XDY XDXd ; XdY ; XDXd ; XdY  Affected heterozygote females >> Half -> sons affected >> Half -> daughters affected. >> XDXd x XdY XDXd ; XDY ; XdXd ; XdY - X-linked recessive:  More males affected -> males with one mutant allele -> hemizygous.  Types of transmission: 1. Transmission -> Female heterozygous carrier  Half -> sons affected  Half -> daughters carriers. >> XAXa x XAY > XAXA ; XAY ; XAXa ; XaY 2. Transmission -> Hemizygous affected male  No sons affected / carriers  All daughters -> carriers  XAXA x XaY XAXa ; XAXY ; XAXa ; XAY 3. Transmission -> Affected female:  All sons -> affected  All daughters -> carriers.  XaXa x XAY XAXa ; XaY ; XAXa ; XaY - > Not present in every generation therefore not autosomal dominant - > No patterns therefore trends of X-linked dominan t& x-linked recessive cannot be applied. - > Therefore must be autosomal recessive.

Question 32

Two genes ‘A’ (with A and a alleles) and ‘B’ (with B and b alleles) are 20 centimorgans apart on the same chromosome. AA BB was crossed with aa bb, and the F1 progeny was crossed with aa bb. In the F2 progeny what percentage are expected to be Aa bb? A. 5% B. 10% C. 15% D. 20%

Answer 32

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Question 33

1. The closest living relation to the whales is: A. cow B. dog C. hippopotamus D. seal

Answer 33

C. hippopotamus • Evidence for Evolution?: 1. Fossils: i) Evolutionary change in trilobites & whales Trilobites: -> incr. no of ribs over 3 million yr periods. Whales: -> Basilosaurus; Hind limbs inherited from ancestors no longer advantageous as aquatic lifestyle adapted. > Closest relative = Hippo. (Give birth & nurse underwater ; multi-chambered stomach ; internal testicles.) ii) Tiktaalik Roseae - > Evidence for evolution of tetrapod limb from fins.  Discovered 2004 – Neil Shubin - Freshwater Sediments – Ellesmere Island.  Tetrapod trackways on Valentia island, Ireland.

Question 34

2. Euphorbs and cacti are an example of: A. climate evolution B. concerted evolution C. convergent evolution D. divergent evolution

Answer 34

C. convergent evolution - Convergent Evolution: Distantly related organisms independently evolve similar traits due to adapting to similar environments. Eg. Placentals -> Marsupials 1. Anteater -> Banded Anteater 2. Flying squirrel -> Sugar glider 3. Mole -> Marsupial mole 4. North American Porcupine -> Crested porcupine ``` - Eg. Succulents (Desert Plants):  Fleshy stems -> water storage  Small surface area of leaves -> reduce water loss  Spines -> deter herbivores  Cacti – South America  Euphorbs – Old world ```

Question 35

3. When did the eukaryotic cell appear first? A. Around 2 billion years ago as a result of a chance event B. Around 2 billion years ago as a result of natural selection C. Around 200 million years ago as a result of a chance event D. Around 200 million years ago as a result of natural selection

Answer 35

A. Around 2 billion years ago as a result of a chance event - Deep Sea -> 3.8 billion years  Proton gradients surrounding alkaline hydrothermal vents Eg. “Lost City” in mid-Atlantic -> possibly supplied energy to fuel first replication. Harbour geological manifestations of both kinds of energy used by life -> chemically reac-tive compounds & natural proton gradients. Remain active for up to 100,000 yrs -> provide constant source of energy over long geo-logical timescales. Proton gradients at Lost City -> same magnitude & orientation as those in modern auto-trophic cells. • Likely that cells existed before life itself.  RNA extremely reactive / unstable -> needs protection/barrier.  In order for chemical reactions to occur & creation of energy -> life; tiny molecules have to react/combine. -> Miniscule therefore requiring contained environment for reaction to oc-cur as molecules otherwise may not have met.  Tiny pores in rocks around hydrothermal vents  Also hypothesised -> lipid protobionts reproduced & metabolised. What were the first replicating molecules? • RNA world may have characterised first life  Basis for both reproduction & biochemical interaction -> acting as enzymes.  Possible combination of Amino acids -> found on comets -> by RNA -> creates proteins. Re-placed RNA enzymes -> wide range of biological functions eg. structural, enzymes, etc. ad-vantageous.  DNA evolved & replaced RNA. Double helix -> more stable for storage of info & identification of errors More resistant to effects of sunlight & cell membrane additional protection. Faster chemical reactions & DNA replication using protein enzymes. Formed LUCA 1. Earliest DNA life traces back to LUCA – Last Universal Common Ancestor. - Life approx. 3.8 billion years ago.  Requires cool temperatures, gravity, water & protection from radiation. 2. Evolution of First prokaryotes  Cyanobacteria -> produced O2  Methanogeneic bacteria consumed this O2 for hundreds of millions of years.  Methanogens extinct -> possibly changes in trace metals in sea. 3. Eukaryogenesis -> Evolution of eukaryotes -> 2 billion years ago.  Eukaryotes can be single-celled ; but to be multicellular -> must be eukaryote.  Likely caused by heterotrophic eubacterium being engulfed by archaebacterium.  This eventually became mitochondria.  Mitochondria enable large complex life due to energy they provide.  Bacterium engulfed by eukaryotic plant ancestor. >> Eukaryogenesis not a product of natural selection -> chance event.  Natural selection acted on chance individuals/populations once event occurred. 4. Ediacaran biota -> 570 million years -> Found in Australia  Spriggina -> Annelid worm/arthropod? – 3cm long  Dikinsonia -> Up to 1m in length  Rangeomorphs -> Fractally branched, fern-like. -> Dark depths of ocean. -> Ecology suggests they were animals.  Erniettmorphs -> modular or quilted. 5. The Cambrian Explosion -> 515-540 million years ago -> Burgess Shale.  Most types of common animal today appeared over 20-30 million years. Eg. Arthropods, chordates, worms. Soft bodies preserved in Burgess Shale (Canada), China, Greenland & Russia  Some can’t be categorised into modern taxonomy Eg. Opabinia -> 7cm long.

Question 36

4. Which of the following factors may be involved in speciation? A. Genetic drift B. Isolation C. Reinforcing selection D. All of the above

Answer 36

D. All of the above • What is speciation? - The evolutionary process in which new, distinct species evolve from an existing population due to reproductive isolation. - Population divides -> populations become isolated from one another -> selection pressures, adaptations, changes in characteristics, mutations -> do not recognise when reintroduced.  Population becomes reproductively isolated following start of speciation due to  Selective pressures  Genetic Drift  Genetic mutations Of each individual population. - Consequential event of specific conditions, involving by-products of subsequent selection. - Occurs over long time periods -> selection not involved. • Genetic Factors in Speciation: - Genetic Drift & Bottlenecks -> long periods of isolation - Natural selection

Question 37

5. Evolution can be defined as: A. a change in allele frequencies in a population B. a change in fitness in a species C. a change in the anatomy, physiology or behaviour of a species D. the appearance of a mutation in an individual

Answer 37

A. a change in allele frequencies in a population • What is Evolution?: The gradual change in allele frequency of a population over time. • Conditions for evolution by natural selection?: 1. Range of characteristics 2. Inherited by offspring 3. Advantageous characteristics enable survival & reproduction.

Question 38

6. Roughly how many human protein-encoding genes are there, and what rough percentage of the human genome do they represent? A. 20,000, 3% B. 30,000, 5% C. 30,000, 10% D. 40,000, 5%

Answer 38

A. 20,000, 3%

Question 39

7. Genetic analysis of a large population of mink inhabiting an island in Michigan in the US revealed an unusually large number of gene loci where one allele was fixed (i.e. allele frequency = 1). Which of the following is the most probable explanation for this genetic homogeneity? A. The population exhibited non-random mating, producing homozygous genotypes B. The gene pool of this population never experienced mutation C. A very small number of mink may have colonised the island, and the founder effect and subsequent genetic drift could have fixed many alleles D. Natural selection has selected for and fixed the best adapted alleles at these loci

Answer 39

C. A very small number of mink may have colonised the island, and the founder effect and subsequent genetic drift could have fixed many alleles ??

Question 40

8. Which event occurs in meiosis I but NOT in meiosis II? A. Alignment of paired homologous chromosomes at the metaphase plate B. Attachment of spindle fibres to the kinetochore C. Chromosome condensation D. Separation of sister chromatids at anaphase condensation

Answer 40

A. Alignment of paired homologous chromosomes at the metaphase plate - Meiosis: Produce -> 4 haploid gametes -> from diploid mother cell 2 cell divisions Occurs -> gonads (ovaries/testes) only. ``` • Meiosis: - G2 phase -> S phase cell cycle - Meiosis I:  Divides pairs of chromosomes  Reductional division (chromosome no. halved)  2 haploid (n) daughter cells - Meiosis II:  Divides sister chromatids  Equational division  4 haploid gametes ```  Meiosis I: >> 2n = 6. >>3 bivalents -> each -> 4 chromatids. Prophase I: Chromosomes condense Homologous chromosomes -> synapsis -> (pairing) Crossing-over -> genetic material -> non-sister chromatids -> each bivalent Metaphase I: Homologous chromosome pairs -> line up -> equator of cell. Anaphase I: Homologous chromosomes -> separate -> opposite poles of cell -->sister chromatids -> still attached Telophase I: One of original pair -> homologous chromosomes -> each pole of cell Cytokinesis I: Cells divide -> 2 haploid daughter cells ->Each daughter cell -> chromosome -> 2 chromatids  Meiosis II: Prophase II: Chromosomes -> attach -> spindle Metaphase II: Individual chromosomes -> line up -> equator Anaphase II: Sister chromatids -> separate -> opposite poles of cell. Telophase II: Each haploid daughter cell -> 1 each type chromosome.

Question 41

9. In an organism with a diploid chromosome number (2n) of 3, how many genetically-different gametes can be produced as a result of independent assortment only? A. 2 B. 4 c. 8 D. 16

Answer 41

c. 8 n= 3 so 2^3 = 2 x2 x2 = 8

Question 42

10. Alfred Knudson’s ‘two hit’ hypothesis can explain: A. incomplete penetrance in familial cancer syndromes B. maternal imprinting of the IGF2 gene C. variable expressivity as the result of insertion of a transposable element D. why women with a BRCA1 mutation are at increased risk for breast and ovarian cancer but not for colon cancer

Answer 42

A. incomplete penetrance in familial cancer syndromes >> 2 hit hypothesis -> familial cancer syndromes -> tumour suppressor genes >Both alleles of tumour suppressor gene -> must be inactivated for tumour formation (loss -> heterozygosity) Eg. >> Neurofibromatosis Type 1 (peripheral nervous system) - Dominant familial cancer syndrome - 1/3500 - Benign neurofibromas under skin & ‘Café au lait’ spotting -> skin - Very varied severity Pearson Twins -> Facial tissue cells -> Adam -> homozygous -> NF1 mutation -> Neil -> heterozygous Loss of heterozygosity -> Adam -> during foetal dev. -> Cell lineage forming face. CT scans -> tumors -> Neils abdomen -> asymptomatic. Incomplete penetrance: adjective Referring to the presence of a gene that is not phenotypically expressed in all members of a family with the gene. A form of penetrance in which not all individuals carrying a deleterious gene express the associated trait or condition. A disease is said to show incomplete penetrance when some individuals express the associated trait while others do not even though they carry the disease-causing gene.

Question 43

11. The SRY gene controls sexual development by: A. activating genes required for testis formation B. encoding the receptor for testosterone C. repressing genes required for ovary formation D. repressing synthesis of testosterone in females

Answer 43

A. activating genes required for testis formation ' • Only one gene of Y chromosome required to determine males  SRY (Sex-determining Region on the Y) • Sex reversal:  Translocation of SRY -> X chromosome >>Found -> Rare XX males • Mutation of SRY gene:  XY females • Confer of gender in SRY gene: - DNA binding protein (transcription factor)  Regulate expression -> genes -> testis formation - Week 4: Genital ridge (Somatic cells) - Week 6: Indifferent gonad (Germ & somatic cells) - Week 8: SRY expression -> Testis formation >Females -> no SRY expression -> Ovary formation.

Question 44

12. A human female is found to have two Barr bodies in each of her skin cells. What is her most likely karyotype? A. 45, X B. 46, XX C. 47, XXX D. 48, XXXX

Answer 44

C. 47, XXX - > 46 chromosomes normally - > she has 2 barr bodies -> indicates 2 inactivated X chromosomes. - >In females (XX) they normally have 1 X chromosome inactivated. Therefore just on extra inactivated. & One extra chromosome. • Individuals -> multiple X chromsomes Eg. XX females, XXY males, XXX females >>Each cell -> expresses only 1 X chromosome Other X chromosomes -> Barr bodies

Question 45

13. Which is the most frequent type of polymorphism in the human genome? A. Non-synonymous base substitutions B. Restriction fragment length polymorphisms (RFLPs) C. Short tandem repeats (STRs) D. Single nucleotide polymorphisms (SNPs)

Answer 45

D. Single nucleotide polymorphisms (SNPs) Polymorphisms -> structures which occur in several different forms. A nonsynonymous substitution is a nucleotide mutation that alters the amino acid sequence of a protein. restriction fragment length polymorphism (RFLP) is a technique that exploits variations in homologous DNA sequences, known as polymorphisms, in order to distinguish individuals, populations, or species or to pinpoint the locations of genes within a sequence.  Single Nucleotide Polymorphisms (SNPs) Non-coding DNA Most common type found -> Human genome Abundant & easy to identify -> gene chip technology 1 Nucleotide difference per 1000 nucleotides -> Any 2 individuals. HapMap Project: >> Gene chip technology -> identification -> common SNPs -> across ethnic groups. SNP genotyping -> GeneChip >> Each individual -> pedigree -> 0.5-1 mill SNPs genotyped -> Each SNP tested -> linakge w/ disease phenotype -> using genetic model. -> Identifies several SNPs in region -> genome where disease must be located. • Haplotype:  Particular combination of SNPs -> small chromosomal region. - 6,000 bp -> 20 of which are SNPs. Each individual -> one of 4 possible haplotypes -> 1 haplotype -> same differences in nucleotide seq. to other haplotypes. Therefore identification of Haplotype can be done using tag SNPs rather than assessing all differences in SNP bases. >>For example -> at particular nucleotide base -> A/G tag -> if A nucleotide -> Haplotype 1/2/4/ At 2nd tag -> T/C -> -> if C nucleotide -> Haplotype 2/4 At 3rd tag -> C/G -> -> if C nucleotide -> Haplotype 4 Therefore identifying haplotype. >>Genotyping 3 tags of 20 sufficient to identify haplotype. - The closer a mutation occurs to a particular SNP on the chromosome, the higher the linkage illustrated by that SNP to the disease in future generations / the higher the no. of future generations with SNPs associated to the disease. - The further away an SNP is located from the original mutation on the chromosome, less likely association will still be present as the no. of future generations/pedigrees incr.  Short Tandem Repeats (STRs) Microsatellite repeats Tandem repeats -> Short, non coding sequences (2-4 nucleotides) Eg. GAGAGA / TTATTATTATTA Longer repeats (>10 nucleotides) -> minisatellite repeats.  Forensic Analysis -> STRs Polymerase Chain Reaction (PCR) >> Amplifies 10 STRs & Gender DNA marker >Separated -> electrophoresis -> DNA profile. Forensic uses -> DNA profiles: >> Each individual excl. identical twins -> unique DNA profile >> Individuals placed at scene -> analysis -> hair, blood, saliva / semen samples. >> DNA profiles of crime -> compared -> suspect / database profiles >> Close matches -> indicate close relatives.

Question 46

14. Which is the most frequent cause of trisomy 21 (Down Syndrome)? A. Non-disjunction in the early cleavage divisions of the embryo B. Non-disjunction in the first meiotic division in the mother C. Non-disjunction in the second meiotic division in the mother D. Translocation between chromosomes 14 and 21

Answer 46

B. non-disjunction occurring during the first meiotic division in the mother Trisomy 21 is caused by a meiotic nondisjunction event. Gamete produced with an extra copy of chromosome 21 (24 chromosomes). When combined with a typical gamete from the other parent, the child now has 47 chromosomes, with three copies of chromosome 21. Trisomy 21 is the cause of approximately 95% of observed Down syndrome, with 88% coming from nondisjunction in the maternal gamete and 8% coming from nondisjunction in the paternal gamete. Mitotic nondisjunction after conception would lead to mosaicism, and is discussed later. Down syndrome, a trisomy of chromosome 21, is the most common anomaly of chromosome number in humans.[2] The majority of cases results from nondisjunction during maternal meiosis I. • Maternal Age & Trisomy  Incr. trisomy  Responsible for incr. miscarriage -> older women.  95% -> trisomy 21  Maternal non-disjunction -> meiosis 1  Human oocytes  Paused -> late meiotic prophase I (diplotene) >> Alongside paired, replicated chromosomes  Begins before birth >> Maintained -> decades -> Until egg matures -> menstrual cycle  Loss of cohesion -> Prophase I  Aneuploidy -> older women >> Premature loss -> cohesion ->> 2 univalents -> segregate independently > Aneuploid (n +1) gametes -> Robertsian Translocation can be due to inheritance or other reasons, however most commonly familial. Translocation Down syndrome is often referred to as familial Down syndrome. It is the cause of about 4.5% of the observed Down syndromes.[4] It does not show the maternal age effect, and is just as likely to have come from fathers as mothers. Mosaic Down syndrome -> some cells are normal & some have trisomy 21, an arrangement called a mosaic. This can occur in one of two ways: A nondisjunction event during an early cell division leads to a fraction of the cells with trisomy 21; An anaphase lag of a chromosome 21 in a Down syndrome embryo leads to a fraction of euploid cells (2n cellsThis is the cause of 1–2% of the observed Down syndromes. ```  Down’s Syndrome:  Charcteristics: - Characteristic facial features - Short stature - Learning disabilities - Higher risk -> Heart defects - Alzheimers & some cancers -> 1/1000 ``` ```  Causes:  Trisomy (2n +1): - Trisomy 21  Robertsian Translocation: - Chromosome 21 & 14  Genetic mosaicism: - Mix of normal & trisomy 21 cells >>Non-disjunction -> early embryonic mitotic divisions ```

Question 47

15. Which of the following is an example of descriptive embryology? A. A catalogue of where and when genes are expressed during undisturbed development B. An assessment of how development changes when a gene is mutated C. An assessment of the effect of transplanting cells or tissues from one place in the embryo to another D. The construction of a specification map at a defined time of development

Answer 47

• Descriptive Embryology:  Doesn’t involve disturbing development  Examination of undisturbed groups of embryological cells Descriptive Embryology: - Observation of the natural development of an embryo  Internal & external observations - Understand mechanisms of growth -> vary with stages & species - Describe different stages of growth by naming them - Use colours to describe types of tissues observing  Ectoderm -> blue -> (outside layer of tissues)  Mesoderm -> red  Endoderm -> yellow -> (inner linings / linings of organs) - Fate maps  Observe natural development over time  Describe development of tissues / cells Eg. How cells & tissues develop / what they develop into. - Describing when (what stage) & where (what part of embryo) genes expressed - Gene expression dynamic -> can be turned on & off

Question 48

16. What are the major advantages of Drosophila melanogaster for the study of development? A. It is a vertebrate, with external development and strong genetics B. A short generation time, strong genetics, imaging, sequenced genome C. It is a mammal with strong genetics D. It is an amphibian, which produces large number of eggs

Answer 48

B. A short generation time, strong genetics, imaging, sequenced genome • Drosphila melanogaster (Fruit Fly) - Short generation time - Genetics - Sequenced genome - Transgenesis - Imaging - Development of segments -> Segmented organisms

Question 49

17. What conclusions could be drawn from the Hilde Mangold and Hans Spemann organizer experiments published in 1924? A. Dorsal mesoderm from the early gastrula can induce ventral ectoderm to become neural tissue and ventral mesoderm to become muscle (somites) B. Dorsal mesoderm is already committed to become notochord by the early gastrula stage C. Ventral ectoderm is competent at the early gastrula stage to become neural tissue D. All of the above

Answer 49

- Evidence of inheritance of cell fate determinants:  Inheritance from parents  Mosaic development > Experiments in tunicates -> (Sea squirts) Mosaic development in embryos >>Egg -> small yellow tinted area of cytoplasm -> inherited by some daughter cells but not all. >>Cells which inherited this developed -> muscles  Yellow cytoplasm became red cell type -> muscle (Purely observation so far) (Then experimental) Isolated cells from env. (rest of embryo) Only cells isolated containing yellow cytoplasm become muscle after isolation >> Only inheritance determines fate - Confirming presence of cytoplasmic determinants: Disrupt cells which inherit cytoplasm & observe development >Inject cytoplasm into cells -> wouldn’t normally inherit it ->Before joined cells divide. ->See if they become muscles –>> they did. - Other evidence for cytoplasmic determinants: > Fruit fly embryo -> Different RNAs present at different poles of embryo -> fate determinants. >> Bicoid mRNA -> Anterior >> Nanos mRNA -> Posterior Injection -> Nanos mRNA into anterior end of cells -> do not normally contain mRNA. >> Mutation in protein (normally asymmetrically present) -> No heads  Inject mRNA in diff place in organism with no anterior mRNA (E.g in middle) > Head structures grow in middle of body >Frog embryo fate map:  Label cell -> blastula stage -> enable development -> C1 cell forms part of gut & notochord Must isolate them from neighbours to see if they know what to become without external influence of neighbours > Specification map: Differences to fate map > At some point later in dev -> needed influence from neighbours to stimulate fate >> Some things similar >> Extra additional parts not present in isolation eg. nervous system, blood & kidney etc. -> Further specification / development of tissues not present. Evidence of regulative development -> Removal of embryo sections > Development of whole embryo still possible from remaining parts of Cell Requires cell-cell determination - Cytoplasmic determinants in early embro:  Notochord doesn’t change between specification & fate maps of tissues  Grey crescent region (pigmented area in embryo) Is regulative development possible in vertebrate embryo: -> Will cell develop lost cell sections destroyed? >> 2 staged frog embryo -> Hot needle used -> kill 1 of 2 cells before 1st cell division -> 2 halves -> Injection of red dye into one of cells -> Dye labels half of embryo so fate map at this stage is visible -> Indicates fate map fits with spec map (As removed one neighbour from one cell) However -> got wrong answer -> Dead half cell not removed & in way Baby hair ligature exp -> Very fine so useful -> Flexible -> can alter embryos using it -> Baby hair used -> separate 2 stage embryo -> 2 cell separated -> Results in dev of 2 normal, fully developed embryo -> Indicates full embryo dev on both sides of cell (Fated to only dev into one half of cell) >> Reforms what was missing >> Regulative dev. - Grey Crescent Area:  If grey crescent not inherited by both halves -> Incomplete development of cell which doesn’t inherit -> only belly piece -> not full embryo >> Indicates grey crescent important in generation of full embryo.  Is grey cresecent area communicating info to env -> Eventually becomes dorsal mesoderm -> Develops into notochord.  If transplant into opposite side -> ventral of other cell -> Siamese twin embryos develop -> 2 nervous systems & 2 embryos in one -> therefore organiser -> organises embryo around it  Is Notochord organising tissue around it or is it actually embryo?  Looked at diff newt species -> some pigmented -> If transplant non-pigmented cytoplasm / cells of side from area with notochord fate into other embryo >> Notochord comes from one embryo >> Rest of tissue ie. organs comes from other embryo  Indicates induction of muscle (somites) & neural tissues by notochord  Dorsal mesoderm determined by early gastrula stage  Ventral ectoderm & mesoderm are competent to become neural & somatic tissue Eg. Notochord transplant -> surrounding tissues (Mesoderm normally blood but responds to notochord to become muscle & skin became spinal chord instead.  Can result in double-headed embryos eventually (deicephaly -> can occur naturally)

Question 50

18. Examples of stem cell therapies in patients include: A. limbal stem cell transplants after corneal injury B. bone marrow transplants C. both A and B D. neither A nor B

Answer 50

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Question 51

19. Embryonic stem cells are derived from: A. The inner cell mass (ICM) of a peri-implantation stage mammalian embryo B. The trophectoderm (TE) of a peri-implantation stage mammalian embryo C. The germ cells (GC) of a peri-implantation stage mammalian embryo D. The neurectoderm (NE) of a peri-implantation stage mammalian embryo

Answer 51

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Question 52

20. Why are ES cells and iPS cells so valuable for the biomedical sciences? A. They can be used to study embryonic development in vitro B. They have the potential of being used for the treatment of genetic diseases in human patients C. They can be used to model genetic diseases in vitro D. All of the above

Answer 52

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Question 53

23. The frequencies of two alleles at the A gene locus are as follows: A1 = 0.2, A2 = 0.8. Assuming the A gene locus is at Hardy-Weinberg equilibrium, what is the expected frequency of A1 A2 heterozygotes in the population? A. 0.16 B. 0.32 C. 0.6 D. 0.89

Answer 53

B. 0.32 (p^2 = A1A1 = (0.2)^2 = 0.04 q^2 = A2A2 = (0.8)^2 = 0.64) Therefore A1A2 = 2pq = 2(0.2)(0.8) = 0.32

Question 54

24. The graph represents the changes in phenotypic frequency over a five year period for a trait under selection. Based on the graph what type of selection is most likely to have been experienced by the population? A. Balancing selection B. Directional selection C. Frequency-dependent selection ' D. Retrograde selection Shows graph with birth weight on x axis & frequency on y-axis. Bell shaped curves for both 2000 & 2005. Same widths & max peaks. However mean of 2005 (x-axis value for peak point of curve has smaller value for birth weight (slightly further to left)

Answer 54

B. Directional selection - > Shape of curve stays the same but just shift s lightly further along x-axis - > hence only variant is the birth weight - >illustrates the mean birth weight has only changed over time therefore shift in mean birth weight- > directional. >>Frequency dependent selection Allele selected against -> high frequency Allele selected for -> low frequency Eg. Scale-eating fish >>Left & right mouthed predators Left -> attack back right flank Right -> attack back left flank ->>If high no. right mouthed -- Prey expecting to be attacked -> left flank -- Less likely to predict attack -> right flank -- Left mouthed have advantage -> incr. no. of left-mouthed individuals. ->>Low no. right mouthed --High no left mouthed -> expect attack on right flank -- Incr. success of right mouthed -> pop. Incr. > Retrograde selection -> backward selection (regressive)?

Question 55

25. This diagram depicts the result of isolating different regions of the blastula embryo and assessing their developmental potential in isolation, as well as when conjugated together. What conclusions can be made from these experiments? A. That the animal region of the blastula embryo is specified to become ectoderm. B. That endoderm can be induced by conjugating the animal region with the vegetal region of blastula stage embryos. C. The mesoderm is committed by the blastula stage. D. All of the above.

Answer 55

D. All of the above. ?? • Determined: - Development of a cell or tissue according to fate, even when transplanted into another site in the embryo / new environment. Eg. Blastophore fated -> Notochord still develops into notochord when transplanted to ventral side of embryo. • Competence: - Range of cell fates acquired by a cell or group of cells, given appropriate conditions. Cell / tissue may be competent to develop into many cell types it would not normally be fated to form. Eg. Cap cells -> Blastula stage of frog embryos  Specified & fated -> Ectodermal tissues only  Competent -> Formation of virtually any cell type in embryo, given appropriate signals. • Induction: - Process in which a cell / group of cells emit signals Influences neighbouring cells to change fate Eg. Induction -> Neural ectoderm by dorsal mesoderm -> Gastrula stage (Organiser experiment)

Question 56

26. In the ascidian embryo, the fate map at the 8-cell stage corresponds to the specification map at the 8-cell stage. Isolation and culturing the blastomeres at the 8-cell stage is an example of: A. regulative development B. experimental embryology C. descriptive embryology D. induction When blastula components isolated on formation: -> Part of blastula labelled animal becomes ectoderm -> Part labelled vegetal becomes endoderm. When blastula components enabled to stay together after formation: -> ectoderm of animal part of blastula & endoderm of vegetal part of blastula can bind to from triploblastic molecule. (Ectoderm, mesoderm & endoderm)

Answer 56

B. experimental embryology Regulative development -> Removal of embryo sections > Development of whole embryo still possible from remaining parts of Cell Requires cell-cell determination

Question 57

What do the two lines, labelled 1 and 2 represent? A. 1. Chance event; 2: natural selection B. 1. Eukaryogenesis; 2: origin of chloroplast C. 1: Origin of plants; 2: fusion between Eukaryotes and Eubacteria D. 1. Origin of plants; 2: origin of photosynthesis - > Illustrates root of tree of evolutionary life. 2 branches branching off to eubacteria & archaebacteria & many sub-branches etc. - > Branches (root -> bottom of page; branches upwards in V shape) - > Line 2 horizontally Connects node on Eukaryote branch (towards end) to end node in eubacteria - > Line 1 connects node near start of archabacterial lineages to node in start/middle of eubacteria lineage & very start of eukaryota lineage.

Answer 57

B. 1. Eukaryogenesis; 2: origin of chloroplast

Question 58

28. Which of the following features might have caused the Cambrian Explosion? 1. Changes in marine trace elements 2. Changes to animal genomes 3. Ecological changes produced by animals 4. Increase in habitable areas produced by continental drift A. 1, 2 and 4 B. 1, 3 and 4 C. 2 and 4 D. 1, 2, 3 and 4

Answer 58

Why did Cambrian Explosion/Life occur? 5. The Cambrian Explosion -> 515-540 million years ago -> Burgess Shale.  Most types of common animal today appeared over 20-30 million years. Eg. Arthropods, chordates, worms. Soft bodies preserved in Burgess Shale (Canada), China, Greenland & Russia  Some can’t be categorised into modern taxonomy Eg. Opabinia -> 7cm long. Theories: - Physiological change -> dissolved O2 levels enabled active lifestyle - Geographical change -> Formation of new seas -> new niches - Geochemical change -> Changing Sea levels led to abundance of trace metals -> used to make exoskeletons. - Biological change -> Incr. zooplankton enabled enabled evolution of new predators -> further incr. selection pressures. - Other geological & biological factors -> Eruptions caused by comet impacts weakened species  Dinosaurs & large reptiles died  Ecological niches of extinct species (pterosaurs, mesosaurs) -> left space for evolution of birds & mammals. (--NB. -> Mammals present for majority of Jurrasic era.) -> Recovery of ecosystem from mass extinction -> 20 million years. • End of Permian -> mass extinction. (250 million years ago) - 95% sea-life extinct -> All trilobites & placoderms (armoured fish) , 99% plankton genera. - 70% all families of land animal extinct - Multiple factors Eg. Volcanism, tectonic plates, asteroids may have caused rapid widespread climate change. - Diapsids (early dinosaurs) & cartilaginous fishes populations able to grow. - Theraspids (our ancestors) just survived.

Question 59

29. What does a Manhattan plot illustrate in genetics? A. The distribution of threshold traits in a population B. The heritability of a quantitative trait C. The probability of association between single-nucleotide polymorphisms (SNPs) and a disease trait D. The variance of a trait in a population

Answer 59

C. The probability of association between single-nucleotide polymorphisms (SNPs) and a disease trait A Manhattan plot is a type of scatter plot, usually used to display data with a large number of data-points with a distribution of higher-magnitude values. commonly used in genome-wide association studies (GWAS) to display significant SNPs .

Question 60

30. The gene encoding insulin-like growth factor 2 (Igf2) is maternally imprinted in the mouse. Loss of function of the Igf2 gene gives rise to a dwarf mouse. A male and female both heterozygous for a loss of function, recessive mutation in the Igf2 gene were crossed. What is the expected phenotypic ratio in the offspring? A. All dwarf B. All normal C. 3 normal : 1 dwarf D. 2 normal ; 2 dwarf

Answer 60

B. All normal - > Maternally imprinted so father's alleles expressed in embryo - > heterozygous -> one mutant allele & 1 normal allele. - > must inherit normal allele from mother. Therefore 50% homozygous for non-mutated allele & 50% heterozygous. - > mutation recessive so a;; normal. • Genomic imprinting: - Paternal:  Paternal allele imprinted & silenced ->***By epigenetic tags  Maternal allele preferentially expressed -> embryo - Maternal:  Maternal allele imprinted & silenced ->***By epigenetic tags  Paternal allele preferentially expressed -> embryo ``` Eg. lgf2 gene:  Insulin-growth-like-factor 2 >>Required -> normal growth >>Only paternal copy of gene expressed >>Maternal copy of gene silenced & imprinted ``` Both mice heterozygous for recessive lgf2 mutant allele >Mouse -> Mutant allele inherited -> mother -> Normal size Maternal genomic imprinting >Mouse -> Mutant allele inherited -> father -> Dwarf size Paternal genomic imprinting >>Parent of origin effect • Parent of origin effect: -->passed to daughter cells.  When the phenotypic effect of an allele depends on whether it is inherited from the mother or father. • Genomic imprinting:  Affects limited no. genes (100 -> mouse)  Many imprinted genes -> involved -> foetal growth Paternally expressed genes -> promote growth Maternally expressed genes -> suppress growth ***-->>> Kinship / Parental conflict theory: Conflict between sexual / reproductive interests -> maternal & paternal genes in foetus. >> Mother -> equally related to all offspring >Wants to divide resources equally >> Father -> likely related to subset of foetuses >Wants to incr. survival chances of his offspring -> promoting their growth.

Question 61

. What is the most likely mode of inheritance for the genetic trait indicated by filled symbols in pedigree above? A. Autosomal dominant B. Maternal inheritance C. X-linked dominant D. X-linked recessive - > No offspring inherit trait passed on from father. - > All offspring inherit trait if passed on from mother.

Answer 61

- > Not autosomal recessive as parents aren't both heterozygotes. - > Unlikely autosomal dominant as unlikely that homozygosity results every time when 50% chance offspring could inherit normal allele from possible heterozygote parent. (Improbable as all offspring inheriting mutant allele.) - > No conditions for X-recessive suit. - > Must be maternal by process of elimination. - ->> Plus, only inherited if mother passes on mutant allele (& is homozygous for mutation) Autosomal dominant:  Phenotype present -> every generation  Equally affects both genders  Homozygous mutant -> sometimes lethal. - X-linked recessive:  More males affected -> males with one mutant allele -> hemizygous.  Types of transmission: 1. Transmission -> Female heterozygous carrier  Half -> sons affected  Half -> daughters carriers. >> XAXa x XAY > XAXA ; XAY ; XAXa ; XaY 2. Transmission -> Hemizygous affected male  No sons affected / carriers  All daughters -> carriers  XAXA x XaY XAXa ; XAXY ; XAXa ; XAY 3. Transmission -> Affected female:  All sons -> affected  All daughters -> carriers.  XaXa x XAY XAXa ; XaY ; XAXa ; XaY - Autosomal recessive:  A genetic condition that appears only in individuals who have received two copies of an autosomal (non-sex chromosome) gene, one copy from each parent.  The parents are heterozygous carriers who have only one copy of the gene  Gene is recessive to its normal counterpart gene.

Question 62

32. These skeletal preparations show the phenotype of knocking out the Hox10 and Hox11 genes in the mouse forelimb and hindlimb. What conclusions can be made from these data? A. That Hox11 genes are required for the development of the radius in the forelimb B. That Hox11 genes are required for the development of the femur in the hindlimb C. Neither A nor B D. Both A and B In controls, radius & femur are long, normal length. In mutated versions, they are barely even present, practically end of bone as soon at it starts st the joint of the limb. - > Removal of Hox11 gene -> arm - > Removal of Hox 10 gene -> leg 33. Mouse Hoxa11 and Hoxd11 are: A. duplicated genes following an ancestral whole genome duplication event B. orthologous genes C. paralogous genes D. both A and C

Answer 62

A. That Hox11 genes are required for the development of the radius in the forelimb D. both A and C - > Exp, knockout genes -> therefore mutations caused because of absence of these genes. - > Hox11 paralogous genes removed from radius & radius mutated after removal therefore must be required for dev of forelimb. - > Hox1o paralogous genes removed from femur & femur mutated after removal therefore must be required for dev of forelimb. ```  Paralogous genes:  Duplicated genes within single chromosome  Orthologous genes:  Same gene present -> diff. organisms  Whole genome duplication: - 2 rounds: (2R Hypothesis) >> Duplication >> Gene Loss >> 2nd Round Duplication >> Gene Loss  Results in Hox Genes ```  Allotetraploidy >> Hybridisation between 2 separate species  Autotetraploidy >> Duplication of genome through improper meiosis

Question 63

34. Chondrodysplasia punctata is caused by an X-linked dominant mutation in heterozygous females but is lethal in hemizygous males (i.e. no hemizygous male births). What is the expected phenotypic ratio in the children of an affected heterozygous female and an unaffected male? A. 1 affected female : 1 normal female : 1 normal male B. 1 affected female : 1 normal female : 1 normal male : 1 affected male C. 1 affected female : 1 normal male D. 1 normal female : 1 affected male

Answer 63

A. 1 affected female : 1 normal female : 1 normal male --> X-linked dominant therefore only require 1 allele to be effected. However therefore lethal in males which are affected. --> X*X x XY -> X*X ; X*Y ; XX ; XY So live offspring => X*X ; XX ; XY Therefore 1 affected female : 1 normal female ; 1 normal male X-linked domiant:  Affected male parents >> All daughters affected >> No sons affected / carriers >> XdXd x XDY XDXd ; XdY ; XDXd ; XdY ``` • Lethal alleles:  Cause skewed phenotypic ratios  Dominant allele  Phenotypic ratio 2:1 instead of 3:1 -> heterozygotes – (AYA & AYA) AYAY genotype -> not produced -> death  3 combinations produced: AYA, AAY, AA  2 phenotypes -> 2 x (AYA) & (AA) >>3 produced -> 2:1  Pleiotropy ```

Question 64

The table below shows two-point map distances between the genes A, B, C and D on the same chromosome. What is the order of the genes on the chromosome? ``` A - B 15 A - C 7 A - D 13 B - C 10 B - D 3 C – D 6 ``` A. CDAB B. DABC C. BCAD D. BDCA

Answer 64

``` Closest -> furthest B - D = 3 C - D = 6 A - C = 7 B - C = 10 A - D = 13 A - B = 15 ``` Start with highest: A-B = 15 These must be on outside A - B Then closest -> B-D So put D inside A & B but closer to B. A -- D-B Then C is in middle ground of both so just slot in A-C-D-B ->> Will see not exact order but that is okay as remember multiple crossovers can make inaccurate.

Question 65

The table below shows two-point map distances between the genes A, B, C and D on the same chromosome. What is the order of the genes on the chromosome? ``` A - B 15 A - C 7 A - D 13 B - C 10 B - D 3 C – D 6 ``` A. CDAB B. DABC C. BCAD D. BDCA

Answer 65

``` Closest -> furthest B - D = 3 C - D = 6 A - C = 7 B - C = 10 A - D = 13 A - B = 15 ``` Start with highest: A-B = 15 These must be on outside A - B Then closest -> B-D So put D inside A & B but closer to B. A -- D-B Then C is in middle ground of both so just slot in A-C-D-B ->> Will see not exact lengths fit order but that is okay as remember multiple crossovers can make inaccurate.