Past Papers Flashcards
- From which stage of the cell cycle do cells begin mitosis and meiosis?
A. G0
B. G1
C. G2
D. S
D. S
- 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
-> 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.
- 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
C. independent assortment ?
- The AB blood type in humans is an example of
A. co-dominance
B. incomplete dominance
C. over-dominance
D. pre-dominance
A. co-dominance
- 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
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.
- 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
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
- 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
? 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
- 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
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
- 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
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
- 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
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.
- The molecular clock can be based on molecular differences in
A. genes
B. genomes
C. proteins
D. all of the above
D. all of the above
- 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.
- 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
??
- > 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.
- 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) - Paraphatic speciation:
Somewhat but not complete geographical isolation -> reproductive barriers when restricted gene flow.
- 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
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.
- 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
??
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.
- All animal eyes are based on molecules called
A. opsins
B. opsitins
C. opticins
D. optins
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.
- Human HOX11A and mouse Hox11A are examples of
A. cytoskeletal genes
B. orthologous genes
C. paralogous genes D. ribosomal genes
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
- 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
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
- 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
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).
- 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
?
- 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
?
- 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)
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?
- 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
?
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
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.
- 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
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
- 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
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
- 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
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.
