Genes, Evo & Dev Blackboard Exam qs

Question 1

The conditions necessary for evolution by natural selection to occur are:

Answer 1

inherited characters that vary and differential survival of individuals

Question 2

The bill shapes of the finches on the Galapagos islands have been subject to natural selection. The best explanation of this process is:

Answer 2

. the specific ecological conditions found on the islands

NOT
changing climatic conditions

Question 3

Molecular phylogenies:

Answer 3

all support the existence of evolution

Question 4

Fitness is:

Answer 4

measured by the number of offspring an individual produces

Question 5

Patterns of lactose intolerance in humans:

Answer 5

are an example of natural selection

Question 6

There are many species of marsupials in Australia because:

Answer 6

Australia was isolated from the rest of the world for a long period

The isolation of Australia allowed radiation of marsupial species in the absence of placental mammals.

Question 7

A proficient engineer can easily design skeletal structures that are more functional than those currently found in the forelimbs of such diverse mammals as horses, whales, and bats. That the actual forelimbs of these mammals do not seem to be optimally arranged is because:

Answer 7

natural selection is generally limited to modifying structures that were present in previous generations and in previous species

Even with very long time periods natural selection may not arrive at the optimal structure because of constraints imposed by the development of the organism. Natural selection is a tinkerer, not a designer.

Question 8

Over evolutionary time, many cave-dwelling organisms have lost their eyes. Tapeworms have lost their digestive systems. Whales have lost their hind limbs. How can natural selection account for these losses?

Answer 8

Under particular circumstances that persisted for long periods, each of these structures presented greater costs than benefits

These complex structures would present a ‘cost’ to the organisms if they were of no adaptive value. Organisms carrying mutations causing loss of these costly structures would be favoured by natural selection.

Question 9

Synonymous base substitutions in a gene:

Answer 9

do not change the amino-acid sequence of the encoded protein

The genetic code is degenerate or redundant: most amino acids can be coded by more than one codon.

Question 10

The molecular clock hypothesis implies that:

Answer 10

the rate of amino-acid substitution in a particular gene is constant over evolutionary time

valid as an approximation but rates of evolutionary change of a particular gene can vary in different lineages. –> The clock runs at different rates for different genes.

NOT
the rate of amino-acid substitution over evolutionary time is the same for all genes
–> The clock runs at different rates for different genes.

Question 11

In the lakes of East Africa more than 1500 species of related cichild fish have evolved over the last 10 million years. These lakes are subject to periodic severe droughts that subdivide the large lakes into more numerous smaller lakes. What mode of speciation is likely to have resulted in this remarkable diversity of species?

Answer 11

Allopatric speciation

Question 12

Sympatric speciation that involves a hybrid between two related species is known as:

Answer 12

allopolyploidy

Question 13

Hybrid infertility is an example of:

Answer 13

a postzygotic reproductive barrier

This is a failure of the hybrid to contribute to future generations.

Question 14

Which is NOT an example of a reproductive isolating mechanism?

Answer 14

hybrid zones

Hybrid zones are regions where two recently derived (or two incipient) species mate and give rise to hybrid offspring.

Question 15

What would happen if all animals were herbivores?

Answer 15

A mutation that led to predatory behaviour would be extremely successful

Question 16

Competition:

Answer 16

occurs within and between species

Competition can be interspecific (between species) and intraspecific (within species)

Question 17

Genetic drift leads to:

Answer 17

a loss of genetic variation within a population

Genetic drift tends to reduce genetic variation within a population because alleles are either fixed or lost by chance.

Question 18

If the relative fitness of genotypes MM, MN, and NN are 0.8, 1.0, and 0.2 respectively, what do you expect to happen to the frequency of the N allele?

Answer 18

It will be maintained in the population because the heterozygote has the highest fitness.
This is an example of heterozygote advantage or overdominance.

Question 19

Inbreeding leads to:

Answer 19

an increase in the frequency of homozygotes in the population

The frequency of homozygotes increases at the expense of heterozygotes, but allele frequencies are not directly affected.

Question 20

allopatric speciation

Answer 20

speciation occurring following geographic isolation

Question 21

altruism

Answer 21

behaviour that decreases the fitness of the actor and increases the fitness of the recipient

Question 22

bottleneck

Answer 22

short term reduction in population size followed by an increase in population size

Question 23

disruptive selection

Answer 23

the result of individuals with more extreme values of a trait having higher fitness

Question 24

Hamilton’s rule

Answer 24

an expression that predicts when altruism should be favoured

Question 25

inclusive fitness

Answer 25

fitness due to an individuals own reproductive success plus the reproductive success of its relatives

Question 26

neutral theory

Answer 26

the hypothesis that the vast majority of base substitutions have no effect on fitness

Question 27

phylogeny

Answer 27

the evolutionary history of a group of species

Question 28

polyandry

Answer 28

a mating system in which some females mate with more than one male

Question 29

postzygotic isolation

Answer 29

reproductive isolation resulting from decreased fitness of hybrids

Question 30

transitional form

Answer 30

a species that exhibits traits common to ancestral and derived species

Question 31

morphospecies

Answer 31

populations designated as separate species based on differences in form

Question 32

The dorsal fins of a porpoise and a salmon

Answer 32

Analogous

Question 33

The bones making up the human arm and the bat wing

Answer 33

Homologous

Question 34

The jointed leg of a ladybird beetle and the leg of a robin

Answer 34

Analogous

Question 35

A rhesus monkey's tail and a human's coccyx

Answer 35

Homologous

Question 36

The giant panda's thumb and a human's thumb

Answer 36

Analogous

Question 37

Scientists investigated the effect of tail length on the reproductive success of male red-collared widowbirds. To do this they trimmed the feathers of one set of males to 20 cm final length; these were called the long-tailed group. The tails of a second set of males were trimmed to 12 cm final length; these were called the short-tailed group. Male widowbirds are territorial and successful males set up nests with several females within their territory. Late in the breeding season, scientists counted the number of active nests in each male's territory. The short-tailed group had a mean of 0.75 nests per male territory whereas the long-tailed group had a mean of 2.8 nests; the difference in reproductive success between the two groups was statistically significant (P < 0.01). In addition the scientists measured the body condition of males throughout the breeding season as shown in the graph. What do you think is the most likely explanation for the higher reproductive success of the long-tailed males? (Line for long-tailed has steeper downward slope -> Body condition decreases faster than short over time)

Answer 37

Females prefer long tailed males as their sexual partner

Question 38

Tail length in male widowbirds is genetically determined. In the part of the population that was not included in the experiments, what is likely to happen to average tail length in males over several generations? (If longer tails illustrated to cause decreased body condition over time , at faster rate. Also have higher mean no. of active nests per square metre.

Answer 38

It will reach an optimum length at which the benefits of long tails for reproductive success are balanced by the cost of mantaining long tails

Question 39

Question 1 In which cell cycle stage does DNA replication occur? G1 G0 G2 S

Answer 39

S

Question 40

Question 2 Maize is a diploid organism with 10 pairs of chromosomes (2n=20). How many chromosomes and chromatids are present in each cell at metaphase 1 of meiosis? 20 chromosomes and 40 chromatids 40 chromosomes and 20 chromatids 10 chromosomes and 20 chromatids 20 chromosomes and 10 chromatids

Answer 40

20 chromosomes and 40 chromatids

Question 41

Question 3 Which of the following events occurs in meiosis but NOT in mitosis? Attachment of microtubules to kinetochores Spindle formation Synapsis Chromosome condensation

Answer 41

Synapsis Pairing of two homologous chromosomes during prophase I of meiosis. Allows matching-up of homologous pairs prior to their segregation, and possible chromosomal crossover between them.

Question 42

Question 4 Being able to curl up your tongue into a U-shape is under the control of a dominant allele (T) at one gene locus. In the homozygous recessive condition (tt) the tongue cannot be rolled. Suppose a woman who can roll her tongue and a man who can also roll his tongue have a child who cannot roll her tongue. What are the genotypes of the parents? TT and TT Not enough information Tt and Tt Tt and TT

Answer 42

Tt and Tt Both parents must have t allele for homozygous recessive child. Both parents do not express recessive phenotype though so must have dominant allele T. Therefore Tt

Question 43

Question 5 Among some ethnic groups, when individuals with straight hair mate with those with curly hair, wavy-haired children are produced. If two individuals with wavy hair mate, what phenotypic ratios would you predict among their offspring? 3 wavy : 1 straight 3 curly : 1 wavy 1 straight : 2 wavy : 1 curly 1 straight : 2 curly : 1 wavy

Answer 43

1 straight : 2 wavy : 1 curly Codominance: Curly hair = HH Straight hair = hh Produces offspring with Hh genotype. -> Wavy hair. Cross of 2 wavy hair: Hh x Hh (Offspring : HH ; Hh ; hH ; hh) -> Curly, wavy, wavy , straight.

Question 44

Question 6 The achoo syndrome (sneezing in response to bright light) and trembling chin (triggered by anxiety) are both dominant traits in humans. What is the probability that the first child of parents who are heterozygous for both the achoo gene and the trembling chin gene (Aa Tt) will have the achoo syndrome but lack the trembling chin? 3/16 1/16 9/16 15/16

Answer 44

3/16 ``` AaTt x AaTt • Dihybrid cross: - Two gene loci  2 alleles per locus - F2 -> 9:3:3:1 phenotypic ratio --> 9 A_T_ --> 3 A_tt --> 3 aaT_ --> 1 aatt ``` -->Achoo -> A allele present -->No trembling chin -> tt -->Therefore 3 A_tt

Question 45

Question 7 Seed capsule shape in the weed "Shepherd's purse" is controlled by two independent genes C and D. The double recessive condition (cc dd) results in ovoid capsule shape. A dominant allele at either locus results in a triangular capsule shape. What phenotypic ratio would result from self pollinating a double heterozygote (Cc Dd)? 1 triangular : 3 ovoid 1 triangular : 15 ovoid 15 triangular : 1 ovoid 9 triangular : 7 ovoid

Answer 45

15 triangular : 1 ovoid --> ccdd -> Ovoid --> Dominant in either allele -> triangular So Ccdd ; CCdd ; ccDd ; ccDD ; CCDD -> All triangular ``` Heterozygous: • Dihybrid cross: - Two gene loci  2 alleles per locus - F2 -> 9:3:3:1 phenotypic ratio Where 1 -> Fully recessive ccdd -> No dominant alleles -> Only possible ovoid ```

Question 46

Question 8 Coat colour in dogs depends on the action of at least two genes. At one locus a dominant epistatic inhibitor of coat colour pigment (I) prevents the expression of colour alleles at another independently assorting locus, thereby producing white coat colour. When the recessive condition exists at the inhibitor locus (ii), the alleles of the second locus will be expressed: ii BB or ii Bb producing black and ii bb producing brown. When dihybrid (double heterozygous) white dogs are mated, what would be the phenotypic ratio in the offspring? 9 : 4 : 3 black:white:cinnamon 9 : 7 black:white 12 : 3 : 1 white:black:brown 1 : 1 : 1 white:black;brown

Answer 46

12 : 3 : 1 white:black:brown I prevents expression if present at all -> Dominant -> does not require double allele for expression. Epistatic allele -> Affects expression of other alleles if present -->Dominant epistasis : 12:3:1 I (White) is dominant so will have most ratio -> I = I___ Black more dominant than b so will have higher ratio B = iiB_ = 3 Brown is recessive so has lowest ratio b = iibb = 1 • 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. • 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 47

Question 9 Which of these statements about the human sex chromosomes (X and Y) is NOT true? The X chromosome is larger than the Y chromosome The pseudoautosomal region is unique to the X chromosome Males are hemizygous for mutations located on the differential region of the X chromosome Females are the homogametic sex

Answer 47

The pseudoautosomal region is unique to the X chromosome ``` • Pseudoautosomal regions: - PAR1 & PAR2 - Shared between X & Y chromosomes  ~30 genes -> Humans - Required -> X-Y pairing -> male meiosis --> Therefore Not unique to X chromosome ``` ``` • Human chromosomes - Male specific region -> Y chromosome (MSY) >>80 genes >> ~60Mb - X – specific region >> >1000 genes >> ~160Mb --> Therefore X larger than Y chromosome ``` • Homogametic Sex:  Females • Heterogametic Sex:  Males Hemizyogous: - Having only one set of alleles instead of two, as in the case of loci on the X chromosome in males - A chromosome in a diploid organism is hemizygous when only one copy is present. Also observed when one copy of a gene is deleted, or in the heterogametic sex when a gene is located on a sex chromosome.

Question 48

Question 10 Duchenne muscular dystrophy in humans is a rare, X-linked, recessive trait. A woman who is unaffected by the disorder has two sons, Bill and Ben, who have the same, unaffected father. Bill has Duchenne muscular dystrophy. Which of the following values most closely estimates the chance that Ben has the same condition? 25% 100% 75% 50%

Answer 48

50% -->Present on X chromosomes --> XY crosses Recessive trait -> Requires double presence for expression. Woman unaffected: -> So at least one unaffected dominant allele M -> X^(M) Offspring male affected -> has X^(M*) Offspring must inherit Y chromosome from father. So must have inherited X^(M*) from mother -->Therefore mother XX* --> XX* x XY cross ->Must inherit Y from father. So inherits X or X* from mother. -> 50% chance of either X -> Producing XY (not affected) ->50% chance producing X*Y (affected) • Sex-linked inheritance: - Involves genes located -> X chromosome >> (X-linkage) - Y-linked genes -> male-specific functions (spermatogenesis) >> Only small % -> sex linkage - Most genes -> X >> Unrelated to sex determination / sex function >> Expressed in males & females - Males >> Hemizygous for genes -> X chromosome ``` • Sex-linked inheritance: - Reciprocal crosses -> same results  - Yellow F x Green M >> 100% Yellow F1 - Offspring cross >> 3 yellow : 1 green  - Green F x Yellow M >> 100% Yellow F1 - Offspring cross >> 3 yellow : 1 green ``` - White mutant -> Drosophila:  Thomas Hunt Morgan -> Spontaneous mutant fly -> white eyes.  Reciprocal crosses with white mutant flies >> No identical Mendelian phenotypic ratios 1. White Male x Red Female >> F1 -> 100% red eyes >>Red dominant 2. F1 Female x F1 Male >> 3 Red : 1 White >> All white eyed flies -> male. >> Red eyed flies -> 2 females : 1 Male 3. White male x F1 Female >> 1 red Male : 1 red Female : 1 white Male : 1 white Female 4. White Female x F1 Male ->> Reciprocal cross >> Red Females & White Males  Parallel genetic data -> cytological observations -> chromosomes of Drosophila. >> Drosophilla -> 4 pairs chromosomes ->> 3 homomorphic pairs > similar in size of chromosomes in chromosomal pair ->> 1 heteromorphic pair > different sizes of chromosomes in chromosomal pair**** >> Females -> All eggs -> one X chromosome >> Males -> Sperm -> 50% -> X & 50% -> Y Hypothesis: White gene on X chromosome 1. X+X+ x XWY Red Female x White Male F1: >> X+XW & X+Y Red Females & Red Males -> 100% red offspring 2. F1 Female x F1 Male >> X+X+ x X+Y x X+XW x XWY 3 Red : 1 White All white -> Male Red flies -> 2 Female : 1 Male ``` • X-linked traits -> Humans - Nearly all -> recessive - More common in males (hemizygous) than females - ~300 known X-linked disease genes ~10% total disease genes ```

Question 49

Which of the following statements about linkage is NOT true? Linked alleles tend to stay together at meiosis Unlinked genes undergo independent assortment at meiosis Crossing-over never occurs between linked genes A linkage map is based on recombination frequencies

Answer 49

Crossing-over never occurs between linked genes - ->All genes undergo independent assortment (alignment on cell equator in metaphase) as all chromosomes undergo independent assortment - ->Crossing over can occur on all chromosomes. incl. chromosomes with linked genes as separate chromosomes cross over hence linked genes on same chromosome can become separated by transfer of sections of chromosome to other chromosome. - > Linkage maps developed using calc. of recombination frequencies to calc. distances between alleles etc. * *IMPORTANT -> Please read all of this info!!*** {Crossing-over DOES occur between linked genes, although the frequency of crossing over decreases as the distance between a pair of genes decreases. } • Genetic Linkage: - Genes -> same chromosome -> linked - > May violate Mendel’s 2nd law -> Independent Assortment - Alleles of genes -> same chromosome > Segregate together in gametes during meiosis >>Unless crossing-over between them occurs. Genes do assort independently. However, it depends on where they lie on the chromosome. Genes on different chromosomes will be sorted separately Genes on the same chromosome, but far apart, will also likely be sorted separately. Linked genes occur when the ratio of parental vs. non-parental siblings are skewed. If the four classes of offspring are not produced in equal numbers, there are linked genes. Linked genes cross depending on where the two genes sit relative to each other. That may be on the same chromosome or a different chromosome. The crossover can unlink genes so that you don’t get the same ratio of offspring. The chance of crossing over happening is dependent on how far away the gene is from each other. Mendel observed that two genes on different chromosomes would have a 50% chance of recombination. 50% represents when: Two genes are on separate chromosomes. Two genes sit on the same chromosome, but far away. As genes sit closer together on the SAME chromosome it becomes less likely they will cross-over.

Question 50

The genes apterous (ap), dumpy (dp),and clot (cl), are all linked on chromosome 2 of Drosophila. In a series of two-point mapping crosses, the following genetic distances were determined: ap-dp, 42 map units; dp-cl, 3 map units; ap-cl, 39 map units. What is the order of the three genes on chromosome 2? ap-cl-dp cl-ap-dp ap-dp-cl

Answer 50

ap-cl-dp Closest -> dp & cl --> dp - cl Next closest -> ap-cl - -> Add ap beside cl - -> dp-cl-ap ``` Furthest distance -> ap-dp -->Ensure ap & dp on furthest possible points away from each other --> makes sense as dp-ap = 42 dp-cl + cl-ap = 3 + 39 = 42 ```

Question 51

Question 3 Copy of CC dd and cc DD individuals were crossed to each other and the F1 generation was backcrossed to a cc dd tester. 953 Cc dd, 947 cc Dd, 46 Cc Dd and 54 cc dd offspring resulted. How far apart are the c and d loci? 10 cM 0. 5 cM 0. 1 cM 5 cM

Answer 51

5 cM >Example of linkage mapping -> Testcross ((Produce double heterozygote ; cross with homozygous recessive test strain >>Double recessive chromosome -> F2 phenotype always corresponds -> genotype of gamete -> F1 double heterozygote. -> Easy classification of offspring as parental or recombinant with respect to double heterozygote gametes.  F2 Phenotype -> Genotype -> gamete -> double heterozygote. )) ** Parental genotypes always higher frequency & recombinant lower frequency. ``` > > CCdd x ccDD -> 100% CcDd --> C & d -> one chromosome --> c & D -> homologous chromosome > CcDd offspring --> Cd & cD chromosomes from parent --> cd from tester parent >> So Cd & cd -> Ccdd cD & cd -> ccDd --> (parent strains) ``` ``` > CcDd x ccdd 953 Cc dd, -> Parent 947 cc Dd, -> Parent 46 Cc Dd 54 cc dd Therefore recombinant -> CcDd & ccdd ``` - Linkage maps: Illustrate relative positioning of genes / markers on chromosomes -> meiotic recombination frequencies (centiMorgan, cM) • Linkage-mapping principles: - Crossing-over -> random positions -> chromosomes Eg. Humans -> 1-2 crossings each chromosome arm -> per meiosis - Frequency -> crossing-over between 2 gene loci > Proportional -> physical distance between them on chromosome. >> Measuring frequency -> crossing-over between 2 genes therefore indicates measure of distance between them. • Frequency -> recombinant gametes  Proportional to frequency of crossing-over & distance apart on same chromosome.  Independent assortment -> genes on diff. chromosomes >> produce 50% recombinant gametes ``` • Recombination frequencies:  50% recombinant gametes indicate ->> Genes -> diff. chromosomes OR ->> Genes long distance apart -> chromosome  <50% recombinant gametes indicate ->> Genes linked -> Same chromosome OR ->> Smaller recombination frequency -> Closer genes. ``` ``` • Genetic Distance = Recombinant Frequency (RF) • RF = [(No. recombinant offspring) x 100] / [Total no. offspring}  Units -> centiMorgans (cM) >> 10% RF = 10cM -->Distance between c & d loci > CcDd x ccdd 953 Cc dd, -> Parent 947 cc Dd, -> Parent 46 Cc Dd 54 cc dd Therefore recombinant -> CcDd & ccdd ``` So distance between c & d loci -> recombinant frequency • RF = [(No. recombinant offspring) x 100] / [Total no. offspring} RF = ((46+54)x100) / (953+947+46+54) = 5% = 5cM

Question 52

The diagram above shows a pedigree of a family affected by familial rickets. What is the most likely mode of inheritance? Male -> all daughters affected, no sons affected Female -> half of sons & half of daughters affected

Answer 52

X-linked dominant ``` • Mendelian Inheritance: - Autosomal recessive:  Phenotype can skip generations  Both parents -> carriers -> pass trait to offspring.  Equally affects both genders. ``` - 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. 2. Transmission -> Hemizygous affected male  No sons affected / carriers  All daughters -> carriers 3. Transmission -> Affected female:  All sons -> affected  All daughters -> carriers. ``` - X-linked domiant:  Affected male parents >> All daughters affected >> No sons affected / carriers ``` --> Not equally shared -> male & females > X-linked. Affected male parents -> Dominant -> >> All daughters affected >> No sons affected / carriers Recessive -> Transmission -> Hemizygous affected male >>No sons affected / carriers >>All daughters -> carriers ``` -> So recessive Therefore X-linked recessive

Question 53

You discover a new variant – the punk mouse – in which mice have spiky fur instead of the usual soft fur. You notice that this trait seems only to be present in males. To investigate this pattern, you cross a spiky fur male with a true breeding female with soft fur, and find that all of the F1 progeny of both sexes have soft fur. You then interbreed the F1 and observe that all of the F2 females have soft fur, but ½ of the F2 males have spiky fur. You conclude that: the spiky allele is autosomal dominant with sex-limited expression the spiky allele is autosomal recessive with sex-limited expression the spiky allele is an X-linked recessive the spiky allele is Y-linked

Answer 53

the spiky allele is an X-linked recessive - -> Only present in males - -> F1 progeny of both sexes have soft fur. - -> F2 females have soft fur, but ½ of the F2 males have spiky fur. - Autosomal recessive:  Phenotype can skip generations  Both parents -> carriers -> pass trait to offspring.  Equally affects both genders. - 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. 2. Transmission -> Hemizygous affected male  No sons affected / carriers  All daughters -> carriers 3. Transmission -> Affected female:  All sons -> affected  All daughters -> carriers. ``` - X-linked domiant:  Affected male parents >> All daughters affected >> No sons affected / carriers -->Cannot be autosomal dominant as not present in every generation (F1 = all smooth fur) -->Cannot be Y-linked as if so, no female offspring could inherit & all male offspring would inherit -->If X-linked recessive: 2. Transmission -> Hemizygous affected male  No sons affected / carriers  All daughters -> carriers >However all F1 offspring had smooth fur therefore F2 gen didn't inherit from Hemizygous male so this does not apply --> Cannot use so try process of elimination. - Autosomal recessive:  Phenotype can skip generations  Both parents -> carriers -> pass trait to offspring.  Equally affects both genders. --> All F1 -> smooth fur >>Evidence of skipping generations. However does not equally effect genders -> unequal in F2: ->> F2 females have soft fur, but ½ of the F2 males have spiky fur. -->Therefore must be X-linked recessive.

Question 54

``` In fruit flies, the recessive pr and cn mutations cause brown and cinnabar (bright red) coloured eyes, respectively (whereas wild-type flies have red eyes). Both genes are on the same chromosome. The double mutant pr pr cn cn combination has orange eyes. Two true-breeding parental flies were crossed to produce an F1. An F1 female who had wild-type eyes was crossed to an orange-eyed male. Their progeny had the following distribution of eye colours: wild-type (red): 8 brown: 241 cinnabar: 239 orange: 12 ``` What were the haplotypes (combinations of alleles on the same chromosome) of the F1 female that was crossed to give rise to these offspring? (pr+ and cn+ indicate the wild type alleles, pr and cn indicate the recessive mutant alleles) pr+ cn on one chromosome, pr cn+ on the other chromosome pr cn on one chromosome, pr+ cn+ on the other chromosome pr cn on one chromosome, pr cn on the other chromosome pr+ cn on one chromosome, pr+ cn on the other chromosome

Answer 54

pr+ cn on one chromosome, pr cn+ on the other chromosome - ->>Example of test cross!!** 1. Produce double heterozygote 2. Cross with homozygous recessive * *Recombinant phenotypes always smaller ratio ``` --> Therefore female => double heterozygote -> pr+pr cn+cn Progeny: wild-type (red): 8 brown: 241 -> Parental cinnabar: 239 -> Parental orange: 12 > Therefore parental genotypes -> Brown -> prcn+ -> Cinnabar -> pr+cn ``` --> so pr+pr cn+cn F1 female passed these genotypes to offspring -> where prcn+ inherited together & pr+cn inherited together So these are combos on haplotypes. - > pr = brown eyes - > cn = cinnabar (bright red) - > wild type -> pr+cn+ = red eyes - > prcn = orange eyes - ->> Therefore - > pr mutation alone gives brown eyes (prcn+) - > cn mutation alone gives cinnabar (bright red eyes) (pr+cn) - > pr and cn mutation gives orange - > no mutations at all give red F1 = pr+_cn+_ -> wild type 2 true-breeding parents (homozygous) -->1 parent must have had pr+pr+cn+cn+ ``` F1 could produce orange offspring therefore has prcn also -> Therefore her genotype F1 = pr+pr cn+cn -> (True-breeding parents) --> 1 parent must have had prprcncn ```

Question 55

Question 7 The endosymbiotic theory was proposed to explain the presence of which structures in the eukaryotic cell? Ribosomes and flagella Ribosomes and chloroplasts Mitochondria and chloroplasts Mitochondria and flagella

Answer 55

Mitochondria and chloroplasts

Question 56

Question 8 If a disease shows higher concordance in monozygotic twins than in dizygotic twins this indicates that the disease: is caused by a recessive mutation does not have a genetic component is affected by environmental rather than genetic factors has a genetic component

Answer 56

has a genetic component >> Twin Heritabilty Studies: 1. Identical (monozygotic twins ; MZ) Share same environment & all alleles Relatedness ; r = 1 2. Non-identical (dizygotic twins ; DZ) Share same environment & half of their alleles Relatedness ; r = 0.5  If MZ twins resemble more than DZ twins ; genes contribute to variation in trait . > Assumes equal environmental effects >> Twin Concordance Studies:  Concordance: Probability a twin is affected by a particular trait if their twin is affected. > Concordance = 1.0 -> Other twin always affected = 0.6 -> 60% chance other twin affected.  Difference in concordance between MZ & DZ twins used to estimate hertitability. Eg. MZ concordance = 0.50 (50%) ; DZ = 0.08 (8%) Heritability = 0.58

Question 57

Question 9 The type of DNA markers typically used in forensic DNA analysis are: Copy number variants Simple tandem repeats (STRs) Single nucleotide polymorphisms (SNPs) Restriction fragment length polymorphisms (RFLPs)

Answer 57

Simple tandem repeats (STRs) {Multiple STRs are used to generate a DNA profile of an individual.} • Difficulties using linkage analysis to study disease genes as requires at least 2 genetic disease genes present in a family at any one time.  Solved by DNA markers. • DNA Markers: - Requirements:  Polymorphic: >> 2 or more alleles -> significant proportion of population >> Easy to assay -> Easy to distinguish alleles. - Common types:  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.  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.

Question 58

Question 10 In human females, meiosis begins: at the start of the menstrual cycle at puberty shortly after birth during foetal development (before birth)

Answer 58

during foetal development (before birth) {Meiosis begins during foetal development but then arrests at Prophase I until puberty.}

Question 59

Match each of the terms (left column) to one of the descriptions (right column). allopatric speciation altruism bottleneck disruptive selection Hamilton's rule inclusive fitness neutral theory phylogeny polyandry postzygotic isolation transitional form morphospecies A. an expression that predicts when altruism should be favoured B. a species that exhibits traits common to ancestral and derived species C. the result of individuals with more extreme values of a trait having higher fitness D. reproductive isolation resulting from differences in mate choice or timing of breeding that prevent mating E. fitness due to an individuals own reproductive success plus the reproductive success of its relatives F. populations designated as separate species based on differences in form G. a mating system in which some females mate with more than one male H. short term reduction in population size followed by an increase in population size I. speciation occurring following geographic isolation J. reproductive isolation resulting from decreased fitness of hybrids K. the hypothesis that the vast majority of base substitutions have no effect on fitness L. speciation occurring between populations in the same area M. a mating system in which some males mate with more than one female N. populations designated as separate species based on reproductive isolation O. behaviour that decreases the fitness of the actor and increases the fitness of the recipient P. the result of individuals with the mean value of a trait having the highest fitness Q. the evolutionary history of a group of species

Answer 59

allopatric speciation speciation occurring following geographic isolation altruism behaviour that decreases the fitness of the actor and increases the fitness of the recipient bottleneck short term reduction in population size followed by an increase in population size disruptive selection the result of individuals with more extreme values of a trait having higher fitness Hamilton's rule an expression that predicts when altruism should be favoured inclusive fitness fitness due to an individuals own reproductive success plus the reproductive success of its relatives neutral theory the hypothesis that the vast majority of base substitutions have no effect on fitness phylogeny the evolutionary history of a group of species polyandry a mating system in which some females mate with more than one male postzygotic isolation reproductive isolation resulting from decreased fitness of hybrids transitional form a species that exhibits traits common to ancestral and derived species morphospecies populations designated as separate species based on differences in form

Question 60

Homologous structures are similarities between species as the result of descent from a common ancestor, even if the structures perform different functions. Analogous structures are similarities that arise as the result of adaptation to a similar environmental problem and are not due to common descent (convergent evolution). Which of the following pairs of structures are homologous and which are analogous? The dorsal fins of a porpoise and a salmon The bones making up the human arm and the bat wing The jointed leg of a ladybird beetle and the leg of a robin A rhesus monkey's tail and a human's coccyx ``` The giant panda's thumb and a human's thumb A. Homologous B. Analogous ```

Answer 60

The dorsal fins of a porpoise and a salmon A. Analogous The bones making up the human arm and the bat wing B. Homologous The jointed leg of a ladybird beetle and the leg of a robin A. Analogous A rhesus monkey's tail and a human's coccyx B. Homologous The giant panda's thumb and a human's thumb A. Analogous

Question 61

The DNA sequences of a gene from two related fruit fly species are shown below. The gene encodes a six amino-acid peptide. The sequences show the coding strand of the DNA (i.e. the strand whose equivalent sequence is found in the messenger RNA of the gene). Drosophila melanogaster gene sequence: 5'-ATGCTTCTTCATCGCGGGTAA-3'; Drosophila simulans gene sequence: 5'-ATGCATCTACATCGTCGGTAA-3'. You are given a table of the genetic code to help with this problem. Remember that a gene always starts with a Met or START codon (ATG in DNA, AUG in RNA) and ends with a STOP codon. How many nucleotide substitutions have occurred in this gene between the two species? Which of the following non-synonymous changes has occurred in the gene in the two fruit fly species? 1. Glu to Gly 2. Leu to Ile 3. His to Gln 4. Gly to Arg 5. Arg to STOP

Answer 61

4 ATG CTT CTT CAT CGC GGG TAA ATG CAT CTA CAT CGT CGG TAA * * * * Gly to Arg Non-synonymous substitution: Change in nucleotide sequence of DNA -> Results in alteration in the amino acid sequence of the protein.

Question 62

Scientists investigated the effect of tail length on the reproductive success of male red-collared widowbirds. To do this they trimmed the feathers of one set of males to 20 cm final length; these were called the long-tailed group. The tails of a second set of males were trimmed to 12 cm final length; these were called the short-tailed group. Male widowbirds are territorial and successful males set up nests with several females within their territory. Late in the breeding season, scientists counted the number of active nests in each male's territory. The short-tailed group had a mean of 0.75 nests per male territory whereas the long-tailed group had a mean of 2.8 nests; the difference in reproductive success between the two groups was statistically significant (P < 0.01). In addition the scientists measured the body condition of males throughout the breeding season as shown in the graph. What do you think is the most likely explanation for the higher reproductive success of the long-tailed males? 1. Females prefer long tailed males as their sexual partner 2. Long-tailed males are more aggressive than short tailed males 3. Long tailed males have larger territories than short tailed males 4. Long tailed males are healthier than short tailed males Trend: Line Graphs: - Both originate from same place approx. on y-axis - y axis -> Body condition - x axis -> Days since territory establishment Short tails: (5, 0.175) - (95, 0) Long tails: (5, 0.5) - (85, -0.175) - short tails have wider disperse of data around line of best fit. - all lines have \ alignment

Answer 62

Females prefer long tailed males as their sexual partner - > No evidence for ling tailed males being more aggressive - ->Even if this was true, and aggressiveness & long tailed genes were inherited together, males with long tails would still demonstrate aggressive behaviour which would skew results - > Results (No. of nests per area standardised so effect of area of a male's territory didn't matter -> nests per male territory) - > Body condition in graph declined more rapidly in those with longer tails & both male groups had same body conditions at start of exp. so can't be confounding factor)

Question 63

Tail length in male widowbirds is genetically determined. In the part of the population that was not included in the experiments, what is likely to happen to average tail length in males over several generations? 1. It will decrease because males with long tails are less healthy than males with short tails 2. It will decrease because males with long tails are more conspicuous to predators such as raptors 3. It will reach an optimum length at which the benefits of long tails for reproductive success are balanced by the cost of maintaining long tails 4. It will increase each generation because males with longer tails have more offspring than short tailed males

Answer 63

It will reach an optimum length at which the benefits of long tails for reproductive success are balanced by the cost of maintaining long tails

Question 64

Match each of the terms (left column) to one of the descriptions (right column). Fixation Founder effect Gene flow Gene pool Genetic drift Overdominance A. when an allele is eliminated from the population B. genetic drift resulting from the establishment of a new population by a small number of individuals C. genetic drift resulting from a population crash D. random changes in allele frequencies caused by sampling error E. when an allele reaches a frequency of 1.0 F. heterozygotes have higher fitness than homozygotes G. all the alleles in a breeding population at one time H. the movement of alleles between populations I. heterozygotes having lower fitness than homozygotes

Answer 64

Fixation . when an allele reaches a frequency of 1.0 Founder effect genetic drift resulting from the establishment of a new population by a small number of individuals Gene flow the movement of alleles between populations Gene pool all the alleles in a breeding population at one time Genetic drift random changes in allele frequencies caused by sampling error Overdominance heterozygotes have higher fitness than homozygotes

Question 65

Question 8 Sea otters in California had been hunted to near extinction up until the early twentieth century. At its lowest point the population of California sea otters numbered fewer than 50 individuals. Since 1911 the sea otters have been protected from hunters and the population has recovered to ~1,500 individuals. Researchers studied the 'PAP' gene locus which has two alleles, A1 and A2. In a sample of 40 sea otters from this population researchers found that there were 20 of the A1 A1 genotype, 8 of the A1 A2 genotype and 12 of the A2 A2 genotype. Match each of the questions (A-E) to one of the possible answers. What is the observed genotype frequency of the A1 A1 genotype? What is the observed genotype frequency of the A1 A2 genotype? What is the observed genotype frequency of the A2 A2 genotype? What is the frequency of the A1 allele? What is the frequency of the A2 allele? ``` A. 0.0 B. 0.1 C. 0.2 D. 0.3 E. 0.4 F. 0.5 G. 0.6 H. 0.7 I. 0.8 J. 0.9 K. 1.0 ``` Using the allele frequencies you have calculated, work out the expected genotype frequencies if the population was at Hardy-Weinberg equilibrium. Compare the expected genotype frequencies with the observed genotype frequencies and then answer the following question. Which statement best describes the results of the study of the PAP gene locus in California sea otters? The observed excess of heterozygotes suggests that the population is under balancing selection at the ‘PAP’ locus. The observed excess of homozygotes suggests that the population has been subject to genetic drift affecting the ‘PAP’ locus. The observed excess of homozygotes suggests that the population is under strong directional selection at the ‘PAP’ locus. The observed and expected genotype frequencies are very similar suggesting that the population is at Hardy-Weinberg equilibrium at the ‘PAP’ locus.

Answer 65

What is the observed genotype frequency of the A1 A1 genotype? F. 0.5 [ 20/40 = 0.5} What is the observed genotype frequency of the A1 A2 genotype? C. 0.2 [ 8/40 = 0.2] What is the observed genotype frequency of the A2 A2 genotype? D. 0.3 [ 12/40 = 0.3] What is the frequency of the A1 allele? G. 0.6 [ (20(2) + 8) / 2(40) = 48/80 = 0.6 ] What is the frequency of the A2 allele? E.0.4 [ (12(2) + 8) / 2(40) = 32/80 = 0.4 ] ``` Observed (A1A1) = 0.5 Expected ; p = 0.6 so p^2 = 0.36 Observed (A2A2) = 0.3 Expected ; q = 0.4 so q^2 = 0.16 Observed (A1A2) = 0.2 Expected ; 2pq = 2(0.6)(0.4) = 0.48 ``` The observed excess of homozygotes suggests that the population has been subject to genetic drift affecting the ‘PAP’ locus.

Question 66

Match each of the terms (left column) to one of the descriptions (right column).

Answer 66

Barr body I. the inactive X chromosome in cells of female mammals centromere C. a region of DNA on a chromosome that becomes attached to the mitotic/meiotic spindle epistasis M. interaction between two or more genes that affect a single phenotype haplotype D. a set of SNPs close together in a small region of a chromosome linkage disequilibrium F. association of particular SNPs with each other in a population monosomy L. where a cell or organism has one chromosome of a homologous pair missing non-disjunction H. aberrant segregation of chromosomes or chromatids during meiosis pseudoautosomal region A. a region of homologous DNA on the X and Y chromosomes of mammals recombinant J. a chromosome, cell or individual that has non-parental combinations of alleles sister chromatids Q. two identical copies of a chromosome derived from replication of the chromosome and held together by cohesin synapsis G. the association of replicated homologous chromosomes that allows crossing-over to occur during prophase I of meiosis

Question 67

Huntington disease is a rare, degenerative human disease determined by a dominant allele, HD. The disorder is usually manifested after the age of 45. A young man in his 20s has learned that his father has developed the disease; his mother does not have the disease and there have been no cases of Huntington disease on her side of the family. Taking into account all the information above, which value most closely estimates the probability that the young man will later develop the disease? The young man (described in Huntington's disease part 1) and his partner, who has no family history of Huntington disease, have decided to start a family. What value most closely estimates the probability that their first child will be affected by Huntington disease as an adult?

Answer 67

Huntington's disease is caused by an autosomal dominant mutation. Mother-> No cases in family & unaffected therefore homozygous & unaffected -> HH Father -> Affected -> Only requires one allele as dominant mutation -> HH* HH x HH* - > Son inherits one unaffected allele from mother -> H - > Son inherits either an unaffected allele or an affected dominant mutated allele from father -> H or H* - > HH ; HH* ; HH ; HH* - > (50% chance dominant allele inherited from father, where if allele inherited, inherited allele from mother is irrelevant) -> Therefore p = 0.5 ii) Partner has no history of mutation -> Therefore homozygous & unaffected -> HH Male -> 50% chance either unaffected or affected -> HH or HH* Probability child affected = Probability male is affected & probability male mutated allele passed to child P(Male affected) = 0.5 P(Mutation inherited): --> If male inherits mutation, alleles inherited from mother irrelevant. (As dominant mutation) -> %0% chance normal allele & 50% mutated allele inherited. Therefore probability child inherits mutation from muatted father is 0.5 --->> Therefore prob child mutated = P(Father mutated) x P(Child inherits mutation) = 0.5 x 0.5 = 0.25

Question 68

In mice, pigment is produced only when the C allele is present. Mice with the cc genotype have no colour (albino). If colour is present, it may be determined by the A and a alleles. AA or Aa genotypes result in agouti colour, whereas aa results in black coats. The C and A genes are not linked. AA CC mice are crossed to aa cc mice. The F1 offspring are then crossed to each other to give F2 offspring. What phenotypic ratio do you expect to find in the F2 offspring?

Answer 68

```  C gene: >>Colour expression  Allele (C): Colour expressed  Allele (c): Albino -> colour not expressed. ``` • 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. ``` • Albino allele:  Epistatic -> all coat colour genes  Mutation  Recessive  Loss of function  Enzyme tyrosinase -> melanin synthesis -> melanocytes > Tyrosine -> Tyrosinase >>Eumelanin (Black) >>Pheomelanin (Yellow) ``` Recessive epistasis -> i) Homozygous cross -> Heterozygous F1 & 9:4:3 ratio in F2 ii) Heterozygous cross -> 9:4:3 ratio in F1 --->>> Phenotypes: 9:4:3 9 Agouti : 4 Albino : 3 Cinnamon

Question 69

Mice in the F 2 with the agouti phenotype can be either of four different genotypes: AA CC, AA Cc, Aa CC or Aa Cc. To identify the genotype of the F2 agouti mice, four female mice with the agouti phenotype were crossed with males of aa cc genotype. The offspring of each of the four female mice are shown in the left column. For each female choose her most likely genotype from the right column. Female 1: 4 agouti, 3 albino offspring Female 2: 4 agouti, 4 black offspring Female 3: 4 albino, 2 black, 1 agouti offspring Female 4: 7 agouti offspring ``` A. Aa CC B. Aa Cc C. AA CC D. AA Cc ```

Answer 69

Combinations; Crosses: 1) AACC x aacc 2) AACc x aacc 3) AaCC x aacc 4) AaCc x aacc All offspring must already have _a_c genotype from father. Albino offspring -> __cc - > So c inherited from mother - -> F1 & 3 Agouti offspring -> A_C_ - > So A & C inherited from mother - -> F1, 2, 3 & 4 Black offspring -> aaC_ - > So a & C inherited from mother - -> F2 & 3 F1 -> A_Cc F2 -> AaC_ F3 -> AaC_ F4 -> A_C_ F1 -> No black offspring (aaC_) >> Cant possess a allele = AACc F2 -> No albino offspring (__cc) >> Cant possess c allele = AaCC F3 -> Must therefore be AaCc F4 -> Must therefore be AACC

Question 70

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?

Answer 70

->Maternal imprinting Maternal allele imprinted & paternal expressed -> Parents heterozygous >> Igf2 Igf2* x Igf2 Igf2* -->> If maternal imprinted, maternal allele inherited irrelevant; -> offspring genotypes: Igf2 _ or Igf2*_ (-> inherited from father ; maternal alleles have no effect so paternal alleles solely determine expression) -> 50% normal (Igf2_) -> 50% dwarf (Igf2*_) So 2 normal : 2 dwarf (Or written 1 normal : 1 Dwarf etc. ) • 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. 2 normal : 2 dwarf

Question 71

The recessive mutations black (black body colour, b), curved (curved wings, cu), purple (purple eyes, pr) and vestigial (crumpled wings, vg) are all on chromosome 2 in Drosophila melanogaster. Bridges and Morgan mapped the 4 genes by a series of two-point crosses. For example, to determine the map distance between black ( b) and purple ( pr) they crossed a black female ( bb pr+pr+) with a purple male ( b+b+ prpr). The F 1 offspring (b+b pr+pr) were crossed with double homozygous recessive flies (bb prpr). The F 2 offspring were classified with respect to their phenotype. They obtained 4 different F 2 phenotypes: (1) wild type body and eyes, (2) black body, (3) purple eyes, (4) black body and purple eyes. (Note that the + added to the gene symbol indicates the wild type allele and a gene symbol without a + indicates the mutant allele). In this experiment what are the genotypes of the recombinant progeny in the F2

Answer 71

??? - >Parental cross ( bb pr+pr+) with a purple male ( b+b+ prpr). - > F 1 offspring (b+b pr+pr) were crossed with double homozygous recessive flies (bb prpr). - > 4 different F 2 phenotypes: (1) wild type body and eyes, (2) black body, (3) purple eyes, (4) black body and purple eyes. Wild type -> b+pr+ Black -> bpr+ Purple eyes -> b+pr Black & purple eyes -> bpr ??? • Frequency -> recombinant gametes  Proportional to frequency of crossing-over & distance apart on same chromosome.  Independent assortment -> genes on diff. chromosomes >> produce 50% recombinant gametes ``` • Recombination frequencies:  50% recombinant gametes indicate ->> Genes -> diff. chromosomes OR ->> Genes long distance apart -> chromosome  <50% recombinant gametes indicate ->> Genes linked -> Same chromosome OR ->> Smaller recombination frequency -> Closer genes. ``` • “Two-point” mapping experiment: - pr -> purple eye - Pr+ -> wild type (red eye) - vg -> vestigial wing - Vg+ -> Wild type (normal wing) ``` F1 genotype -> Pr+pr Vg+vg ** -> But occasional crossing over Possible genotypes: >> Pr+Vg* >> Pr+vg -> (If crossing occurs) >> prVg+ -> (If crossing occurs) >> prvg ```  Pr+pr Vg+vg ** x prpr vgvg  Offspring: >> Pr+pr Vg+vg -> 1339 -> Phenotype > Pr+Vg+ -> WT -> Parental genotype >> Pr+pr vgvg -> 154 -> Phenotype > Pr+vg -> Vestigial -> Recombinant genotype >> prpr Vg+vg -> 151 -> Phenotype > prVg+ -> Purple -> Recombinant genotype >> prpr vgvg -> 1195 -> Phenotype > prbg -> Purple vestigial -> Parental genotype  Parental genotypes: Pr+Vg+ & prvg  Recombinant genotypes: Pr+vg & prVg+  RF= [(151 + 154) x 100] / [2839] = 10.7% >>Map distance = 10.7cM • Correctly identifying the parental and recombinant progeny:  Parental cross version 1: Pr+Pr+ Vg+Vg+ x prpr vgvg ->> Pr+pr Vg+vg double heterozygote Parental types: Pr+Vg+ and prvg Recombinant types: Pr+vg and prVg+  Parental cross version 2: Pr+Pr+ vgvg x prpr Vg+Vg+ ->> Pr+pr Vg+vg double heterozygote Parental types: Pr+vg and prVg+ Recombinant types: Pr+Vg+ and prvg

Question 72

The “odds-ratio” test for genome-wide associations with disease:

Answer 72

indicates whether an allele increases or decreases the susceptibility to disease

Question 73

Question 1 Cell grafting experiments have helped elucidating how cells become different. This experiment reveals that: The grafted cells were: 1. induced 2. specified 3. determined 4. none of the above - > Dorsal lip cell of pigmented blastophore donated to non-pigmented recipient blastophore. - > Primary cell develops primary non-pigmented neural tube & notochord at one pole of cell & secondary non-pigmented neural tube & pigmented notochord at opposite pole.

Answer 73

3. determined • 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) • 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. • Specification: - First stage in differentiation. Specified cells can have their commitment reversed while the determined state is irreversible. There are two main types of specification: autonomous and conditional. >> Autonomous specification: Development into a specific fate based upon cytoplasmic determinants with no regard to the environment the cell is in. >> Conditional specification: Development into a specific fate based upon other surrounding cells or morphogen gradients. - Cytoplasmic determinants in early embro:  Notochord doesn’t change between specification & fate maps of tissues  Grey crescent region (pigmented area in embryo) - 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) • What generates grey crescent & why is it important? - Cortical rotation breaks in radial symmetry  Unferitilised egg -> Radially symmetrical  Following fertilisation -> Zygote becomes bilaterally symmetrical Darker pigment / grey crescent rotates 30*  Cleavage occurs -> new region of grey crescent area (part of 30*)  Future dorsal side of cell  Development of blastophore - Nieuwkoop Centre  Donor -> Embryo -> normal development  Recipient -> Embryo -> Twinned & duplicated axis. - 4 signal model of induction events -> Early Frog Embryo:  General mesoderm inducer -> Vegetal pole  Nieuwkoop centre -> Dorsal vegetal pole  Induces organiser  Organiser -> Induces neural ectoderm -> Dorsalises ventral mesoderm (Induces somites -> ventral mesoderm)  Ventral mesoderm antagonises dorsal mesoderm Enables establishment -> dorsal-ventral polarity & patterning

Question 74

Question 2 The grafted cells (cells of one blastophore from which cells were added to another) normally induce (via signalling) the development of the... 1. dorsal side 2. ventral side

Answer 74

1. dorsal side - Cytoplasmic determinants in early embro:  Notochord doesn’t change between specification & fate maps of tissues  Grey crescent region (pigmented area in embryo) - 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) • What generates grey crescent & why is it important? - Cortical rotation breaks in radial symmetry  Unferitilised egg -> Radially symmetrical  Following fertilisation -> Zygote becomes bilaterally symmetrical Darker pigment / grey crescent rotates 30*  Cleavage occurs -> new region of grey crescent area (part of 30*)  Future dorsal side of cell  Development of blastophore - Nieuwkoop Centre  Donor -> Embryo -> normal development  Recipient -> Embryo -> Twinned & duplicated axis. - 4 signal model of induction events -> Early Frog Embryo:  General mesoderm inducer -> Vegetal pole  Nieuwkoop centre -> Dorsal vegetal pole  Induces organiser  Organiser -> Induces neural ectoderm -> Dorsalises ventral mesoderm (Induces somites -> ventral mesoderm)  Ventral mesoderm antagonises dorsal mesoderm Enables establishment -> dorsal-ventral polarity & patterning

Question 75

Question 3 The grafted cells... (cells of one blastophore from which cells were added to another) 1. cannot change the fate of their neighbours 2. can change the fate of their neighbours

Answer 75

2. can change the fate of their neighbours Cytoplasmic determinants in early embro:  Notochord doesn’t change between specification & fate maps of tissues  Grey crescent region (pigmented area in embryo) - 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) • What generates grey crescent & why is it important? - Cortical rotation breaks in radial symmetry  Unferitilised egg -> Radially symmetrical  Following fertilisation -> Zygote becomes bilaterally symmetrical Darker pigment / grey crescent rotates 30*  Cleavage occurs -> new region of grey crescent area (part of 30*)  Future dorsal side of cell  Development of blastophore - Nieuwkoop Centre  Donor -> Embryo -> normal development  Recipient -> Embryo -> Twinned & duplicated axis. - 4 signal model of induction events -> Early Frog Embryo:  General mesoderm inducer -> Vegetal pole  Nieuwkoop centre -> Dorsal vegetal pole  Induces organiser  Organiser -> Induces neural ectoderm -> Dorsalises ventral mesoderm (Induces somites -> ventral mesoderm)  Ventral mesoderm antagonises dorsal mesoderm Enables establishment -> dorsal-ventral polarity & patterning

Question 76

Question 4 The diagram on the left depicts the skeletal morphology of a normal mouse and the expression patterns of the Hox 10 paralogs (light blue box) and the Hox 11 paralogs (dark blue box). At the top are the cervical vertebrae C1-C7, which have no ribs. Next are the ribbed thoracic vertebrae, T1-T13 with T1-T7 wrapped around to connect to the sternum. Then come the lumbar vertebrae, L1-L6, the sacral vertebrae S1-S4 (which articulate with the pelvis--> Fused sacral wings), and the many small caudal vertebrae that form the tail. The diagram in the middle depicts the skeleton of mutant mouse that has had all the Hox10 paralogs knocked out [deleted] (homozygous mice). The diagram on the right depicts the skeleton of a mutant mouse that has had all the Hox11 paralogs knocked out (homozygous mice). Normal Mouse: - > Hox 10 -> Sacral vertebrae 1-4 & 6 susbsequent caudal vertebrae. - > Hox 11 -> Lumbar vertebrae 1-6 & sacral vertebrae 1-4. Hox 10 Mutated Mouse: -> Sacral vertebrae 1-4 & 6 susbsequent caudal vertebrae. >> All have ribs attached (& are highlighted blue?) >> Sacral vertebrae have orange outline?? >> Sacral vertebrae ribs attached between them & the fused sacral wings Hox11 -> No sacral vertebrae >> Replaced by lumbar vertebrae >> 5 subsequent caudal vertebrae also missing, replaced by lumbar vertebrae. -> No fused sacral wings at all >> All lumbar vertebrae highlighted in blue? --> Everything else not mentioned regarding structure of Hox 10 & 11 Mice is the same as Normal mouse. Below is an analysis of these results. For each stem, choose the correct ending. The Hox 10 mutant displays: 1. ectopic cervical and thoracic vertebrae. (In wrong place) 2. missing ribs on the lumbar and sacral vertebrae. 3. ectopic ribs on the lumbar and sacral vertebrae. (In wrong place) 4. ectopic ribs on the cervical and thoracic vertebrae. (In wrong place) 5. missing ribs on the cervical and thoracic vertebrae. 6. ectopic lumbar and sacral vertebrae. (In wrong place)

Answer 76

3. ectopic ribs on the lumbar and sacral vertebrae.

Question 77

Question 5 The diagram on the left depicts the skeletal morphology of a normal mouse and the expression patterns of the Hox 10 paralogs (light blue box) and the Hox 11 paralogs (dark blue box). At the top are the cervical vertebrae C1-C7, which have no ribs. Next are the ribbed thoracic vertebrae, T1-T13 with T1-T7 wrapped around to connect to the sternum. Then come the lumbar vertebrae, L1-L6, the sacral vertebrae S1-S4 (which articulate with the pelvis--> Fused sacral wings), and the many small caudal vertebrae that form the tail. The diagram in the middle depicts the skeleton of mutant mouse that has had all the Hox10 paralogs knocked out [deleted] (homozygous mice). The diagram on the right depicts the skeleton of a mutant mouse that has had all the Hox11 paralogs knocked out (homozygous mice). Normal Mouse: - > Hox 10 -> Sacral vertebrae 1-4 & 6 susbsequent caudal vertebrae. - > Hox 11 -> Lumbar vertebrae 1-6 & sacral vertebrae 1-4. Hox 10 Mutated Mouse: -> Sacral vertebrae 1-4 & 6 susbsequent caudal vertebrae. >> All have ribs attached (& are highlighted blue?) >> Sacral vertebrae have orange outline?? >> Sacral vertebrae ribs attached between them & the fused sacral wings Hox11 -> No sacral vertebrae >> Replaced by lumbar vertebrae >> 5 subsequent caudal vertebrae also missing, replaced by lumbar vertebrae. -> No fused sacral wings at all >> All lumbar vertebrae highlighted in blue? --> Everything else not mentioned regarding structure of Hox 10 & 11 Mice is the same as Normal mouse. Together with the endogeneous expression pattern of the Hox 10 paralogs, the mutant data suggest that the normal Hox 10 paralogs function is to: 1. Prevent rib formation on the lumbar and sacral vertebrae. 2. Give thoracic and lumbar vertebrae their identity 3. Give cervical and thoracic vertebrae their identity 4. Activate rib formation on thoracic vertebrae

Answer 77

1. Prevent rib formation on the lumbar and sacral vertebrae. - -> Ribs present on lumbar & sacral where they should not be. - > Thoracic, lumbar & cervical vertebrae already seem to have their identity as they all are the same as normal mouse. Eg. Thoracic vertebrae already have ribs present.

Question 78

The diagram on the left depicts the skeletal morphology of a normal mouse and the expression patterns of the Hox 10 paralogs (light blue box) and the Hox 11 paralogs (dark blue box). At the top are the cervical vertebrae C1-C7, which have no ribs. Next are the ribbed thoracic vertebrae, T1-T13 with T1-T7 wrapped around to connect to the sternum. Then come the lumbar vertebrae, L1-L6, the sacral vertebrae S1-S4 (which articulate with the pelvis--> Fused sacral wings), and the many small caudal vertebrae that form the tail. The diagram in the middle depicts the skeleton of mutant mouse that has had all the Hox10 paralogs knocked out [deleted] (homozygous mice). The diagram on the right depicts the skeleton of a mutant mouse that has had all the Hox11 paralogs knocked out (homozygous mice). Normal Mouse: - > Hox 10 -> Sacral vertebrae 1-4 & 6 susbsequent caudal vertebrae. - > Hox 11 -> Lumbar vertebrae 1-6 & sacral vertebrae 1-4. Hox 10 Mutated Mouse: -> Sacral vertebrae 1-4 & 6 susbsequent caudal vertebrae. >> All have ribs attached (& are highlighted blue?) >> Sacral vertebrae have orange outline?? >> Sacral vertebrae ribs attached between them & the fused sacral wings Hox11 -> No sacral vertebrae >> Replaced by lumbar vertebrae >> 5 subsequent caudal vertebrae also missing, replaced by lumbar vertebrae. -> No fused sacral wings at all >> All lumbar vertebrae highlighted in blue? --> Everything else not mentioned regarding structure of Hox 10 & 11 Mice is the same as Normal mouse. The Hox 11 mutant displays: 1. ectopic lumbar and sacral vertebrae (in wrong position/location) 2. missing ribs on the lumbar and sacral vertebrae 3. ectopic lumbar vertebrae and missing sacral vertebrae (in wrong position/location) 4. ectopic lumbar vertebrae (in wrong position/location)and missing all sacral and some caudal vertebrae 5. ectopic lumbar vertebrae and caudal vertebrae (in wrong position/location) 6. ectopic ribs on the thoracic vertebrae (in wrong position/location)

Answer 78

4. ectopic lumbar vertebrae and missing all sacral and some caudal vertebrae

Question 79

.The diagram on the left depicts the skeletal morphology of a normal mouse and the expression patterns of the Hox 10 paralogs (light blue box) and the Hox 11 paralogs (dark blue box). At the top are the cervical vertebrae C1-C7, which have no ribs. Next are the ribbed thoracic vertebrae, T1-T13 with T1-T7 wrapped around to connect to the sternum. Then come the lumbar vertebrae, L1-L6, the sacral vertebrae S1-S4 (which articulate with the pelvis--> Fused sacral wings), and the many small caudal vertebrae that form the tail. The diagram in the middle depicts the skeleton of mutant mouse that has had all the Hox10 paralogs knocked out [deleted] (homozygous mice). The diagram on the right depicts the skeleton of a mutant mouse that has had all the Hox11 paralogs knocked out (homozygous mice). Normal Mouse: - > Hox 10 -> Sacral vertebrae 1-4 & 6 susbsequent caudal vertebrae. - > Hox 11 -> Lumbar vertebrae 1-6 & sacral vertebrae 1-4. Hox 10 Mutated Mouse: -> Sacral vertebrae 1-4 & 6 susbsequent caudal vertebrae. >> All have ribs attached (& are highlighted blue?) >> Sacral vertebrae have orange outline?? >> Sacral vertebrae ribs attached between them & the fused sacral wings Hox11 -> No sacral vertebrae >> Replaced by lumbar vertebrae >> 5 subsequent caudal vertebrae also missing, replaced by lumbar vertebrae. -> No fused sacral wings at all >> All lumbar vertebrae highlighted in blue? --> Everything else not mentioned regarding structure of Hox 10 & 11 Mice is the same as Normal mouse. Question 7 Together with the endogeneous expression pattern of the Hox 11 paralogs, the mutant data suggest that the normal Hox 11 paralogs function is to: 1.. Give lumbar vertebrae their identity 2. Prevent the formation of lumbar vertebrae and induce the formation of sacral vertebrae and some caudal vertebrae 3. Induce the formation of the sacral and some caudal vertebrae 4. Prevent the formation of the sacral and some caudal vertebrae

Answer 79

?? 2. Prevent the formation of lumbar vertebrae and induce the formation of sacral vertebrae and some caudal vertebrae ? -> Too many lumbar vertebrae & present where not supposed to be -> possibly mistaken / no identity -> Would lumbar vertebrae form if sacral & caudal vertebrae present? >> Didn't form where actual caudal vertebrae present so this suggests not. Prevention of formation of lumbar vertebrae may also prevent formation of any; not just the misplaced ones; which would not be correct. -> Inducing formation of sacral & some caudal vertebrae may be sufficient to solve problem,

Question 80

The diagram on the left depicts the skeletal morphology of a normal mouse and the expression patterns of the Hox 10 paralogs (light blue box) and the Hox 11 paralogs (dark blue box). At the top are the cervical vertebrae C1-C7, which have no ribs. Next are the ribbed thoracic vertebrae, T1-T13 with T1-T7 wrapped around to connect to the sternum. Then come the lumbar vertebrae, L1-L6, the sacral vertebrae S1-S4 (which articulate with the pelvis--> Fused sacral wings), and the many small caudal vertebrae that form the tail. The diagram in the middle depicts the skeleton of mutant mouse that has had all the Hox10 paralogs knocked out [deleted] (homozygous mice). The diagram on the right depicts the skeleton of a mutant mouse that has had all the Hox11 paralogs knocked out (homozygous mice). Normal Mouse: - > Hox 10 -> Sacral vertebrae 1-4 & 6 susbsequent caudal vertebrae. - > Hox 11 -> Lumbar vertebrae 1-6 & sacral vertebrae 1-4. Hox 10 Mutated Mouse: -> Sacral vertebrae 1-4 & 6 susbsequent caudal vertebrae. >> All have ribs attached (& are highlighted blue?) >> Sacral vertebrae have orange outline?? >> Sacral vertebrae ribs attached between them & the fused sacral wings Hox11 -> No sacral vertebrae >> Replaced by lumbar vertebrae >> 5 subsequent caudal vertebrae also missing, replaced by lumbar vertebrae. -> No fused sacral wings at all >> All lumbar vertebrae highlighted in blue? --> Everything else not mentioned regarding structure of Hox 10 & 11 Mice is the same as Normal mouse. Question 8 When considered together, the expression patterns of the Hox 10 and Hox 11 paralogs suggest that: 1. Hox 10 and 11 work together to form the fused wings of the sacral vertebrae and Hox 11 works alone for the lumbar identity 2. Hox 10 and 11 work together to give the sacral vertebrae their identity and Hox 11 works alone for the caudal identity 3. Hox 10 and 11 work together to prevent lumbar vertebrae formation and Hox 11 works alone for the caudal identity 4. Hox 10 and 11 work together to prevent sacral vertebrae formation and Hox 11 work alone to prevent the caudal vertebrae formation

Answer 80

?? 3. Hox 10 and 11 work together to prevent lumbar vertebrae formation and Hox 11 works alone for the caudal identity ? -> Fused wings of sacral vertebrae not present at all when Hox 11 knocked-out, indicating this is solely responsible for their formation. So 1 untrue. -> Sacral vertebrae present when Hox 10 gene removed & nothing wrong with it otherwise as rib formation caused by other factor & not identity of scaral vertebrae. Therefore only Hox 11 responsible for identity of Sacral vertebrae. So 2 untrue. ***** NEED TO REVIEW

Question 81

Match each of the answers with the letters on the figure. Trophectoderm Blastocoel Primitive endoderm Contains pluripotent stem cells ``` 1. A 2. D 3. C 4. B ``` -> Type of cell >> Empty in middle / no components & doesn't even appear to be cytoplasm >> Circle of green cells / sections forming circular ring >> At top pole of cell; cluster of blue cells attached to inner lining of green circular ring >> Yellow band of cells enclosing the cluster of blue cells to the green circular ring & separating from apparent hole. A points to one of the green cells B points to the blue cells C points to the yellow cells D points to the open space/ vacuole.

Answer 81

``` Trophectoderm 1. A Blastocoel 2. D Primitive endoderm 3. C Contains pluripotent stem cells 4. B ``` • Early Mammalian Embryos/ Structure of Mammalian Blastocysts: - Trophectoderm: >> Outer layer of cells of a blastocyst. >> Provide nutrients -> the developing embryo, facilitate attachment to the uterine lining and become part the placenta. - Blastocoel: >> Fluid-filled, inner cavity of a blastocyst resulting from cleavage of the oocyte after fertilization. - Epiblast: (Primitive mesoderm) >> Forms embryo proper >> Cell layer -> develops from inner cell mass (accumulation of cells attached to inner lining of trophoblast) of blastocyst. >> Differentiates -> Ectoderm, mesoderm & endoderm >> Develops -> Extraembryonic membranes. - Hypoblast (Primitive endoderm) >> Forms inner cell mass >> Layer of cells which enclose epiblast / accumulation of cells attached -> trophoblast & separate from blastocoel. >> Develops -> Extraembryonic membranes

Question 82

Diagram -> Type of cell >> Empty in middle / no components & doesn't even appear to be cytoplasm >> Circle of green cells / sections forming circular ring >> At top pole of cell; cluster of blue cells attached to inner lining of green circular ring >> Yellow band of cells enclosing the cluster of blue cells to the green circular ring & separating from apparent hole. A points to one of the green cells B points to the blue cells C points to the yellow cells D points to the open space/ vacuole. Question 10 Gives rise to extraembryonic membranes 1. A & D 2. A & C 3. A & B 4. B

Answer 82

?? • Early Mammalian Embryos/ Structure of Mammalian Blastocysts: - Trophectoderm: >> Outer layer of cells of a blastocyst. >> Provide nutrients -> the developing embryo, facilitate attachment to the uterine lining and become part the placenta. - Blastocoel: >> Fluid-filled, inner cavity of a blastocyst resulting from cleavage of the oocyte after fertilization. - Epiblast: (Primitive mesoderm) >> Forms embryo proper >> Cell layer -> develops from inner cell mass (accumulation of cells attached to inner lining of trophoblast) of blastocyst. >> Differentiates -> Ectoderm, mesoderm & endoderm >> Develops -> Extraembryonic membranes. - Hypoblast (Primitive endoderm) >> Forms inner cell mass >> Layer of cells which enclose epiblast / accumulation of cells attached -> trophoblast & separate from blastocoel. >> Develops -> Extraembryonic membranes

Question 83

Diagram -> Type of cell >> Empty in middle / no components & doesn't even appear to be cytoplasm >> Circle of green cells / sections forming circular ring >> At top pole of cell; cluster of blue cells attached to inner lining of green circular ring >> Yellow band of cells enclosing the cluster of blue cells to the green circular ring & separating from apparent hole. A points to one of the green cells B points to the blue cells C points to the yellow cells D points to the open space/ vacuole. Question 11 The inner cell mass includes 1. A 2. B & D 3. B & C 4. A & C

Answer 83

3. B & C Cell mass incl. - Hypoblast (Primitive endoderm) - Epiblast (Primitive mesoderm) • Early Mammalian Embryos/ Structure of Mammalian Blastocysts: - Trophectoderm: >> Outer layer of cells of a blastocyst. >> Provide nutrients -> the developing embryo, facilitate attachment to the uterine lining and become part the placenta. - Blastocoel: >> Fluid-filled, inner cavity of a blastocyst resulting from cleavage of the oocyte after fertilization. - Epiblast: (Primitive mesoderm) >> Forms embryo proper >> Cell layer -> develops from inner cell mass (accumulation of cells attached to inner lining of trophoblast) of blastocyst. >> Differentiates -> Ectoderm, mesoderm & endoderm >> Develops -> Extraembryonic membranes. - Hypoblast (Primitive endoderm) >> Forms inner cell mass >> Layer of cells which enclose epiblast / accumulation of cells attached -> trophoblast & separate from blastocoel. >> Develops -> Extraembryonic membranes

Question 84

Question 12 Match model organism with statements Dictyostelium discoidum Arabidopsis thaliana Caenorhabditis elegans Drosophila melanogaster Ciona intestinalis Danio rerio Xenopus laevis ``` Mus musculus A. basal chordate B. Transparent embryos, external development C. Strong genetics, mammal D. Genetically tractable insect model E. can be used to study social behavior, cheating and altruism F. genetically tractable plant model G. self-reproduction in hermaphrodites H. Good system for transplantation experiments, external development ```

Answer 84

Dictyostelium discoidum E. can be used to study social behaviour, cheating and altruism Arabidopsis thaliana F. genetically tractable plant model Caenorhabditis elegans G. self-reproduction in hermaphrodites Drosophila melanogaster D. Genetically tractable insect model Ciona intestinalis A. basal chordate Danio rerio B. Transparent embryos, external development Xenopus laevis H. Good system for transplantation experiments, external development Mus musculus C. Strong genetics, mammal ``` • Caenorhabditis elegans - Short Generation time - Sexual & self-reproduction - Males & hermaphrodites Hemaphrodites produce eggs & sperm --> Can self-fertilise & perform genetic cross -> One organism. - Genetics - RNAi knockdown technology - Sequenced genome - Study -> Aging - Lineage of all cells known - Imaging - Provided most knowledge -> Cell death ``` • Arabidopsis thaliana (Thale Cress) - Small - Easy to grow - Short generation - Flowering plant - Embryology - Good imaging - Genetics - Sequenced genome - Trasngenesis - Agrobacterium -> Make transgenic plants ``` • Dictyostelium discoidum - Chemotaxis -> Moves towards signals sent by other molecules (CAMP) to aggregate & form fruiting body under stressful conditions. - Genetics - Multicellularity -> Single amoeba cell which can aggregate to form fruiting body understressful conditions. -> Low food conditions -> form spores. - Good imaging - Sequenced genome - Transgenesis - Social behaviour - Cheating & altruism - Chemotaxis toward cAMP ``` • Drosphila melanogaster (Fruit Fly) - Short generation time - Genetics - Sequenced genome - Transgenesis - Imaging - Development of segments -> Segmented organisms • Ciona intestinalis: - Basal chordate - Transgenesis - Genetics - Sequenced genome - Imaging - Develop notochord in dev. after 1 day • Danio rerio: (Zebrafish) - Genetics - Sequenced genome - Transgenesis - Imaging - Transparent embryos - Regeneration capacity - External development - Vertebrate Xenopus: - Embryos produced -> Large nos - External development - Rapid development - Cell division - Transparent tadpoles - Easy -> production of explants & transplants Can cut embryos -> albino & pigmented. Attach different halves -> rapid effective healing -> produce half albino & half pigmented individuals. - Generation of transgenic embryos -  Promoter elements / regulatory proteins of other organisms -> Mouse elastase-GFP transgenic embryo -> insertion of other organism -> Mouse Tie2-GFP transgenic tadpole -> Circulatory system - Rapid, effective wound healing Imaging - Regeneration of appendages -> tadpoles - Tetrapods • Mus musculus: (Mouse) - Mammal - Strong genetics - Sequenced genome - Mammal - Pluripotent embryonic stem cells - Short generation time

Question 85

Question 13 The Waddington landscape states that: A. Cell fate becomes more restricted with time B. Cell fate decisions are irreversible C. Cell fate decisions are reversible D. Potency increases with time E. Potency decreases with time None of the above B, D A, C, E A, B, E

Answer 85

A, B, E A. Cell fate becomes more restricted with time B. Cell fate decisions are irreversible E. Potency decreases with time Waddington Landscape Hypothesis:  Totipotent cells become differentiated > Once differentiation begins, it cannot be reversed. > Potency decreases over time.

Question 86

Question 14 Match person(s) with experiment Hans Spemann Briggs and Kings John Gurdon Campbell and Wilmut Shinya Yamanaka Hans Driesch A. First to clone an animal B. First to successfully transplant nuclei into eggs C. Used baby hair to constrict embryos and show that blastomeres in salamander embryos up are totipotent up to 16 cell stage D. First to induce pluripotency in adult cells without nuclear transplantation E. First to clone a mammal F. First to show each cell of a two-cell stage sea urchin is totipotent

Answer 86

Hans Spemann C. Used baby hair to constrict embryos and show that blastomeres in salamander embryos up are totipotent up to 16 cell stage Campbell and Wilmut E. First to clone a mammal Shinya Yamanaka D. First to induce pluripotency in adult cells without nuclear transplantation Hans Driesch F. First to show each cell of a two-cell stage sea urchin is totipotent Briggs and Kings B. First to successfully transplant nuclei into eggs John Gurdon A. First to clone an animal

Question 87

Question 15 John Gurdon and Shinya Yamanaka's experiments showed that: A. Adult cells can be reprogrammed to an embryonic pluripotent state B. Adult cells are unable to reprogram to an embryonic pluripotent state C. Differentiated cells retain all the genes required for the development of an organism D. Differentiated cells loose some of the genes during their differentiation programme. B & C A & D B & D A & C

Answer 87

 A & C A. Adult cells can be reprogrammed to an embryonic pluripotent state C. Differentiated cells retain all the genes required for the development of an organism Cloning Protocol: (John Gurdon)  Extraction -> Terminally different pigmented cells -> Tadpole epithelia. > Transplanted into denucleated wild type (pigmented) eggs -> (Denucleated via UV radiation)  Serial transplantations -> Produced blastula stage embryo >> Died.  Transplant nuclei -> Dying blastula stage embryo -> Into Albino type Eggs > Repeat several times.  Eventually developed fully formed identical frogs • Reprogramming Differentiated cells: (Shinya Yamanaka)  Differentiated cells reprogrammed -> return to Pluripotent Embryonic state >> Without use of nuclear transplantation. >> Production -> Induced Pluripotent Stem (iPS) cells  Process: - Various solutions of gene mixtures added -> Fibroblasts >> Investigate which mixtures reprogramme cells > Produce Pluripotent Embryonic cells  Result: - Generation of clones -> Induced Pluripotent Stem (iPS) stem cells >> Act like Embryonic stem cells -> iPS > Cocktail : Klf-4, Sox-2, Oct4, Myc  Uses / Benefits: - Can use any cell Eg. Skin -> Produce Pluripotent stem cells >> Embryonic stem cells not required -> Ethics - Produce pluripotent stem cells >> Determine / manipulate development -> specific required cells. - Produce genetically modified organism - Used to generate human iPS cells - Taken from those with disease >> Model diseases in vitro for investigation rather than patient - Use in transplants >> Production of cells / organs / tissues which are not immunologically rejected. - Produce iPS cells from patients >> Cell replacement therapies / regenerative medicine. Eg. Mouse -> Sickle Cell Anaemia

Question 88

Question 16 Mechanisms of gene duplication include: A. Tandem duplications B. Translocations C. Whole genome duplications D. Inversion E. Segmental duplications A only D only C only A, C, E only A, C only B, D only

Answer 88

Selected Answer: A, C, E only -> Tandem duplications, whole genome duplications, Segmental duplications ``` • Evolutionary Developmental Biology: - Gene Duplication Events  Establishes gene redundancy  Enables maintenance of some functions -> One of duplicated genes >> Other can acquire new functions -> without loss of ancestral function. -->> Development -> Novel function.  Methods:  Tandem Gene Duplication  Segmental Duplication  Whole Genome Duplication ```  Tandem Gene Duplication: >> Creation of gene clusters  Unequal crossing-over > Mis-pairing of chromosome during meiosis --> Possible cause -> Repeat DNA seq.  Leads to:  Paralogous genes:  > Duplicated / same genes -> common ancestor -> within single chromosome  Orthologous genes: > Same gene present -> common ancestor -> diff. organisms  Subfunctionalisation: > Acquisition of new function -> duplicated gene -> following multiple tandem gene duplications: > Other function not lost -> multiple other duplicate genes still code for this function -> No effect on organism. > Methods: 1. Duplication 2. Divergence > Removal of duplicated gene if new function not acquired. >> 1. Duplication  Alteration -> Protein sequence ->> Similar proteins & structures ->> New binding properties / slightly diff. functions etc.  Can also happen -> transcription factors Eg. Hox genes ->> Same derived ancestral hox gene -> Different functions -> dependent on location of body in which expressed. 2. Divergence  Alteration -> Time / place of expression ->> Duplication -> Cis-regulatory elements ->> Mutations -> Regulatory regions > New expression domains > Change -> expression / timing > Altered level of expression.  Segmental duplication:  Similar -> Giant tandem duplication >> Affects whole sections of chromosome  Elongation -> existing gene clusters >> (Formed -> Tandem gene duplication) >> Localised, duplicated genes already exist -> related to one another. -> Uneven cross-over > Further incr. number of duplicated genes in this region.  Whole genome duplication:  Duplication of single genome duplication event.  Failure of meiosis -> Results in diploid germ cells (sperm/egg)  Fertilisation -> tetraploid organism.  Types: 1.  Allotetraploidy >> Hybridisation between 2 separate species -->> Closely related enough for chromosomal cross-over -->> Not sufficiently closely related for proper meiosis >> Meiosis failure ->> However 4 chromosomes not identical -> Not from same species. 2.  Autotetraploidy >> Duplication of genome through improper meiosis >> Meiosis failure ->> 4 identical chromosomes. ```  Formation -> Vertebrate Lineage: 2 rounds: (2R Hypothesis) >> Duplication >> Gene Loss >> 2nd Round Duplication >> Gene Loss  Results in Hox Genes 1st Event: -> Lamprey Lineage (Jawless Fish) -> Most basal fish 2 Hox Clusters 2nd Event: -> More advanced fishes 4 Hox Clusters 3rd Event: -> Bony Fish 8 Hox Clusters ```

Question 89

Question 17 Match definition with term Orthologous genes Paralogous genes Homologous genes Homeotic mutation ``` A. Same genes in different organisms B. Duplicated genes within an organism C. Genes that share a common ancestor D. Mutation that leads to the transformation one body part into another ```

Answer 89

Orthologous genes A. Same genes in different organisms Paralogous genes B. Duplicated genes within an organism Homologous genes C. Genes that share a common ancestor Homeotic mutation D. Mutation that leads to the transformation one body part into another  Paralogous genes:  > Duplicated / same genes -> common ancestor -> within single chromosome  Orthologous genes: > Same gene present -> common ancestor -> diff. organisms • Homeotic mutation: - Transformation of one body part into another >> Development of correctly developed structure in incorrect location of body. Eg. Legs instead of antennae -> Head Wings instead of halteres. -> 2nd thoracic segment of insects. Homologous genes -> Genes derived from a common ancestor, shared by a group of organisms