Genes, Evo & Dev Blackboard Exam qs Flashcards

Question 1

Q

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

Answer

A

inherited characters that vary and differential survival of individuals

Question 2

Q

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

Answer

A

. the specific ecological conditions found on the islands

NOT
changing climatic conditions

Question 3

Q

Molecular phylogenies:

Answer

A

all support the existence of evolution

Question 4

Q

Fitness is:

Answer

A

measured by the number of offspring an individual produces

Question 5

Q

Patterns of lactose intolerance in humans:

Answer

A

are an example of natural selection

Question 6

Q

There are many species of marsupials in Australia because:

Answer

A

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

Q

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

A

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

Q

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

A

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

Q

Synonymous base substitutions in a gene:

Answer

A

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

Q

The molecular clock hypothesis implies that:

Answer

A

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

Q

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

A

Allopatric speciation

Question 12

Q

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

Answer

A

allopolyploidy

Question 13

Q

Hybrid infertility is an example of:

Answer

A

a postzygotic reproductive barrier

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

Question 14

Q

Which is NOT an example of a reproductive isolating mechanism?

Answer

A

hybrid zones

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

Question 15

Q

What would happen if all animals were herbivores?

Answer

A

A mutation that led to predatory behaviour would be extremely successful

Question 16

Q

Competition:

Answer

A

occurs within and between species

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

Question 17

Q

Genetic drift leads to:

Answer

A

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

Q

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

A

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

Q

Inbreeding leads to:

Answer

A

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

Q

allopatric speciation

Answer

A

speciation occurring following geographic isolation

Question 21

Q

altruism

Answer

A

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

Question 22

Q

bottleneck

Answer

A

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

Question 23

Q

disruptive selection

Answer

A

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

Question 24

Q

Hamilton’s rule

Answer

A

an expression that predicts when altruism should be favoured

Question 25

Q

inclusive fitness

Answer

A

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

Question 26

Q

neutral theory

Answer

A

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

Question 27

Q

phylogeny

Answer

A

the evolutionary history of a group of species

Question 28

Q

polyandry

Answer

A

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

Question 29

Q

postzygotic isolation

Answer

A

reproductive isolation resulting from decreased fitness of hybrids

Question 30

Q

transitional form

Answer

A

a species that exhibits traits common to ancestral and derived species

Question 31

Q

morphospecies

Answer

A

populations designated as separate species based on differences in form

Question 32

Q

The dorsal fins of a porpoise and a salmon

Answer

A

Analogous

Question 33

Q

The bones making up the human arm and the bat wing

Answer

A

Homologous

Question 34

Q

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

Answer

A

Analogous

Question 35

Q

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

Answer

A

Homologous

Question 36

Q

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

Answer

A

Analogous

Question 37

Q

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

A

Females prefer long tailed males as their sexual partner

Question 38

Q

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

A

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

Q

Question 1
In which cell cycle stage does DNA replication occur?

G1

G0

G2

S

Question 40

Q

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

A

20 chromosomes and 40 chromatids

Question 41

Q

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

A

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

Q

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

A

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

Q

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

A

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

Q

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

A

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

Q

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

A

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

Q

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

A

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

Q

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

A

The pseudoautosomal region is unique to the X chromosome

•	Pseudoautosomal regions:
-	PAR1 &amp; PAR2 
-	Shared between X &amp; 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

Q

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

A

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
&raquo_space;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: &raquo_space; X+XW & X+Y
Red Females & Red Males
-> 100% red offspring

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

Q

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

A

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
&raquo_space;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

Q

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

A

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 &amp; dp on furthest possible points away from each other
--> makes sense as 
dp-ap = 42
dp-cl + cl-ap = 3 + 39 = 42

Question 51

Q

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

5 cM
1 cM

5 cM

Answer

A

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 &amp; d -> one chromosome
--> c &amp; D -> homologous chromosome
> CcDd offspring 
--> Cd &amp; cD chromosomes from parent
--> cd from tester parent
>> So Cd &amp; cd -> Ccdd
           cD &amp; cd -> ccDd
   --> (parent strains)

> CcDd x ccdd
953 Cc dd, -> Parent
947 cc Dd,  -> Parent 
46 Cc Dd
54 cc dd
Therefore recombinant -> CcDd &amp; 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.
&raquo_space; 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 &amp; d loci 
> CcDd x ccdd
953 Cc dd, -> Parent
947 cc Dd,  -> Parent 
46 Cc Dd
54 cc dd
Therefore recombinant -> CcDd &amp; 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

Q

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

A

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 &amp; 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

Q

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

A

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

Q

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

A

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

Q

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

A

Mitochondria and chloroplasts

Question 56

Q

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

A

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 &amp; DZ twins used to estimate hertitability.     Eg. MZ concordance = 0.50 (50%) ; DZ  = 0.08 (8%) 
      Heritability = 0.58

Question 57

Q

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

A

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)
&raquo_space; Amplifies 10 STRs & Gender DNA marker
>Separated -> electrophoresis -> DNA profile.
Forensic uses -> DNA profiles:
&raquo_space; Each individual excl. identical twins -> unique DNA profile
&raquo_space; Individuals placed at scene -> analysis -> hair, blood, saliva / semen
samples.
&raquo_space; DNA profiles of crime -> compared -> suspect / database profiles
&raquo_space; 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:
&raquo_space; Gene chip technology -> identification -> common SNPs -> across
ethnic groups.
SNP genotyping -> GeneChip
&raquo_space; 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

Q

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

A

during foetal development (before birth)

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

Question 59

Q

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

A

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

Q

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

A

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

Q

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

A

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

Q

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

A

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

Q

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

A

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

Q

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

A

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

Answer 64

A

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.

Answer 65

A

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

Answer 66

A

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

Answer 67

A

	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

Answer 68

A

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_)
&raquo_space; Cant possess a allele
= AACc

F2 -> No albino offspring (__cc)
&raquo_space; Cant possess c allele
= AaCC

F3 -> Must therefore be AaCc
F4 -> Must therefore be AACC

Answer 69

A

->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 &amp; 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.
&raquo_space; Mother -> equally related to all offspring
>Wants to divide resources equally
&raquo_space; Father -> likely related to subset of foetuses
>Wants to incr. survival chances of his offspring -> promoting
their growth.

2 normal : 2 dwarf

Answer 70

A

???

> 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

Answer 71

A

indicates whether an allele increases or decreases the susceptibility to disease

Answer 72

A

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
&raquo_space; 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
&raquo_space; Notochord comes from one embryo
&raquo_space; 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

Answer 73

A

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
&raquo_space; 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
&raquo_space; Notochord comes from one embryo
&raquo_space; 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

Answer 74

A

2.
can change the fate of their neighbours

Cytoplasmic determinants in early embro: 	Notochord doesn’t change between specification &amp; 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
&raquo_space; 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
&raquo_space; Notochord comes from one embryo
&raquo_space; 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

Answer 75

A

ectopic ribs on the lumbar and sacral vertebrae.

Answer 76

A

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.

Answer 77

A

ectopic lumbar vertebrae and missing all sacral and some caudal vertebrae

Answer 78

A

??

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,

Answer 79

A

??

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

Answer 80

A

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

Answer 81

A

??

• 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

Answer 82

A

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

Answer 83

A

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 &amp; self-reproduction 
-	Males &amp; hermaphrodites
Hemaphrodites produce eggs &amp; sperm 
   --> Can self-fertilise &amp; 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 &amp; 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 &amp; 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

Answer 84

A

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.

Answer 85

A

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

Answer 86

A

 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.
&raquo_space; 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

Answer 87

A

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 &amp; 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
&raquo_space; Affects whole sections of chromosome
 Elongation -> existing gene clusters
&raquo_space; (Formed -> Tandem gene duplication)
&raquo_space; 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
&raquo_space; Hybridisation between 2 separate species
–» Closely related enough for chromosomal cross-over
–» Not sufficiently closely related for proper meiosis
&raquo_space; Meiosis failure
-» However 4 chromosomes not identical
-> Not from same species.
2.  Autotetraploidy
&raquo_space; Duplication of genome through improper meiosis
&raquo_space; 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

Answer 88

A

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

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