Inheritance Flashcards

Question

What can novel genes and functions result in?

Answer 1

o Novel genes and functions, although rare, can have a significant impact

Answer 2

o Flower colour  Purple- dominant  White-recessive o Flower position  Axial- dominant  Terminal- recessive o Seed colour  Yellow- dominant  Green- recessive o Seed shape  Round-dominant  Wrinkled- recessive o Pod shape  Inflated- dominant  Constricted- recessive o Pod color  Green-dominant  Yellow- Recessive o Stem length  Tall- dominant  Dwarf- recessive

Answer 3

 Controlled by basic-helix-loop-helix (bHLH) transcription factor • bHLH switches on expression of genes involving in making anthocyanin (purple) pigments o bHLH binds to promoter of gene which leads to transcription of protein that makes anthocyanin pigments • When it isl mutant, flower is white o When there is G->A mutation in bHLH gene, bHLH transcription factor is not produced which means that can no longer bind to promoter of gene and protein that makes antocyanin pigment is not produced, turning the flower white

Answer 4

 Controlled by magnesium dechelatase gene (stay-green gene): • Positive regulator of the chlorophyll degrading pathway • When it is mutant, seed colour is green

Answer 5

 Controlled by starch-branching enzyme 1 (sbe I) • Converts amylose to amylopectin, the branched form of starch (starch is a polymer of glucose subunits, some of which are linear and some are branched) • When it is mutant, peas are wrinkled o In round peas, starch branching enzyme links all linear glucose molecules together to produce starch, which is a rigid structure making the peas appear round o The starch branching enzyme is wrinkled in peas, leading to reduced concentrations of amylopectin, the branched form of starch as glucose polymers not joined together  Higher sugar content means less water in peas making it wrinkled

Answer 6

 Controlled by gibberellin 3 beta-hydroxylase: • Gibberellin is a growth hormone that stimulates cell elongation and cases plants to grow taller • When it is mutant, plants are dwarves

Answer 7

• Dominance relationships between multiple alleles are relative

Answer 8

Context and what you measure

Answer 9

a heterozygote expresses the phenotype of both alleles simultaneously

Answer 10

 EG. ABO blood antigens • IA and IB allele produce A and B antigens and these alleles are co-dominant with each other, leading to an AB phenotype • Dominant to i allele (basis of O blood) that produces nothing

Answer 11

A heterozygote is intermediate between the two homozygous phenotypes

Answer 12

cross between a red lower and white flower gives pink flower

Answer 13

 Decrease or complete loss of functional gene product • Partial loss of gene function- leaky or hypomorphic mutation • Complete loss of gene function- null mutation  These mutations are usually recessive -Include haplo-sufficient and haplo-insufficient mutations

Answer 14

 Haplo-sufficient- one wild type allele provides enough normal gene product to produce a wild-type phenotype • The mutation is recessive

Answer 15

• Example- albinism o Albinism results from loss of melanin pigmentation in skin cells o Several genes control the synthesis of melanin in melanocytes  Tyrosinase gene takes tyrosine (unpigmented) and converts it into melanin (brown pigmentation)  Tyrosinase loss of function mutants are recessive  Sufficient enzyme produced in heterozygotes for normal pigmentation  Wild-type allele is haplo-sufficient

Answer 16

 Haplo-insufficient- a single wild-type allele in a heterozygote cannot provide the amount of gene product needed • The mutation is dominant

Answer 17

 Increase in functional gene product or new cellular function  These mutations are usually dominant

Answer 18

 EG- drosophila malenogaster • In drosophila melanogaster, the Antp gene is a homeotic gene (developmental regulator) that promotes leg growth • Antennapedia mutants arise from the mis-expression of the Antp gene causing legs to grow where antennae should be • Mutation is dominant because the gain-off-function cannot be compensated by the wild-type allele and heterozygotes show the phenotype

Answer 19

• While diploid individuals can only contain 2 alleles per locus, genes can have many different alleles within a population

Answer 20

• Alleles form an order of dominance or an allelic series, according to phenotypes expressed in the heterozygotes o Have to figure out relative dominance between multiple alleles

Answer 21

o The agouti in mice has more than 100 different alleles, including the black a allele, the yellow A^Y allele and the agouti A^w-J allele, which forms an allelic series of A^y>A^w-j>a

Answer 22

the agouti locus produces a molecule that regulates the synthesis of black pigment eumelanin  Different alleles of the agouti locus modify this pattern of bands on a hair shaft in various ways, with pheomelanin appearing when eumelanin production is blocked

Answer 23

 In dogs, the ay allele blocks synthesis of eumelanin so is dominant to all other alleles  Allele a^y>a^w>a^t>a  a lacks the ability to control distribution eumelanin, giving a completely black coat

Answer 24

o The C gene contains multiple alleles including the C allele (resulting in black rabbits), the c^ch allele (resulting in chinchilla grey rabbits), the c^h allele (resulting in himalayan rabbits) and the c allele, resulting in white rabbits o The allelic series is C>c^ch>c^h>c o The c,c^ch and c^h recessive alleles result in a decrease or complete loss of functional gene product  Chinchilla is partial loss of gene function (leaky or hypomorphic mutation)  White is complete loss of gene function (null mutation)  Himalayan is conditional mutation (mutant phenotype is temperature sensitive) • Enzyme active at low temperatures (black ears, nose…) • Enzyme inactive at high temperatures (white cores)

Answer 25

• Conditional mutants- mutation causes a phenotype only under certain environmental conditions (e.g. temperature, chemical exposure, resource availability)

Answer 26

o Restrictive condition- mutant phenotype (restricts growth of individual) o Permissive condition- wild-type phenotype (allow growth of individual/normal phenotype)

Answer 27

• Incomplete penetrance- not every individual with the mutant genotype displays a mutant phenotype

Answer 28

o Penetrance- the proportion of individuals of a specific genotype that exhibits the corresponding phenotype

Answer 29

o e.g. polydactyly  Single dominant gene, causing growth of extra fingers and toes  Incomplete penetrance- at least 1 in 4 people with the mutation have 5 digits (penetrance=0.75: ¼ of people have normal phenotype and ¾ of people have mutant phenotype)

Answer 30

alleles that cause the death of the organism that carries them

Answer 31

o Example: the agouti AY allele in mice  In mice, A alleles produce wild-type (agouti): dark hair with yellow band  The AY allele in mice is a gain of function mutation resulting in ubiquitous overexpression of the agouti protein (mutants produce solid yellow hair)  Homozygous mutants are embryonic lethal • Not possible to generate a pure-breeding yellow line

Answer 32

 Dominant lethal alleles • Homozygous dominant mutant will die • Offspring ratio= 2:1

Answer 33

 Recessive lethal alleles • Homozygous recessive mutant will die • Offspring ratio: 1 (all same dominant phenotype)

Answer 34

• Pleitropic genes- a single gene that affects more than one phenotype or process

Answer 35

o Example- the agouti AY allele affects both coat colour (yellow dominant), weight (obese dominant), tumor susceptibility, hyperinsulemia (leading to embryonic lethality)  Too much agouti protein results in increased antagonism of melanocortin receptors

Answer 36

• Mutants with the same phenotype could be alleles of the same gene, or mutations in two different genes

Answer 37

 A gene (agouti)- determines distribution of pigment along the hair  B gene- determines the colour of pigment  C gene- permits colour expression  D gene- controls the intensity of pigment  S gene- controls distribution throughout the body (spotting)

Answer 38

• The complementation test- a functional test between recessive mutants with the same phenotype Process: o Two homozygous recessive mutants with the same phenotype are mated o F1 offspring are analysed o If the F1 is a wild-type, the two mutations complement each other and represent two different genes o If the F1 is a mutant, the two mutations fail to complement and represent two different alleles of the same gene

Answer 39

• Genes rarely act in isolation • Can use genotypic and phenotypic ratios of some crosses to determine how many and what type of gene interactions are occurring o If deviation from expected ratios, means some kind of gene interaction is occurring

Answer 40

- No interaction - Complementary gene interaction - Duplicate gene interaction - Dominant gene interaction - Epistasis - ----Recessive epistasis - ----Dominant epistasis - ----Dominant suppression epistasis

Answer 41

 Two genes, no interaction |  Get a 9:3:3:1 ratio

Answer 42

 Two genes act sequentially to produce a phenotype. Both are required  Get a 9:7 ratio

Answer 43

 Two genes perform the same function (redundancy). Either are required  Get a 15:1 ratio

Answer 44

 Two genes with the same phenotype interact additively (causes enhanced/additive phenotype)  Get a 9:6:1 ratio

Answer 45

 One gene stops/masks the phenotype of another gene

Answer 46

• Recessive epistasis- homozygous recessive alleles for one gene masks the phenotypic expression of alleles at a second gene. Most common epistasis version. o Get a 9:3:4 ratio

Answer 47

• Dominant epistasis- dominant allele of one gene masks the phenotypic expression of another gene o Get a 12:3:1 ratio

Answer 48

• Dominant suppression epistasis- dominant allele of one gene suppress the expression of the dominant allele of another gene o Get a 13:3 ratio

Answer 49

• Chromosomes visualised as early as 1842

Answer 50

• Process of division described later (1873-82)• Process of division described later (1873-82)

Answer 51

o Mitosis- Somatic cells: Chromosomes replicate and divide, and chromosome number is maintained in daughter cells o Meiosis- Germline cells: Chromosomes replicate and undergo 2 rounds of division: chromosome number is halved in daughter cells

Answer 52

• Complete set of chromosomes is called a karyotype

Answer 53

o Complete set of human chromosomes is 23 pairs (46 chromosomes)

Answer 54

• A homologous chromosome pair has similar DNA sequence and encodes the same genes

Answer 55

Chromosome theory of inheritance states that individual genes are found at specific locations on particular chromosomes, and that the behavior of chromosomes during meiosis can explain why genes are inherited according to Mendel’s laws

Answer 56

• At the time it was proposed, the chromosome theory of inheritance lacked experimental proof and was wildly controversial o Morgan and drosophila melanogaster (1910)  Experimental evidence came from one of the first identified mutants in Drosophila melanogaster: a male with white eyes which is unusual as the wild type usually have red-eye colour  Morgan crossed his white type male to his wild type female, and got all wild type offspring (F1 individuals)  Morgan then crossed all his F1 individuals together but found that none of the females were white eyed, but half the males had red eyes and the other half had white eyes (not the 3:1 ratio expected)

Answer 57

o This is because chromosomes come in homologous pairs, and one member of the pair comes from the mother and the other from the father o The members of a homologous pair separate in meiosis, so each sperm or egg receives one- This process matches the segregation of alleles into gametes in Mendel’s law of segregation o Different chromosome pairs sort into gametes independently of one another in meiosis, like as predicted for genes in Mendel’s law of independent assortment

Answer 58

o ABC transporters-ATP-Binding cassette transporters which are transmembrane proteins that use ATP to transport molecules across membranes

Answer 59

• The white gene is an ABC transporter required for eye pigment production • White gene occurs upstream of many pigmentation genes and delivers pigment precursors guanine and tryptophan to the eye to make the brick red pigment o Trypotophan, aided or regulated by many genes, eventually makes zanthommatin brown pigment o Guanine, aided or regulated by many genes, eventually makes drosopterin, orange pigment • Xw mutation blocks the transport of pigment precursors guanine and tryptophan, resulting in white eyes

Answer 60

• Nettie Stevens (1905) used the mealworm beetle to show that: o Male gametes had 10 large chromosomes, or 9 large chromosomes and 1 small chromosome o Female gametes all had 10 large chromosomes • Observed sex-based differences in chromosomes from a range of insects

Answer 61

• In many organisms, sex is determined by the presence of sex chromosomes but some organisms do not have sex chromosomes

Answer 62

• Heterogametic sex- the sex with different sex chromosomes

Answer 63

• Homogametic sex- the sex with homologous sex chromosomes

Answer 64

• Hemizygote- Only one copy of an allele is present at a locus instead of two. The phenotype is present regardless of dominance

Answer 65

Heterogametic (XY)

Answer 66

Homogametic (XX)

Answer 67

o Sex chromosomes in humans were originally derived from a pair of homologous autosomes (non-sex chromosomes) 1. The Y chromosome acquired a sex determining locus • Accumulated genes with functions in sex and male fertility 2. Differentiation of X and Y specific regions, with small regions of homology retained in pseudo-autosomal regions (which help X and Y chromosomes to pair up during cell division) 3. Y chromosome lost genes and accumulated mutations 4. Dosage compensation 5. Final heteromorphic chromosomes differ greatly in size and morphology -Over evolutionary time, they have changed in size and morphology

Answer 68

o The X chromosome  165 Megabases, about 1000 genes  Highly conserved in placental mammals

Answer 69

o The Y chromosome  59 Megabases, about 200 genes  Specialised male fertility genes  Many residual and non-functional sequences  Many human sex-linked traits predominantly affect males  Y chromosome has changed due to gene loss over time  Sex determining region on the Y chromosome

Answer 70

o Pseudoautosomal region- regions of chromosomes that have maintained homology

Answer 71

X-linked genes

Answer 72

Y-linked genes

Answer 73

• Parental cross of homozygous wild-type female with hemizygous mutant male results in a 2:2:1: ratio when trait is sex linked o 2 female wild type: 2 male wild-type: 1 mutant male

Answer 74

• When trait is sex-linked, parental cross of a heterozygous wild-type female with hemizygous mutant male results in a 1:1:1:1 ratio when trait is sex-linked o 1 female wild type:1 female mutant:1 male wild-type: 1 male mutant

Answer 75

• Criss-cross inheritance- the trait goes from mother to son, mirroring the inheritance of the X chromosome o Mothers will only pass on X chromosomes to their sons, and fathers will only pass on X chromosomes to their daughters

Answer 76

o Mothers that are carriers of the trait are assumed based on the phenotype of sons  Unaffected mothers with affected sons must be heterozygotes  Mothers who are not affected when their sons are is a way of guessing that it is a sex-linked trait

Answer 77

o Trait occurs in males only o Occurs in all sons of the affected males o Daughters of affected males are normal and do not have affected offspring

Answer 78

 Trait is present in males hemizygous for the allele  Trait is present in females homozygous or heterozygous for the allele  Each individual who has the disease has at least one affected parent • Affected heterozygous females mated to normal males transmit the defect to ½ of their sons and ½ of their daughters • Affected hemizygous males transmit to all their daughters • Every affected has at least one affected parent, except in the case of de-novo mutations

Answer 79

 Trait is present in males hemizygous for the allele  Trait is present in females homozygous for the allele  The defect or disease may skip generations • Normal parents with a female carrier will give no affected females and ½ affected males • Affected males mated to normal females will have no affected offspring, but all daughters are carriers • Affected females mated to normal males will have all sons affected and no affected daughter, but all daughters are carriers

Answer 80

o Autosomal recessive inheritance  Individuals who have the disease are often born to parents wo do not  If only one parent has the disorder the risk that a child will have it depends on the genotype of the other parent  If both parents have the disorder, all children will have it  The sex ratio of affected offspring is expected to be equal  The disease is not usually seen in each generation but if an affected child is produced by unaffected parents, the risk to subsequent children is ¼  If the disease is rare in the population, unaffected parents of affected children are likely to be related to one another

Answer 81

 Each individual who has the disease has at least one affected parent  Males and females are affected in equal numbers  Either sex can transmit the disease allele  In crosses where one parent is affected and the other is not, approximately half the offspring have the disease  Two unaffected parents will not have any children with the disease  Two affected parents may produce unaffected children

Answer 82

``` o Instead of lower case upper case allelic notations, wild-type alleles are indicated by a + superscript.  For a recessive trait with allele e • Wild type allele: e+  For a dominant trait with allele B • Wild type allele: B+  For a sex-linked trait with allele w: • Wild type allele: Xw+ o Dominant mutations are given a capital letter and recessive mutations are lower case ```

Answer 83

o Drosophila melanogaster have 4 chromosomes o 1 X chromosome and 3 autosomes o Each chromosome has a left and right arm at either side of the centromere except for chromosome 4

Answer 84

• Sex determination in Drosophila depends on the number of X chromosomes, not on the presence of a Y chromosome: two X chromosome is a female and one X chromosome is a male o XX or XXY= female o XY or XO= male

Answer 85

* Used to determine sex-linkage: two crosses are performed, where the genotypes of the male and female parents are swapped * If F1 offspring ratios differ, it indicates that this trait is sex-linked

Answer 86

* Pedigrees, or family trees, are a way of tracing the inheritance of traits * They are useful to understand the mode of inheritance of a trait when offspring numbers are low, but family records are available

Answer 87

- Males: square - Females: Circle - Unaffected: Empty square/circle - Affected: Filled square/circle - Heterozygous: Dot inside square/Circle - Deceased: Diagonal line (from bottom left to top right inside shape) - Unspecified sex: Diamond - Generation: Vertical line - Parents: Horizontal line - Parents related by blood: 2 horizontal lines - Adoption: Dotted vertical line - Siblings: Horizontal line joining vertical lines - Identical twins: Two lines joined by horizontal line - Fraternal twins: Two lines - Generations: Indicated by roman numberals - Individuals in a generation: Indicated by arabic numerals Timestamp: 3:40 on the 13/08/2019

Answer 88

one mutant allele is sufficient for trait or disease

Answer 89

two mutant alleles result in trait or disease

Answer 90

hemizygous males express the trait, regardless of dominance

Answer 91

* Thousands of human genetic traits and diseases are caused by mutations of single genes which follow mendelian patterns of inheritance * Can occur on autosomes or sex chromosomes

Answer 92

• Many deleterious alleles causing genetic diseases are recessive (autosomal or X-linked) o Detection of carriers is important as carrierxcarrier matings will give affected offspring in autosomal genes at ¼ frequency • Can do molecular test where alleles of individuals are examined and detect whether the offspring has inherited 2 mutant alleles o Can also result in employment of personalised medicine where treatment is targeted to a person’s specific genomes and susceptibility

Answer 93

• Commercial genetic testing can report one’s traits and genetic characteristic but results should always be interpreted with a medical background as well, not just a genetic background and commercial genetic testing has many limitations.

Answer 94

• Many species exhibit sexual dimorphism- physical differences between males and females. For example: o Colour dimorphism o Size dimorphism

Answer 95

• There are lots of different mechanisms for sex determination- o Both genetic and environmental sex determination mechanisms exist

Answer 96

• Monoecious plants- male and female flowers on the same plant such as zucchini

Answer 97

o Hermaphrodites- contain both male and female sex organs

Answer 98

 94% of plants have both male and female organs, often within the same flower

Answer 99

 About 5% of animals (some worms, snails and fish (such as clownfish)) are hermaphrodites

Answer 100

• Dioecious plants- male and female flowers on different plants o Only 6% of plants have split into different sexes, including papaya and asparagus

Answer 101

genetic elements specify whether individuals are female or male o Individuals release some genetic signal during embryonic development to tell to develop as male or female

Answer 102

 Heteromorphic sex chromosomes- the presence of specific sex chromosomes that are different and visually distinct from each other, and distinct from autosomes • Heterogametic sex- the sex with different sex chromosomes • Homogametic sex- the sex with homologous sex chromosomes • For example, mammals, birds, some reptiles and fish, many invertebrates  Sex-determining genes on homomorphic chromosomes • Sex-determining/mating type genes are located on autosomes • Species do not have sex chromosomes • Amphibians, some reptiles and fish, some invertebrates o E.g. swordtails: a mating-type gene with 2 alleles determines sex. Females are homozygotes and males are heterozygotes  Chromosome ploidy: • Males develop from unfertilised eggs and are haploid (n) • Females develop from fertilized eggs and are diploid (2n) • Invertebrates (all bees, wasps, ants, most mites and ticks)

Answer 103

• Environmental sex determination- external stimuli control sex determination o Favoured when specific environments are more beneficial to one sex

Answer 104

1. Reptiles and birds: o Birds: ZW, Reptiles: ZW, XY and environmental o Sex chromosomes have no homology to mammalian XY (come about on different chromosomes present in these organisms). 2. Mammals o XY(or variations) o Monotremes: earliest offshoots of the mammalian lineage

Answer 105

• Sex determination is quite robust-genetic sex determination that can transition in environmental mechanism and vice-versa

Answer 106

• Australian bearded dragons use both sex chromosomes and temperature to determine sex o Normal temperature:  Heteromorphic sex chromosomes and genetic sex determination • ZW females and ZZ males o High temperatures higher than 32  Male dragons with ZZ sex chromosomes undergo sex reversal and develop as female • W chromosome eliminated from population  Must be advantages to having more females as females are favoured in high temperature zones o Transition to environmental sex determination

Answer 107

- XY systems - XO systems - ZW systems - Haplo-diploidy

Answer 108

 Female: XX • The absence of SRY allows expression of genes needed for ovary differentiation (the default pathway)  Male: XY • In mammals, the presence of the Y chromosome determines if an offspring develops as a male • Male sex depends on a master sex-determining locus, the SRY (sex determining region Y) gene • SRY encodes a protein, the testis-determining factor (TDF) which leads to testis differentiation and male sexual development

Answer 109

 Monotremes have 10 sex chromosomes • Females have 10 X chromosomes, males have 5X and 5Y chromosomes • X1 chromosome has homology to other mammal X chromosomes

Answer 110

 Female: XX, XXY  Male: XO, XY  The number of X chromosomes determines sex in Drosophila and many other invertebrates • More than one X chromosome- female • One X chromosome- male  In many nematodes and other insects, there is no Y chromosome- XX are female and XO are male e.g. C elegans

Answer 111

 The drosophila chromosome is not involved in sex determination but contains genes required for male fertility. An XO is male but sterile. • Sex-lethal (Sxl) is the master sex-determining locus in Drosophila o XX embryo: High sex-lethal expression, which triggers production of the female isoform of the gene doublesex (dsx) via alternate splicing, leading to female development and results in no hypertranscription of X o XY or XO embryo: low Sxl expression which leads to production of the male isoform of dsx and results in hypertranscription of X

Answer 112

* XX means that master sex-determining locus xol-1 is off, which makes a hermaphrodite * XO means that the master sex-determining locus xol-1 is on, which makes a male.

Answer 113

 Female: ZW  Male: ZZ  All birds, some reptiles, some insects and fish  Males are the homogametic sex (ZZ) while females are the heterogametic sex (ZW) • Z chromosome: large, gene rich • W chromosome: Small, few genes  ZW heteromorphic chromosomes are structurally similar to XY (large vs small) but have completely different gene content • Dose-dependent Z (need multiple copies of Z to become a male)

Answer 114

• Master sex-determining gene: dmrt1

Answer 115

 Female: Diploid |  Male: Haploid

Answer 116

Haplo-diploidy  E.g. honey bees • Females have 32 chromosomes (16 pairs) whilst males have 16 unpaired chromosomes • Males can only have daughters (if produce diploid malformed males are killed by the colony at development) • Sons do not have fathers • Sex is determined by the complementary sex determiner (csd) gene o Males are haploid and hemizygous for csd o Females must be heterozygous at the csd locus to develop normally o Homozygous diploid eggs develop into inviable males o There are hundreds of different csd alleles o Honey bee queens mate with over 20 different males to ensure that they have high diversity of alleles at this gene  Will have sperm stored in their ovaries to make sure they can fertilise eggs with lots of csd alleles to make sure they have lots of female heterozygous offspring

Answer 117

• Environmental systems o Temperature of egg incubation in reptiles (such as turtles)  Female: Warmer temperatures  Male: Cooler temperatures o Photoperiod in marine amphipods o Social factors in many coral reef-dwelling fish

Answer 118

o Gynandromorphs- embryos develop as part male and part female, often leading to spectular phenotypes  Chimeras- fusing of two embryos in early development  Polyspermy- multiple sperm fertilise egg, different lineages develop  Errors in mitotic division early in development (sex chromosome loss)

Answer 119

each cell decides its own sexual fate based on its sex chromosome constitution

Answer 120

o Although the SRY gene is considered a master sex-determining locus, other genes are involved in the sex determination cascade  Other genes involved in production of testosterone, some acknowledge testosterone (such as testosterone receptors)  If genes in these other pathways are interrupted, can have changes in sex determination cascades o Sex-reversal can occur when genetic males develop as females and vice versa due to mutations in key genes in the pathway, for example:  Translocation of the SRY gene to the X chromosome male development with 2X chromosomes  Mutations in genes downstream of SRY XY genetic male but female development

Answer 121

• Species where sex determination is cell nonautonomous (mammals) o Sex determining gene controls the fate of the gonads early in development: ovaries or testes o Gonads direct establishment of phenotypic sex through the production of hormones o Hormones direct development of sex-specific characteristics throughout the body  Gynandromorphs not possible

Answer 122

• In species with heteromorphic sex chromosomes, XX and XY individuals would express their X-linked genes at different levels, unless a special mechanism exists to control the expression of these genes

Answer 123

o In the absence of a special mechanism, XX animals and XY animals would have different levels of X-linked transcription o Dosage compensation mechanisms have evolved that balance the level of X-linked gene products between the sexes  Up-regulation in the expression of X-linked genes in males; in Drosophila  Down-regulation of X-linked genes in females: in the worm C.elegans  Complete inactivation of one of the two X chromosomes in females: in mammals

Answer 124

* Drosophila upregulate genes on the X chromosome in males: XY flies have an increased level of X-linked transcription from their single X chromosome (2 fold increase) compared to XX flies * Dosage compensation complex (DCC), a complex of proteins and RNA molecules, is recruited to the X chromosome in males and is responsible for the increased transcription of X-linked genes

Answer 125

* XX worms (females) have a lower level of transcription of each X chromosome compared to XO worms (males) * The dosage compensation complex (DCC) is recruited to both X chromosomes in XX worms, and both X chromosomes are transcribed at a lower level

Answer 126

condensed, inactivated X chromosome found in female nucleus

Answer 127

o Visible under the light microscope, with a stain that detects condensed chromatin

Answer 128

• Only one X chromosome is transcribed in XX individuals- the other X chromosome is condensed into heterochromatin and is no longer transcribed • Only one X chromosome is expressed in any given cell • X-linked genes are expressed at the same level in males and females through X-chromosome inactivation dosage compensation o Most genes are inactivated but not all o X chromosome inactivation requires Xist  Xist RNA coats the inactive X chromosome and recruits proteins that silence gene expression and result in chromosome condensation * Inactivation is random (happens in 50/50 ratio) * Once an X chromosome has been inactivated, it remains inactive through subsequent cell divisions

Answer 129

• X chromosome inactivation in mammals happens early in development as cells divide: one of the two X chromosomes becomes condensed and localised to the nuclear membrane where it is not transcribed

Answer 130

 In placental mammals, the X-inactivation-specific-transcript (Xist gene) is an RNA gene on the X chromosome  Xist is transcribed from the X-inactivation centre of the future inactive X to produce a 17kb long non-coding RNA

Answer 131

• Example- calico cats o Cats have an X-linked gene B that affects coat color o In XBXb heterozygotes, some cells have the chromosome with the B allele inactivated, while other cells have the chromosome with the b allele inactivated  XBXb heterozygotes are a mosaic, with patches of black fur and patches of orange fur (calico)

Answer 132

• Females that are carriers of X linked disease: half of their cells will have inactivated the mutant X, while half have inactivated wild-type • In females that are ‘mosaic’ for a disease, the products of both alleles are expressed: sufficient to prevent full expression of X-linked recessive alleles in a heterozygous female o Will have enough expression of wild-type X chromosome to take over effect of X-linked recessive allele o Often asymptomatic or only minor symptoms of a given disorder

Answer 133

• X-inactivation skewing: sometimes a non-random proportion of X’s of one type are inactivated

Answer 134

• Genes occupy physical locations (loci) on chromosomes

Answer 135

• Chromosomes have a characteristic morphology that can be observed under a microscope o Heterochromatin and Euchromatin o Telomere, centromere, short (p) and long (q) arms

Answer 136

• Chromosomes occur in species-specific sets (karyotypes) o Species have characteristic numbers of chromosomes with distinct sizes and morphology o All eukaryotes have a defined number of chromosome sets int their genome o Different species have evolved to have a different number of chromosomes

Answer 137

o DNA is wrapped around chromosomal proteins called histones to form nucleosomes o Nucleosomes form compact chromatin  Histone H1 helps nucleosomes condense into 30nm fibre o Chromatin is folded into higher-order structures that form the chromosome

Answer 138

 Each nucleosome contains 146 bp DNA wrapped around histones (H2A H2B, H3 and H4).  There is an electrostatic interaction between histones and DNA< as histone proteins are positively charged while DNA backbones are negatively charged  This structure needs to be unfolded to allow separation of the DNA strands during gene expression or DNA replication

Answer 139

o Telomeres-Mix of protein and DNA sequences at either end of the chromosomes to protect the ends from further degradation o Centromeres- Spindle microtubules bind to the kinetochore which is formed on centromeric DNA-essential to allow chromosomes to line up during division and not get lost during cell division  Chromosomes will line up on a metaphase plate and will be attached by a spindle formed out of microtubules o P (short) and Q (long) arms

Answer 140

• Types of chromosomes- o Metacentric chromosome-  Centromere in middle  P and Q arms are the same o Acrocentric chromosomes-  P arm is the short arm  Q arm is the long arm  Centromere is near the top of the chromosome o Telocentric chromosomes  There is only the q arm  Centromere is at the top of the chromosome

Answer 141

 Smallest number of chromosomes: Jack jumper ant: 2 for females, 1 for males

Answer 142

 Largest number of chromosomes: agrodiae butterfly: 268 chromosomes

Answer 143

o Number of chromosomes in a basic set is the monoploid number (n)

Answer 144

o Ploidy- the number of chromosome sets in an organism’s genome

Answer 145

 Haploid (n)- one set of every chromosome  Diploid (2n)- two sets of every chromosome  Triploid (3n)- three sets of every chromosome

Answer 146

o Chromosomes have a characteristic banding pattern  Euchromatin contains most of the active genes • Represented by a light stain G-banding pattern achieved through Geimsa staining  Heterochromatin is more condensed • Represented by a dark stain G-banding pattern achieved through Geimsa staining due to AT rich regions in that DNA that the Giemsa stain stains

Answer 147

• Mitosis- somatic cells: chromosomes replicate once and divide once-chromosome number is maintained

Answer 148

• Meiosis-germline cells: chromosomes replicate once and undergo 2 rounds of division to make haploid gametes: chromosome number is halved

Answer 149

• Sister chromatids- basically identical to each other (except for mutations) and both replicated from the original version-DNA replication has occurred

Answer 150

• Homologous pair- matching pair of chromosomes that have been inherited from each parent o Chromosomes are identical in size and characteristics, have a lot of differences in their DNA sequence

Answer 151

o Chromosomes become progressively more condensed and visible during prophase I: chromosomes are synapsed (homologous chromosomes are paired together) and crossing over occurs o Chromosomes are aligned across the middle during metaphase I o In Anaphase I, homologues separate but sister chromatids remain together o In Telophase I, cells separate o Process starts again in meiosis II, but there is no pairing, no crossing over and the sister chromatids separate to produce 4 haploid products.

Answer 152

 Crossing over- occurs between chromatids of the two homologous chromosomes: sequences from one homologous are now on the other-transfer of genetic material

Answer 153

No.  Every pair of homologues crosses over at least once  The larger the chromosome is, the more chance that recombination events occurs (multiple events can occur on one chromosome) • Can have recombination at sister homologues, but it wouldn’t be seen because they are identical sequences

Answer 154

 Homologues are used as the template to repair double strand breaks • Double stranded breaks important for keeping chromosomes paired up correctly during meiosis • DNA synthesis to repair a double stranded DNA break • The newly synthesised DNA is attached to the DNA molecule on the other homologue • Gaps are repaired • One sister chromatid has DNA sequences from each homolog • Recombination important as it allows the chromosomes to remain together at the right times

Answer 155

 Crossing over occurs at chiasmata • Chiasmata- the sites where homologue strands exchange DNA/cross over, which occurs during prophase I • Homologues are held together by chiasmata while pulled towards opposite poles during chromosome separation at anaphase I o Allow separation once double stranded breaks have been repaired

Answer 156

• Mendel’s 1st law, the segregation of alleles, is upheld as homologues separate during meiosis I and sister chromatids separate during Meiosis II: a gamete only has a haploid chromosome number and offspring will have diploid chromosome number

Answer 157

• Each pair of chromosomes segregates independently during meiosis, which is the cellular basis for Mendel’s 2nd law of independent assortment

Answer 158

sperm production

Answer 159

egg production

Answer 160

Spermatogenesis- continuous meiosis | Oogenesis- suspended meiosis between first and second stage

Answer 161

 First stage of meiosis occurs during embryoic development where all the eggs to be used in lifetime are made, while second phase only starts occurring when the female reaches sexual maturity  Females are born with all of their future ova present and arrested at prophase I  During ovulation, primary oocytes complete meiosis I and arrest at metaphase II  Fertilised 2nd oocytes complete Meiosis II o Only 1 of 4 products becomes an ovum, the rest become polar bodies -one precursor cell will yield one mature ovum

Answer 162

• Genetic testing- take polar bodies and test their genotype o If we know different alleles that are segregating in the mother, can determine whether functional oocyte inherited the good alleles if we can test that the polar body inherited the bad ones o Polar body from Meiosis II should be the same as her functional egg o The polar body from Meiosis 1 and the secondary oocyte (future ovum) will have different alleles

Answer 163

• Caused by problems in meiosis and gamete formation

Answer 164

• Polyploidy occurs when there is variation in the number of sets of chromosomes - Autopolyploid - Allopolyploid

Answer 165

o Autopolyploid- multiple copies of the same genome: single species chromosome duplication  Occurs when the first division of meiosis fails and gives gametes without reduction in chromosome number • Unreduced (2n) gametes fuse • Results in either triploid (3n) or tetraploid (4n)  Common in plants- polyploids have larger cells, hence larger plants, leaves and fruits which is desired in agriculture, but these plants are sterile

Answer 166

o Allopolyploid- contain two (or more) different genomes after a hybridisation event: the fusion of two or more diploid genomes from different species  Wheat is an allopolyploid of three species (which all have 7 chromosomes)

Answer 167

• Aneuploidy occurs when the number of chromosomes of a particular pair is unbalanced o Monosomy- missing one member of a pair o Trisomy- with one extra chromosome in a pair

Answer 168

• Non-disjunction- failure of chromosomes to segregate normally

Answer 169

o Results in daughter cells with abnormal chromosome numbers (aneuploidy)

Answer 170

o Can occur at both meiosis I and meiosis II  Non-disjunction at meiosis I results in one gamete that has both homologues, and one that has neither  Non-disjunction at meiosis II yields gametes that are homozygous

Answer 171

o Non-disjunction can occur for any chromosome, but is more common in sex chromosomes

Answer 172

 Monosomy-one chromosome missing | • E.g. XO: Turner syndrome (2n=45)

Answer 173

```  Trisomy- one chromosome is duplicated (2n=47) • XXX-triple X syndrome: Female o Two barr bodies • XXY-Klinefelter syndrome: Male o One barr body • XYY: Male  Trisomy-21: Down syndrome (2n=47) ```

Answer 174

 Rare trisomy- • Rare: trisomy 13 (Patau syndrome) • Rare: trisomy 18 (Edward syndrome) • Trisomy 13 and 18 don’t commonly live beyond the age of 1. • Other trisomy types aren’t really a thing as most would be terminated before birth

Answer 175

• Incidence of some trisomies increase with maternal age as there is increased non-disjunction during oogenesis as developing eggs are held in meiosis for a long period of time o Division only finishes when fertilisation occurs o The longer the eggs are paused at prophase I, the more likely that non-disjunction occurs

Answer 176

 Trisomy-21: Down syndrome (2n=47) • Most common type of trisomy- 1/2000 live births • During oogenesis, chromosome 21 usually has a single crossover o If no chiasma forms, both homologues segregate to same pole at Meiosis I o Ocasionally, chiasma at extreme ends does not provide enough resistance to tension from spindle, leading to premature pulling to the poles- recombination provides an extra timeframe to prevent premature pulling • In 5% of cases of trisomy 21, part or all of chromosome 21 is attached to another chromosome

Answer 177

o Translocations-partial rearrangements where an individual that carries them could be normal but can have partial sterility (due to unbalanced meiotic products)

Answer 178

 When segments of chromosomes break off, and attach to or exchange with other non-homologous chromosomes. Different types of translocation include: - Reciprocal - Non-reciprocal - Robertsonian translocations

Answer 179

o A segment from one chromosome is exchanged with a segment from another, non-homologous chromosome, creating two translocation chromosomes  Inviable gametes due to imbalance of genetic material- When translocated and normal chromosomes are mixed  Viable gametes- when two translocated chromosomes or two normal chromosomes are in a gamete

Answer 180

o One piece of chromosome attaches to the other with no exchange

Answer 181

o Long arms of two telocentric chromosomes fuse, forming a single meta- or acrocentric chromosome o Translocation of chromosome 29 to chromosome 1 is relatively common in European breed cattle and results in reduced fertility o Human chromosome 2 is a result of end-to-end fusion of 2 ancestral chromosomes.  All members of Hominidae (great apes) except humans, Neanderthals and denisovans have 24 pairs of chromosomes

Answer 182

-Translocation -Inversions • Individuals carrying chromosome rearrangements may be phenotypically normal, but have partial sterility o Produce unbalanced meiotic products that cause zygotes to die

Answer 183

* Genes located on the same chromosome (called syntenic genes) * Genes that are close together on a chromosome such that their alleles are often inherited together are considered to be linked genes

Answer 184

-When they are on different chromosome or are far apart enough on one chromosome

Answer 185

Independently

Answer 186

o Linked genes are not assorted independently

Answer 187

o First genetic map was Drosophila melanogaster X chromosome map created by Sturtevant (Morgan’s student)

Answer 188

* Recombination (crossing over) seperates alleles of linked genes * Recombination increases genetic variation by producing new alleleic combinations * Crossing over creates parental and recombinant gametes after segregation * This occurs more frequently between genes that are far apart

Answer 189

* Occurs during prophase I of meiosis * Crossing over occurs when chromatids between two homologous chromosomes exchange DNA * This occurs more frequently between genes that are far apart

Answer 190

• Make a test cross to figure out if genes are linked or unlinked

Answer 191

o Dihybrid test cross- F1 heterozygous X homozygous recessive individual  Only one type of gamete is produced by the double recessive individual- the same recessive alleles are passed on to all offspring  Four types of gametes are produced from the heterozygous parent phenotype of the test cross progeny are derived from the alleles in the gametes of the heterozygous parent • Two gametes are parental and two are recombinant

Answer 192

• Higher ratio of parental phenotypes (resemble the parents) and a lower ratio of recombinant phenotypes (different phenotypes to the parents) if the two genes are linked o Linked genes don’t have 9:3:3:1 ratio in dihybrid cross, and no 1:1:1:1 ratio in a test dihybrid cross o The further two genes are apart, the more likely there will be more recombination between them and thus more recombination progeny

Answer 193

• If two genes are unlinked, expect 1:1:1:1 equal ratio in test cross

Answer 194

Recombination rate | • Recombination frequency can be used to determine the arrangement of elements along a chromosome

Answer 195

• Recombination frequency in map units (m.u) or centiMorgans (cM)= (number of recombinant offspring/Total number of offspring)*100

Answer 196

o Genes >50 map units apart have so much recombination between them that they are effectively unlinked an assort independently • Cannot distinguish recombination frequencies of more than 50% from genes that are assorting independently on separate chromosomes

Answer 197

1. Generate a heterozygote 2. Perform a test cross 3. Identify parental and recombinant offspring 4. Calculate recombination frequency

Answer 198

o Genetic linkage map- estimation of relative locations of phenotypic or molecular markers along a chromosome

Answer 199

• Use genetic mapping to link phenotypes back to sequence changes in the genome • Connect genetic linkage maps to physical maps (DNA sequence) o Map the gene using genetic techniques then use molecular techniques to find physical DNA sequence responsible

Answer 200

 Molecular markers preferred as there are only so many genes that we can reliably characterise a visible phenotype for -pad out the map more

Answer 201

o Identifying genes and biological processes underlying inherited traits o Functional genomics to discover why the gene causes the phenotype o Genetic tests for diseases and potential cures  Linkage mapping is important in understanding human disease because can find genetic causes of familial traits that are passed on through generations • BRCA1 genetic mapping of breast cancer on q arm of chromosome 17  If know genetic cause, can use that information to find a cure o A scaffold or map to inform the assembly of whole genome sequences using genetic markers

Answer 202

* Diagram of linked markers along a chromosome * Map distances are additive * Larger values calculated as a sum of shorter intervals

Answer 203

• Recombination frequency cannot exceed 50%

Answer 204

• For greater accuracy, use shorter intervals (more markers) and more offspring o Many thousands of molecular makers can be used to fill the gaps between the limited number of visible markers due to limited number of genes that provide usable phenotypes for mapping

Answer 205

o Are sites of silent DNA variation not associated with any measurable phenotypic variation o Work the same way as visible markers- can be heterozygous, have recombination between them o Are critical for understanding human variation and mapping positions of disease genes  Amount of variation present without any phenotypic variation is particularly high in humans compared to other organisms

Answer 206

• In bacteria, vast majority of genome codes for protein- if there is mutations in bacterial DNA not as common to find one that doesn’t provide a phenotypic variation o Most of the genome is exon, but a change could be a third base change, conservative substitution or redundant information -Relatively useless

Answer 207

``` • In worms/flies, only 35% of genome is exon (codes for protein), but a change could be: o A third base change o A conservative substitution o Gene family o Duplicated pathways o Recessive alleles -Relatively useful ```

Answer 208

``` • In humans, only 2% of the genome is exon, but a change could be: o A third base change o A conservative substitution o Gene family o Duplicated pathways o Recessive allele -Extremely useful ```

Answer 209

``` o SNPs (single nucleotide polymorphisms) o RFLPs (restriction fragment length polymorphisms) o Microsatellites (short tandem repeats) ```

Answer 210

 Single nucleotide differences in DNA sequence

Answer 211

 Most common type of molecular change between individuals

Answer 212

• Differences in DNA sequence may or may not be associated with differences in protein structure or phenotypes

Answer 213

* Sequencing (e.g. whole genome sequencing) but is expensive * Microarray chip (SNP array)-most common method

Answer 214

o Microarray chip containing spots of DNA oligonucleotide probes, to detect sites of known SNP variation in the genome 1. Attach a single stranded oligonucleotide (40-60 bp in length) corresponding to one variant to one feature on the microarray and single-stranded oligonucleotide corresponding to the other variant to another feature on the microarray 2. Prepare genomic DNA from the individual being tested 3. Label with fluorescent dye 4. Hybridise to the microarray (sites of known variation) o If there is DNA binding between oligonucleotide and microarray , then there is fluorescence-means that SNPs match up and we know what SNPs the individual has by the number of fluorescent spots they have

Answer 215

 Nucleotide differences that generate a change in a restriction enzyme cutting site  SNP variations can destroy or create restriction enzyme recognition sequences

Answer 216

 Differences in DNA cutting sites results in different sized bands after a restriction digest on agarose gel-> can identify whether SNPs occurred or not based on size of restriction digest

Answer 217

 2-5 nucleotides tandemly repeated in a DNA sequence

Answer 218

 These repeats occur in conserved regions of the genome so can use PCR primers to detect this region of the genome but within the microsatellite the repeat sequences will differ in number  Number of repeats in microsatellite DNA sequences are variable within the allele of individuals  Amplify microsatellite alleles with PCR • Size of PCR product produced changes with number of microsatellite repeats o Can be used to detect paternity

Answer 219

o Step 1: Determine if the genes are linked  8 possible phenotypic combination  If not linked, then 1:1:1:1:1:1:1:1  If partially linked, see 4 big number categories and 4 small ones  If linked, then there is deviation in that ratio • Three categories o Highest frequency category- parental test cross phenotypes (because recombination is rare event) o Intermediate frequency category-Single recombination event o Lowest frequency category-Double recombination event (middle marker has switched chromosomes independent of the ages) o Step 2: Determine gene order  Look for double crossover classes- the classes with the fewest recombinants (the lowest number)  The marker that is different from the parental configuration is the middle marker o Step 3: Look at markers in pairwise configurations  Factor in double recombination events between flanking markers • Double recombinants added in calculation twice o Don’t do this if there is no clear double-recombinant class  Unlinked genes give 50 or more map units o Step 4: Put the map together

Answer 220

o Determine if 3 genes are linked, the gene order and the map distance

Answer 221

 Notation in exams- • Even if notation is linkage location: o Do not assume all genes are linked o Do not assume that the order of writing is the order of the genes on the chromosome

Answer 222

o Chromatin structure o Sex differences o Chromosome aberrations

Answer 223

 Hotspots-high levels of recombination | • Greater genetic distance relative to physical distance

Answer 224

 Coldspots-low levels of recombination | • Shorter genetic distance relative to physical distance

Answer 225

o DNA in coldspots doesn’t cross over so well due to the proteins surrounding that DNA e.g. genes separated by a centromere ------Pericentric heterochromatin  May be due to the feasibility of DNA in those regions to initiate crossing over

Answer 226

 Pericentric heterochromatin-condensed chromatin at the centromere which lacks standard meiotic recombination (no recombination here)

Answer 227

 Recombination rates differ between males and females for most animals

Answer 228

• In male fruit flies, there is no crossing over at all o Unusual prophase I of meiosis in which no chiasmata develop presumably due to lack of expression of recombination proteins which usually help this recombination to occur

Answer 229

• Genetic distance and maps on human male and female chromosomes can look different o Females have slightly more recombination than males o Probably because males vs females have different tissues and different gamete production rates

Answer 230

 Inversion- a type of chromosome aberration, a structural variation in a chromosome leading to rearrangement of chromosome segments • Piece of chromosome breaks up and flips around and inserts in the opposite orientation of its original orientation

Answer 231

 Heterozygous individuals for inversion chromosomes may be phenotypically normal, but have partial sterility • Produce unbalanced meiotic products if recombination occurs within the inversion

Answer 232

 Inversions can be paracentric (include centromere) or pericentric (no centromere)

Answer 233

• During meiosis, for genes to align together, a loop needs to form (all markers paired up with each other) • Products of single recombination mean that there are ½ viable and ½ inviable chromomes o Dicentric-Have two centromeres o Acentric-Have no centromeres • Double recombination events can restore viability to a chromosome

Answer 234

• Physical map- position of markers in the DNA sequence (in megabases)

Answer 235

• Genetic map- position of markers in linkage group (in map units)

Answer 236

• Recombination rate (megabases per map unit)- differs along chromosomes o Linear line- no recombination hotspots/coldspots in chromosome o Non-linear line: recombination results in high rate whilst no recombination results in stationary points

Answer 237

• Genetic mapping of traits uses natural genetic variation of individuals in a population

Answer 238

• Haplotype- a group of alleles that are inherited together throughout generations o A block of alleles in a set of linked loci that are inherited together o These alleles will tend to be passed on together during meiosis, unless separated by recombination o May consist of any type of genetic variation:  SNPs  Phenotypic markers  RFLPs  Microsatellites

Answer 239

• A causal mutation that leads to trait: o Inherited with other nearby variants on the chromosome o Recombination can generate blocks of variation that tend to be inherited with a mutation of interest o Recombination reduces large blocks transmitted through the pedigree o Variation that happened to be linked to a causative mutation will still be associated with the mutant many generations later

Answer 240

o Ancestry haplogroups- A group of similar haplotypes that share a common ancestor  Indicate likely similar origin or line of descent  Genome-wide SNP analysis can determine ancestry haplogroups

Answer 241

 Positional cloning |  Genome Wide Association Studies (GWAS)

Answer 242

• Technique (used a lot in the 1980s) to locate position of a gene along the chromosome • Use genetic mapping to connect between the gene that affects the phenotype and the DNA sequence o DNA sequence for a gene can be identified based on its genetic map position • Mapping done by observing individuals in a pedigree, checking suite of molecular markers for these individuals and finding area of the genome which always comes up with same markers that are variable in mutated individuals. Then, can look at region of DNA between them

Answer 243

• Step 1: Gene of interest is mapped between two cloned markers o Map the phenotype by recombination between known molecular markers • Step 2: The DNA between the markers is isolated and analysed • Step 3: The compatible candidate gene is identified based on o Expression pattern (expression at right tissues at right time) o Evolutionary comparisons-compare with other organisms to see if gene has the same function in those other organisms o Sequence analysis-see if it differs from wild type individual

Answer 244

• First example of positional cloning is mapping of cystic fibrosis gene in 1989  Cystic fibrosis was mapped to a region on 7q31  A region that was conserved in mammals was isolated  A cDNA from this region was isolated  This cDNA was used to probe for transcripts from tissues affected in CF patients  Found that gene expressed in all tissues affected by cystic fibrosis and became candidate for causing the disease

Answer 245

o Common autosomal recessive disorder, affecting lungs and other organs and tissues that produce sweat, digestive fluids or mucus o Extensive pedigrees available o Gene located on chromosome 7 in 1000bp interval

Answer 246

* Sequencing is more accessible * SNPs used * Compare genome-wide SNP variation in of thousands of affected vs unaffected individuals * Relative frequencies of each SNP and haplotype are calculated and plotted among affected and unaffected individuals * Can identify SNPs associated with a disease, but they cannot specify which genes are causal * SNPs and haplotypes that are more common among affected individuals identify candidate locations for causative alleles for the trait * Genomic locations can be surveyed for candidate causative genes based on gene function, expression patterns, similarity to mutants in other animals…

Answer 247

• Interpreting GWAS-Manhattan plots o Each dot represents a haplotype, a region containing variable SNPs o Each number on the X axis represents a chromosome o Each colour indicates when chromosome changes o Probability that a haplotype is associated with the trait is located on the y axis o SNPs/haplotypes with a significant association with the trait are highlighted in a different colour (green)

Answer 248

o Expression pattern- from transcript analysis such as RNA sequencing and microarrays  Which genes are expressed in the appropriate time and location o Evolutionary comparison  Has the gene been identified with a phenotype in other closely related organisms? o Mutations and polymorphisms- Do people affected by the trait share a DNA sequence variation in one of these genes?

Answer 249

• Modern agricultural practices rely on chemical methods of pest control • Compounds that act as insecticides have diverse chemical structures with different modes of action but a lot of them target nervous system of insects o E.g. pyrethrins, neonicotinoids, DDT, dieldrin  Disrupt nervous system of arthropods, causing paralysis and death

Answer 250

• Problem of insecticide use- o Puts pest species under intense selective pressure o Variants containing mutations in genes that provide resistance (ability to tolerate insecticide) are favoured o Resistance is selected for in a population and becomes widespread

Answer 251

o First used in 1950s o Insect-specific neurotoxic chemical o Dieldrin- a synthetic chemical with high insecticidal activity o Resistant populations developed quickly  Present in pests of medical and agricultural significance  Crop pests and vectors of disease

Answer 252

o Unknown cause of the resistance phenotype  Resistant homozygotes survive much higher doses (4000 fold higher doses) than wild-type homozygotes  Heterozygotes survive at intermediate levels

Answer 253

o Resistance to this chemical shows temperature sensitivity  When exposed to high temperatures (380C), resistant homozygotes (RR) exhibit higher sensitivity than wild-type homozygotes (SS) • Heterozygotes (RS) have intermediate levels of temperature sensitivity  Resistant flies will go into heat shock (go to sleep in a while and eventually die)  Recovery times of resistant mutants is much longer (10 minutes) than wild type flies (5 minutes)

Answer 254

• Backcross and linkage mapping with 2 markers o Bristle phenotype- stubble (position 58.5) o Lyra wings (position 40.5) • Rdl (resistance to dieldrin) was mapped on position 25 on chromosome 3  Further refining of location of causal mutation with positional cloning • Screened lots of different markers along chromosome to lead to identification of region containing the resistance gene • Resistance mapped to a locus containing the first identified insect ligand-gated chloride channel gene

Answer 255

* Rdl (resistance to dieldrin) was mapped on position 25 on chromosome 3 * Resistance mapped to a locus containing the first identified insect ligand-gated chloride channel gene

Answer 256

 Normal- • Ligand-gated chloride channel with large extracellular domain that binds to GABA (an excitatory ligand) and allows for chloride ion influx to flow through o Five Rdl subunits join together to make a chloride channel pore in the membrane

Answer 257

• Insecticide dieldrin blocks chloride influx

Answer 258

• Single nucleotide polymorphism (SNP)-> G to T mutation which results in an amino acid change in transmembrane region of the protein (alanine to serine substitution at position 301) • Mutant receptor no longer binds to dieldrin o Chloride channel is not blocked by insecticide o Insects survive

Answer 259

o Mutation screening-see how many populations around the world have this mutation  G to T SNP that leads to resistance generates an RFLP and removes HaeII recognition site  PCR performed on fragment  Restriction enzyme HaeII cuts twice in wild type individuals but in resistance mutants only gets cut once: therefore different banding pattern can be seen • Uncut fragment- 268 bp • Homozygous resistant- 205 and 63 bp • Homozygous susceptible mutant- 3 bands • Heterozygote- all bands present in either allele  HaeII method used to recognise mutants around the world

Answer 260

• Tasmanian devil- o World’s largest carnivorous marsupial, distribution now restricted to Tasmania o Has 7 different chromosomes

Answer 261

• Tasmanian devil face tumour disease- o Transmissible cancer that can spread like a contagious disease o Spread by the transfer of living cancer cells by biting o Spread throughout Tasmania since first identified in 1996

Answer 262

o Fatal transmissible cancer is decimating the devil population in Tasmania  All DFTD cells are genetically identical to each other and genetically distinct from their hosts own cell • Seen by using genetic markers and karyotyping

Answer 263

 Cancer originated from a single individual and spread, rather than arising repeatedly • First strain in 1996 (female origin) • New strain identified in 2015 (male origin) o Tumor cells must have arisen from a female because they contain X-chromosome fragments

Answer 264

o There are translocations, deletions, rearrangements… o Chromosome 1 and sex chromosomes missing from tumor karyotypes  Translocations and joining from different chromosomes o Chromosome 5 aneuploid o Four additional marker chromosomes (M1-M4) are present, made up of chromosome fragment translocations, probably inversions and other chromosome abnormalities  Not additional chromosomes-just have been formed from bits of pre-existing chromosomes

Answer 265

o Fluorescence labelling of chromosomes highlights karyotype differences  Label then look under a microscope

Answer 266

• Genome-wide association study and functional characterisation of novel genes involved in human skin pigmentation • Studies often focus on European lineages • Large diversity of pigmentation phenotypes within Africa o Melanin index- skin tone measurements o Erythemal dose- level of radiation required to turn skin red • Compared populations to each other using those phenotypes • Look at Africans as baseline for rest of population o Measure phenotypic diversity (pigmentation levels) o From blood, isolate DNA  Analysed 1,570 DNA samples of Africans with diverse pigmentation phenotypes o Genotyping using SNP array  Illumina Omni5M exome array about 4.3 million SNPs • SNPs from international HapMap and 1000 genomes projects o Association between genetic variation and the trait  Manhattan plot of GWAS  Then functionally characterised each gene to make sure they had function for skin pigmentation-functional gene experiments necessary to confirm causative genes once associations have been found

Answer 267

• GWAS identified novel variants associated with skin pigmentation • Significant variants clustered in 4 genomic regions that together account for about 30% of the phenotypic variation • Four regions identified account f about 30% of variation in skin pigmentation phenotype o SLC24A5, MFSD12, DDBI and HERC2 • Proved phenotype of candidate genes with functional experiments in model organisms o MFSD12- codes for a protein that modifies pigmentation in human melanocytes  Decreased MFSD12 expression associated with darker pigmentation  Genetic knockouts of MFSD12 orthologues affect pigmentation in mice • Identified previously uncharacterised genes associated with skin pigmentation in ethically diverse Africans

Answer 268

o Gene interactions possibly involved:  Complementary gene interactions (both genes needed for phenotype)  Duplicate gene interaction (either gene needed for phenotype/redundancy)  Dominant gene interaction (genes with same phenotype interact additively)  Epistasis (one gene stops/masks the phenotype of another) • These genes have diverse functions o Repairing UV damage: melanogenesis in skin cells

Answer 269

* Gene expression analysis * Molecular function of candidate gene * Functional experiments in model organisms

Inheritance Flashcards

Emily Remnant lecture series