Complex Traits Flashcards

Question

Affect of Recessive Lethal Alleles

Answer 1

Lead to death when homozygous --> Lethal Alleles can lead to missing or reduced #s of genotypes In monohybrid cross with recessive lethal alleles = have fewer offspring overall

Answer 2

Y = yellow -- dominant to y y = No yellow Yy = Yellow yy = no yellow YY = death Yy X Yy GR -- 1 YY: 2 Yy: 1 yy PR -- 2:1 -- 2 yellow: 1 non-yellow because YY will never develop

Answer 3

Can take out the one that dies = change the probability of all others Ff X Ff 1 FF -- dead 2 Ff -- short 2 ff -- tall Because 1 FF is dead = only 3 pffspring -- 1/3 is tall F = dominat for achodrolasia BUT recessive for lethal (becaise need 1 F for short BUT need two F for dead) ***Doesn't matter which you assign F or f -- gives same results F = dominant for acondroplasia -- only need 1 copy (one copy in Ff = have achsdrolplasia) F = recssive for lethal because need two copies (FF = dead)

Answer 4

Y = lethal Y = dominant to y for agouti -- because Yy is yellow (heterozygout has phenotype of Y) BUT Y = recessive for lethality because need 2 of them -- only die if YY Yy = alive --> y is dominant for lethality Yy = alive and yy = alive BUT YY is dead = y is dominat for lethality because Yy is like yy

Answer 5

S = truncated tails and is dominat to s SS = leads to spindal development probelms and is embryonic lethal

Answer 6

Organelles contain mtDNA or cpDNA that can influence a trait - Unpredictable organelle inheritance may lead to differences in trait expression In mitosis -- different quanityoes of organelle is ingerited = mtDNA and cpDNA is not inherited according to mendelian ratios = not predictable

Answer 7

IN mammals egg cells (female gametes) pass on all cytoplasmic components -- mtDNA is inherited through mothers Paternal mtDNA is activley elimanted in zygote

Answer 8

Unpredictable organelle inheritance may lead to differences in trait expression

Answer 9

1. Present in Males and Females 2. Usually inherited from one parent -- usually the maternal parent 3. Reciprical crosses give different results 4. Exhibit extemsive phenotypic varaition even within a single family

Answer 10

Called Paternal leakage -- sometimes --> rare exception NOT general rule - 17 examples of paternal mtDNA in children Possibilities for how -- defects in the ability to eliminate paternal mtDNA

Answer 11

Found genes that encode regulatory RNAs that are passed to the offspring in the cytoplasm of the egg RULE: Mother's genotype detemrines the offspring's phenotype

Answer 12

Moms envirnment to offspring --> envirnment can alter gene expression in offspring that givs rise to phenotype that are reuslt of the materal envirnment Different envirnment can alter phenotypic ratios in offspring BUT the mosther's genotype detemrines the offsrpings phenotype

Answer 13

BOTH -- caused by genes on autosomes = both males and females can transmitt the trait but the trait is only expressed on one sex - Can be on sex chromosmes (then it will also be sex linked and sex-influenced/sex limited) OR can be on autosomes Sex limit = trait is ONLY expressed in one sex Sex influence = trait is mostly expressed on one sex Can be caused by genes on autsomes -- sometimes they are also sex-linked - Sex linked if they are on sex chromsomes

Answer 14

P = Early puberty -- dominant mutations in autosomal gene encoding hormone receptor can lead to early puberty in boys --> lead to early puberty in boys only p = normal puberty Mutation in AUTOSOME -- both males and females can pass down but only have trait in males pp (normal female) X Pp (Early puberty male) - Gives Normal female, Normal Female, Early male, andnormal male (SEE IN IMAGE) - Get gametes with pX from mom, and PX PY, pX, pY from dad (because get P on auatsome and get sex chromsomes) OR pp (normal male) X Pp (Normal female) - Daughters and half sons have normal pubey, Half sons have early puberty Shows both males and femals can pass down trait but only males show the trait NOT sex linked because when do reciprical cross = we get the same ratios (get 2 normal females, one early male, and one normal male) --> because same ratios = NOT sex linked traits = both can transmit the trait but only males present the trait

Answer 15

Mutations in an AUTOSOMAL locus on chr. 21 can lead to hair loss in men, and to a lesser degree women. sex influenced (but not sex-linked) Sex influneced because can see trait in women but not sex linked because on autosome Males: HH = have hair Hh = have balding hh = bald Females (have the same alleles BUT do not see hair loss to full degree): HH = normal hh = have thinning of hair BUT not bald in same way that males will be ***Inherited on autosomes = not sex linked

Answer 16

Varaints on gene on X chromosome leads to hair loss in men but not wome (Sex linked because on X chrosmomes AND sex limited because only in men) Xh = hair loss in men xH = no hair loss (full hair) xhxH or xHxH = full hairs Sex linked -- on sex chromosome vs. Sex infulenced is on autosomes

Answer 17

Influence mendelian ratios --> Allels that ONLY show phenotype in certain envirnments - Only present their phenotype under certain envirnmments - Conditional on envirnment Ex. Siemmese cats -- temperature sensitive alleles cs = non-functional at high temperatures but functional at low temp CS = WT -- can produce melasis = get full color CS is dominant to cs cs at low temp expressed = can produce pigment but at high temp not expressed = cannot produce pigment --> expression is dependent on envirnment Siemesse cats = cscs --> can make tyrosnine to get pigment at low temperature but not at high --> when they are born they are born all white because at high tempeture in the womb = no tryosinase = no pigemnt --> as they grow their expremities cool but their body stays warm = extremeisties get opigment because of cooling and body stays white because warm (warm areas stay white) This is conditional allele because expression of the allele depends on the envirnmen cs = Temperature sensative allele - The extremities = get cold = have color - The core stays warm = no color - ALL cells have the alelle but its not expressed in all - When first born = all white because wamr in the womb = no color

Answer 18

Lethal --> means that they are lethal only under certain conditions = have conditional lethal allele

Answer 19

X-linked recessive mutation --> mutation prevents G6P dehydrogenase - Mutation in Glu-6-P dehydrogenase = reduced glutathione = RBCs are not protected against oxydants = cells undergo hemylysis = can lead to death - Leads to low glutathione, an antioxidant that protects red blood cells - Hemolysis can occur if oxidant loads are too high Trigger for increase in oxidants = fava beans + antimalarial drugs + some antibiotics + henna Most people homozygous for mutation are fine BUT only have health issues if oxidants level increases = conditional lethality

Answer 20

Offers protection against malaria - Individuals that have favism allele = protected against malaria - Increase in malarua alleles = increase in allele frequncey of favism vraints May explain why some cultures use chickpeas instead of fava beans

Answer 21

At certain times during embryonic development both X chromsomes are needed - finally a cell with select 1 at random for X chromsome inactivation ALSO sometimes there is expresion of certain genes from teh condensed X chromosome

Answer 22

Androgen insensitivity syndrome Low response to testostrone

Answer 23

XX with congential Adrenal Hyperplasia -- genetic variants can cause an overproduction of testosterone XX individuals with CAh can be: 1. Typicaly female 2. Typical male 3. Cmbined male and female charachteristics

Answer 24

XX can develope as males if SRY targets are activated errenously -- if have recombination between X and Y chrosmoems that adds SRY to X chromsome - Recombination doesn't normally happen between X and Y chrosmomes but sometimes mistakes happen

Answer 25

Not always X and Y Ex. Lots of repriles and bird and some invertabrates Female = ZW (heteroganetic --> can make gamated with Z or W chromsomes) Male = ZZ (homogametic --> can only make one type of gamete) Still have sex chrosmomes but not called X and Y called Z and W Ex. 2 -- Some Invertabrates Female = XX Male = XO (1 X only)

Answer 26

Some species take sex chromosomes to the extreme Ex. Platapus Female = XX, XX, XX, XX, XX Male = XY, XY, XY, XY, XY Have 5 sets of sex chromosomes

Answer 27

Some rely on ploidy rather than specific sex chromsomes Ex. Bees Males = Haploid -- basically unfertalized gamete becomes male Females = diploid Haploids = develope as malel Diploid develop as female

Answer 28

A single gene determines mating types in some organisms -- the allele of that gene determines Sex - No sex chromsomes Ex. Yeast - Has two mating types -- A and Alpha --> mate to form diploid --> controled by a single gene - The gene = flanked by silent copies of both mating types --> chnages alleles at mating type locus = daughter cells can change mating type Mating types in yeast: Saccharomyces cerevisiae: 2 mating types Tetrahymena thermophila: 7 mating types --> have pheranomes + pheramone receptors that allow this to happen Cryptococcus neoformans: 2 mating types but ~20 different mating receptors

Answer 29

Have pheranomes + pheramone receptors that allow this to happen

Answer 30

Female = XX Male = XY + 3 pairs of autosomes *The Y chromosome does not determine sex --> Not a gene on Y chromosome that makes male/female *The ratio of X to autosomes (X:A) determines the expression of Sxl, a gene that signals the female sex determination pathway - Female XXAA -- X makes yellow protein and A makes red protein (have no X inactivation -- both are active) --> then the proteins make a heterodimer of Yellow and red THEN the left over yellow make a homodimer -- the yellow homodimer promotes transcription = female IF not enogh yellow = no Homodimer = no transcription = male IF more X = more yellow = homodimer = transcriotion = get gene = female MALE XYAA - X still makes yellow and A makes red BUT male makes less yellow because only one X --> NOW there is no yellow for homodimer = no transcription = male SXL expression = female development NO SXL = male development

Answer 31

Some species -- individuals are both male and female = hermaphradism Ex 1 -- C. Elengans -- self-fertile females (hermaphradites as male and females for repredouction) Ex 2 -- Flowers -- hermaphrodites - They prefer to out mate BUT will self-fertalize (self mate) --> it is better to make seeds than to not make seeds at all

Answer 32

Model system for development -- are hermaphrodites

Answer 33

Sex can be triggered by envirnment Ex. Lizards + Aligators - Dveelopment of sex in egg = based on temperatire egg is reared at - Warmer = get more males; colder = get more females HERE = have G X E --> All genes are there to make males and females its just what gets activated Ex 2 - Clownfish -- sequntial hermaphradies

Answer 34

They are seuqntial hermaphridites -- they switch --> NOT always male and female -- same indiviodual can concert to female * Switch between male and female * The largest fish in the school is female, all others are male * If something happens to the female, the largest male turns into a female

Answer 35

Unisexual -- all female - They steal sperm from other species --> don't use genetics just activates to make eggs - Eggs have varaible ploidies - Sometimes DNA from sperm is present - This is how they reproduce To make eggs, the female steal sperm from other salamander species --> This triggers the females to make eggs (with variable ploidy- mostly 2, but 3 and 4 sets sometimes observed) -- Sometimes, the DNA from the sperm is also present

Answer 36

Incomplete dominance -- because see neither purple or white -- heterozygoues is a blend of the hoozygous Co-dominance = see both homozygous

Answer 37

You need to assign one phenotype as h = 1 and one phenotype has h = 0 Ex. Red h = 1 White h = 0 IF Aa is pink --> h = 0.5 If Aa is closer to red --> h > 0.5 Need to define one as h = 1

Answer 38

Incomplete penetrance = refers to pedigree when people with genotype don't show the trait

Answer 39

NO -- individuals do not have expresivity -- traits have expresivity

Answer 40

S = spinal defromation s = normal Ss = no tail SS = dead in utero -- recessive lethal S = spinal defect over s but it is recesisve for lethality because 1 copy is not lethal bit 2 copies is lethat - Dom for spinal development but recessive for lethal - Spinal development = Ss more like SS than ss = S is dominnat BUT for SS need two copies (Ss is NOT like SS it is like ss for lethality) = S is recessive (need two copies) for lethality

Answer 41

Expressed only in one sex but inheritance is controlled bu sutosomes - Males and females have the allele but the trait is only expressed in one Ex. Early puberty in males -- Females = have the allele but it is only expressed in males

Answer 42

Because for sex limited/sex associated the recipircal cross gives you the same ratio For Sex linked the reciprical cross gives you two different ratios

Answer 43

Complex traits demonstrate continuous phenotypic variation -- do not fit into tidy discrete categories - NOT ALWAYS BELL CURVE ***Complex traits do not behave according to Mendelian Genetics predictions

Answer 44

Help us understand mechanism pathways

Answer 45

1 -- Sex linked (X chrosmome) and Sex limited (ONLY in men) 2 -- Sex assicated/sex influenced 3 -- Sex limited (ON autosome BUT only in boys) 4 -- Sex limited 5 -- Sex linked + Sex limited (women do not have postsates) Sex linked = ON sex chromosomes (if on X or Y = Sex linked) Sex limited = can be on autosomes OR sex chromoeomes

Answer 46

Maternals Genotype determined the offspring phenotype -- MEANS that no matter the phenotype of mom or genotype of kid the kid will have the phenotyp assicated with genotype of the mom - Kid genotype doesn't matters -- mom genotype is what matters Example -- D = Dextral d - Sinistral IF Mom is D_--> All kids are dextra (Because mom genotype has D) If mom is dd --> ALL kids are Sinistral EVEN if they have Dd genotype - Because based on mom genotype and mom has dd genotype = kids phenotype is Sinistral even if Dd genotype

Answer 47

ALL kids are Dextral STILL have 1 DD: 2 Dd: 1 dd BUT the dd kid is still dextra because mom is Dd - Since mom is Dd = all kids are dextral because kid phenotype us based on mom genotype Mom = Dd = all offsrping have dominant trait even if they have dd --> genotype of mom gives information on how offspring develope The mom's genotype is Dd (gioves dextral phenotype) so the offsrping will show the omindnat trait even if they inherit recssive alleles - They still show the dominant traiyt even if dd because mom is Dd (gives dominant phenotyype) and its based on mom genotype

Answer 48

Encode RNAs found in the egg --> Stuff that ends up in egg (not in the polar bodies) includes Proteins + mitocondria + RNAs that have regulatory function --> egg has RNAs to turn on development - Matters what is theer to actrivate egen expressuon ***Genotype of mom = affects regulatpry RNAs in egg = affects the phenotype in offspring - Mom puts regulatory RNA in egg BUT she also can only put her own genotype (If she is gg she can only give g NOT G) -- all of the offspring get the regulatory RNA that mom can make - The reguoatory RNA turns on expression during development - Regulatory RNA made from DNA of mom THE offspring can then make regulatory RNA from their own genotype to give to kids (if offspring is dd then they can make regulatory RNA from their offspring) --> They can give different regulatory RNA to their own offspring than they got from their mom FOUND 11 Maternal Effect genes in humans

Answer 49

Phenotype ratio 4 Sinstral: 0 --> Mom is dd = mom genotype gives sinistral AND the mom genotype dictates the phenotype of kids (if mom is dd genotype kids all have phenotype associated with dd regardless of their own genotype) - Since mom is dd --> ALL offsrping show recssive phenotype even though mom shows the Dominnat phenotype BEACAUSE based on mom genotype NOT mom phenotype

Answer 50

G = Grey g = White B = Black b = Red G_bb --> Grey and red Answer: 3/16

Answer 51

Expected Genotype/phenotype ratio of Dihybrid cross where each indepent gene influencves seperate traits (2 genes = 2 traits) GR = 9:3:3:1 PR = 9:3:3:1 Example: R = Round r = Wrinckled Y = yellow y = green RRYY --> Only produces RY gametes rryy --> Only produces ry gametes RRYY X rryy --> Get RrYy (Heterozygous -- all show dominant traits) RrYy = can make 4 types of gametes (RY, Ry, rY, ry) --> because of law of Independent assortment = expect equal frequency of each type of gamete THEN can use punnet square to find expected genotype and phenotype frequnceies of F2 - Build a punnet square of 4 Gametes by 4 gametes of each parent across tops and sides --> THEN fill in the sopace by combing gametes for each parent END: 9R_Y_: 3R_yy: 3 rrY_: 1 rryy When 2 genes infleunce 2 seperate traits phenotype ratio = 9:3:3:1

Answer 52

The genotype ratio always stays the same -- expect genotype ratios are always the same Phenotype ratio = Can shift if not independent (if genes are not indepent if they affect one trait) - Change ratio if you have two genes that affect one traut and are not independent from each other - If those 2 genes influence 2 seperate traits expected phenotype rations = 9:3:3:1 - Ways genes interact with each other = can change phenotypic ratio

Answer 53

These genes can intercat in different ways to produce novel phenotypes

Answer 54

Different types of genetic interactions can sometimes reveal the pathways or molecular mechanisms underlying a trait - Different types of genetic interactions + how they predict phenotype ratios in a Dihybrid cross can sometimes reveal the pathways or molecular mechanisms underlying a trait

Answer 55

Polygenic -- two genes influencing the same trait (two genes control one trait)

Answer 56

Two genes control one trait -- Color in peppers Two loci - Y and C loci control the production of color Y+ is dominant to y C+ is dominnat ti c Y+_C+_ = red Y+_cc (dominant at one and recessive at the other) = Peach colored yyC+_ (dominant at one and recessive at the other) = Orange yycc (recessive for both) = white IF Y+Y+C+C+ X yycc F1 -- ALL Y+yC+c --> ALL RED F2 -- Y+yC+c X Y+yC+c F2 expected Genotype Frequncey (Doesn't change): 9Y+_C+_: 3Y+_cc: 3yyC+_: 1yycc Expected Phenotype Rato: 9 Red: 3Peach: 3Orange:1 White HERE -- 2 genes infleunce the same trait in indendent fashion -- expected PR = 9:3:3:1 - Here two Loci are contributing independently because still 9:3:3:1 --> Here they are additive = indepedent = no epistasis

Answer 57

ANSWER: 3/16 STILL 9:3:3:1 ration but now 2 genes are for the same (for one) trait 9Y+_C+_: 3Y+_cc: 3 yyC+_: 1 yycc Orange = yyC+_

Answer 58

Genetic interactions where one gene influences (masks) the effect if another gene - Allele at one gene masks effect at another gene

Answer 59

The gene that does the masking (the gene you end up seeing)

Answer 60

The gene whose affect is masked (the gene that you do not see)

Answer 61

Also refers to instances where two genes do not contribute equally to a trait -- Non-additive effects - Interacts in a non-additive way

Answer 62

1. Recessive Epistasis 2. Dominnat Epistasis 3. Duplicate Recessive Epistasis 4. Duplicate Dominant Epistasis 5. One domiant, One recesisve Epistasis 6. Non-additive epistasis

Answer 63

When recessive alleles at one locus mask the effects of the alleles at a second locus Phenotype ratio -- 9:4:3 Example -- Labs Coat color is influenced by 2 labs coat color is influenced by 2 genes (Have B locus and Extension locus) - Extension locus (E locus) -- allows color to extend to fur B = Black (Dom to b) b = Brown E = Extends color = get color in fur (Dom to e) e = no color in fur (ee = yellow) - ALL ee = yellow lab because can't extend pigment into coat Genotypes: B_E_ -- Black - Black = BB or Bb AND Must have EE or Ee --> B_E_ bb = chocolate lab AND must have EE or Ee (bbE_) B_ee -- Yellow --> The recessive e allele is epistatic to B and b (Bceause if have ee then you can't know the genotype at B -- no mater what is ee will be yellow no matter what is at B locus) - Yellow lab = _ _ ee --> Can be BB, Bb, bb --> no matter what if ee then it will be yellow bbE_ -- Brown bbee -- yellow Recessive epistasis because e (which is the recessive allele) makes affects at B -- the recessive e allele is epistatsic to B and b - Recessive epistasis because ee masks phenotype at B locus - e = epistastic allele - B loci = hypostastic gene 9B_E_: 3B_ee: 3bbE_: 1 bbee 9 black (Black + can extend): 3 yellow (Black but no extend = yellow): 3 brown (Brown and can extend): 1 yellow (NO extension) --> 9black: 4 yellow: 3 brown - Different phenotypic ration (9:4:3)

Answer 64

When dominant alleles at one locus mask the effects of the alleles at a second locus - The dominant Alleles do the masking Expected Phenotypic ratio -- 12:3:1 Example -- Compund A (White) --> Compund B (Green) --> Compound C (Yellow) - A --> B -- need enzyme 1 -- W_ = encodes an inhibitor = blocks synthesis of Enzyme 1 = get white -- get build up of A (white squash) - B --> C have enzyme 2 -- Y Produces enzume; - Mutation along the oath = blocks synthesis of compund C = build up of green or white W = produces an inhibitor to enzyme 1 = No A --> B - Dominant W = inhibits enzyme 1 = dominant allele masks affect w = get A --> B Y = produces an Enzyme 2 y = No enzyme 2 (No B --> C -- unable to get enzyme 2 = build up of compund B = stays green) W_YY, W_yy, W_Yy = ALL stay white because inhibits Enzyme 1 (W blocks A --> B -- because block in early step = doesn't matter what allele you have at Y locus) - Means that W (dominant W) masks affects at Y locus (If have dominant W = don't know genotypes at Y locus = Dominant epistasis) wwyy = green squash --> Need to make compund B = need ww + can't make B --C = they need yy wwY_ = yellow squash -- need ww to get A --> B and need 1 Y to get B --> C WwYy X WwYy 9 W_Y_(White) : 3W_yy (White) : 3 wwY_ (Yellow): 1 wwyy (green) 12 white (W_Y_ or W_yy): 3 Yellow: 1 Green Overall: Dominnat W blocks enzyme 1 = blocks affect of Y Locus (If have W = don't know genotype at Y Locus -- white can be W_Y_ or W_yy -- no matter what W means white) --> W shows dominant epistasis to Y gene - Because if stop A --> B = B --> C doesn't matter = masks affect of Y locus = dominant epistasis to Y locus

Answer 65

When recessive alleles at one locus mask the effects of the alleles at a second locus, and the recessive alleles at the second locus mask the effects of the alleles of the first locus - When the recessive allele of both genes mask the effects of the other - Effects of 2 genes on the same trait -- both demonstarte recessive epistasis Predicted Phenotyp Ratio -- 9:7 Occurs often if have 2 genes in the SAME linear biochemical pathway --> frequentley observed when 2 genes produce enzymes for the SAME linear pathways Example -- Snail shells (Image) NEED Compound A --> Compund B --> Comound C (compound C = brown) - A --> B -- have enzyme (coded for by A locus) -- A = make enzyme for A --> B - B --> C -- have enzyme 2 codeed for by B Locus (B makes Enzyme 2 -- Makes B --> C -- get Brown) aa = no A --> B (no enzyme for A --> B = no A --> B - stays white) - aa = white -- masks the affects if alleles at B locus (Masks influence of B -- aa is recessive epistatic to B gene) bb = No B --> C (no enzyme for B --> C -- no compound C = stays white) - bb = white -- makes the affect of any genotype at A (Masks influence of A locus = bb is recessive epistatic to A gene) A_B_ = Brown A_bb = Whote aaB_ = whote aabb = white ***If have either aa or bb or both aabb = will be white Need enzyme 1 and 2 to make color Epistasis because if you have white snail you don't know the genotype of a or b --> could be aa or bb or aabb = don't know genotype at the B locus if have aa and don't know genotype at A locus if have bb (because if have bb white no matter what and if have aa then white no matter what = don't know genotypes at the other locus + since they are both recssive that does the masking = duplicate recssive epistsis Since both are recessive that do the making = duplicate recessive epistasis Since if have bb don't know genotype at A locus and if have aa don't know genotype at B locus = duplicate epistasis - aa = masks B (Can be BB, Bb, bb -- still white) - bb = Makes A (Can be AA, Aa, aa -- still white) - Two genes have epistasis = duplicate epistasis AaBb X AaBb 9A_B_: 3 A_bb: 3 aaB_: 1 aabb - A_B_ = brown - A_bb = white (Have Enzyme 1 BUT no enzyme 2 because bb = white) - aaB_ = white (No enzyme 1 = stays white) - aabb= white (No enzyme 1 or 2 = white) Phenotypic Ratio 9Brown: 7 White

Answer 66

When dominant alleles at one locus mask the effects of the alleles at a second locus, and the dominant alleles at the second locus mask the effects of the alleles of the first locus - When the dominant allele of each gene masks the effect of the other gene - Both act in dominant epistatic manner Common ratio = 15:1 Example -- Wheat Precursor + Gene A --> Pigmemnts Procursor + Gene B --> Pigmenyt Get pigmeny by A or B -- A and B have redundent functions (A or B = get kerneel color) aabb -- recessive = can't make precursur to colored kernerl = stays white A_B_ or A_bb or aaB_ = colored AaBb X AaBb 9A_B_ (Colored): 3A_bb (Colored): 3aaB_ (Colored): 1 aabb - Because just need 1 A or 1 B to get color 15 Color: 1 White

Answer 67

When dominant alleles at one locus mask the effects of the alleles at a second locus, and the recessive alleles at the second locus mask the effects of the alleles of the first locus Common ratio = 13:1 Example Gene B = produces feather color B = colored b = white Gene B produces feathers color -- the B allele produces colored feathers and sis dominant to b = get white feathers Gene A = encodes inhibitor that blocks color formation A = dominant to a and epistatic to B locus bb genotypes = don't make colored feathers -- you can't determine the genotype at the the A locus if have bb = recessive epistasis _ _ bb = all not colored = don't know genotype at A locus if have bb = recessive epistasis A_ genotypes = don't make colored feathers --> You can't determine the genotype at the B locus = dominant epiostasis A_ bb, A_BB, A_Bb = all no color = can't know genotype at B locus = dominant epistasis Colored feather = need B_ and aa (to not have inhibitor --> aaB_) White feather = can encode inhibitory (A_B_) OR Can encode inhibitor and no color (A_bb) OR not encode inhibitor and no color (aabb) AaBb (Make color but also have inhibitor -- white because have inhibitor even though they can make color) X AaBb 9A_B_ (White): 3A_bb (White): 3aaB_ (Colored): 1 aabb (White) PR: 13 White: 3 Color

Answer 68

When alleles at different loci do not contribute equally to a single trait

Answer 69

Answer: B_E_ bbee --> Yellow bbE_ --> Brown B_ee --> Yellow

Answer 70

Answer: __ee --> ee masks effect at B locus = can't know the genotyoe at B locus if yellow lab (Could be BB, Bb, bb -- no matter what is ee then it will be yellow) The recessive e allele is epistatic to B and b (Bceause if have ee then you can't know the genotype at B -- no mater what is ee will be yellow no matter what is at B locus) Recessive epistasis because e (which is the recessive allele) makes affects at B -- the recessive e allele is epistatsic to B and b

Answer 71

Answer: 4/16 (3 + 1 = 4) 9B_E_:3B_ee:3bbE_:1bbee 9 black: 3 yellow: 3 brown: 1 yellow --> 3 + 1 = 4 Phenotype ratio: 9 black: 4 yellow: 3 brown

Answer 72

Answer: B --> gene that does the masking -- masks the effects if color locus Epistatic gene = gene that does the masking

Answer 73

Beacause the rcessive allele does the maksing (e allele is recssive to E = recssive allele --> and the e allele is the allele that masks B locus) e allele (ee) = prevents you from knowing genotyope at B locus -- since e is the recssive -- it is the recessive that does the maksing = recessive epistasis - e = recessive to E NOR recessive to B locus ***Recessive epistasis when the recessive alelle does the maksing Recessive = refering to E to e (e is recssive to E and does the masking) e is recssive to E and does the making = recessive epistasis

Answer 74

Answer: Both are white bb = get A --> B but no B --> C (Stays white -- B is white) aa = no A --> B = white

Answer 75

Common in genes with the SAME linear biochemical pathway

Answer 76

Alleles that are unable to perform the normal function of a gene - Can't produce the function Loss of function is NOT always epistasis

Answer 77

Recessive -- Because in diploid might have mutation in one copy BUT then you still have the normal copy --> if have loss of function still have the other version that can make the gene - In diploid = have a backup -- other copy can do the job

Answer 78

Frequntley observed when genes have redundant functions

Answer 79

Example A+ and A- = incompletley dominant B+ and B- are incompletley dominant where the degree of dominance is h = 0.5 1/16 =

Answer 80

Complex traits can involve polygenic + Additive + Epistatic genes whose relationships can change depending on the environment

Answer 81

A goal of quantitative (complex trait) genetics is to study phenotypic variation in a population and try to parse these effects into the basic compents of dominance, additive, epistatic relationship of one or more genes, and g x e interactions

Answer 82

Most often a single trait is influenced by multiple genes --> genes can interact in different ways to produce novel phenotypes - Interactions = produce novel traits

Answer 83

RrYy = can make 4 types of gametes (RY, Ry, rY, ry) --> because of law of Independent assortment = expect equal frequency of each type of gamete

Answer 84

Can use punnet square to find expected genotype and phenotype frequencies - Build a punnet square of 4 Gametes by 4 gametes of each parent across tops and sides --> THEN fill in the sopace by combing gametes for each parent

Answer 85

IF get 9:3:3:1 Phenotypic ratio --> 2 genes infleunce the same trait in indendent fashion - Here two Loci are contributing independently because still 9:3:3:1 --> Here they are additive = indepedent = no epistasis Still 9:3:3:1 = additive = they contribute independently = no epistasus Ex. In pepper color

Answer 86

Mom genotype = dd --> mom genotype determines the offspring phenotype = all offspring have p phenotype from dd = all sinistral MEG= always 4:0 ratio

Answer 87

Durring Oogeeniss = make gametes BUT inreality only get 1 viable cell and 3 polar bodies If mom Has Dd genotype --> The MEG = makes regulatary RNA - The Regulatory RNA is based on mom genotype --> turns on if she will be Dexttral; or Sinistral (which way it goes is controlled by expression of gene -- contriolled by regulatory RNA) -- regulatopry RNA needs to be in egg cell - The gamete will get that regulatpry RNA made based on mom DNA NOW get two eggs -- one can have D and one can have d - D egg in development is regulated by regulatpory RNA to trin on spinal (based on reg DNA from mom) - d egg = still has regulatpry RNA -->? stull on on dominnat dextral phenotype ebven though it has d -- because of regulatpry RMA -- even if genptype ends up being dd --> but has the regulatpry RNA to make it dextral -- if the regulatpry RNA is there = get Dextral phenoytype BUT the daughter cells from these daighter cells = can't amke regulatpry RNA -- they have enough at first (when daughter cell sget from mom) --> enough at first at the start of development to get dextral but can't pass down to offspring THEN if dd offspriunf that is dextral because mom id Dd --> dd is dextral but has dd = can't make regulatpry RNA for dextral ohenotype like their mom gave them = can make dd Regualtopry RNA in kids --> kids get dd regulatpry RNA = kid is all sinstral Maternal effect = always 4:0 ratio

Answer 88

Dd = all dextral ***MEG = always 4:0 ratio

Answer 89

When genomic DNA is altered so that the expression of a gene is permanetley prevents -- detele a gene or alter promister region (reguilatory region) = can't make that gebe from that copt of the chromosome - Gene is there but never tirns on A gene knockout = A loss of function allele

Answer 90

A gene knockout = a loss of function allele

Answer 91

Because we have 2 chromsomes -- if have loss of function in 1 chrosmomes --> can be active on the other chromsomes -- have other one as a backup

Answer 92

Loss of function alleles can be dominant Example = image Two wild type -- Get normal phenotyp 1 Wt and 1 mutant (1 loss of function allele) - Normal = can make regulatlry RNA + protein BUT not as miuch -- get 1/2 the amount of protein --> might not be enough Here = end with obesity -- there is not enough to get normal phenoyyp -- get obsesity 2 loss of function -- 2 copies of Knockoiut --> have obese mice Here -- Obesity is cominant because heteropzygous shows obesisty SHOWS -- halpoinsifcincey -- when 1 cope of a gene is not enough - Aa -- that one cope of Wt allele = not enough for W-T phenotype --> shows that Loss of function can be dominant

Answer 93

When 1 cope of a gene is not enough - Aa -- that one cope of Wt allele = not enough for W-T phenotype --> shows that Loss of function can be dominant

Answer 94

Example -- Wheat (Red X white wheat) A+ and A- are incompletley domiant; B+ and B- are incompletley dominant where h = 0.5 - A+A- = middle - B+B- = middle --> A+A-B+B- = middle END = see 5 catagpries of color --> NOT 4 like before -- We know it is not epistasis because in epistasis it is max 4 catagpries HERE they thought - what if each + allele contributes 1 unite of color - A-A-B-B- --> no units of color - A+A+B+B+ -- 4 units of color - A+A-B+B- -- 2 units of color = in the middle 1/16 = lihghtests color (light purple) 1/16 = darlest colro (dark purple) - - ***Very common in biology

Answer 95

get bell curve becasue additive -- if contonous --> bell curve disyrbution

Answer 96

2 genes = 5 categories --> based off the # of additive alleles that are there 1A+A+B+B+ -- 4 units of color 2A+A+B+B -- 3 units of color A+A-B+B+ -- 3 units of color 4 A+A-B+B- -- 2 units of color 1 A+A+B-B- -- 2 units of color 1A-A-B+B+ -- 2 units of color More genes = start to see continous distrubution of traits (more genes = more catagoreies)

Answer 97

The proporition of indvioduals at eaitehr end of dirbution = (1/4)^n n = number of genes - look at number of indiviuals at the more left or most right caratgories Ex from wheat 1/16 = white (no alleles); 1/16 = red (4 alleles) - 1/16 at the ends 1/16 = (1/4)^4 n = 2 Can get the total number of genes that control a trait if you now the number of individuals at the end and the total number of individuals ***If the end numbers are different = pick one end -- can solve for both

Answer 98

First need to look at how many out of total (how many out of 256) has extreme phenotype -- look at the two ends --> look at the tails (look at the two extremes) Here ends = 1 indivual (1/256) 1/256 = (1/4)^n n = 4 --> 4 genes

Answer 99

Start -- subtrarct both homozygous varlues Here -- 20- 0 = 20 Then divide that value by the total number of alleles --> 20/10 = 2 Means each additive allele contributes +2 HERE -- have 5 genes A+A+B+B+C+C+D+D+E+E+ = 20% (most we can make is 20%) aabbccddee = 0% most we can make is 20 and least is 0 --> 10 allles total --> 20/10 = 2 00> each allele contributes 2% oil Each adds 2% = for 8% oil = we need 4 additive alleles Range = 0- 20 20 - 0 = 20 20 (difference between highest and lowester) /10 (10 allels) = 2

Answer 100

Can be any combo of 4 alleles -- all would give the same phenotype for additive alleles

Answer 101

bb = recesive eepistasis because b allele is recessive allele and that recessive b allles masks teh effects of A -- don't know genetype of A A_ = dominant that blocks B locus -- dominant A = masks B locus