Complex Traits Flashcards
Relationship between alleles at one locus
Can lead to non-mendelian genetics patterns
Mendle
Discovered how alleles at one gene can be dominant or recessive
Did Aa X Aa – go t phenotypic ratio in offspring were predictable
3:1 Phenotypic ratio
3:1 F2 phenotypic ratio in Monohybrid cross
Pattern of inheritance = Mendelian Inheritance –> extends to corsses with heterozygosity at 2 loci
F2 phenotypic ratio in Dihybrid cross AaBb X AaBb
Get phenotypic ratio 9:3:3:1
Non-mendelian inheritance
When phenotypic ratios deviate from 3:1
Non-Mendeian inheritnace at one locus (At one gene)
- Incomplete dominance
- Co-dominance
- maternal Effects + Cytoplasmic Inheritance
- Sex Influence + Sex-linked
- Lethal Alleles
- Conditional Allele
Complex (Quanatative traits)
Demonstrate continuous phenotypic variation + do not behave according to Mendelian Genetics
Mendelian Inheritance
Monohybrid (Aa X Aa)
GR – 1:2:1
PR – 3 Dom: 1 recessive
Incomplete dominance
Relationship between 2 alleles were Aa is intermediate between either homozygote
Example:
P = Purple
p = White
PP = Purple
pp = white
Pp = Violat – mid between purple and white
Pp X Pp – PR –> 1 Purple: 2 Violate: 1 White = deviation from 3:1 = Non-mendelian inheritance
Degree of Dominace (h)
Describes the degree to which the phenotype of the heterozygote resembles the phenotype of a homozygous parent
Range from 0 - 1
Meaning/finding Degree of dominance
You need to define one homozygous phenotype as h = 1
Ex. Flowers
Red – H = 1
White – h = 0
h= 0.5 – midpoint pink
h > 0.5 – Darker pink
h < 0.5 – lighter pink
***need to define one of the homozygotes as h = 1
Answer: C – Incomplete dominance where h = 0.7
Black – h = 1 (given in question)
White – h = 0
Dark grey = closer to black = h >0.5 –> h = 0.7
Expressivity
When phenotypes vary between individuals with a particular genotype
- There is a wide range of phenotypes for a single genotype
High expresivity = there is a BIG range of phenotypes
Expressivity example – Polydactalism
A = dominant to a
AA or Aa = extra digits
aa = normal # of digits
Aa genotypes – have some with just extra small nub and some with many more fingers = wide range of phenotypes but individuals all have the same genotype
- There is a wide range of phenotypes for a single genotype
A = have polydactalism
aa = normal
A = extra digits
Degrees of polydactalism can vary
Autosomal Dominant with incomplete penetrance
Extect II - 1 to show the traut because offspring is affected but they don’t
Expressivity + Incomplete dominance
Expressivity (range of phenotypes) can sometimes explain incomplete penetrance – if you have high expresivity = have very wide range of phenotypes –> if have wide range then you can have a very low end phenotype – since it is very low end it might not be detected but then can have offspring with detected trait
- High expresivity is one way to explain incomplete penetrance
Penetrance
The proportion of individuals with a particular genotype that also express the associated trait
Incomplete penetrance
When some individuals with a particular genotype express the trait while others do not
Calculating penetrance example
Here 90% – because 9 out of 10 individuals with the genotype have the trait
To solve:
Need to find the genotypes of individuals – look at who NEEDS to have the genotype then look at how many out of those actually have the trait
with trait/ # with genotype
Co-dominance
relationship where Aa includes phenotypes of AA and aa
***Example – Red + White –> Red and white spots
Example – Blood types
AA –> Type A – Have A antigen on RBC surface
BB –> Type B – Have B antigen on RBC surface
AB – Type AB (CO-DOMINANT) –> produce BOTH A and B antoigen (Have both AA and BB phenotype NOT a intermediate one)
AB X AB –> Get 1 AA:2AB :1BB Phenotypeic ratio –> Non-mendelian ratio
Blood type adding in i allele
KNOW – AB is co-dominant BUT both are dominant to i
i = no antigen – ii = type O
Type A = AA or Ai
Type B = BB or Bi
Have multiple allles in gene = can make it harder to predict outcome of cross = interupts mendelian inheritance pattern
Having multiple alleles per gene + mendelain inheritance
Have multiple alleles in gene = can make it harder to predict outcome of cross = interupts mendelian inheritance pattern
Example –
Type A = AA or Ai
Type B = BB or Bi
Without knowing the genotypes (AA or Ai) = don’t know outcome of the cross
Allele context
Allelic relations = need context –> Allele is only dominant to another allele BUT if comparing to another allele that might change
Blood transfusions
Type A = doesn’t have B antigen = sees B as a foreign = have immune response producing anti-B antigen = essential that the person with Type A doesn’t get blood with B allele
- Can’t get B or AB
- Can only get A or O
Type B = recognizes A as foreign
- Requires B or O
Type AB = have A and B = no immune response to Type A, Type B ot Type O
- Can accept blood from all blood types –> Universal accpetor
Type O = No Antigen = reocgnizes A and B as foreign
- All other types can accept Type O = universal donor
- Can only get type O
Affect of Recessive Lethal Alleles
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
Example Recessive lethal – Agouti in mice
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
Findoing probability in recessive lethal
+ Example
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)
Why ressive lethal if the allele is also dominant
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
Recessive lethal phenotypic ratio in monohybird
2:1
Recessive Lethal Example 2 – Tail in cats
S = truncated tails and is dominat to s
SS = leads to spindal development probelms and is embryonic lethal
Organelle inheritance
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
What passes on cytoplasmic componenets
IN mammals egg cells (female gametes) pass on all cytoplasmic components – mtDNA is inherited through mothers
Paternal mtDNA is activley elimanted in zygote
Affect of organelle inheritance
Unpredictable organelle inheritance may lead to differences in trait expression
Characteristics of cytoplasmically inherited traits
- Present in Males and Females
- Usually inherited from one parent – usually the maternal parent
- Reciprical crosses give different results
- Exhibit extemsive phenotypic varaition even within a single family
Can there be paternal inheritance of mtDNA
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
Maternal affects Genes
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
Maternal effects
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
Sex limits vs. Sex influences traits
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
Example Sex limited – Precocious puberty
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
Example of sex influenced – male pattern baldness
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
Sex linked male pattern baldness
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
Conditional alleles
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
Conditional alleles can be
Lethal –> means that they are lethal only under certain conditions = have conditional lethal allele
Example conditional lethal allele – favism
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
Allele frequencey for favism varaint
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
Given X-inactivation – why do Turner (XO) or Klinfelder (XXY) syndromes exist
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
Common mechanism in XY females
Androgen insensitivity syndrome
Low response to testostrone
Overproducing testosterone
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
XX males
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
Sex chromomomes across biology
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)
Extreme sex chromsomes
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
Sex + Ploidy
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
Genic Mating System
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
How are there multiple mating types in yeast
Have pheranomes + pheramone receptors that allow this to happen
Drosphila sex determination
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
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
Maternal Effects example
D = Dextral
d = Sinistral
MOM Dd X Dad Dd
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
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
Rh Factor
G = Grey
g = White
B = Black
b = Red
G_bb –> Grey and red
Answer: 3/16
Menelian Dihybrid Cross Genotype and Phenotype Ratios (walks through Dihybrid cross steps)
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
Example of Polygenic trait (Two genes one trait)
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: 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+_
Dominant Epistasis
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
Duplicate Recessive Epistasis
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
Duplicate Dominant Epistasis
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
One Dominant, One Recessive Epistasis
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: B_E_
bbee –> Yellow
bbE_ –> Brown
B_ee –> Yellow
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: 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: B –> gene that does the masking – masks the effects if color locus
Epistatic gene = gene that does the masking
Black, brown, yellow lab example – why is it recessive epistasis
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: Both are white
bb = get A –> B but no B –> C (Stays white – B is white)
aa = no A –> B = white
Expected phenotypic ratios – summary
Additive traits
Example
A+ and A- = incompletley dominant
B+ and B- are incompletley dominant where the degree of dominance is h = 0.5
1/16 =
Distrbution of Addition Traits
Calculating the number of loci involved in additive traits
Practice –
Two genes infleuncing trait independently
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
Mom genotype = dd –> mom genotype determines the offspring phenotype = all offspring have p
phenotype from dd = all
sinistral
MEG= always 4:0 ratio
How do Maternal Effect Genes work?
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
Dd = all dextral
***MEG = always 4:0 ratio
Haploinsufficinecey
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
Additive traits
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)
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***Very common in biology
Distribution of additive traits
get bell curve becasue additive – if contonous –> bell curve disyrbution
Example – finding the amount of genes that control a trait
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
Finding the genotype example
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
Can be any combo of 4 alleles – all would give the same phenotype for additive alleles
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
