Linkage Flashcards
What is the goal of linkage mapping
To identify genes that contribute to a trait
Who discovered linkage/Linkage mapping
Nancey Wexler –> Gene Hunter
- Idetofied gene for Huntingtons Disease
Discovered the locus causing huntington’s disease
What do Unlinked loci follow
Unlinked Loci follow Independent assortment
- Alleles on different chromosomes assort independently
- When genes are not linked they sort independently (easiest to see on different chrosmomes)
- Loci that are on separate chromosomes will assort independently
- DNA sequences far apart on the same chromosome assort independently due to recombination
Independent assortment on the same chromosome
DNA sequences far apart on the same chromosome assort independently due to recombination
- 2 genes on the same chromosome that are unlinked = follow independent assortment
- When genes are not linked they sort independently
- Unlinked genes = generate same freuqmncey of recombined and non-recombined gametes after meiosis
IMAGE – because A and B are far apart and recombination happens = have IA
- When far enough apart = have equal representation of all genotypes
- A and B are on the same chromsomes but are far enough (at different ends) = recombination occurs = IA
- During prophase corssover can occur = get chromsomes with recombined version (chromatid has A and b and other has a and B) –> continue in meiosis and end with gametes AB, ab, aB, Ab
Same arrangment of alleles that occurs on parental genotype = same alleles that occur in the original chrosmomes = non-recombined (AB and ab) = non-recombinant/parental genotyopes
aB and Ab = result of recombination = recombined gametes
Linked loci during Mitosis
Linked loci travel together during Mitosis
- If close together = won’t assort independently – assort together
- Things close to gene = carried along
IMAGE – recombination rarely occurs between A and B = gametes are not going to have an equal combination –> Get AB and ab more = not even frequcey (NOT 1:1:1:1)
Recombination Frequency
The measure of how frequently recombination occurs between 2 loci
- How often does recombination occur between 2 loci
RF of completley linked vs. partial linkage
Completely linked – RF = 0% –> No recombinant gametes produces
Partial linkage – 0 < RF < 50% (RF is between 0-50) –> Some recombinant gametes are produced but fewer than you would expect
Unlinked vs. Linked RF vs. Compltley linked
Unlinked – RF = 50%
Linked – RF = <50%
Compltley linked – RF = 0%
Genes = in complete linkage when no recombinant genotypes are made during meiosis
Genes = linked if recombination between them occurs <50% of the time
> 50% Recombination Frequency
Means the genes are in repulsion – never want to be together
When does recombination occur
Prophase I
Homologous Recombination (Process)
During Prophase I homologous chromosomes are paired together tightly through synapsis
- HC = held together by cohesion proteins
During Synapsis Homologous regions of chromostis are cut (restricted) and Pasted (ligated) back together –> Results in exchange of DNA between homologous chromosomes
MINE – Homologous regions are cut and ligated back together – cut strands and paste back together –> One cut and one exchange
Synapsis
Process that pairs HC together tightly during Prophase I
Restricted vs. Ligated
Restricted = cut
Ligated = pasted back together
One crossover event = one cut/one exchange/one ligation
Recombinant vs. non-recombinant chromosomes after crossover
Crossing over
Crossing over = Homologous Recombination
Where does Recombination occur between
Recombination occurs between identical sequences on homologous chromosomes
DNA = cut at the same position on homologous chromosomes – cut in same place on both chromatids
- Chromatids exchange ends and glue them back together
***NO DNA LOST OR GAINED
Identical DNA needed for recommendation
All species have different amounts of identical DNA needed for recombination
Frequency of crossover
Recombination occurs 1 - 3 times per arm per chromosome per meiosis
Parents vs. you – Parent have 1/2 recombination events –> pass chromosome to you – reason have grandma and grandfather on same chromosome
ANSWER: 1 Recombination event
Cut both HC at once – one cute = 1 recombination event
***Two chromosomes but still 1 recombination event
Counting number of recombination events
1 cut = 1 recombination event
Even if 2 chromosomes –> if one cut = one recombination event
ANSWER: 2
2 cuts = 2 recombination events –> Only way to have different chromosome in the middle
Single crossover vs. double crossover
Single crossover – one recombination event (1 cut)
Double crossover – 2 recombination events (2 cuts)
Gametic phase
Phase represents the allele combinations that were inherited from each parent
Example: AaBb
Two ways to get AaBb individual
1. Mom AB X Dad ab
PHASE = AB/ab
- AB = comes from one parent
- ab = comes from other parent
- Dad Ab X Mom aB
Ab/aB
PHASE = Ab/aB
- Ab = comes from one parent
- aB = comes from other parent
GET the same genotype BUT different phase
***Both get the same genotype but different arrangement = different phase – allelic combinations inherited from one parent vs. other
***A and B can be on different chromosomes –> Just tells you the phase – doesn’t mean you know genes are linked
Phase + Linkage
Phase DOESN’T mean linkage – the genes you are looking at can be on different chromosomes –> just know the phase doesn’t mean you know genes are linked
If two alleles are in phase it DOES NOT mean they are linked
ANSWER: E
BOTH C and D are right – there is no convention on which to put on the top or bottom
Phase = alleles from mom/alleles from dad
Finding phase if don’t know parent genotype
LOOK at genotype of offspring – see which one is in highest abundance – see what is linked –> put the linked ones together
***Look at practice problems
Gametes can have allele combinations that are
- Parents
- Non-parental
Example
AB/ab –> Can make AB, ab, aB, Ab gametes
- AB, ab = parental gametes (same phase as parents) –> on phase/non-recombinant/parental genotype
n Ab, aB = recombinat/non-parental
Parental gametes
Parental/non-recombinant/in phase –> the phase in the gametes that matches the phase in the parent
NO change in phase
Ex. AB/ab –> AB, ab = parental
Non-parental genotypes
Non-parental/recombinant/in repulsion/uncoupled/out of phase –> The phase in gametes is different than the phases in the parent
- Have a recombination event to make gamtes that are different
Ex. AB/ab –> Ab, aB = non-parental
ANSWER: Ab and aB –> because to get Ab or aB = need to have recombination events
Need phase of parents to know if offspring have recombinant or non-recombinant
Answer: AB and ab
What do you need to know if offspring have recombinant or non-recombinant
Need phase of parents to know if offspring have recombinant or non-recombinant
Beginning studies of linkage
Studies of linkage began in Fruit flies –> Thomas hunt Morgan
Morgan = understood linkage – understood that there was not an equal number of genotypes in offspring if genes are linked
Harreit creighton and Barbra Mcclintock
Unsung heros –> discovered recombination
- Discovered transposable elements + she also documented linkage in plants –> Followed chromosomes with a Knob through meiosis – looked at end to know it was linked to another gene –> Found recombination events
How do we know if two genes are linked
Linkage analysis – test cross for linkage
Linkage analysis – test cross for linkage Steps
- Create heterozygoes with Known phase
- Preform a test cross –> Determine the genotypes and phase of offspring classes – identify which contain parental or recombinant gametes
- Score progeny
- If progeny containing parental genotypes > progeny with non-parental gametes – you have discovered linked gemes
- Use statsics to support your claim –> Chi square test for independence
- Calculate the genetic distance between genes
First step of linkage anlysis
Create heteozygotes with known linkage –> Take a WT homozygous dominant ( know only can have dominant = phase is CV+Y+/CV+Y+) X Homozygous recesive (Know phase must be cvy/cvy)
END = get heterozygous – know the phase MUST be
CV+Y+/cvy –> Know phase (one parent ONLY can give dominant alleles and one parent can only give recessive alleles)
Test Cross in linkage analysis
Cross geterozygotes with known phase to homozygous recessive
- Heterozgous = can make 4 types of gametes
CC+Y+/cvy X cvy/cvy
THEN – Determine the genotypes and phase of offspring classes – identify which contain parental or recombinant gametes
Here =
***IN ALL one of the phase MUST be cvy – because can only get cvy from cvy/cvy parent
1. CV+Y+/cvy – parental (in phase – non-recombinant)
2. cvY+/cvy – out of phase/recombinant/non-parental
3. CV+y/cvy – out of phase/non-parental
4. cvy/cvy – parental (Same phase/non-recombinant)
Discovering linked genes
If progeny conatining parental genotype > progeney with non-parental gametes – you have discivered linked genes
EXAMPLE–
63 CV+Y+/cvy – parental
28 cvY+/cvy – non-parental
33 CV+y/cvy – non-parental
77 cvy/cvy – parental
HERE = more parental = looks linked (67 + 77 > 28 + 33) –> more parental = likely linkage
Linked genes
If linked –> recombinants shouldn’t happen as frequncetley
- # of offspring in parental classes should occur more often if genes are linked
- Unlinked = equal numbers across all
- Linked = have more of parental
Calculating genetic distance between linked genes
Genetic distance = recombination frequencey
Recombination frequency = (# of recombination events/total progeny) X 100
- # of recombination events = # of individuals in recombination classes
Example
63 CV+Y+/cvy – parental
28 cvY+/cvy – non-parental
33 CV+y/cvy – non-parental
77 cvy/cvy – parental
RF = 28 + 33/ 201 X 100 = 61/2-3 = 30% –> 30 RF = 30cM = 30 mu
- 30% frequncey = 30 mu = 30 cM
This number should be less than 50 because we think they are linked –> see more parental than recombinant = think linked = logical that it should be <50
Genetic distance
RF = mu = cM
ANSWER:20 cM
ANSWER: 17 cM
ANSWER: 10%
Because have 2 catagories for recombinat and 2 catagories for non-recombinant
20cM = 20% recombinant (20% non-parental)
Have 2 catagories of reocmbinant –> each catagory = 10%
Types of mapping
- Cytogenic Map
- Genetic map
- Physical map
Cytogenic map
Show positions on chromosomes based on cytological features
- Banding pattern + where centromere is
Ex. 4P2.2
Genetic Map
Shows relative positions of genes (or SNPs) based on how frequently recombination between them occurs
RF –> gives genetic map
Physical Map
Based on DNA sequences – # of BP in between 2 genes –> often correlated with recombination frequcney (not always)
- More BP between = more likley recombine
Example Independent Assortment
RrYy – R and Y are on separate chromosomes
In meiosis – chromsomes can align in two ways (image)
Get RY, ry gametes OR rY, Ry gametes
Takes 2 seperate meiosis events to produce ALL 4 tyoes of gametes and each miotic event is equallly likley = each genotype is equally likley = Unlinked independentley assorting genes
What do unlinked genes produce
Unlinked genes = generate same frequency of recombined and non-recombined gametes after meiosis – same genotype frequencey as of the loci were on different chromosomes
Genetic Linkage
Tendency of DNA sequences that are close together on a chromosomes to be inherited together following meiosis
***Genes = in complete linkage when no recombinant genotypes are made during meiosis
IMAGE – A and B on the same chromsomes close together = unlikely that recombination event that may occur will happen between the 2 loci = gametes made contain orginal parental non-recomcombined versions
Recombination between genes close together
Can have recombination between genes that are close together but target region for recombined is small = likelihood of recombination will separate two alleles decreases when genes are close together
Genes = linked if recombination between them occurs <50% of the time
- DNA sequences close together = inherited together
- Genes are linked of thete ate more parental genotypes than recombinat genotypes (see more parental genes than non-parental genotypes in gametes)
When are genes linked
Genes are linked of there ate more parental genotypes than recombinant genotypes (see more parental genes than non-parental genotypes in gametes)
Testing for linkage (overall)
Test for linkage by preforming crosses and following genotypes of offspring produced
Testing for linkage (Drosphilla example)
CV+ = non-curved
cv = curved
Y+ = non yellow
y = yellow
WT = CV+Y+
Question = are the Y and CV loci linked?
Step #1 – Set up cross to see if recombination occurs
- To set up a test cross = first need Homozygous dominnat X homozygous recessive to get F1 heterozygous with known phase
- To make Heterozygous = CV+CV+Y+Y+ X cvcvyy –> Get CV+cvY+y
- Can write CV+cvY+y as genetic compenent from one parent/genetic component from other parent = CV+Y+/cvy (know CV+Y+ came from one parent and cvy came from other)
CV+cvY+y = CV+Y+/cvy (Know the phase because know parents = parents could only give those)
CROSS = Heterozygous (with known phase) X homozygous recessive
- Heterozygous gives – CV+Y+, CV+y, cvY+, cvy
- Homozygous = can only give cvy
After corss = get 4 offspring types
1. CV+Y+/cvy – non- recombinant
2. CV+y/cvy – recombinant
3. cvY+/cvy – recombinant
4. cv/cvy – non-recombinant
CV+Y+ and cvy = parenat genotype = non-recombinannt
CV+y and cvY+ = recombinant/non parental gametes from heetrozygous
END – observed
63 CV+Y+/cvy – non- recombinant
33 CV+y/cvy – recombinant
28 cvY+/cvy – recombinant
77 cv/cvy – non-recombinant
***If not linked expect equal frequncey of all gametic combinations (equal frequncey of all phenotype classes)
Which gametes in test cross tell us if recombination occurs
Gametes from heteozygous tell us if recombination in gametes occurs or not
Males vs. females in drosphila
Males don’t undergo recombination – in test cross the homozygous recessive needs to be male and females are heterozygous
Chi square test for independence
Overall: testing for linkage (also corrects for differential survival)
- Write genotypes of cross testing – CV+Y+/cvy X cvy/cvy
- State Null – Deviations from expected numbers of each phenotypic class for sample size of ____ is due to random chance/sampling error
- Determine expected value based on sample size – and find X^2 value
Example – X^2 = 30.7 - Find Degree of freedom – df = (# rows - 1) X (# columns -1)
- State alpha (P) value for hypothesis – at P = 0.05 – X^2 for df = 1 = 3.814
- Look at the table
- State whether H0 can be rejected – compare X^2 vs. X^2 CV
X^2 = 30.7 –> 30.7 > 3.814 (X^2 > CV) = reject the null hypothesis
- State overall conclusion – deviations form expected values assuming Independent assortment are NOT due to random chance or sampling error = due to someplthing else –> Most likley that Y and CV genes are linked
ONCE you know that two things are linked you can then look how linked they are – how close they are
What does Chi square test for indepence resolve
For unlinked genes we expect a 1:1:1:1 ratio of phenotype classes BUT this assumes that all genotypes are equally likley to survive amd are equally represented in porgencey –> The Chi square test for Independence corrects for differential survivoship
Example – lethal alleles/partially lethal alleles could screq up the phenotypic ratio of classes (this would not be related to linkage)
How do you find expected value for Chi square TOI
Use a contingencey table
Do gene BY gene of the gamtes of heteozygotes (break doiwn gamete contribution of heterozygote) –> then fill in # of each of the indiviuals with that genotype
- Fill out the table by considering phenotypic catagories corresponding to gamete contributeion of heterozygote (63 CV+Y+ from heterozygote or 28 cvY+ from heterozygote)
Expected for each genotype = Row total X column totak/grand total
Grand total = Row total + row total OR column total + column total
Example – 96 + 105
Example
CV+Y+ – 96 X 91/201 = 43.5
CV+y = 96 X 110/201 = 52.5
cvY+ = 91 X 105/ 201 = 47.5
cvy = 105 X 110/ 201 = 57.5
THEN do Chi square –> (O-E)^2/E –> sum all of the values
(63 - 43.5)^2/43.2 + (28 - 47.5)^2/47.5 + (33 - 52.5)^2/52.5 + (77 - 57.5)^2/57.2 = 30.7
Determining dF for contigencey table
df = (# rows - 1) X (# colums -1)
Example (in ours) – (2-1) X (2-1) = 1
Steps after Ch-square TOI
Can know Y and CV are linked BUT THEN can ask how linked are they –> can know how close they are
Genetic distance
Genetic distance = Recombination frequncey – get distance by looking at how common recombination occurs
Genetic distance + probabilities
We can think of genetic distance as probabilities
EX. RF = 30.7 –> 0.307 percent chance that there will be recombination event during meiosis between these 2 loci
Phase coupling + recombination
Heterozygote = can have 2 phases of their alleles
AaBb – Can be AB/ab OR aB/Ab
- Get AB from one parent and ab from other parent OR aB from one parent and Ab from other parent
Can test for linkage using either type of zygote just need to know which is parental genotype and which is recombinant genotype
If two possibilities for phases which should you choose?
Can test for linkage using either type of zygote just need to know which is parental genotype and which is recombinant genotype
Example – AB/ab or Ab/aB
AB/ab – give parental gametes of AB or ab and recombinat gametes of Ab and aB
Ab/aB – gives parental gametes of Ab and aB and recombinant gamteres if AB and ab
Parental arrangment + phase
Parental arrangment = in phase
Example AB/ab –> AB is in phase/couples; ab = in phase/coupled
- PHASE DOES NOT MEAN THAT THEY ARE LINKED – JUST MEANS THAT RECEIVED SAME AS PARENT
recombinant for AB/ab –> Ab and aB = out of phase/in repulsion
