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
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
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
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)
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%
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
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
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
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
