Linkage Flashcards

Question

Answer 1

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

Answer 2

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

Answer 3

1. Parents 2. 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 4

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

Answer 5

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 6

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 7

Answer: AB and ab

Answer 8

Need phase of parents to know if offspring have recombinant or non-recombinant

Answer 9

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

Answer 10

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

Answer 11

Linkage analysis -- test cross for linkage

Answer 12

1. Create heterozygoes with Known phase 2. Preform a test cross --> Determine the genotypes and phase of offspring classes -- identify which contain parental or recombinant gametes 3. Score progeny 4. If progeny containing parental genotypes > progeny with non-parental gametes -- you have discovered linked gemes 5. Use statsics to support your claim --> Chi square test for independence 6. Calculate the genetic distance between genes

Answer 13

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)

Answer 14

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 15

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

Answer 16

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

Answer 17

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

Answer 18

RF = mu = cM

Answer 19

ANSWER:20 cM

Answer 20

ANSWER: 17 cM

Answer 21

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%

Answer 22

1. Cytogenic Map 2. Genetic map 3. Physical map

Answer 23

Show positions on chromosomes based on cytological features - Banding pattern + where centromere is Ex. 4P2.2

Answer 24

Shows relative positions of genes (or SNPs) based on how frequently recombination between them occurs RF --> gives genetic map

Answer 25

Based on DNA sequences -- # of BP in between 2 genes --> often correlated with recombination frequcney (not always) - More BP between = more likley recombine

Answer 26

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

Answer 27

Unlinked genes = generate same frequency of recombined and non-recombined gametes after meiosis -- same genotype frequencey as of the loci were on different chromosomes

Answer 28

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

Answer 29

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)

Answer 30

Genes are linked of there ate more parental genotypes than recombinant genotypes (see more parental genes than non-parental genotypes in gametes)

Answer 31

Test for linkage by preforming crosses and following genotypes of offspring produced

Answer 32

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)

Answer 33

Gametes from heteozygous tell us if recombination in gametes occurs or not

Answer 34

Males don't undergo recombination -- in test cross the homozygous recessive needs to be male and females are heterozygous

Answer 35

Overall: testing for linkage (also corrects for differential survival) 1. Write genotypes of cross testing -- CV+Y+/cvy X cvy/cvy 2. State Null -- Deviations from expected numbers of each phenotypic class for sample size of ____ is due to random chance/sampling error 3. Determine expected value based on sample size -- and find X^2 value Example -- X^2 = 30.7 4. Find Degree of freedom -- df = (# rows - 1) X (# columns -1) 5. State alpha (P) value for hypothesis -- at P = 0.05 -- X^2 for df = 1 = 3.814 - Look at the table 6. 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 7. 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

Answer 36

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)

Answer 37

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

Answer 38

df = (# rows - 1) X (# colums -1) Example (in ours) -- (2-1) X (2-1) = 1

Answer 39

Can know Y and CV are linked BUT THEN can ask how linked are they --> can know how close they are

Answer 40

Genetic distance = Recombination frequncey -- get distance by looking at how common recombination occurs

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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