5.1-5.4 Flashcards
5.1 Gene Linkage and Recombination
most chrosmoems carry many genes
recognition that that many genes lay on a crhsome asks how do genes on same chromsomes assort
some genes on same chrosome dont assort indpendtly - linked
x-linke dropsophila genes
females have two x so two alleles of an x linked gene while males only have one alelle
the two x-lnked genes that dtermine fruit fly eye colour and body colour
these genes are syntenic bc they are lcoated on the same chrosmome
the white gene- w+ is red and w is white eyes
the alleles for body colour are y+ brwon and y yellow
in a cross between feamle with white eye nand brown body (wy+/wy+) and male with red eyes and yellow body (w+y/Y)
the f1 offpsring are evenly brown body female with normal red eyes (wy+/w+y) and brown bdoiss males with mutant alleles (wy+/Y)
male proegny look like mum bc phenotype reflects the geenotype of teh single x chrosme they got from here
with tow alles of x linked enes, one from each parent, the domiance relations of each pair fo alleles detmines the female phenotype
if these two drosphila genes for eye and body coloru aossrt indepdnetly then dihybrid f1 females should make 4 kinds of gamtes with 4 diff combos of alleles: wy+,w+y,w+y+ and wy - they should occur in 1:1:1:1 ratio so that half of teh gametes are of the two parental types (wy+ or w+y like parents)
the remang half of gametes would be two recombinant types - w+y+ or wy allele combos not seen in P gen
by looking at male progeny you can see if tehres teh 1:1:1:1 ratio bc they get x-linked genes form maternal gamete
the breeding results showed a 99% of paental genotype to 1% recomb
the two genes failed to assort idnepdnetly - must be related to fact that both genes are x-linekd and thus syntenic
there are two psosobilities why its the parental that are large majority:
- the wy+ and w+y combos might be prefered bc inistinc chemical affinity
or these combos might show up bc they are parnetla types
that is the f1 female inehrted w and y+ toeghgtehr and w+ and y toegtehr from father
the f1 female is tehn mroe likley to pass on tehse patenral combo of alleles, ratehr than recomb to their progeny
a second set of crosses with diff arangement of alles show why dirhbrid f1 dont produce 1:1:1:1 ratio in teh four possible types of gametes
in second set, the orignal aprental gen was red ey ebrown females (w+y+/w+y+) and white eye yellow male (xy/Y) os f1 females are w+y+/wy dihyrbids
in the sons of f1 cross the w+y/Y and wy+/Y recomb account for little more than 1% and parnetal recombs are almost 99%
clearly theres no prefered asociation of w_ and y or of y+ and w
the frequency of the various types of progeny dpend on how the arrangeemnt of alleles in f1 female orgnated
the parental classes - combo presnet in P gen that show up most frqntely in f2
the reshfuled recomb classes occur less often
deisgnaton of parnetal and reocmb gametes or porgeny of doubly ehrtogous f1 female is detrmiend by partcualr set of alleles taht feamle received erom each of her parents
when alleels of genes assort indepdntly, number of parnetal and recomb in f2 are equal bc doubly hetroyzgous f1 produce equal numebrs of all 4 agmetes
but when two genes are linked whne the number of f2 proegny with parnetal geneotypes exceeds the number of f2 proegny with recom genotypes
instead of asosrting indepndtly, partcular allels behave as if tehy are connected to each otehr much of the time
linkage isnt always as tight
ex. mutation for mini wings and red eye colours - the genotype of f2 gen is 67% parnetal and 33% recomb
this prepondeance of parnetal types among f2 males rveals tha the two genes are linked
but comapred to 99% linked for eye coloru and body colour this linakeg isnt that tight
(33 out of 100 rather than 1 out 100)
linkage of automsal traits
linked uatomal genes arent inehrted according to 9:3:3:1 ratio for two idepndtly assorting genes
if there was linkage and f1 dihyrbids were AB/ab - the 9/16 and 1/16 classes wouldve increased at expense of the 3/16 classes
but if allels of dihhrbid f1 were Ab/aB then the two 3/16 clases would inc
linkage undoes the 9:3:3:1 ratio
no more euqally likely fetrliation
testcrosses simplfy detction of linkage
geneticst found it difucl to inetrpet corsses involing autosomal genesbc it was impsoble to tarce which aleles came from what parent
by setting up testcrosses in which one parent was homzygous recieve for both genes, genetcis tcan analyze eaisly the gene combo transmited by the otehr doubly hetrozygous parent
in furit flies ex. they carry an autosmal gene for body colour; wild type brown and mutant black
a second gene under same autsome detrmiens shape of wing - theres c+ straight and c curve
in breeding of two pure breeds - balck body straight wing and brwon body with curve all the f1 dihrid are bc+/b+c and phenotypcialy wild type
in testcorss of f1 females with bc/bc males, all offspirng receive recieve b and c from father
the pgenotpes of offspring thus idncate the kidns of gametes rceived from the mom
for ex. a black fly with normal wings would be geneotyep bc+/bc ; bc we know b c was fro father, the bc+ was from mom
rouggly 77% of testcorss proegny receive paretal gene combo while 23% were recombs
bc the parnetal classes outnumber recomb classes, we can conclude that automsla genes for balck body and curve wings are linked
5.2 Recombination: A result of Crossing over during Meiosis
its easy to udneftsand that genes connected on same chrosome can be transmitetd togethr and show linakge but how do linked genes still show some recomb if enough progeny is made?
Janssens descirbed the strcutures happening in phosphase 1 as chisamata - they might have to do iwth recomb genes
chiasmata seem to reprsent regions in which nosister chormatdis of homologous chrosomes cross over with eachother
Morgan sugegsted that the chiasmata observed trhough light microsocpe were istes of chrosome breakage and exchange reuslting in genetic recomb
reciprocal exchanges between homologs are the physical basis of recomb
morgans idea that the phsycial brekaing and rejoining of chromsomes during meisis was baiss of gentic recomb was reasonbale
but even though janssens chiasmata could be inetrpested as signs of the process, there wasnt clear cut evdience that crossing over between homologs actually occurs
the ID of physical markers or cytologcially visible abnormalties in chrosmme strcuture that can be follwoed from one gen to the next enabled rsearchers to provde morgans conjecture
ppl studied corn and drosophila showing that egentic recomb does depdn on reciprocal exchnage of parts between maternal and paternal homologs
Stern bred female flies with two diiff X chrosmmes, each containing a distinct physical marker ner one of the ends
these same females were also doubly heteroyzgous for two X linked genetic markers- alleles of genes that can serve as points of reference in dtemring whetehr progeny were result of recomb
one x chromsome carried mutation to produce carnation eyes (car) that were kidney shaped (bar) - teh chromse was marked phsycially by visible disconintunt which resulted whne the end of the x chromsome was broekn off and attached to autsome
the other x had wild type alleels and phscial marker was part of Y chromsome that became connected to the x chrosmome centromere
all sons showing a pheontype dtermined by one or the otehr parnetal combo of genes (car Bar or car+ Bar+) had an x chrosmome that was strcally indistingusble from one of the orginal x chrosmomes in mom
in recomb sons hwoever like canration eyes and normal shape, an exchange of the abnroal features markng the ends of homolohs chromsoems accomapeind the recomb genes
evdience thus tied genetic recomb to corssing over of specifckaly marked parts of chromsomes (egentic recomb involves reciprocal exchange of segments between homologs during emeiss)
why recomb?
recomb provides genetic divsierty bc reshufling of alleels
but also it ensures that chrmsomes segrgate proerly when tramsitted betwen parent and progeny
other nondisjunction during meisis 1 would be common - and species could never retain the same number of chromsomes in succesive generations
issue is that proper chromsome segration reuires pair homologs to be conected to spindle fibers and so they they are pulled to opp poles
key element for this bipoalr orienattion is the phsycial tension that meisis 1 spindle exters on rpeprly connected chrosmoems
tension is created only when homologs are pulled in opp direction but remain joined by a phaysical link
in absence of tension centreoemre send signal that prevent anaphase
more than synaptoenmal coplex or recomb noules are needed to link homologs
but homologs are stileld conencted after synapotenma complex and recomb nodules disolve after prophase 1
the physal linkage still remains til onset of anaphase
its bc the actual crososver events marked by chistamata where nonsister chromatids of homolgs exchange places
second are moelcular complexes called cohesin that make connection between sister chromatids soon after chromsomes are replicated
after crossover, cohesin complexes distal to the crossover point that keep the homolgs toegterh at metapahse plate and ensure proepr chrosmome segregation
cohesion plays many roles
importance of crossing over to proepr chrosme segrgation is that bivalent has at least one recomb nodule or chiasma
a mechanism called inetrfence ensures thate ahc chormose pair udnergoes at least oen crossoevr prevent nonjdisnction of any chrosme except when rare mistake happenes
recomb freq reflects distance bewteen two genes
Morgan intuited that chiasmata rep sites of phsycal crossing over which can result in recomb
led to: diff gene pairs exhibit diff linakge frequencies bc genes are arrnaged in a line along a chrosmoem
close toegtehr on chrosome, smaller their chance to be sperated by recomb
proposed that the percentage of total progeny that were recomb types, the recombination frequency (RF_ could be used to gauge phsycal distance seprating two genes on a chromsome
one RF percent point as unit of measure along chrosmoe
theres also unit centimorgan or map unit
we used centimorgans (all inetrchanable tho)
ex 1.1 m.u means super clsoely linked traits - recomb is unlikely - recomb percepnt is 1.1
if genes A and B clsoe on chromsme, emisis takes place with no crossover betwene genes - no cross over meisis
in dihyrbid, the outcome of NCO meisis are four parental gametes
if single crossoevr meisis then two recomb and two parnetal gametes
these crosover are rare bc most of the lenth of chromsome lies outside region between A and B
as distance between genes inc, the freq of single corss oevr meisisi inc and so does chance of recomb gametes
thtas why two type of parnetal gametes are produced at equal rates and why recomb appear in aprox same numebes to each otehr aslo
whenevr two nonsister chromatids dont cross over between genes A and B, one eof each type of parental chrosoome is generated
but if non sister chomaatids do crossover then both recipocal recombs are produced
the map unit if idnex of recomb proabilities
recomb freq between two genes never exceeds 50%
if defiination of linkage is that the proporton of recomb classes is smaller than that of parental classes, a recomb freq of less than 50% indicates linkage
there can be recomb freq of 50%
genes on diff genes - non homolog chrosmomes obey mendel law for two reaosn
first, the two chrosomes can line up on spindle during emeisi in eitehr two equally likley configs
as a result whne summd over teh rpoducts of many meisis, the number of parental and recomb gametes are equal
second, if a crososevr happens betwnee any one gene and the centromere, that emisis will produce two parental and two recomb gametes
thus if genes A and B are on nonhomologs a dihurbid can produce all four types of gametes - AB, Ab, aB and ab
expeimrnets have established that genes located far apart on same chrosmome also show recomb freq of approx 50%
consider the dif fmeisis that contirbute to poll of gamtes counted in testcross experiment
when two genes are close, only two kinds of meisis can happen - no crosseoevr betwen the two genes or rare meisis with single corssover betwen the gene
a no crossover yeidls only parental awhile a single crososevr prdcues 50% recomb gametes - tehse clsoely spaced genes are linekd bc some meisis (no crossoevr) dont make any recomb gametes
when the two genes are fartehr apart, single crosover is freq and in some two crosoevrs between A and B (doible cross over meisis) can happen
doubel emisis can be one of four diff types: two, three or all four non sister chromatdis may crosses over
bc teh four double crosoevr events are equaly freq, the avg fraction of recomb is 50%
the same is true for triple corssover etc
even if the two genes are far apart that at least one crososver happens, the pool of gamtes by dihyrbid would only be 50% recomb type
even though corsses between two genes laying far aprt on chromsome can show no linkage, you demosnatte they are on same chrosmme if u can tie each of the widley sperated genes to one or more comon intermediares
linked genes: parental more than recomb (recomb freq less than 50%) - linked genes have to be suffenctly clsoe so alleles dont asosrt idnepdnetly
unlined genes: parental = recomb (recomb freq = 50%) - happens when two genes are on diff chrosome or whne they are far aprt on same chrsoome that at least one crosoevr happens betwen them in every emeisis
5.3 Mapping: Locating Genes along a Chrosmme
maps are images of the relative psitions of objects in space
maps that assign genes to specific locations on aprtcialr chrosomes are called loci
by transofming gentic data into spatiala rrnagemnets, maps sharpen the ability to predict inehrtance patterns of specific traits
knowing a genes location gives sicnetist ability to idenify segment of chromsal dna corespendonung to the gene
the recombination frequency RF is a measure of the distance seperating two genes along a chroomsome
we now examine how data from many corsses following two and three genes at a time can be compled and comapred to generate acurate and comprehsive gene/chrosome maps
compariosn of two point croses establish relative gene positions
Sturtevant asked whetehr data obtained form a larg enumber of two point crosses (crosses tracing two genes at a time) would support the idea that genes form a deifnit linear series along a chrosomoe
he began by looking at x linked genes in drosphila
rmbr that the distance bewteen two genes that yeilds 1% recomb progeny - an RF of 1% is 1 m.u.
consider three genes w,y and m
if these genes are arranged in a line then one of them must be in the mdile, flanked on eitehr side by the otehr two
the greatest gentic distance should seprate the two genes on the outside and this value should roughly equal the sum of the distances seprating the middle gene from each outside gene
the data sturtvent obtaiend are consinet with this idea implying the w lies between y and m
left to right orientation of map was selcted at random - could be y on right and m on left
by following the same procedure for each set of three genes cosindered in pairs, sturtvenat established a self consisetn orde rfor all the genes he investiagtes on the drosophila x chromsome
the fact that the recombo datayeild a simple linear map of gene position supports idea that genes reside in unique linear order on chromsoes
limiations of two point crosses
the pairwise mapping of genes has shortcoming that limit usefulenss
first in crosses inviling only two genes at a time, it might be dififcult to detemrine gene order if some pairs lie close together
ex. in maping y, w and m - 34.3 mu seps the outside genes y and m while nearly as great a distance (33.8 mu) seprates middle w from outside m
before being able to consued that y and m are furtehr apart than w and m - that small difference isnt a reuslt of a sampling erorro - u have to do many crosses and subejct dat ato statsical test like chi squaer test
second prob is tha the actual distances in his mapping dont always add up even aproximately
ex. the locus of the y gene on far left is position 0 - the w gene then would lie near psotion 1 and m would be near 34 mu
but what abt the r gene named for mutation - based solely on its distance from y, we would palce it at 42.9
but if we calc the psotion as sum of all inetrvening distances infered from data (that is as the sum of y-w plus w-v plus v-m and m-r) the local of r is 55
what explains teh difference?
three point crosses provide more accurate mapping
the simultanous analyssi of three markers reveals the sources of discrepancies between two maps and makes it clear whcih is acurate
lets look at three linked genes on dropsphila autsome
homozygous female with vg and b and pr was mated to wild type of each trait
all the triply hetroyzgous f1 progeny had normal phenotypes, indicating tha the mutation are automsal receisve
in a tesctrsos of the f1 females with male homozygotes for receive three genes, the progeny were of eight diff phenotypes, reflecting 8 diff genotpes
we are deducing gene order/mapping
ina anlzying the datt we look at two genes at a time bc recomb freq is always a function of a pair. of genes
for the pair vg and b the parnetal combo are vg b and vg+ b+ - the recomb are vgb+ and vg+b
to detrmine whetehr a partcular class of porgeny is parnetal or recomb for vg and b, we dont carry abt the pr gene - so any progeny with vg+b or vgb+ are all added and divided by total progeny to get the distance between the two genes
these recomb frquency show that vg and b are seprated by large distance and have to be outside genes, flanking pr in mdidle
but as with x-linked y and r genes, the distance speraein outside vg and b genes (17.7) doenst equal sum of the two inetrvening distances (18.7) - reason for this discpency is double crossoevrs
correction of double crossovers
homologs of autsome of f1 females are hertozygous for three genes vg, pr and b
a close examination of the chromsoem reveal the kinds of crososvers that must have occured to genrate the classs and numebrs observed
region 1 is space bewteen vg and pr and space between pr and b is region 1
recall that teh progeny form the trhee pont testcross perfomred ealier fall into eight groups
the two largest ones carry config of genes as the p gen - parnetal classes
the next two groups - vg+bpr and vgb+pr+ are recomb that must be the reciprocal products of corssoevr inr egion 1 between vg and pr
while vg+ bpr+ and vgb+pr must have reuslted from reocmb in region 2 betwene pr and b
what abt two smallest groups: vgbpr+ and vg+b+pr - most likley they result fromdouble cross meisis - crosover in region 1 and 2
these double recomb gamtes still have the parnetal config for outside genes vg and b even thought not one but two exchnages mustve occured
bc the esietsnce of double crossoevrs, the vg-b distance of 17.7 calced doesnt reflect all the recomb ebents prpducing the gametes that give rise to the observed progeny - to correct, need to adjust teh recomb frquency by adding the double crososvers twice bc each indivdual in the double crososver groups is the result of two exchnages between vg and b - so new distance ends up being 18.7 mu
this value makes sense bc u have accounted for all corsovers that happen in regin 1 and 2
as a reuslt the corrected value is now exactly the sum of distances bewteen vg and pr (region 1) and between pr and b (region 2)
whne sturtventa organilly mappes the x-linked genes in dropshila by two point crosses, the locus of the udimentary wings (r) gen was aminghous
a two point cross invlving y and r gave a recomb freq of 42.9 but sum of all intervening distances was 55
the discrepcny couldve been bc of two point cross - all chromatids that cross over twice in large inetrval were undetctable
the data. summing the smaller intervening disatnces accounted for at least some of these double crossesover by ctaching recomb involing other genes located between y and r
moreoevr, each smaller diatcne is less likely to encompass a double crososver than a alrger distance so each number for a smalelr distacne is inherntly more accurate
even a three point cross ignores psinility of two recomb ebent happening in place like region 1 - its alway sbest to constrcyt map using many genes seprates by short distances
interference - fewer double crossovers than expected
in a three point cross following three linked genes, of the ieght pissible genotypic classes, two parnetal classes conatin most progeny while the two double recomb classes from double crosso oevr are always smallest
double crososver are rare bc of probability fo occurence
since exchange in one region doesnt affect ecnage in other, probability they occur simulatneosuly is product of seprate proabilities - so if its 10% and then 20% - gives 2% chance
if eight classes of prgeny are obtained in a three point cross, the two clases conating fewest rpoegny are from double crososver
the umerical freq of obserevd double progeny is almost never conincing with expcttaion from product rule
proabulity fo single crosover bewteen vg and pr is 0.123 (12.3 m.u.) and the proability fo single coss between pr and b is 0.064(6.4 mu) - product is 79% but the osrved proportion was 13+9/4194 so 0.53%
observed doublecrossover less than expected so occurance of one crossover redcues likelihood of another - crosovers arent indepdnet so its chromosomal inetrfence
inetrfence likely exists to ensure evry pair of homologs udnergoes at elast one crossover
bc proteins involved in recombo ar epresnet in samll amount, a limited amount of crosovers can happene - inetrfence helps distribute crosovers which reduce number of crossovers on large chrosmome
interfence was also seen whne bivalents needed at least one recomb nodule
most human sperm chromsomes only undegro a single crosover
interfence is mediated likley by synaptonemal complex
interfence isnt uniform and may even vayr for diff regions of the same chrosmome
inevstiatros can obtain a qunativtve measure of the amount of inetrfence in diffi chrosmomal inetrvals by frist calculating a coef of coincidence - a ratio betwene actual freq of double crossovers obersvred and number expected on basis of proabilities - then do 1- coef of conincde to get interfecne
if interfence is 0 the observed frequence of obsrved double crosver prog equal expectation (coef is 1) and crosovers happen indently
if inetrfecne is complete (interfence is 1), no double crosover occur (coef is 0) bc one exchnage prevents another in region examined
determing if gene in the mdidle
smallest of the clases in prgeny are the two double recomb types
its psisble to use the compositin of alleles in tehse double crsosver classes to detmrine which of the trhee genes lies in the mdidle even without calculating any recomb freq
testcross the gave double cross vgpr+b and vg+prb+
the alleles of vg and b retain parental associateions (vg b and vg+b+)
while pr gene has recombines to both other genes
the same is true for all three point crosses
treu fro all trhee point - gene whose alle have recombined to parnetal config of otehr two has to be the middle
how genetic maps correlate with phsycial reality
order of genes revelaed by recomb mapping reflects the order of those same genes along DNA moelcule
in constastthe physical distances between genes, that is the amount of dna sperating them doesnt always correspond liearly to egentic map distances
underestimation of phscial distances between gnes by recomb freq
certain progeny resulting from doubel cross overs may go undeteced as recombs reuslting in undercounting of number of crossovers and tehrefore an undestimatin of the distance between them
this isnt probt when two genes are close togetehr that doubel crosover is infrequent
hwoever whne far apart, double and multple crosoevrs can occur often to affect the relationship between recomb freq and map distance
this relationship cant be linear bc recomb freq for two point cross cant exceed 50% regardless how far aprt egnes are on the same chromsome
when gene A and B are close - most emisis are no crssover and some time single
single cosover produces two recoob gamtes and so a perfect linear corresndce exists between the number of crossovers and number of recomb gamtes (1:2)
but when far aprt, the double cross oer can happen
only one of the four equally frequency double crsos oevr preserves the linear relationship between cross and recoomb gametes - two cross in a 4 strand doible corssover and 4 recomb gamtes (2:4_
in other types of double cross over the result in fewer than four recomb gamtes
theres a discepncy between number of crososvers and observed recomb freq
when phsycal distance betwene gene inc, the two lines will differ - recomb freq becomes less precise
geneticst have dveloped mapping functions that relate RF measured in testcross to actual number of crosoer that took place between two genes
however the corrections for large distances made by mapping fucntins are at best imprecise bc tehse equations are based on simplfying asusmptions
map is best by summing smaller intervals, lcoating wide sperateged genes through their linakge to common inetrmediaries
nonuniofrm crossover freq
recoomb isnt random
in dna most crssovers happen in recomb hotspots - small regions of DNA where freq of recomb is much higher than average
genes with hotspots between them (A and B_ are more distanc from eachotehr (mu) than another gen pair without hot spot between them even though the phsycial distance (base pairs) seprating the geme is the same
hotspots are relatvly frequent - recomb freq is neverthelss a reasonable estimation of phsycal distance between most genes
frequencies of recomb may differ in species
1% RF can be diff in DNA exapnses ind iff organisms
recomb freq sometimes vayr even betwene the two sexes of a single species
in humans the freq of crossoevrs is abt twice higher in female egrm line than males
the same two genes will appear roughly twice as far on gentic map genated by recomb freq in feamle meisis than if corsing over during male meisis was measured
drosphila are etxreme example - no recomb in meisis of males
multiple factor crosses help establish linkage groups
genes chained togetehr by linkage relationship are known as linakge group
whne enough egnes are asisgned to chromsome, the terms chromsome and lnakge group are synonmous
if a linked to b, b to c so and so ofrth u can sya they are syntenic genes
when egnetci map is so dense that u can show any gene on chromsome is linked to anotehr gene on same chromsoe, the number of linkage groups is equal to homoog pairs of chromsome in species
total egentic distance along a chrosome which is obtaiend by adding short distance between genes may be more than 50 m.u
eqns
recomb freq can be added to get the distance betwene two genes (.+.+.+. so on)
or:
from the distance given from data of the two genes (gievn value)
thre point cross gives more accurate mapping
disacne between two genes from three point data can be done by looking at allles recomb freq over the total progeny - so any porgeny with vgb+ or vg+b (pr or pr+ doesnt matter) all over total gives the % which u can multiply by 100 to get mu or distance between vg and b
(applies to recomb gene pairs - cant be of parent genotype)
above id fif than sum of two intervening distances (diff bc double crossovers)
in the addition of porgeny with genotype, you double the indivduals in the double crossover genotypes
probability of double crosover is the product of the proability of each single corss over
observed double crososoevr are aomunt in progeny observed over totla
coef of coindicden = freq observed/freq expected
interference = 1-coef of concidnece
5.4 The Chi Square Test and Linkage Analysis
how do u know from expiemrnet that two genes assort indepdnet or are linked
u would think that for unlinked, a dihbrid female produces four types of gametes in equal - so in test cross, one half progeny ar eparnetla class and other half are recomb
in constarst for linked the two parental calsses would outnumber the two types of recomb classes in progeny of testcrooss
the porb is that there are chance events - so if there is deviation from euqla numbers of parental and recomb expected of indepdnet asostrment, how can u rlly conldue they are linked? could it have been chance?
its also imprnta to consider linked genes that give recomb classes of almost 50% bc not tightly linked
the chisquare test evaluates the signficance of differences between predicted and observed values
the chi square test - for goodness of fit
measures how well observed results conform to predicted ones and it is designed to account for the fact that the size of an expeirmental population (the sample size) is an importnat components of statiscal signficance
two consepts emerge from the toin coss ex:
a comparison of percentage or ratios alone doesnt allow u to dtermine whetehr or not observed data are signficnatly diff than predicated values
second the aboslute numbers obtained are impronat bc they reflect size of expeirmnet
the larger the size, the closer observed precentages can be expected to match the values predicted by expeirmntal hypothesis
so chi square test happens with numebrs from data
chi square test cant prove a hypothesis but can allow researchers to reject a hypothesis
a crucial prereq of chi square test is the frmaing of a null hypothesis: statement that potentially can be nullified
null hypothesis are formulated as statements that no sig differences exist between the observed data and a partcular moel that leads to clear cut numerical predictions
most use chi squae test to disocver whether data from expeirmnt provide evidence for or against hypothesis that genes are linked
but saying that gene A and B are linked allows for no numerical prediction for what to expect in terms of results - bc the frequency of recombs vary withe ach linked pair
but saying A and B arnet linked gives rise to prediction that alleles of diff genes will assort idnepdntly and produce a 1:1:1:1 ratio of porgeny (or 9:3:3:1 in dihyrbid)
so they would tets the null hypotehsis that no sig diff exist between observed data and 1:1:1:1 rati
if test shows that observed data differ sig from those expected for indepdnet assortment - they differ enough to not reasonably attribute it to chance - then researcher can reject null hypothesis of no linkage and accept the alternative of linkage between the two genes
yhe final result of calculation is the numerical probility - p value- that a set of observed expiemrnetal results represents a chance of deviation from the predicted values by hypotehssi
if p value is high, the data are a good fit to the mdoel and the observed deviation from epxectd results is considered insignifcant - if mdoel being tested is no linkage, a high p means that the genes might be unlinked
if the p value is low the observed deviation from expected results is signficnat and model cant be rejected - usually data means it supports linkage
null hypotehsi of no linkage (recomb freq of 50%) rather than a null hypotehsis of a certain degree of linkage bc the test alllows u to reject hypotehsis not prove - knowing that recombfreq is diff than 38% meaning rejcction of null, doesnt tell u much but rejected that two genes are unlikned says its likely they are syntenic and close enough to be linked
how to do chi square test for goodness of fit
- use data to answer: total number of offspring analzyed? how many classes of offpsirng? how many offpsirng in each class?
- calc how many offpsirng would be expected for each class if null is correct - multiply the fraction predicted by null by total number of offspring
- to calc the square begin with one class of offspring - subtract the expected number from the observed to obtain the deviation, then square the result and divide by the expected number - do it for all classes and sum the indivdual results - this is chi square or X^2
- consider the degres of freedom, a mesure of the number of indepdnetly varying parameters in the expeirment - its on less than number of classes so df = N-1
- use chi square nad df to detrmine p value: the porbability that a deviation from the predicted numbers at least as large as that observed in expeirment occured by chance - use the table values
- p value is the proability that null is true - value greater than 0.05 idnictaes that more than 1 in 20 repitions of expeiment of same size, the osberved deviations could be obtained by chance, even if null is true; data are therefore not signifcnat to reject null - but a p alue lower than 0.05 means data isnt good fit to the model, the deviation is signficnat and you can rject the null
