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Frekvencije rekombinacije

Frekvencije rekombinacije



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Učio sam o frekvencijama rekombinacije, ali i dalje postajem pomalo zbunjen unatoč tome što sam pregledao mnoge veze u Googleu u vezi s njima. Zanimalo me može li netko provjeriti je li sljedeće točno.

  • Da biste izračunali frekvencije rekombinacije i na taj način stvorili genetsku kartu, morate imati prave roditelje za razmnožavanje za dva ili više različitih gena.
  • Morate prekrižiti te roditelje, tako da imate generaciju F1 koja se sastoji samo od heterozigota
  • Zatim morate ukrstiti osobe F1 s bilo kojim od roditelja. Ako su svi ti geni smješteni na različitim kromosomima, samo ćete dobiti neovisnu segregaciju kromosoma i stoga 50/50 šanse da se svaki alel pojavi s drugim u potomstvu. Dok ako se pojave na istom kromosomu, tada će se u F1 heterozigotima aleli svakog roditelja različito odvojiti u gamete samo ako dođe do rekombinacije.
  • Za izračun frekvencija rekombinacije uzimamo stoga broj potomaka za koje se kombinacija ova dva alela razlikuje od one uočene kod roditelja- nazivamo ih rekombinantima.

Je li to točno?


Ovaj odgovor je uglavnom kopiran iz mojih odgovora ovdje i ovdje.

Važnost polimorfizma

Da biste razumjeli koju generaciju želite prijeći unatrag, važno je razumjeti zašto su nam potrebni polimorfni lokusi.

Ako je barem jedan lokus homozigotan

Ako se rekombinacijski događaj dogodi između dva lokusa gdje je barem jedno homozigotno, tada nećete vidjeti ništa. Razmotrimo, na primjer, sljedeće nizove nizova kod diploidnog pojedinca

----- A ----- B ---- ----- a ----- B ----

Bilo da se dogodio rekombinacijski događaj ili ne, dva moguća kromosoma prenesena na potomstvo jesu

----- A ----- B ---- ----- a ----- B ----

Stoga ne možete reći je li došlo do rekombinacije.

Ako su oba lokusa heterozigotna

Sada razmislite o sljedećoj osobi

----- A ----- B ---- ----- a ----- b ----

Ako se nije dogodio rekombinacijski događaj između dva interesna lokusa, tada će se prenijeti dva moguća kromosoma

----- A ----- B ---- ----- a ----- b ----

S druge strane, ako se dogodio rekombinacijski događaj između dva interesna lokusa, tada će se prenijeti dva moguća kromosoma

----- A ----- b ---- ----- a ----- B ----

Stoga možete reći je li došlo do rekombinacije ili ne.

Statistika rekombinacije

Ispod je mali dio matematike. Ove su jednadžbe uglavnom za znatiželju jer se može razumjeti odgovor bez razumijevanja matematike koja stoji iza toga.

Definicije $ r $ i $ M $

Zbunjujete se između dvije različite statistike

  • Stopa rekombinacije $ r $ između dva lokusa
    • $ r $ je vjerojatnost da dvije sekvence pronađene na dva lokusa ostanu u istoj gameti nakon što se dogodila rekombinacija. Ta vjerojatnost ne može biti veća od 0,5 ($ 0 le r le 0.5 $).
  • Udaljenost u Morganima $ M $ (ili češće u centiMorganima) između dva lokusa
    • $ M $ je očekivani broj križanja koji se javlja između dva lokusa.

Morgani i centiMorgani

Primijetit ćete da više govorim u Morganima nego u centimorganima, što je neuobičajeno u literaturi, ali pomaže u prenošenju intuicije o tome što to znači. ako je $ M = 150 $ centiMorgans $ = 1,5 $ Morgans, tada je očekivani broj križanja između dva lokusa 1,5. Ispod je još nekoliko objašnjenja ove dvije definicije s nekim crtežom :)

Studija slučaja s lokusimaAiB

Iako su $ M $ i $ r $ blisko povezani, oni nisu potpuno ista stvar. Razmotrite sljedeći slijed s lokusimaAiB

--- [A] ------------------ [B] ---

Pretpostavimo da su dva lokusa jako udaljena i $ M = 2 $. Vjerojatnost da će doći do točno $ k $ crossovera stoga je data Poissonovom distribucijom sa stopom $ M = 2 $

$$ P (k) = frac {e^{-M} M^k} {k!} $$

Recimo za dati slučaj da je $ k = 1 $ (došlo je do jednog ukrštanja). Ovaj prijelaz označen je s "/" ispod

--- [A] -------------/---- [B] ---

Ovdje jasno dva niza na lokusimaAiBbit će razdvojeni. Recimo sada da je $ k = 2 $ (došlo je do dva križanja).

--- [A] ---/-----/-------- [B] ---

Ovdje, čak i ako je došlo do križanja, dvije sekvence na mjestimaAiBostat će zajedno. Samo niz između dva križanja doći će iz homolognog kromosoma.

Odnos između $ r $ i $ M $ - riječima

Možda ćete vidjeti da dolazi iz prethodnog odjeljka. Vjerojatnost $ r $ ova dva lokusaAiBda se odvoji rekombinacijom vjerojatnost je da se između njih dogodi neparan broj rekombinacijskih događaja (znajući da je $ M $ očekivani broj ukrštanja).

Odnos između $ r $ i $ M $ - u jednadžbi

Izračunajmo najprije vjerojatnost $ p_ {even} $ da će doći do parnog broja ukrštanja. Ova vjerojatnost je pravedna

$$ p_ {even} = sum_ {k = 0}^{ infty} {e^{-M} M^{2k} over (2k)!} $$

, gdje sam upravo dodao konstantu $ 2 $ prije $ k $ i u brojniku i u nazivniku. Pomoću neke algebre i triga to se može pokazati

$$ p_ {par} = sum_ {k = 0}^{ infty} {e^{-M} M^{2k} over (2k)!} = e^{-M} sum_ {k = 0}^{ infty} {M^{2k} over (2k)!} = $$

$$ e^{-M} : cosh (m) = e^{-M} lijevo ( frac {e^{M} + e^{-M}} {2} desno) = {1 + e^{-2M} preko 2} $$

As, $ r = p_ {odd} = 1 - p_ {even} $,

$$ r = 1 - {1 + e^{ - 2M} preko 2} = {1 - e^{ - 2M} preko 2} $$

Idemo! Imamo naš odnos između $ r $ i $ M $! Nacrtajmo to grafički

Odnos između $ r $ i $ M $ - na grafikonu

Upravo sam prikazao gornju jednadžbu u Mathematici (Zemljište [y = (1 - Istek [-2 M])/2, {M, 0, 5}])

Ovdje je isti grafikon, ali zumiran na nižim vrijednostima $ r $ i $ M $ (Zemljište [y = (1 - Exp [-2 M])/2, {M, 0, 0.1}])

Iz grafikona jasno vidimo da za niske vrijednosti od $ M $, kako se $ M $ povećava $ r $, kvazi linearno se povećava ($ M≈r $). Za veće vrijednosti od $ M $, $ r $ se i dalje povećava, ali sve sporije sve dok ne dosegne asimptotu / plato na $ frac {1} {2} $. $ r $ je doista ograničeno između 0 (kada je $ M = 0 $) i $ frac {1} {2} $ (kada je $ M = infty $).

Imajte na umu da je činjenica da je zbroj vjerojatnosti svakog parnog $ k $ u Poissonovoj distribuciji uvijek manji ili jednak 0,5 zanimljiva matematička činjenica sama po sebi!


Da, tako to funkcionira koliko ja znam

Budući da je teorijski dio prilično jasan. Budući da se geni nalaze na istom kromosomu, manja je vjerojatnost da budu odvojeni u segregaciji tijekom mejoze

Razmotrimo jednostavan primjer

---------A B C D E----------------- ---------------- f

Neka ovo bude uže sa abcdef od 6 točaka Ako ovdje napravim nasumičan rez, vjerojatnost da će se a, b odvojiti mnogo je manja od a, f… jer su a i f toliko udaljeni da svaki presjek u sredina će osigurati njihovo razdvajanje. Ako ga ponovno pogledate ... šansa da se d, e razdvoji izrazito je mala

Ovo je princip LINKING

Dopustite mi sada da kažem da gen a označava visokog, b označava plavo, a f grmoliko ... (nemojte suditi o likovima)

Ako je ovo dio kromosoma na roditelju, a dolazi do mejoze Svi znamo da je i prelazak ovdje fenomen

Sada se tijekom prelaska ... a i f sigurno razdvajaju, pa će biti neovisno razvrstani, poštujući zakon Mendela

No, kada govorimo o a i b, oni se mogu, ali i ne moraju razdvojiti u različitim spolnim stanicama

Što su bliže, manja je šansa za njihovo razdvajanje, a time i veće šanse da se pojave u istoj gameti, pa stoga i veće šanse da budu izražene u istom potomstvu (pod pretpostavkom da su dominantni)

pretpostavimo da prelazimo visokog plavog (dominantno) sa zelenim patuljkom (recesivno) Kao što znate f1 će biti visok plavo (bez nepotpune ili sudominacije, molim)

Ako uzmem ovo plavo visoko i prekrižim ga s roditeljskim zelenim patuljkom

Imat ćete vrlo veliki broj plavih visokih i zelenih patuljaka (zapamtite, postoje aleli ISTOG RODA ... I ZAPOSLITE ISTI LOCI) (dakle ... velika vjerojatnost da se oboje pojave zajedno u istoj gameti)

Stoga će broj rekombinanata koje dobijete biti vrlo nizak (plavi patuljak i visok zeleni ... jer je šansa da ti likovi uđu u različite spolne stanice vrlo mala)

Stoga možemo zaključiti da je veći broj rekombinacija ... veća vjerojatnost odvajanja gena što podrazumijeva veću udaljenost između gena)

Stoga veća frekvencija rekombinacije implicira veću udaljenost između 2 povezana gena

Nadam se da ste razumjeli koncept i da se ne trudite da ga shvatite kao silu Hvala


7.6: Genetsko mapiranje

  • Doprinijeli Todd Nickle i Isabelle Barrette-Ng
  • Profesori (biologija) na Sveučilištu Mount Royal i Sveučilištu amp u Calgaryju

Budući da je učestalost rekombinacije između dva lokusa (do 50%) približno proporcionalna kromosomskoj udaljenosti između njih, možemo koristiti frekvencije rekombinacije za izradu genetskih karata svih lokusa duž kromosoma i na kraju u cijelom genomu. Jedinice genetske udaljenosti zovu se jedinice karte (mu) ili centiMorgani (cM), u čast Thomas Hunt Morgan od svog učenika, Alfred Sturtevant, koji je razvio koncept. Genetičari rutinski pretvaraju frekvencije rekombinacije u cM: frekvencija rekombinacije u postocima približno je jednaka udaljenosti karte u cM. Na primjer, ako dva lokusa imaju frekvenciju rekombinacije od 25%, za njih se kaže da su

25 cm udaljenosti na kromosomu (slika ( PageIndex <9> )). Napomena: ova aproksimacija dobro funkcionira na malim udaljenostima (RF & lt30%), ali postupno ne uspijeva na većim udaljenostima jer RF doseže maksimum na 50%. Neki kromosomi su dugi> 100 cM, ali lokusi na vrhovima imaju samo RF od 50%. Metoda za mapiranje ovih dugih kromosoma prikazana je u nastavku.

Imajte na umu da udaljenost karte samo dva lokusa ne govori nam ništa o orijentaciji tih lokusa u odnosu na druge značajke, poput centromera ili telomera, na kromosomu.

Slika ( PageIndex <9> ): Dvije genetske karte u skladu s frekvencijom rekombinacije od 25% između A i B. Zabilježite mjesto centromere. (Izvornik-Deyholos-CC: AN)

Udaljenost karte uvijek se računa za jedan par lokusa odjednom. Međutim, kombiniranjem rezultata višestrukih proračuna u paru, a genetska karta mogu se stvoriti mnogi lokusi na kromosomu (slika ( PageIndex <10> )). Genetska karta prikazuje udaljenost karte, u cM, koja razdvaja bilo koja dva lokusa, i položaj tih lokusa u odnosu na sve ostale mapirane lokuse. Udaljenost genetske karte približno je proporcionalna fizičkoj udaljenosti, tj. Količini DNA između dva lokusa. Na primjer, u Arabidopsis, 1,0 cM odgovara približno 150 000 bp i sadrži približno 50 gena. Točan broj baza DNA u cM ovisi o organizmu, a o posebnom položaju u kromosomu neki dijelovi kromosoma (& ldquocrossover vruće točke & rdquo) imaju veće stope rekombinacije od drugih, dok su druge regije smanjile križanje i često odgovaraju velike regije heterokromatina.

Slika ( PageIndex <10> ): Genetske karte za područja dva kromosoma iz dvije vrste moljca, Bombyx. Ljestvica na lijevoj strani prikazuje udaljenost u cM, a položaj različitih lokusa označen je na svakom kromosomu. Dijagonalne linije koje povezuju lokuse na različitim kromosomima pokazuju položaj odgovarajućih lokusa u različitim vrstama. To se naziva regijama očuvana sintenija. (NCBI-NIH-PD)

Kada se novi gen ili lokus identificira mutacijom ili polimorfizmom, njegov približni položaj na kromosomu može se odrediti ukrštanjem s prethodno mapiranim genima, a zatim izračunavanjem frekvencije rekombinacije. Ako novi gen i prethodno mapirani geni pokažu potpunu ili djelomičnu vezu, učestalost rekombinacije pokazat će približan položaj novog gena unutar genetske karte. Ove su informacije korisne za izolaciju (tj. Kloniranje) specifičnog fragmenta DNA koji kodira novi gen, putem procesa koji se naziva kloniranje temeljeno na karti.

Genetske karte također su korisne za praćenje gena/alela u uzgoju usjeva i životinja, za proučavanje evolucijskih odnosa među vrstama, te za utvrđivanje uzroka i individualne osjetljivosti nekih ljudskih bolesti.

Slika ( PageIndex <11> ): Dvostruko križanje između dva lokusa proizvest će gamete s roditeljskim genotipovima. (Izvornik-Deyholos-CC: AN)

Genetske karte korisne su za prikaz redoslijeda lokusa duž kromosoma, ali udaljenosti su samo približna. Korelacija između frekvencije rekombinacije i stvarne kromosomske udaljenosti točnija je za kratke udaljenosti (niske RF vrijednosti) nego za velike udaljenosti. Uočene frekvencije rekombinacije između dva relativno udaljena markera imaju tendenciju podcjenjivanja stvarnog broja križanja koja su se dogodila. To je zato što se s povećanjem udaljenosti između lokusa povećava mogućnost ponovnog (ili više) križanja između lokusa. Ovo je problem za genetičare, jer s obzirom na lokuse koji se proučavaju, ovi dvostruke skretnice proizvode gamete s istim genotipovima kao da se nije dogodio rekombinacijski događaj (slika ( PageIndex <11> )) & ndash imaju roditeljske genotipove. Stoga će se dvostruki crossover činiti roditeljskim tipom i neće se računati kao rekombinantni, unatoč tome što ima dva (ili više) crossovera. Genetičari će ponekad koristiti posebne matematičke formule za podešavanje velikih frekvencija rekombinacije kako bi se uzela u obzir mogućnost višestrukih križanja i tako dobili bolju procjenu stvarne udaljenosti između dva lokusa.


Rekombinacija gena: učestalost i spolni kromosomi (sa dijagramom)

Tijekom prelaska preko homolognih kromosoma međusobno izmjenjuju dijelove kromatida. Rezultat je rekombinacija povezanih gena.

Ovaj proces je poznat kao rekombinacija. Uzgojni eksperimenti provedeni na kukuruzu pokazali su da čak ni povezani geni ne ostaju uvijek zajedno tijekom nasljeđivanja.

Ponekad se odvajaju razmjenom alela zbog razmjene dijelova između kromatida homolognih kromosoma u vrijeme mejoze. To rezultira stvaranjem nove ili ne-roditeljske kombinacije starih gena.

Ovaj fenomen nove kombinacije starih općenitih poznatih kao rekombinacija gena i potomaka s novom kombinacijom znakova poznati su kao rekombinantni tipovi. Rezultati oplemenjivačkog pokusa kukuruza pokazali su da je nova kombinacija ili roditeljska kombinacija 3,6%. To je postotak rekombinacije gena.

Učestalost (ili postotak) rekombinacije:

Učestalost rekombinacije ili ukrštanja bilo koja dva gena je broj križanja nastalih između njih. Izravno je proporcionalna udaljenosti između dva gena. Učestalost prelaska se koristi kao indeks relativnih udaljenosti između gena na kromosomu.

Učestalost prelaska ili postotak rekombinacije može se odrediti:

(i) Brojenjem broja hiasmata nastalih tijekom diplotena u profazi 1 mejoze pod mikroskopom.

(ii) Provođenjem kontroliranih pokusa i izračunavanjem učestalosti roditeljskih i rekombinantnih vrsta potomaka.

Učestalost rekombinacije:

Br. Rekombinanata u potomstvu test križa /ukupno br. potomstva test križa × 100

Primjer. Potpuno heterozigotna siva ženka, normalna krilata ženka F] Drosophila križana je s crnim tijelom i ružičastim krilatim mužjakom.

Dao je sljedeće rezultate:

Spolni kromosomi:

U jezgri viših životinja prisutne su dvije vrste kromosoma. Oni su autosomi i spolni kromosomi. Autosomi nose gene za druge likove osim spola, dok spolni kromosomi ili alosomi određuju spol životinje. Dok su spolni kromosomi općenito formirani od para kromosoma, preostali kromosomi su označeni kao autosomi. U čovjeka, od ukupno 23 para kromosoma, postoje 22 para autosoma i jedan par spolnih kromosoma [Sl. 5.12 (A)].

Njemački biolog Henking (1891.) prvi je put otkrio X kromosom (X za nepoznato) kod određenih insekata. Mc Clung (1902), američki zoolog, pokazao je da se X kromosom bavi određivanjem spola u košu za travu. Kasnije je Stevens (1905.) primijetio da diploidna stanica ženke ima par X kromosoma.

Daljnje studije pokazale su da se kod mužjaka nekih vrsta X kromosom tijekom mejoze upari s drugim kromosomom (različite veličine i oblika). Ovaj kromosom je nazvan Y kromosom. Budući da su kromosomi X i Y povezani s određivanjem spola pojedinca, nazvani su spolni kromosomi. Kasnije su otkrivene i druge vrste spolnih kromosoma.

Vrste spolnih kromosoma:

Spolni kromosomi različito se nazivaju X i Y kromosomi (čovjek i Drosophila), Z i W kromosomi (ptice i moljci), neparni kromosomi, idiosomi, heterosomi ili alosomi. Geni koji određuju spol nalaze se na tim kromosomima. Dva člana ovog para često su različiti kod muškaraca i predstavljeni su kao X i Y kromosomi ili kao Z i W kromosomi.

(a) XX ženski i XY muški tipovi:

Ova vrsta spolnog mehanizma nalazi se u Drosophili (voćna muha) i većini sisavaca. U ovom tipu ženka je homogametična (XX), a muški heterogametičan (XY) koji se sastoji od dva različita kromosoma X i Y (pojedinosti se odnose na određivanje spola).

(b) ZW ženski i ZZ muški tip:

Kod leptira i ptica ženka je heterogametna s različitim Z i W kromosomima, dok je mužjak homogametičan sa sličnim ZZ kromosomima (konvencija je da se ženka označi kao ZW umjesto XY, a mužjak kao ZZ umjesto XX). Ovdje je situacija samo obrnuta od prve vrste.

(c) XX ženski i XO muški tip:

Ova vrsta spolnog mehanizma uočena je kod kukaca, osobito kod skakavaca. Kod muškaraca nema para za X kromosom, pa je stoga dano ime XO, nema Y kromosoma. U ženki postoje dva slična ili homomorfna spolna kromosoma XX.

Od dva spolna kromosoma kod ženki sisavaca, jedan se prikazuje u obliku tamno obojenog tijela u međufaznoj jezgri. Zove se kao tijelo barra u muškom tijelu barra se ne promatra. Tako Barr tijelo pomaže u određivanju spola embrija. Dok se većina gena povezanih sa spolom nalazi na X kromosomima, nekoliko gena je prisutno na Y-kromosomima.

Neki od ovih gena alelni su za neke od gena X kromosoma i nazivaju se nepotpuno spolno povezani geni. Također postoji nekoliko gena u nehomolognom dijelu Y kromosoma muškaraca (poznatih kao Y-povezani geni) koji pokazuju vrlo neobičan način nasljeđivanja.


Kako izraditi kartu kromosoma iz uzastopnih frekvencija

Rekombinacija: Tijekom križanja (profaza I Mejoze), geni na kromosomima mijenjaju mjesta. Križanje je nasumično, ali vjerojatnost da će se ukrštanje dva gena povećati ako su ti geni udaljeniji. Bliže međusobno povezani geni vjerojatnije će se "spojiti zajedno" i neće mijenjati mjesta.

Karte povezivanja gena: Koristeći ukrštene frekvencije, možete konstruirati kartu za predstavljanje udaljenosti između gena.

Ova karta prikazuje kromosom #2 od Drosophila melanogaster. Udaljenost između gena može se zapisati kao postotak ili kao JEDINICA MAPE. Gen za boju tijela i veličinu krila udaljeni su 17 jedinica karte.

S obzirom na učestalost ukrštanja svakog od gena na karti, sastavite kartu kromosoma.

Gen Učestalost crossovera
A-C 30%
PRIJE KRISTA 45%
B-D 40%
OGLAS 25%

Korak 1: Počnite s genima koji su najdalje udaljeni: B i C udaljeni su 45 jedinica karte i bili bi udaljeni.

Korak 2: Riješite ga poput zagonetke, olovkom odredite položaje drugih gena.

Korak 3: Oduzimanje će biti potrebno za određivanje konačnih udaljenosti između svakog gena.

1. U Drosophile, oči u obliku šipke (B), ošiljena krila (S), križna krila (W) i boja očiju (C) nalaze se na X kromosomu. Učestalost rekombinacije svakog gena navedena je na tablici. Napravite kartu kromosoma.

Gen Učestalost crossovera
Ž-B 2.5%
ZAHOD 3.0%
PRIJE KRISTA 5.5%
B-S 5.5%
Z-J 8.0%
C-S 11.0%

2. Sljedeći grafikon prikazuje frekvencije ukrštanja gena na autosomu Armor Plated Squirtlesaura. Napravite kartu kromosoma.

Gen Učestalost crossovera
P-Q 5%
P-R 8%
P.S 12%
Q-R 13%
Q-S 17%

3. Napravite kartu s obzirom na sljedeće podatke.

Gen Učestalost crossovera
A-B 24%
A-C 8%
CD 2%
A-F 16%
F-B 8%
D-Ž 6%

/> Ovo je djelo licencirano pod Creative Commons Attribution-NonCommercial-ShareAlike 4.0 međunarodnom licencom.


24.2: Rekombinacija

  • Doprinijeli Todd Nickle i Isabelle Barrette-Ng
  • Profesori (biologija) na Sveučilištu Mount Royal i Sveučilištu amp u Calgaryju

Izraz & ldquorecombination & rdquo koristi se u nekoliko različitih konteksta u genetici. Vezano za nasljedstvo, rekombinacija definira se kao svaki proces koji rezultira gametama s kombinacijama alela koje nisu bile prisutne u gametama prethodne generacije (vidi sliku ( PageIndex <2> )). Uterkromosomska rekombinacija događa bilo kroz nezavisni asortiman alela čiji su lokusi na različitim kromosomima (6. poglavlje). Utrakromosomska rekombinacija javlja se kroz skretnice između lokusa na istim kromosomima (kako je dolje opisano). Važno je zapamtiti da je u oba ova slučaja rekombinacija proces koji se događa tijekom mejoze (mitotička rekombinacija može se pojaviti i kod nekih vrsta, ali je to relativno rijetko). Ako mejoza rezultira rekombinacijom, za produkte se kaže da imaju rekombinantni genotip. S druge strane, ako se tijekom mejoze ne dogodi rekombinacija, proizvodi imaju svoje izvorne kombinacije i za njih se kaže da su rekombinantni, ili roditeljski genotip. Rekombinacija je važna jer doprinosi genetskoj varijaciji koja se može primijetiti među pojedincima unutar populacije i na koju se može djelovati selekcijom kako bi se proizvela evolucija.

Slika ( PageIndex <2> ): Kad su dva lokusa na nehomolognim kromosomima, njihovi će se aleli odvojiti u kombinacijama identičnim onima prisutnim u roditeljskim gametama (Ab, aB) i u rekombinantnim genotipovima (AB, ab) koji se razlikuju od roditeljskih spolnih stanica. (Izvornik-Deyholos-CC: AN)

Kao primjer međukromosomske rekombinacije, razmotrite lokuse na dva različita kromosoma kako je prikazano na slici ( PageIndex <2> ). Znamo da ako su ti lokusi na različitim kromosomima, nema fizičkih veza među njima, pa jesu nepovezan i odvojit će se neovisno kao i osobine Mendela & rsquosa. Segregacija ovisi o relativnoj orijentaciji svakog para kromosoma u metafazi. Budući da je orijentacija slučajna i neovisna o drugim kromosomima, svaki raspored (i njihovi mejotički produkti) jednako je moguć za dva nepovezana lokusa kako je prikazano na slici ( PageIndex <2> ). Točnije, postoji 50% vjerojatnost za rekombinantne genotipove i 50% vjerojatnost za roditeljske genotipove unutar gameta koje proizvodi mejocit s nepovezanim lokusima. Doista, kad bismo ispitali sve gamete koje bi ova jedinka mogla proizvesti (koje su produkti više nezavisnih mejoza), primijetile bismo da bi otprilike 50% gameta bilo rekombinantno, a 50% roditeljsko. Učestalost rekombinacije (RF) je jednostavno broj rekombinantnih gameta, podijeljen s ukupnim brojem gameta. Učestalost od približno 50% rekombinacije stoga je definirajuća karakteristika nepovezanih lokusa. Stoga je najveća očekivana rekombinantna frekvencija


Rekombinacija: definicija, mehanizam i vrste | Mikrobiologija

U ovom članku ćemo raspravljati o:- 1. Definiciji rekombinacije 2. Mehanizmu rekombinacije 3. Vrste.

Definicija rekombinacije:

Najvažnije značajke organizama su prilagodba u okolišu i održavanje njihove DNK sekvence u generaciji stanica generacijama uz vrlo male izmjene. Dugoročno preživljavanje organizama ovisi o genetskim varijacijama, ključnoj značajci putem koje se organizam može prilagoditi okolišu koje se mijenja s vremenom.

Ova varijabilnost među organizmima javlja se kroz sposobnost DNK da podliježe genetskim preuređenjima što rezultira malom promjenom u kombinaciji gena. Preuređenje DNA događa se genetskom rekombinacijom.

Dakle, rekombinacija je proces stvaranja novog rekombinantnog kromosoma kombiniranjem genetskog materijala iz dva organizma. Novi rekombinanti pokazuju promjene u fenotipskim obilježjima.

Većina eukariota pokazuje potpuni spolni životni ciklus uključujući mejozu, važan događaj koji rekombinacijom stvara nove alelne kombinacije. To je omogućeno kromosomskom izmjenom koja je rezultat križanja između dva homologna kromosoma koji sadrže identične genske sekvence.

Do 1945. godine mnogo se radilo na eukariotskoj genetici koja je postavila temelje klasične genetike. Rad na genetici bakterija obavljen je između 1945. i 1965. godine koji je unaprijedio razumijevanje mikrobne genetike na molekularnoj razini.

Mehanizam rekombinacije:

U osnovi, postoje tri teorije, a to su: lom i ponovno okupljanje, lom i kopiranje te potpuni izbor kopija koji objašnjavaju mehanizam rekombinacije (slika 8.23).

(i) Prekid i ponovno okupljanje:

Dva homologna dupleksa polaganja kromosoma u uparenom obliku lome se između lokusa gena a i b, i a + i b + (slika 8.23A). Slomljeni segmenti spajaju se poprečno i daju rekombinate koji sadrže segment a i b +, te segment + i b. Ova vrsta rekombinacije ne zahtijeva sintezu nove DNA. Ovaj koncept je korišten za objašnjenje genetske rekombinacije.

(ii) Razbijanje i kopiranje:

Jedna spirala uparenog homolognog kromosoma (ab i a + b +) lomi se između a i b (slika 8.23B). Segment b zamijenjen je novo sintetiziranim segmentom kopiranim iz b + i pridružen odjeljku. Tako rekombinanti sadrže i ab + i a + b +.

(iii) Potpuni izbor kopije:

Godine 1931. Belling je predložio ovu teoriju za rekombinaciju kromosoma kod viših životinja. Međutim, nekoliko je radnika to dovelo u pitanje. Stoga ima samo povijesnu važnost.

Prema ovoj teoriji, dio jednog roditeljskog lanca homolognog kromosoma djeluje kao predložak za sintezu kopije njegove molekule DNA. Proces kopiranja prelazi na drugu roditeljsku nit. Dakle, rekombinanti sadrže neke genetske podatke o jednoj roditeljskoj niti i onoj drugoj (Slika 8.23 ​​C).

Vrste rekombinacije:

U mikroorganizmima se javljaju mnoge vrste rekombinacija.

Oni se u osnovi dijele na sljedeće tri skupine:

(ii) Ne-recipročna rekombinacija, i

(iii) Rekombinacija specifična za mjesto.

(i) Opća rekombinacija:

Opća rekombinacija događa se samo između komplementarnih lanaca dviju homolognih molekula DNA. Smith (1989.) je pregledao homolognu rekombinaciju u prokariota. Opća rekombinacija u E. coli vođena je interakcijama uparivanja baza između komplementarnih nizova dviju homolognih molekula DNA.

Dvostruka spirala s dvije molekule DNA puca, a dva slomljena kraja spajaju se sa svojim suprotnim partnerima kako bi se ponovno spojili u dvostruku spiralu. Mjesto razmjene može se pojaviti bilo gdje u homolognoj nukleotidnoj sekvenci gdje lanac jedne molekule DNK postaje baza uparen s drugom niti kako bi se dobio heterodupleks samo između dvije dvostruke spirale (slika 8.24).

U heterodupleksu se nukleotidne sekvence ne mijenjaju na mjestu razmjene zbog događaja cijepanja i ponovnog spajanja. Međutim, heterodupleksni spojevi mogu imati mali broj neusklađenih parova baza.

Opća rekombinacija poznata je i kao homologna rekombinacija jer zahtijeva homologne kromosome. U bakterijama i virusima opća rekombinacija provodi se proizvodima rec gena kao što je RecA protein. Protein RecA je vrlo važan za popravak DNK, stoga je rekombinacija ovisna o recA.

Hollidayov model za opću rekombinaciju:

Holliday (1974.) predstavio je model koji prikazuje opću rekombinaciju (slika 8.25). Prema ovom modelu, rekombinacija se odvija u pet koraka, kao što su pucanje niti, uparivanje niti, invazija/asimilacija niti, stvaranje hijazme (prelazak), lom i ponovno okupljanje i popravak neusklađenosti.

Sl.8.25: Hollidayov model za recipročnu opću rekombinaciju.

Do opće rekombinacije dolazi ukrštanjem uparivanjem između komplementarnih pojedinačnih niti DNK dupleksa (a). Dvije homologne regije dvostruke spirale DNA podliježu reakciji izmjene.

Homologno područje sadrži dugi niz komplementarnih baza koje se uparivaju između niti jedne ili dvije originalne dvostruke spirale i komplementarne niti s druge. Međutim, nije poznato kako se homologna regija DNA međusobno prepoznaje.

Popis rekombinacijskih gena i njihova funkcija dan je u tablici 8.2. Za rekombinaciju u E. coli potrebni su RecBCD proteini recBCD ili recJ gena. Ovaj protein ulazi u DNK s jednog kraja dvostruke spirale i putuje duž DNK s dvostrukom spiralom, brzinom od oko 300 nukleotida u sekundi.

On stvara petlju ssDNA duž putujuće DNA (b). Koristi energiju dobivenu hidrolizom molekula ATP -a. Posebno mjesto prepoznavanja (a) slijed od osam nukleotida raspršenih po kromosomu E. coli (b) uklješteno je u putujućoj petlji DNA koju tvori RecBCD protein.

Tablica 8.2: Rekombinacijski (rec) geni i njihova funkcija.

(b) Uparivanje niti:

Proteini RecBCD djeluju kao DNA helikaza jer hidroliziraju ATP i putuju duž spirale DNA. Dakle, proteini RecBCD rezultiraju stvaranjem jednolančanih brkova na mjestu prepoznavanja koji su pomaknuti od spirale (c). Time se započinje interakcija uparivanja baze između dvije komplementarne sekvence dvostruke spirale DNA.

(c) Invazija/asimilacija niti:

Jedan lanac (brčić) nastao iz jedne dvostruke spirale DNK napada drugu dvostruku spiralu (d). U E. coli recA gen proizvodi RecA protein koji je važan za rekombinaciju između kromosoma, poput jednolančanih (SSB) proteina, RecA protein se čvrsto veže za jednolančanu DNK i tvori nukleoproteinski filament.

Roca i Cox (1990) pregledali su strukturu i funkciju proteina RecA. RecA protein potiče brzu renaturaciju komplementarne ssDNA koja hidrolizira ATP u tom procesu. Protein RecA ima nekoliko veznih mjesta, stoga može vezati ssDNA, a zatim i dsDNA. Protein RecA veže se prvo za ssDNA, a zatim traži homologiju između donorske niti i molekule primatelja.

Zbog prisutnosti ovih mjesta RecA protein katalizira višestupanjsku reakciju (zvanu sinapsa) između homologne regije ssDNA i dvostruke spirale DNA. E. coli SSB protein pomaže Rec proteinu u provođenju ovih reakcija. Kada se područje homologije identificira početnim uparivanjem baze između komplementarnih sekvenci, dolazi do ključnog koraka u sinapsi.

In vivo pokusi su pokazali da se nekoliko vrsta kompleksa stvara između ssDNA prekrivene proteinom RecA i spirale dsDNA. Prvo se formira nebazni upareni kompleks koji se pretvara u trolančanu strukturu (ssDNA, dsDNA i RecA protein) kada se pronađe homologna regija.

Ovaj kompleks je nestabilan i izvlači DNK heterodupleks plus istisnutu ssDNA iz izvorne spirale. Nakon što se nađu homologna područja i ssDNA i dsDNA budu kompleksirane, nastaje stabilna D-petlja (d).

Sljedeći korak je asimilacija vezivanja niti i nick -a (e). Donatorski lanac postupno istiskuje primateljski lanac koji se naziva migracija grane. Nakon stvaranja sinapse, heterodupleksna regija se povećava migracijom grana usmjerenih na proteine ​​katalizirane proteinom RecA.

Migracija grane usmjerene prema proteinu RecA odvija se ujednačenom brzinom u jednom smjeru zbog dodavanja više proteina RecA na jedan kraj RecA proteinske niti na ssDNA. Migracija grane može se dogoditi u bilo kojoj točki u kojoj dvije pojedinačne niti sa nizom pokušavaju upariti s istom komplementarnom niti.

An unpaired region of the other single strand resulting in movement of branch point without changing the total number of DNA base pairs. Special DNA helicases that catalyse protein directed branch migration are involved in recombination. In contrast, the spontaneous branch migration proceeds in both the directions almost at the same rate. Therefore, it makes a little progress over a long distance.

(e) Chiasma or crossing over formation:

Exchange of a single strand between two double helices is a different step in a general recombination event. After the initial cross strand exchange, further strand exchanges between the two closely opposed helices is thought to proceed rapidly. A nuclease cleaves and partly degrades the D-loop at some points.

At this stage possibly different organisms follow different pathways. However, in most of the cases an important structure called cross-strand exchange (also called Holliday Juncture or chi form or chiasmas, is formed by the two participating DNA helices (g). A chi form of single stranded connections in the cross over region has also been observed under the electron microscope by Dressier and Potter (1982).

The chi form of two homologous helices that initially paired and held together by mutual exchange of two of the four strands where one strand originates from each of the helices (g).

The chi form has two important properties, (i) the point of exchange can migrate rapidly back and forth along the helices by a double branch migration, and (ii) it contains two pairs of strands, one pair of crossing strands and the other pair of non-crossing strands.

(f) Breakage and reunion:

The chi structure can isomerize several rotations (h). This results in alteration of two original non-crossing strands into the crossing strands, and the crossing strands into the non-crossing strands. In order to regenerate two separate DNA helices, breakage and reunion in two crossing strands are required.

If breakage and reunion occur before isomerization the two crossing strands would not occur. Therefore, isomerization is required for the breakage and reunion of two homologous DNA double helices resulting from general genetic recombination.

Breakage and reunion occur either in the vertical or horizontal plane. If breakage occurs horizontally the recombinants would contain genotype ABlab with a little change in base sequences at the inner region (i).

However, if breakage occurs vertically the recombinants would contain Ab/aB (J). The RurC protein and RecG protein expressed from ruvC and recG genes respectively are thought to be alternative endonucleases specific for Holliday structure.

(g) Mismatch Repair (Mismatch Proof Read­ing System):

It is such a repair system which corrects mismatched base pairs of unpaired regions after recombination. This system recognises mis­matched function of DNA polymerase. The mecha­nism involves the excision of one of the other mismatched bases along with about 3,000 nucleotides. This RecFJO is involved in the repair of short mismatch either in the initial stage or at the end of recombination.

The two proteins MutS and MutL are present in bacteria and eukaryotes. The MutS protein binds to mismatched base pair, whereas MutL scan the DNA for a nick (Fig. 8.26).

When a nick is formed MutL triggers the degradation of the nicked strand all the way back through the mismatch, because the nicks are largely confined to the newly replicated strands in eukaryotes, replication errors are selectively removed. In bacteria the mechanism is the same except that an additional protein MutH nicks the un-methylated GATC sequences and begins the process.

It has been demonstrated in yeast and bacteria that the same mismatch repair system which removes replication errors as in Fig. 8.26 also interrupts the genetic recombination events between imperfectly matched DNA sequences. It is known that homologous genes in two closely related bacteria (E. coli and S.typhimurium) generally will not recombine, even after having 80% identical nucleotide sequences.

However, when mismatch repair system is inactivated by mutation, the frequency of such interspecies recombination increases by 100-fold. This mechanism protects the bacterial genome from sequence changes that would be caused by recombination with foreign DNA molecules entering in the cell.

(ii) Non-reciprocal Recombination (Gene Conversion):

The fundamental law of genetics is that the two partners contribute the equal amount of genes to the offsprings. It means that the offsprings inherit the half complete set of genes from the male and half from the female. One diploid cell undergoes meiosis producing four haploid cells therefore, the number of genes contributed by male gets halved and so the genes of female.

In higher animals like man it is not possible to analyse these genes taking a single cell. However, in certain organisms such as fungi it is possible to recover and analyse all the four daughter cells produced from a single cell through meiosis.

Occasionally, three copies of maternal allele and only one copy of paternal allele is formed by meiosis. This indicates that one of two copies of parental alleles has been altered to the maternal allele. This gene alteration is of non-reciprocal type and is called gene conversion. Gene conversion is thought to be an important event in the evolution of certain genes and occurs as a result of the mechanism of general recombination and DNA repair.

Non-reciprocal general recombination is given in Fig. 8.27. Kobayashi (1992) has discussed the mechanism for gene conversion and homologous recombination.

This process starts when a nick is made in one of the strands (a). From this point DNA polymerase synthesizes an extra copy of a strand and displaces the original copy as a single strand (b). This single strand starts pairing with the homologous region as in lower duplex of DNA molecule (b). The short unpaired strand produced in step (b) is degraded when the transfer of nucleotide sequence is completed. The results are observed (in the next cycle) when DNA replication has separated the two non-matching strands (c).

(iii) Site-Specific Recombination:

Site specific recombination alters the relative position of nucleotide sequences in chromosome. The base pairing reaction depends on protein mediated recognition of the two DNA sequences that will combine. Very long homologous sequence is not required.

Unlike general recombination, site specific recombination is guided by a recombination enzyme that recognises specific nucleotide sequences present on one of both recombining DNA molecules. Base pairing is not involved, however, if occurs the heteroduplex joint is only a few base pair long.

It was first discovered in phage λ by which its genome moves into and out of the E. coli chromosome. After penetration phage encoded an enzyme, lambda integrase which catalyses the recombination process (Fig. 8.28). Lambda integrase binds to a specific attachment site of DNA sequence on each chromosome.

It makes cuts and breaks a short homologous DNA sequences. The integrase switches the partner strands and rejoins them to form a heteroduplex joint of 7 bp long. The integrase resembles a DNA topoisomerase in rejoining the strands which have previously been broken.

Site specific recombination is of the following two types:

(a) Conservative site-specific recombina­tion:

Production of a very short heteroduplex by requiring some DNA sequence that is the same on the two DNA molecules is known as conservative site-specific recombination. The detail procedure is described in Fig. 8.28.

(b) Trans-positional site-specific recombi­nation:

There is another type of recombination system known as trans-positional site-specific (TSS) recombination. The TSS recombination does not produce heteroduplex and requires no specific sequences on the largest DNA.

There are several mobile DNA sequences including many viruses and transposable elements that encode integrates. The enzyme integrates by involving a mechanism different from phage λ insert its DNA into a chromosome. Each enzyme of integrates recog­nises a specific DNA sequence like phage λ.

K. Mizuuchi (1992a) reviewed the mecha­nism of trans-positional recombination based on the studies of bacteriophage Mu and the other elements. The enzyme integrase was first purified from Mu. Similar to integrase of phage λ, the Mu integrase also carries out of its cutting and rejoining reactions without requirement of ATP. Also they do not require a specific DNA sequence in the target chromosome and do not form a joint of heteroduplex.

Different steps of TSS recombinational events are shown in Fig. 8.29. The integrase makes a cut in one strand at each end of the viral DNA sequences, and exposes the 3′-OH group that protrudes out. Therefore, each of these 3′-OH ends directly invades a phosphodiester bond on opposite strands of a randomly selected site on a target chromosome.

This facilitates to insert the viral DNA sequence into the target chromosome, leaving two short single stranded gaps on each side of recombinational DNA molecule.

These gaps are filled in later on by DNA repair process (i.e. DNA polymerase) to complete the recombination process. This mechanism results in formation of short duplication (short repeats of about 3 to 12 nucleotide long) of the adjacent target DNA sequence. Formation of short repeats is the hall-marks of a TSS recombination.

Fig. 8.29 : Mechanism of trans-positional site-specific recombination SDR, short direct repeats of target DNA sequence.


13.1 Chromosomal Theory and Genetic Linkages

U ovom odjeljku istražit ćete sljedeće pitanje:

Povezivanje za AP ® tečajeve

Proposed independently by Sutton and Boveri in the early 1900s, the Chromosomal Theory of Inheritance states that chromosomes are vehicles of genetic heredity. As we have discovered, patterns of inheritance are more complex than Mendel could have imagined. Mendel was investigating the behavior of genes. He was fortunate in choosing traits coded by genes that happened to be on different chromosomes or far apart on the same chromosome. When genes are linked or near each other on the same chromosome, patterns of segregation and independent assortment change. In 1913, Sturtevant devised a method to assess recombination frequency and infer the relative positions and distances of linked genes on a chromosome based on the average number of crossovers between them during meiosis.

The content presented in this section supports the Learning Objectives outlined in Big Idea 3 of the AP ® Biology Curriculum Framework. The AP ® Learning Objectives merge essential knowledge content with one or more of the seven Science Practices. These objectives provide a transparent foundation for the AP ® Biology course, along with inquiry-based laboratory experiences, instructional activities, and AP ® exam questions.

Velika ideja 3 Živi sustavi pohranjuju, preuzimaju, prenose i reagiraju na informacije bitne za životne procese.
Enduring Understanding 3.A Heritable information provides for continuity of life.
Osnovno znanje 3.A.2 In eukaryotes, heritable information is passed to the next generation via processes that include the cell cycle and mitosis or meiosis plus fertilization.
Znanstvena praksa 7.1 Učenik može povezati pojave i modele na prostornim i vremenskim ljestvicama.
Cilj učenja 3.10 The student is able to represent the connection between meiosis and increased genetic diversity necessary for evolution.
Osnovno znanje 3.A.3 The chromosomal basis of inheritance provides an understanding of the pattern of passage (transmission) of genes from parent to offspring.
Znanstvena praksa 1.1 Student može stvarati prikaze i modele prirodnih pojava ili sustava koje je stvorio čovjek u domeni.
Znanstvena praksa 7.2 The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understandings and/or big ideas.
Cilj učenja 3.12 The student is able to construct a representation that connects the process of meiosis to the passage of traits from parent to offspring.

Podrška učitelja

Introduce genetic linkage using visuals such as this video.

Students can read about corn genetics in this review article.

Students can read about linked genes and Mendel’s work in this article.

Have students work through inheritance scenarios where genes are linked and where they are on different chromosomes using the following activity sheet.

Teacher preparation notes for this activity are available here.

Pitanja izazova iz znanstvene prakse sadrže dodatna testna pitanja za ovaj odjeljak koja će vam pomoći u pripremi za AP ispit. Ova pitanja zadovoljavaju sljedeće standarde:
[APLO 3.2][APLO 3.11][APLO 3.14][APLO 3.15][APLO 3.28][APLO 3.26][APLO 3.17][APLO 4.22]

Long before chromosomes were visualized under a microscope, the father of modern genetics, Gregor Mendel, began studying heredity in 1843. With the improvement of microscopic techniques during the late 1800s, cell biologists could stain and visualize subcellular structures with dyes and observe their actions during cell division and meiosis. With each mitotic division, chromosomes replicated, condensed from an amorphous (no constant shape) nuclear mass into distinct X-shaped bodies (pairs of identical sister chromatids), and migrated to separate cellular poles.

Chromosomal Theory of Inheritance

The speculation that chromosomes might be the key to understanding heredity led several scientists to examine Mendel’s publications and re-evaluate his model in terms of the behavior of chromosomes during mitosis and meiosis. In 1902, Theodor Boveri observed that proper embryonic development of sea urchins does not occur unless chromosomes are present. That same year, Walter Sutton observed the separation of chromosomes into daughter cells during meiosis (Figure 13.2). Together, these observations led to the development of the Chromosomal Theory of Inheritance , which identified chromosomes as the genetic material responsible for Mendelian inheritance.

The Chromosomal Theory of Inheritance was consistent with Mendel’s laws and was supported by the following observations:

  • During meiosis, homologous chromosome pairs migrate as discrete structures that are independent of other chromosome pairs.
  • The sorting of chromosomes from each homologous pair into pre-gametes appears to be random.
  • Each parent synthesizes gametes that contain only half of their chromosomal complement.
  • Even though male and female gametes (sperm and egg) differ in size and morphology, they have the same number of chromosomes, suggesting equal genetic contributions from each parent.
  • The gametic chromosomes combine during fertilization to produce offspring with the same chromosome number as their parents.

Despite compelling correlations between the behavior of chromosomes during meiosis and Mendel’s abstract laws, the Chromosomal Theory of Inheritance was proposed long before there was any direct evidence that traits were carried on chromosomes. Critics pointed out that individuals had far more independently segregating traits than they had chromosomes. It was only after several years of carrying out crosses with the fruit fly, Drosophila melanogaster, that Thomas Hunt Morgan provided experimental evidence to support the Chromosomal Theory of Inheritance.

Genetic Linkage and Distances

Mendel’s work suggested that traits are inherited independently of each other. Morgan identified a 1:1 correspondence between a segregating trait and the X chromosome, suggesting that the random segregation of chromosomes was the physical basis of Mendel’s model. This also demonstrated that linked genes disrupt Mendel’s predicted outcomes. The fact that each chromosome can carry many linked genes explains how individuals can have many more traits than they have chromosomes. However, observations by researchers in Morgan’s laboratory suggested that alleles positioned on the same chromosome were not always inherited together. During meiosis, linked genes somehow became unlinked.

Homologous Recombination

In 1909, Frans Janssen observed chiasmata—the point at which chromatids are in contact with each other and may exchange segments—prior to the first division of meiosis. He suggested that alleles become unlinked and chromosomes physically exchange segments. As chromosomes condensed and paired with their homologs, they appeared to interact at distinct points. Janssen suggested that these points corresponded to regions in which chromosome segments were exchanged. It is now known that the pairing and interaction between homologous chromosomes, known as synapsis, does more than simply organize the homologs for migration to separate daughter cells. When synapsed, homologous chromosomes undergo reciprocal physical exchanges at their arms in a process called homologous recombination , or more simply, “crossing over.”

To better understand the type of experimental results that researchers were obtaining at this time, consider a heterozygous individual that inherited dominant maternal alleles for two genes on the same chromosome (such as AB) and two recessive paternal alleles for those same genes (such as ab). If the genes are linked, one would expect this individual to produce gametes that are either AB ili ab with a 1:1 ratio. If the genes are unlinked, the individual should produce AB, Ab, aB, i ab gametes with equal frequencies, according to the Mendelian concept of independent assortment. Because they correspond to new allele combinations, the genotypes Ab and aB are nonparental types that result from homologous recombination during meiosis. Parental types are progeny that exhibit the same allelic combination as their parents. Morgan and his colleagues, however, found that when such heterozygous individuals were test crossed to a homozygous recessive parent (AaBb × aabb), both parental and nonparental cases occurred. For example, 950 offspring might be recovered that were either AaBb ili aabb, but 50 offspring would also be obtained that were either Aabb ili aaBb. These results suggested that linkage occurred most often, but a significant minority of offspring were the products of recombination.

Visual Connection

  1. Yes, the predicted offspring frequencies range from 0\% to 100\%
  2. No, the predicted offspring frequencies cannot be higher than 30\% .
  3. Yes, the predicted offspring frequencies range from 0\% to 60\% .
  4. No, the predicted offspring frequencies range from 0\% to 50\% .

Science Practice Connection for AP® Courses

Razmisli o tome

A test cross involving F1 dihybrid flies produces more parental-type offspring than recombinant-type offspring. How can you explain these observed results?

Podrška učitelja

The question is an application of Learning Objective 3.12 and Science Practices 1.1 and 7.2, and Learning Objective 3.10 and Science Practice 7.1 because students are explaining how meiosis can result in gametes with genetic variation in turn, these gametes can introduce variation in offspring.

Odgovor

More parental type offspring are produced because the genes that are being examined in the dihybrid cross are linked. Genes whose loci are nearer to each other are less likely to be separated onto different chromatids during meiosis as a result of chromosomal crossover. Therefore, there will be more offspring with the parental phenotype than the recombinant phenotype.

More information about linked genes can be found at the following resources:

Everyday Connection for AP® Courses

Genetic Markers for Cancers

Scientists have used genetic linkage to discover the location in the human genome of many genes that cause disease. They locate disease genes by tracking inheritance of traits through generations of families and creating linkage maps that measure recombination among groups of genetic “markers.” The two BRCA genes, mutations which can lead to breast and ovarian cancers, were some of the first genes discovered by genetic mapping. Women who have family histories of these cancers can now be screened to determine if one or both of these genes carry a mutation. If so, they can opt to have their breasts and ovaries surgically removed. This decreases their chances of getting cancer later in life. The actress Angelia Jolie brought this to the public’s attention when she opted for surgery in 2014 and again in 2015 after doctors found she carried a mutated BRCA1 gene.

  1. Genes responsible for temperament are on the same chromosome as genes responsible for certain facial features.
  2. A single gene codes for both temperament and certain facial features, such as jaw size.
  3. Genes responsible for mild temperament are only expressed when genes encoding a cute face are also present.
  4. The products of genes encoding temperament interact with the products of genes encoding facial features.

Genetic Maps

Janssen did not have the technology to demonstrate crossing over so it remained an abstract idea that was not widely accepted. Scientists thought chiasmata were a variation on synapsis and could not understand how chromosomes could break and rejoin. Yet, the data were clear that linkage did not always occur. Ultimately, it took a young undergraduate student and an “all-nighter” to mathematically elucidate the problem of linkage and recombination.

In 1913, Alfred Sturtevant, a student in Morgan’s laboratory, gathered results from researchers in the laboratory, and took them home one night to mull them over. By the next morning, he had created the first “chromosome map,” a linear representation of gene order and relative distance on a chromosome (Figure 13.4).

Visual Connection

Which of the following statements is true?
  1. Recombination of the red/brown eye and long/short aristae alleles will occur more frequently than recombination of the alleles for wing length and body color.
  2. Recombination of the body color and red/cinnabar eye alleles will occur more frequently than recombination of the alleles for wing length and aristae length.
  3. Recombination of the body color and aristae length alleles will occur more frequently than recombination of red/brown eye alleles and the aristae length alleles.
  4. Recombination of the gray/black body color and long/short aristae alleles will not occur.

As shown in Figure 13.4, by using recombination frequency to predict genetic distance, the relative order of genes on chromosome 2 could be inferred. The values shown represent map distances in centimorgans (cM), which correspond to recombination frequencies (in percent). Therefore, the genes for body color and wing size were 65.5 − 48.5 = 17 cM apart, indicating that the maternal and paternal alleles for these genes recombine in 17 percent of offspring, on average.

To construct a chromosome map, Sturtevant assumed that genes were ordered serially on threadlike chromosomes. He also assumed that the incidence of recombination between two homologous chromosomes could occur with equal likelihood anywhere along the length of the chromosome. Operating under these assumptions, Sturtevant postulated that alleles that were far apart on a chromosome were more likely to dissociate during meiosis simply because there was a larger region over which recombination could occur. Conversely, alleles that were close to each other on the chromosome were likely to be inherited together. The average number of crossovers between two alleles—that is, their recombination frequency —correlated with their genetic distance from each other, relative to the locations of other genes on that chromosome. Considering the example cross between AaBb i aabb above, the frequency of recombination could be calculated as 50/1000 = 0.05. That is, the likelihood of a crossover between genes A/a i B/b was 0.05, or 5 percent. Such a result would indicate that the genes were definitively linked, but that they were far enough apart for crossovers to occasionally occur. Sturtevant divided his genetic map into map units, or centimorgans (cM) , in which a recombination frequency of 0.01 corresponds to 1 cM.

By representing alleles in a linear map, Sturtevant suggested that genes can range from being perfectly linked (recombination frequency = 0) to being perfectly unlinked (recombination frequency = 0.5) when genes are on different chromosomes or genes are separated very far apart on the same chromosome. Perfectly unlinked genes correspond to the frequencies predicted by Mendel to assort independently in a dihybrid cross. A recombination frequency of 0.5 indicates that 50 percent of offspring are recombinants and the other 50 percent are parental types. That is, every type of allele combination is represented with equal frequency. This representation allowed Sturtevant to additively calculate distances between several genes on the same chromosome. However, as the genetic distances approached 0.50, his predictions became less accurate because it was not clear whether the genes were very far apart on the same chromosome or on different chromosomes.

In 1931, Barbara McClintock and Harriet Creighton demonstrated the crossover of homologous chromosomes in corn plants. Weeks later, homologous recombination in Drosophila was demonstrated microscopically by Curt Stern. Stern observed several X-linked phenotypes that were associated with a structurally unusual and dissimilar X chromosome pair in which one X was missing a small terminal segment, and the other X was fused to a piece of the Y chromosome. By crossing flies, observing their offspring, and then visualizing the offspring’s chromosomes, Stern demonstrated that every time the offspring allele combination deviated from either of the parental combinations, there was a corresponding exchange of an X chromosome segment. Using mutant flies with structurally distinct X chromosomes was the key to observing the products of recombination because DNA sequencing and other molecular tools were not yet available. It is now known that homologous chromosomes regularly exchange segments in meiosis by reciprocally breaking and rejoining their DNA at precise locations.

Veza do učenja

Review Sturtevant’s process to create a genetic map on the basis of recombination frequencies here.

  1. Chromosomal crossover is a specific, non-random process during which chromosomes are linked together and exchange DNA, which contributes to the genetic diversity.
  2. Chromosomal crossover occurs during meiosis when chromosome pairs are linked and exchange DNA. Thus, crossover increases the variance of genetic combinations in the haploid gamete cell.
  3. Chromosomal crossover results in the inheritance of enhanced genetic material by offspring, and the subsequent recombination event is not variable in frequency or location.
  4. Chromosomal crossover occurs during the mitotic process when chromosomes are linked together and recombination takes place, increasing the variance of genetic combinations in the haploid mitotic cells formed from mitosis.

Mendel’s Mapped Traits

Homologous recombination is a common genetic process, yet Mendel never observed it. Had he investigated both linked and unlinked genes, it would have been much more difficult for him to create a unified model of his data on the basis of probabilistic calculations. Researchers who have since mapped the seven traits investigated by Mendel onto the seven chromosomes of the pea plant genome have confirmed that all of the genes he examined are either on separate chromosomes or are sufficiently far apart as to be statistically unlinked. Some have suggested that Mendel was enormously lucky to select only unlinked genes, whereas others question whether Mendel discarded any data suggesting linkage. In any case, Mendel consistently observed independent assortment because he examined genes that were effectively unlinked.

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    • Autori: Julianne Zedalis, John Eggebrecht
    • Izdavač/web stranica: OpenStax
    • Naslov knjige: Biologija za AP® tečajeve
    • Datum objave: 8. ožujka 2018
    • Mjesto: Houston, Texas
    • URL knjige: https://openstax.org/books/biology-ap-courses/pages/1-introduction
    • Section URL: https://openstax.org/books/biology-ap-courses/pages/13-1-chromosomal-theory-and-genetic-linkages

    © 12. siječnja 2021. OpenStax. Sadržaj udžbenika koji proizvodi OpenStax licenciran je pod licencom Creative Commons Attribution License 4.0. Naziv OpenStax, logotip OpenStax, naslovnice knjiga OpenStax, naziv OpenStax CNX i logotip OpenStax CNX ne podliježu licenci Creative Commons i ne smiju se reproducirati bez prethodnog i izričitog pisanog pristanka Sveučilišta Rice.


    Genetic Problems Solutions Campbell Ch14

    Genetics Problems Campbell 1. A man with hemophilia (a recessive , sex-linked condition has a daughter of normal phenotype. She marries a man who is normal for the trait. What is the probability that a daughter of this mating will be a hemophiliac? A son? If the couple has four sons, what is the probability that all four will be born with hemophilia?

    2. Pseudohypertropic muscular dystrophy is a disorder that causes gradual deterioration of the muscles. It is seen only in boys born to apparently normal parents and usually results in death in the early teens. (a) Is pseudohypertrophic muscular dystrophy caused by a dominant or recessive allele? (b) Is its inheritance sex-linked or autosomal? (c) How do you know? Explain why this disorder is always seen in boys and never girls.

    3. Red-green color blindness is caused by a sex-linked recessive allele. A color-blind man marries a woman with normal vision whose father was color-blind. (a) What is the probability that they will have a color-blind daughter? (b) What is the probability that their first son will be color-blind? (Note: the two questions are worded a bit differently.)

    4. A wild-type fruit fly (heterozygous for gray body color and normal wings was mated with a black fly with vestigial wings. The offspring had the following phenotypic distribution: wild type, 778 black-vestigial, 785 black-normal, 158 gray-vestigial, 162. What is the recombination frequency between these genes for body color and wing type.

    5. In another cross, a wild-type fruit fly (heterozygous for gray body color and red eyes) was mated with a black fruit fly with purple eyes. The offspring were as follows: wild-type, 721 black-purple, 751 gray-purple, 49 black-red, 45. (a) What is the recombination frequency between these genes for body color and eye color? (b) Following up on this problem and problem 4, what fruit flies (genotypes and phenotypes) would you mate to determine the sequence of the body color, wing shape, and eye color genes on the chromosomes?

    6. A space probe discovers a planet inhabited by creatures who reproduce with the same hereditary patterns as those in humans. Three phenotypic characters are height (T = tall, t = dwarf), hearing appendages (A = antennae, a = no antennae), and nose morphology (S = upturned snout, s = downturned snout). Since the creatures were not “intelligent” Earth scientists were able to do some controlled breeding experiments, using various heterozygotes in testcrosses. For a tall heterozygote with antennae, the offspring were tall-antennae, 46 dwarf-antennae 7 dwarf-no antennae 42 tall-no antennae 5. For a heterozygote with antennae and an upturned snout, the offspring were antennae-upturned snout 47 antennae-downturned snout, 2 no antennae-downturned snout, 48: no antennae-upturned snout 3. Calculate the recombination frequencies for both experiments.

    7. Using the information from problem 6, a further testcross was done using a heterozygote for height and nose morphology. The offspring were tall-upturned nose, 40 dwarf-upturned nose, 9 dwarf-downturned nose, 42 tall-downturned nose, 9. Calculate the recombination frequency from these data then use your answer from problem 6 to determine the correct sequence of the three linked genes.

    8. Imagine that a geneticist has identified two disorders that appear to be caused by the same chromosomal defect and are affected by genomic imprinting: blindness and numbness of the limbs. A blind woman (whose mother suffered from numbness) has four children, two of whom, a son and daughter, have inherited the chromosomal defect. If this defect works like Prader-Willi and Angelman syndromes, what disorders do this son and daughter display? What disorders would be seen in their sons and daughters?

    9. What pattern of inheritance would lead a geneticist to suspect that an inherited disorder of cell metabolism is due to a defective mitochondrial gene?

    10. An aneuploid person is obviously female, but her cells have two Barr bodies. what is the probable complement of sex chromosomes in this individual?

    11. Determine the sequence of genes along a chromosome based on the following recombination frequencies: A-B, 8 map units A-C, 28 map units A-D, 25 map units B-C , 20 map units B-D, 33 map units.

    12. About 5% of individuals with Downs syndrome are the result of chromosomal translocation. In most of these cases, one copy of chromosome 21 becomes attached to chromosome 14. How does this translocation lead to children with Down syndrome?

    13. Assume genes A and B are linked and are 50 map units apart. An individual heterozygous at both loci is crossed with an individual who is homozygous recessive at both loci. (a) What percentage of the offspring will show phenotypes resulting from crossovers? (b) If you did not know genes A and B were linked, how would you interpret the results of this cross?

    14. In Drosophila, the gene for white eyes and the gene that produces “hairy” wings have both been mapped to the same chromosome and have a crossover frequency of 1.5%. A geneticist doing some crosses involving these two mutant characteristics noticed that in a particular stock of flies, these two genes assorted independently that is they behaved as though they were on different chromosomes. What explanation can you offer for this observation?


    Recombination frequencies - Biology

    Recombination frequency question. The recombinant frequency of two genes A and B located on the same chromosome is 38%. Parents heterozygous for genes A and B were test crossed, and 250 offspring resulted. How many of the offspring would have the parental genotypes ofr A and B?

    I was told that you have to use this formula : recombination frequency= # of recombinant types / total number of offspring x 100%. Ja sam ovo učinio. I went 0.38= # of recombinant types / 250 offsprings x 100%. i got rid off the 100% to get 0.38 i took 0.38 x 250 i ogt 95. is my answer correct?

    Answered By Olena V

    Yes, your answer is correct.

    The number of offspring that would have the parental genotypes A and B would be 95. The 38% lets you know how often this occurs in a group of offspring. So if you know that your total amount of offspring is 250 you then multiply it by .38 to get your answer of 95 offspring.

    Basically this question lets you know that unlike the traditional punett square where if you combine two heterzogous individuals (AB and AB) where the possibility of the offspring to be AB as well is 50%, the chance of that is actually lower (38%).


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