Gregor Johan Mendel (1822 1884)

Clugr austriac

A experimentat cu plantele demazre

Susinea c factorii ereditari(genele) pstreaz individualitateadin generaie n generaie

1865 - Versuche ber PflanzenHybriden von Gregor Mendel

Experiments in Plant Hybridizationby Gregor Mendel

Hibridare - Tipuri Fenotip Genotip Homozigot Heterozigot

x +

P F1, F2 Gena Alela Dominant Recesiv Caractere alelomorfe

Mendel aplic ipoteza puritii gameilor

A iniiat experimente cu 34 specii de Pisum sativum

Structura florii exclude polenizarea strin ntmpltoare

Perioad scurt de vegetaie Caractere bine conturate Multe varieti

Dup 2 ani a ajuns la 22 linii pure

Caractere analizate de Mendel (7):

Stnga -

dominante

Dreapta

recesive

Forma alternativ a genelor alel

Fiecare caracter este determinat de 2 gene (una de la mam, una de la tat)

Gameii conin cte o singur alel

Cnd o alel se exprim, iar cealalt nu are un caracter notabil, aceast alel este dominant

Experimentele lui Mendel:

Plante cu caractere distinctede difereniere

Pe perioada nfloririi hibriziide plante trebuie izolaireproductiv pentru a excludeptrunderea polenului strin

Hibrizii i descendena nutrebuie s sufere nici omodificare n fertilitatea sa

Monohibridarea uniformitatea fenotipic a F1

Genotipuri n F1 4/4 Ss

Fenotipuri n F1 4/4 netede

Se ncrucieaz 2 indivizi heterozigoi (Ss) din F1 i se obin generaii cu raport de segregare fenotipic.

Genotipuri n F2 1/4 SS 1/2 Ss 1/4 ss

Fenotipuri n F2 3/4 netede 1/4 rugoase

RSF 3:1RSG 1:2:1

S bob neted

s bob rugos

Y bob galben

y bob verde

Genotipuri n F1 4/4 SsYy

Fenotipuri n F1 4/4 netede galbene

RSF 9:3:3:1 dup forma bobului

3:1

dup culoarea bobului 3:1

RSG 1:2:2:4:1:2:1:2:1

La ncruciarea formelor parentale carese deosebesc prin 2 sau mai multeperechi de caractere, n F2 se produceo segregare fenotipic independent

(3:1) n numrul de caractere urmrite

8 tipuri de gamei

27 genotipuri diferite

8 fenotipuri diferite

RSF 27:9:9:9:3:3:3:1

Nr. perechi alele

Gamei n F1 Segregarea n F2

Tipuri genetice de

gamei

Combinaii posibile ntre

gamei

Nr. clase fenotipice

Nr. clase genotipice

1 2 4 2 3

2 4 16 4 9

3 8 64 8 27

4 16 256 16 81

5 32 1024 32 243

10 1024 1048576 1024 59094

n 2n 4n 2n 3n

Legea segregrii i asortrii independente este valid dac:

Gameii i zigoii sunt viabili i viguroi

Gameii purttori de alele se unesc randomizat

Genele sunt localizate pe cromozomi diferii

Genele sunt localizate pe autozomi

Bazele citologice ale segregrii mendeliene:

Fiecare pereche de cromozomi omologi dintr-ocelul somatic conine un cromozom matern i unulpatern

Fiecare pereche de gene analizate este localizatpe cromozomi diferii

Orientarea cromozomilor omologi n cadruldiviziunii meiotice este ntmpltoare

Gameii obinui pot conine diferite combinaii degene

Neted Neted Neted Rugoas Rugoas Normal Normal Pitic Normal Pitic

Repartizarea randomizat a genelor

Gameii paterni:

ncruciarea dihibrid SB Sb sB sb

SB

Sb

sB

sb

SSBB SSBb SsBB SsBb

SSBb SSbb SsBb Ssbb

SsBB SsBb ssBB ssBb

SsBb Ssbb ssBb ssbb

Rezultatul: O celul diploid = patru celule haploide

n ncrucirile efectuate de Mendel, fenotipuldescendenilor hibrizi poate fi prevzut dingenotipul prinilor tiind care sunt geneledominante i care sunt cele recesive.

Deci, cunoscnd genotipul prinilor putemcaracteriza i anticipa fenotipul descendenilor.

Se pune ntrebarea dac se poate procedai invers se poate determina genotipulunui individ pornind de la fenotipul su?

Aceast problem este deosebit deimportant i, chiar decisiv, pentrulucrrile de ameliorare cnd este necesars cunoatem dac indivizii sunthomozigoi sau heterozigoi pentru uncaracter urmrit.

Datorit fenomenului de dominan, judecnd dupfenotip, nu se poate spune dac un individ este homozigotsau heterozigot pentru o anumit alel dominant.

Forme

parentaleAA x aa

F1 Aa Aa

n acest scop, s-a recurs la numeroase ncruciri i s-a stabilit c ceamai adecvat este ncruciarea ntre un membru al generaiei hibride F1i un membru al unei linii parentale backcross sau retroncruciare.

Forme

parentale

Aa

(individ de analizat)x

AA

(forma parental)

Gamei A a A A

FB AA AA Aa Aa

Uniformitate fenotipic

Cnd se ncrucieaz un individ de analizat Aa, cu forma parental AA, toidescendenii obinui vor fi identici din punct de vedere fenotipic.

Toi gameii formei parentale AA vor fi de un singur tip, purttori ai geneidominante A.

Hibridul de analizat Aa va produce 2 tipuri de gamei: cu gena dominant A icu alela recesiv a.

Ca rezultat al combinrii ntmpltoare a acestor gamei, n descendena acesteincruciri segregarea fenotipic nu are loc datorit prezenei alelei dominante A.

Genotipic, ns, vor fi diferii, 50% fiind AA i 50% Aa.

n cazul acestei ncruciri, necunoscnd genotipul individului de analizat, lipsasegregrii fenotipice nu ne dezvluie formula genotipic.

Forme

parentale

AA

(individ de analizat)x

AA

(forma parental)

Gamei A

AA A

FB AA AA AA AA

Uniformitate fenotipic

Acelai rezultat s-ar obine i dac individul de analizat ar fi AA.

FB descendena unui backcross

Cu totul alt rezultat se obine atunci cnd se ncrucieaz un individ deanalizat Aa cu forma parental homozigot recesiv aa.

Descendena va segrega fenotipic n 2 categorii n proporie de 1:1, ocategorie fenotipic reprezentat de gena dominant A i o categoriedeterminat de aa.

Baza acestei segregri este faptul c toi gameii formei parentale aa vor fide un singur tip (a), iar hibridul Aa va forma 2 categorii (A i a).

n urma fecundrii, jumtate din descendeni vor fi purttori ai alelei A ijumtate purttori de a n doz dubl.

FT descendena unui testcross

Forme

parentale

Aa

(individ de analizat)x

aa

(forma parental)

Gamei A a a a

FT Aa Aa aa aa

50% 50%

Segregare

fenotipic1 : 1

Uniformitate fenotipic

Acest tip de retroncruciare saubackcross a unui organism hibrid cu oform parental homozigot recesiv,soldat cu segregarea fenotipic adescendenilor i dezvluirea naturiiheterozigote a organismului analizat senumete ncruciare de analiz sautestcross.

Faptul c testcross-ul este ncruciareacare asigur dezvluirea structuriigenotipice a unui individ reiese i dinncruciarea dintre un individ de analizatAA i forma parental aa.

Forme

parentale

AA

(individ de analizat)x

aa

(forma parental)

Gamei A A a a

FT Aa Aa Aa Aa

Uniformitate fenotipic

n concluzie:

Dac dintr-un testcross rezult uniformitatefenotipic, individul de analizat este AA.

Dac dintr-un testcross rezult segregare fenotipicde 1:1, individul de analizat este Aa.

Testcross backcross efectuat ntre un individcu genotip necunoscut i un individ cunoscut cahomozigot recesiv, cu scopul de a determinadac individul n discuie este heterozigot sauhomozigot pentru o anumit alel.

n cazul analizei genotipice a dihibrizilor, ncruciareantre indivizi diheterozigoi cu forme parentalehomozigote recesive duce la realizarea n FB a unui raportde segregare fenotipic de 1:1:1:1. Apar 4 categoriifenotipice corespunztoare celor 4 tipuri de gameiprodui din dihibridul de analizat.

Forme

parentale

AaBb

(individ de analizat)x

aabb

(forma parental)

Gamei AB Ab aB ab ab ab

FB AaBb Aabb aaBb aabb

RSF 1 1 1

O problem foarte important a studiilorexperimentale este aceea a siguranei obineriiacelorai rezultate n cazul repetrii experienei.

Statistica matematic, prin metodele ei, ne ajut laevidenierea caracteristicilor eseniale, astfelnct la observarea unui mare numr de cazuriindividuale, care fiecare n parte prezint deosebiride caracter ntmpltor sau nu, s decoperim legilecrora se supun.

Testul X2 (chi-ptrat) asigur estimareaprobabilitii de repetare a fenomenului observatprin msurarea diferenei ntre 2 serii de variaii.

Procedeul chi-ptrat simplific multmsurarea acestor diferene.

Prin acest procedeu aflm dac exist odiferen esenial ntre 2 sau maimulte serii de variaii sau diferenaeste doar ntmpltoare.

Formula pentru calcul este:

X= d/e

d diferena ntre valoarea calculat i cea observat e valoarea calculat sau ateptat a fenomenului

O valoarea observat E valoarea calculat sau ateptat a fenomenului

n valoarea lu X se folosesc numai cifre absolute.

n cazul a 2 perechi de gene alelomorfe Aa i Bb, cu genele A i Bdominante i alelele a i b recesive, ele segreg independent n timpulncrucirilor.

ncruciarea ntre 2 heterozigoi (AaBb) va produce 4 clase fenotipicentr-un raport de 9 (AB):3(Ab):3(aB):1(ab).

Dac ntr-o polpulaie n care exist cele 2 perechi de gene alelomorfe,cifrele observate sunt 1080 AB, 210 Ab, 200aB i 110 ab, se punentrebarea n ce msur cifrele observate concord cu cifrele calculatesau ateptate pentru segregarea fenotipic n dihibridare.

Date observate

AB Ab aB ab

1080 210 200 110

Date ateptate 9/16 x 1600=900 3/16 x 1600=300 3/16 x 1600=300 1/16 x 1600=100

Diferena d 180 -90 -100 10

d/e 36 27 33,33 1,0

X = 36 + 27 + 33,33 + 1 = 97,33

Valoarea lui se compar cu valorile lui din tabelulspecial Fisher innd cont de gradele de libertate.

n acest exemplu GL = 4-1 = 3

Din tabelul Fisher, cu ajutorul valorii lui Xi agradelor de libertate, se afl valoarea probabilitii(P).

Valoarea P reprezint probabilitatea de obinere aunei diferene tot att de mari sau mai mari dectcea obinut din experiment, prin pur ntmplare.

Cu ce posibiliti ne putem ntlni dinconfruntarea datelor?

P < 0,05 (5%)

diferenele nu se datoresc pe deplin ansei, decidiferenele nu sunt ntmpltoare i ipoteza se respinge.

P > 0,05

datele sunt conforme destul de satisfctor cu ipoteza iaceasta este acceptat.

n cazul de fa:

3 grade de libertate X = 97,33 P < 0,05 i anume la mai puin de 0,01 (1%),

Ipoteza segregrii fenotipice n raport de 9:3:3:1se respinge

probabilitatea repetrii fenomenului, n condiiiledate este de mai puin de 1%. P fiind mic seconchide c diferenele nu se datoresc n ntregimentmplrii i ipoteza se respinge.

594 insecte normale - vg*vg*e*e* 196 aripi vestigiale i culoarea corpului

normal - vgvge*e* 200 aripi normale i culoarea neagr a

corpului - vg*vg*ee 71 aripi vestigiale i culoarea neagr a

corpului - vgvgee

Pentru a verifica dac aceast segregare arecaracter ntmpltor sau este vorba de olegitate mendelian a segregrii, se aplictestul .

X = 0,0132 + 0,0432 + 0,0057 + 0,3317 = 0,3938

La 3 grade de libertate , valoarea lui se plaseaz ntre P=0,90i P=0,93, deci probabilitatea repetrii fenomenului n condiiiledate este ntre 90 i 95% i prin urmare se poate considera cabaterile observate se datoresc ntmplrii.

Ipoteza c segregarea se produce n raport de 9:3:3:1 esteadevrat, se admite.

Date observate

vg*vg*e*e* vgvge*e* vg*vg*ee vgvgee

594 196 200 71

Date ateptate9/16 x 1061=

596,82

3/16 x 1061=

198,33

3/16 x 1061=

198,33

1/16 x 1061=

66,32

Diferena d -2,82 2,93 1,07 4,68

d/e 0,0132 0,0432 0,0057 0,3317

This dominant trait is also called thephoto sneeze reflex. If, whensuddenly exposed to light, you sneeze(usually two or three times) you havethe genes for achoo syndrome.

Next time you go to a movie, exit thedark theater through a door thatleads directly outside. It's fun towait outside and watch the peopleemerge from the movie. Some willsneeze as soon as they are exposedto light.

Dominant (have it) Recessive (don't have it)

A prominent cleft in thechin is due to the bondstructure whichunderlies the Y-shapedfissure of the chin.

Females appear to beless conspicuouslyaffected than males.

Dominant (have it) Recessive (don't have

it)

Nearsightedness, ormyopia, is a complextrait with at least 4gene loci involved,however theheritability of myopia isvery high and shows adominant pattern.

Dominant (have it) Recessive (don't have

it)

A dominant allele causes thelast joint of the little fingerto dramatically bend inwardtoward the 4th finger.

Lay both hands flat on a tablerelax your muscles, and notewhether your have a bent orstraight little finger.

Dominant (have it) Recessive (don't have it)

If you aren't sure ifyou have them, smile!

Dimples are easiest tosee when smiling.

With dominantphenotype, you mayhave a dimple only onone side, or on both.

Dominant (have it) Recessive (don't have

it)

We're kind of cheating here. Eye color, as well as hair and skincolor, is a complex trait; not a case of simple inheritance.

The main pigment is melanin, and the more melanin, the darkerthe color.

Although the genetics of eye color is complex, alleles for theproduction of melanin dominate those for lack of melanin.

So if we evaluate eye color as being blue (recessive) or non-blue(dominant) we can treat it as a characteristic of simple

inheritance.

Dominant (have it) Recessive (don't have it)

The dominant trait is forlobes to hang free, a bit oflobe hanging down prior tothe point where the bottomof the ear attaches to thehead.

With the recessivephenotype, the lobes areattached directly to the

head.

Dominant (have it) Recessive (don't have it)

Clasp your hands together(without thinking about it!).Most people place their leftthumb on top of their rightand this happens to be thedominant phenotype.

Now, for fun, try clasping yourhands so that the oppositethumb is on top. Feels strangeand unnatural, doesn't it?

Dominant (have it) Recessive (don't have it)

Some people have hair on the second (middle) joint of one or more of their fingers, while others don't.

Having any hair at all means that you have the dominant phenotype.

Complete absence of hair is recessive.

Dominant (have it) Recessive (don't have it)

If you have the abilityto roll the sides of yourtongue upwards to forma closed tube, you havethe dominant phenotypefor this motor skill.

Those who are notdominant for this traitcannot roll their tongue,no matter how hardthey may try.

Dominant (have it) Recessive (don't have it)

This trait isreportedly dueto a singlegene;

the presenceof freckles isdominant

the absence offreckles isrecessive

Early geneticistsreported that curlyhair was dominantand straight hairwas recessive.

More recent studiessuggest that morethan one gene maybe involved

While allergic reactions are induced bythings a person comes in contact with,such as dust, particular foods, andpollen, the tendency to have allergies isinherited.

If a parent has allergies, there is a onein four (25%) chance that their childwill also have allergy problems. This riskincreases if both parents have allergies.

Colorblindness is due to a recessive allele locatedon the X chromosome.

Women have two X chromosomes, one of whichusually carries the allele for normal color vision.

Therefore, few women are colorblind.

Men only have one X chromosome, so if they carrythe allele for colorblindness, they will exhibit thistrait.

Thus, colorblindness is seen more frequently in menthan in women.

Is reportedly due to a single gene: widows peak

dominant straight hairline

recessive.

Widows Peak Hairline

Straight Hairline

For some people the chemical PTC (phenylthiocarbamide) tastes verybitter. For others, it is tasteless.

the ability to taste PTC shows dominant inheritance and is controlledby a gene on chromosomes 7.

This gene codes for part of the bitter taste receptor in tongue cells.One of its five alleles (forms) causes a lack of ability to sense bittertastes;

the other four alleles produce intermediate to fully sensitive tasteabilities. Approximately 75% of people can taste PTC while theremaining 25% cannot.

PTC-like chemicals are found in the Brassica family of vegetables, suchas cabbage, brussels sprouts, and broccoli. People who can taste PTCoften do not enjoy eating these vegetables, since they taste bitter tothem.

Non-tasters tend not to notice bitter tastes and therefore may bemore likely to become addicted to nicotine (which is bitter).

PTC-tasting ability has also provided information related tohuman evolution. Populations in Sub-Sahara Africa, and peoplewho are descended from this area, contain at least five formsof the gene.

Some of these forms confer a PTC-tasting ability that isintermediate between taster and non-taster. However, withonly a few exceptions, only two forms taster and non-taster are found in populations outside of Africa and theirdescendents. This is consistent with the out-of-Africahypothesis of modern human origins.

Some scientists think that tasters have fewer cavities,suggesting that there might be a substance in the saliva oftasters that inhibits the bacteria that cause cavities to form.Others think that PTC tasting may be in some way connectedwith thyroid function.

PTC tasting was a chance discovery in 1931.