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.