Ca'm nang ti; 6n luygn thi Sinh hge - Phung Duy D8'ng
3. Mot cap vd chong sinh difrtc mot diJa con bi thieu mau hong cau Mnh liem bi chet luc 2 tuoi. Kieu gen ciia cap vd chong do nhif the nao? bi chet luc 2 tuoi. Kieu gen ciia cap vd chong do nhif the nao?
4. Ba noi va ba ngoai deu bieu hien benh thieu mau hong cau hinh Hem nhc, ong ngoai va ong noi c6 kieu hinh binh thu'dng, bo me sinh diTcfc 2 ngiTdi nhc, ong ngoai va ong noi c6 kieu hinh binh thu'dng, bo me sinh diTcfc 2 ngiTdi con, diJa thiJ nhat che't v i benh hong cau hinh liem, du'a thiJ hai bi benh hong cau hinh liem nhe.
- M o i du'a con da thijfa hiTdng gen cua bo me nhu' the' nao?
- Neu bo me tic'p tuc sinh con nu"a thi kha nang c6 the xuat hien duTa tre c6 kieu hinh binh thu'dng khong? Giai thich? kieu hinh binh thu'dng khong? Giai thich?
HUdng dan
1. DiJa con c6 bieu hien benh thieu mau hong cau hinh \\idi liem nhe. Kieu gen ciia du'a con do la: Hb^Hb' nhan mot giao i\i cua bo va mot giao tuf cua me. gen ciia du'a con do la: Hb^Hb' nhan mot giao i\i cua bo va mot giao tuf cua me.
Kieu gen cua bo'me c6 the nhtf sau: ,,
a) P: Hb'Hb^ (me) x Hb'Hb^ (bo) b) P: Hb'Hb^ (me) x Hb^Hb^ (bo) b) P: Hb'Hb^ (me) x Hb^Hb^ (bo) c) P: Hb^^Hb' (me) x H b ' H b ' (bo)
(khong CO kieu gen Hb'^Hb^ vi bi benh thieu mau nang thiTdng chft't tru'dc 2 tuoi).
2. Du'a con c6 kieu hinh binh thu'dng.
oj Kieu gen cua du'a con d6 la Hb'Hb\n cua moi bo me giao tur Hb'. Do do: Kieu gen cua bo me c6 the nhU' sau:
a) P: H b ' H b ' x H b ' H b ' (Kieu hinh bo me binh thifdng).
b) P: Hb^Hb' x Hb'Hb' (Mot ngu'di binh thiTdng mot ngiTdi thieu mau nhe).
c) P: Hb^Hb'; Hb'Hb" (Ca 2 deu thieu mau nhe)
3. Con b i thieu mau hong cau hinh liem bi chet luc 2 tuoi, c6 kieu gen la
H b ' H b ' . Do do b d me deu phai c6 gen Hb^ => P: Hb^Hb'x Hb^Hb^
4. Ong noi binh thu'dng c6 kieu gen Hb'Hb' truyen cho b d gen Hb'. Ong ngoai binh thu'dng c6 kieu gen Hb^Hb' truyen cho me gen Hb'. DuTa con chet c6 kieu gen binh thu'dng c6 kieu gen Hb^Hb' truyen cho me gen Hb'. DuTa con chet c6 kieu gen Hb^Hb^ nhan cua moi bd me 1 giao tur Hb^. V i vay bd me deu c6 gen Hb^.
Do do kieu gen cua P: Hb^Hb'xHb^Hb'
Du'a con thiJ hai bi thie'u mau hong cau hinh liem nhe c6 kieu gen Hb^Hb' nhan giao tuf Hb^ hoac cua me hoac cua bd.
5. Neu b d me tiep tuc sinh con ni^a thi c6 kha nang xuat hien tre c6 kieu hinh binh thu'dng Hb*Hb' xac sua't 25% theo sd do sau: binh thu'dng Hb*Hb' xac sua't 25% theo sd do sau:
, , B d m e : H b ' H b ' x Hb^Hb* ,,^
G : H b ^ H b ' H b ' . H b '
^^u.|J,,^ : I H b ' H b ^ 2 H b ^ H b M H b W . '
I , ( M l ' i f M
Vi du 4: 6 ruoi dain gen n tren NST g i d i tinh X quy dinh dang canh binh
thu'dng, gen dot bien N tren NST gidi tinh X quy dinh canh c6 mau nho.
K h i lai ruoi diTc hoang dai c6 canh binh thiTdng (X"Y) vdi rudi cai dot bien (X^X") canh CO nhiyng mau nho thi:
1. F| xua't hien kieu hinh theo t i le: - ruoi di/c c6 cdnh binh thu'dng, - ruoi
3 3
c^i c6 canh binh thu'dng va ^ rudi cdi c6 canh c6 nhffng mau nho.
2. T i le giffa rudi difc va rudi cai d F, la 1 diTc : 2 cai. Cf F2 la 3 di/c : 4 cai.
3. T i le dang hoang dai va dang dot bien d F, la 1 dot bien : 2 hoang dai. 6 F2
la: 1 dot bie'n : 6 hoang dai.
Hay giai thich cac hien tU'(?ng tren. '
HUdngddn . .
1. Lai ruoi difc hoang dai (X"Y) vdi ruoi cai dot bien (X^X") P : X " Y X X ^ X " P : X " Y X X ^ X "
G(P) : X " , Y X ^ X "
F, : 1 X ^ X M X " X M X " Y : IX'^Y (chet) * '
Rudi diTc X^Y bi chet, nen Fj chi xuat hien 3 loai hinh:
^ X ^ X " (rudi cai c6 mau) (1 dot bien) _^
^ X " X " (ruoi cai canh binh thifdng) (hoang dai) i v ^ X " Y (ruoi diTc canh binh thiTdng) (hoang dai)
2. Vii CO ti le 1 difc : 2 cai. 1 dot bie^n : 2 hoang dai. '
3. 6 Fi xay ra 2 phep lai:
a) F, : X"Y x X^X" / , , •^'::v:^>>;,.: G ( F , ) : X " , Y x^x" ' G ( F , ) : X " , Y x^x" ' F2 : ^ X ^ X " : ^ X " X " : ^ X " Y (X'^Ybjchet) y . b) F, : x"Y X x-x" ; /; ; .''^ G ( F , ) : X " , Y X " •/-%fp/,;'' -I'^ii'-jy' F2 : I X " X " : 1X"Y ; '
Tinh C h u n g cho ca 2 phep lai ta c6: ; • ; >
3X"X" ( ? hoang dai): 3X"Y {S hoang dai): I X ^ X " ( $ dot bien)
D o d 6 l i l e d F 2 l a : 3 d i r c : 4 c a i
cam nang tij on va Iuy6n Ihi Sinh hpc - Phting Duy Pgng
Vi du 5: 6 tarn dau (2n = 28) gidi tinh cai difdc quy dinh bdi cap NST X Y , gidi tinh diTc dtfcJc quy dinh bdi cap NST X X . Gen quy dinh mau sac trijrng nam tren cap NST so 10. M e n A xac dinh trtfng mau trang, alen a quy dinh tru-ng mau
xam den. Tit 2 noi tam triJng trang va triJng xam den, ngu'di' ta da tao diTOc 2 noi tam ma khi lai vdi nhau se cho ^ so triJng mau xam den, nd toan tam cai va -
so triJng c6n lai cd mau trang nd ra toan llm d\ic. (Ken tam dufc cd nang sua't td
tang trung binh 30% so vdi ken tam cai vi vay ngiTdi ta chi suf dung ken tam di/c de siin xua't td). Hay giai thich ngiTdi ta da tao ra 2 dong tam nay nhu" the nao?
HMngddn
+ Hal noi tam ma khi cho lai vdi nhau cho: - so trtog mau trang nd ra toan
2
tam du-c va - con lai mau xam den, nd ra toan tam cai. Chtfng td tinh trang mau
2
sac triJng lien ke't vdi gidi tinh => gen xac dinh nd nam tren NST gidi tinh.
+ T i le 1 : 1 nay la ket qua cua phep lai phan tich.
P : $ X ^ Y X cJX" X" G(P) : - X ^ , - Y X "
G(P) : - X ^ , - Y X "
+ V i gen quy dinh mau sac trtfng n^m tren NST thtf 10. Do vay, de cd diTdc 2 noi tam tren, ta phai gay dot bien chuyen doan NST mang gen xac dinh mau
sac tru-ng tiT NST thu" 10 sang NST gidi tinh X, nhif sau:
1. Doi vdi noi tam tri^ng trang:
N O — — gay d6l bien ^ 2^ _ 2^ ' ^ Y chuyd'n doan ^ A Y
Tam cai dot bien nay se cho triJng mang NST Y va NST so 10 khong
mang gen A ( - Y )
A X g a y doi bia'ii ^ _ _ X A
b) = = gay noi Dieii ^
^ ^ c h u y e n doan
Tam difc dot bien nay se cho tinh trung mang NST X ^ va NST so 10 khong mang gen A (-X'*').
c) Cho tnJng ( - Y ) ket hdp vdi tinh trung ( - X'^) se tao thanh hdp t^ = X ^ Y Y phat trien thanh con cai cd chtfa cap NST so 10 khong mang gen quy dinh m^u sac, ma cd NST X mang gen A quy dinh mau trtfug trang.
18
Cty T N H H M T V D W H Khang Vigt
2. Do'i vdi noi tnJng mau x i m den.
d6t bien X''
a) 9 ^ ^
^ Y chuyfin doan Y
TCf loai ca the nay se cho triJng mang NST X ' va NST so 10 khong mang en a ( - X").
i.\ 2^ dot bie'n X" -
b) c5 ;—'-r~, > —=•
a X chuytn doan ^ ^
TuT loai ca the nay se cho tinh trung mang X" va NST so' 10 khong mang en a (-X").
v a
c) Cho tnJng (-X") ket hdp vdi tinh trilng (-X") se tao thanh hdp tu^ = — X" X"
iii tren thanh con diTc cd NST so 10 khong mang gen quy djnh mau s^c ma cd
T X mang gen a quy dinh mau xam den.
P3. Cho lai noi tam cai trtfng trang dot bien X
A A
vdi noi tam di/c trtfug
den dot bien se cho — so' triJng mau xam den nd ra toan tam cai va — so'
2 2 g mau xam den nd ra toan tam difc nhu" sd do lai d tren. g mau xam den nd ra toan tam difc nhu" sd do lai d tren.
Vidu6:
Mot loai cay cd bo NST Itfdng boi 2n = 24. Ng^di ta phdng xa hat cua loai y do khi trong phoi mam cac NST trong te bao con chtfa ti/ nhan doi.
Quan sat diTdi kinh hien vi cac phien trung k i cua Ian phan bao dau tien cua
bao phoi mam ngu-di ta thay so lifdng NST Idn hdn 24, tuy nhien ham Itfdng
p N trong te bao khong doi so vdi te bao chiJa bo NST lu-dng boi 2n. Hay giai
•
ch hien tu-dng tren.
1^ HUdngddn
O loai nay, nhieu phan tren mot NST cd cMc nang tu-dng tiT nhU" tam dong.
phdng xa lam dut NST thanh nhieu doan nhd, cac doan nho cd chiJa chi^c
'ng nhu- tam dong trd thanh NST doc lap, lam cho so Itfcfng NST trong te bao
Ig ma khong lam thay doi ham lu-dng A D N trong te bao do. J T A P V A N D V N G
Hay van dung cac phu-dng phap giai cua tijfng loai toan da de cap de'n d tren, i ciic bai tap sau:
2
i 1: Mot gen chi huy tong hdp chuoi pohpeptit gom 498 axit amin cd — = - .
G 3 bie't dot bie'n xay ra khong lam thay doi so nucleotit cua gen. bie't dot bie'n xay ra khong lam thay doi so nucleotit cua gen.
1. Mot dot bie'n xay ra lam cho gen sau dot bien cd ti le — = 66,48%. Dot G G
In thuoc dang nao cua dot bie'n gen?
d m nang tij 6n luygn thi Sinh hpc - Phung Duy D6ng
2. M o t dot bien xay ra lam cho gen sau dot bien c6 ti Ic — = 66,85% • Dpt G G
bien noi tren da lam cho cau true cua gen bi thay ddi nhU' the" nao, thupc dang nao cua dot bien gen?
Hudng dan
1. Gen' ban dau chi tong hdp chuoi polipeptit gom 498 axit amin, chiJng to
gen do CO so'nucleotit la:
[(498 X 3) + 3 (nucleotit cua bo ba m d dau) + 3 (nucleotit cua bo ba ket thijc]
X 2 = 3000 So'nucleotit tufng loai cija gen: So'nucleotit tufng loai cija gen:
Ta CO he phu"cfng trinh:
A + G=3000:2 = 1500 ' " ''^^ A _ 2 A _ 2
- G ~ 3
Giai he phiTdng trinh ta c6: A = T = 600 nucleotit G = X = 900 nucleotit
- Dot bien gen da cho t i le: ^='^=66,61% giam xuong thanh 66,48% v i so
G 3
It/dng nucleotit trong gen khong thay ddi nen so loai nucleotit loai A giam cung bang so' nucleotit loai G tang.
Gpi X la so nucleotit loai A giam di do dot bien, ta c6 phiTdng trinh:
A l ^ = ^ = 66,48% = 0,6648 G + x 900 + x
Giai phufdng trinh ta c6 x = 1. 1 ''
Nhir vay dot bien da lam thay ddi 1 cap A - T bang 1 cap G - X. Dot bien dang thay the 1 cap nucleotit nay bang 1 cap nucleotit khac.
2. Dot bien lam cho t i le — tiT 66,67% tang len 66,85%. Chi?ng to so G G
nucleotit loai A tang. Tat nhien so nucleotit A tang len bang so nucleotit G giam di, nen ta c6 phu'dng trinh:
..oK ft. X ; • 6 0 0 + x
= 66,85% = > x - l 9 0 0 - x
' Nhir vay dot bien da lam thay ddi mot cSp G - X bang mot cSp A - T. Dpt bien dang thay the" cap nucleotit nay bkng cSp nucleotit khac.
Bai 2: Dot bien xay ra trong gen B lam mat axit amin thu" 2 cua chuoi pohpeptit do gen do tdng hpp. Hay xac dinh dang dot bien va vi tri dot bien xay ra trong gen do.
Cty TNHH MTV D W H Khang V i g t ^
HUdng ddn
- Dot bien xay ra trong gen B lam ma't mot axit amin chtfng to dot bien dang mat 3 cap nucleotit ke tiep nhau chiJa trpn trong 1 bp ba ma hoa.
