Download Slide 1

Dihybrid Cross
A cross between two true-breeding parents that
possess different forms (alleles) of two genes
true-breeding plant with
round yellow seeds
X
true-breeding plant with
wrinkled green seeds
Non linked genes
HOMOLOGOUS PAIRS
Y
R
r
DNA REPLICATION
R
Y
R
Y
y
r
Genotype =
y
RrYy
r
y
Non linked genes
Meiosis 1
Gametes
R
Y
R
Y
r
y
RY
ry
r
y
R
y
Ry
R
y
r
Y
rY
r
Y
Original cross
RRYY
gametes
All RY
X
R = allele for round
rryy
r = allele for wrinkled
F1
All ry
Y = allele for yellow
y = allele for green
All RrYy
Second cross
gametes
RY
RrYy
x
Ry rY ry
RrYy
RY
F1 Self-fertilised
Ry rY
ry
F2
RY
Ry
rY
ry
RY
RRYY
RRYy
RrYY
RrYy
Ry
RRyY
RRyy
RryY
Rryy
rY
ry
(phenotypic
ratio)
9 round yellow
3 round green
3 wrinkled yellow
rRYY
rRYy
rrYY
rrYy
rRyY
rRyy
rryY
rryy
1 wrinkled green
9:3:3:1 ratio
Recombination
• In a dihybrid cross two of the F2
phenotypes resemble the
original parents
• Two display new combinations
• The process by which new
combinations of parental
characteristics arise is called
recombination
• The individuals possessing
them are called recombinants
Original parents
RRYY
rryy
Recombinants
RRyy
rrYY
Mendel’s second law
• The principle of independent assortment
During gamete formation, the alleles of a gene
segregate into different gametes independently of
the segregation of the two alleles of another gene
• Let R = Red let r = white
• Let S = straight let s =
curly
RrSs x RrSs
• Possible gametes
• RS,Rs,rS,rs x
RS,Rs,rS,rs
• 9/16 red straight
• 3/16 red curly
• 3/16 white straight
• 1/16 white curly
• 800 offspring
• red straight = 450
• red curly = 150
• white straight = 150
• white curly = 50
Dihybrid cross 2
X
RS
Rs
rS
rs
RRSS
RRSs
RrSS
RrSs
Rs
RRSs
RRss
rS
RrSS
RrSs
rs
RrSs
RS
Rrss
RrSs Rrss
rrSS
rrSs
rrSs
rrss
• Let H = hairless, h = hairy
• Let T = tall, t = dwarf
HhTt x HhTt
• Possible gametes
• HT,Ht,hT,ht x
HT,Ht,hT,ht
• 9/16 hairless tall
• 3/16 hairless dwarf
• 3/16 hairy tall
• 1/16 hairy dwarf
• 1280 offspring
• Hairless tall = 720
• Hairless dwarf = 240
• Hairy tall = 240
• Hairy dwarf = 80
Dihybrid cross 3
X
HT
Ht
Ht
HHTT
HHTt
HhTT HhTt
HHTt
HHtt
HhTt Hhtt
hT
HhTT
ht
HhTt
HhTt
Hhtt
hT
ht
HT
hhTT
hhTt
hhTt
hhtt
• Let P = purple, p = cut
• Let N = normal, n = twisted
PpNn x PpNn
• Possible gametes
• PN,Pn,pN,pn x
PN,Pn,pN,pn
• 6400 offspring
• Homozygous (4/16) =
1600
• Purple (12/16) = 4800
• Cut (12/16) = 1600
• Twisted(4/16) = 1600
Dihybrid cross 4
X
PN
Pn
PN
pN
pn
PPNn
PpNN PpNn
PPnn
PpNn Ppnn
pN
PpNN
PpNn
ppNn
pn
PpNn
Ppnn
PPNN
Pn
PPNn
ppNN
ppNn ppnn
Dihybrid Cross 4
•
•
•
•
•
•
•
•
Let R = Red let r = white
Let S = straight let s = curly
rrSS x RRss
Gametes
rS x Rs
F1
100% RrSs
100% Red straight
X
Rs
rS
RrSs
Dihybrid Test cross 1
• Let H = hairless, h =
hairy
• Let T = tall, t = dwarf
HHTT x hhtt
• Gametes
• HT x ht
• F1 100% HhTt
• F1 100% hairless tall
X
ht
HT
HhTt
Dihybrid test cross 2
•
•
•
•
•
•
•
•
•
•
Let P = purple, p = cut
Let N = normal, n = twisted
PpNn x
ppnn
PN,Pn,pN,pn x pn
1000 offspring
in 1:1:1:1ratio
Purple normal (¼) = 250
Purple twisted (¼) = 250
Cut normal (¼) = 250
Cut twisted (¼) = 250
X
PN
pn
PpNn
Pn
Ppnn
pN
ppNn
pn
ppnn
Linked genes 1
HOMOLOGOUS PAIR
R
r
DNA REPLICATION
R
Y
R
Y
r
y
r
y
Y
y
Genotype = RrYy
Linked genes 2
Gametes
Meiosis 1
R
Y
R
RY
Y
X
ry
r
y
r
y
Crossing over
X
Ry
rY
Large
numbers
of parental
gametes
Small
numbers of
recombinant
gametes
Linked genes 3
HOMOLOGOUS PAIR
H
h
DNA REPLICATION
H
t
H
t
t
T
Genotype = HhTt
h
T
h
T
Crossing over
Frequency of crossing over
Chiasmata can occur at any point along a chromosome
More crossing over (recombination) occurs between two
distantly located genes than two that are close together
Only cross over 1
would break the
link between A and
B or a and b
Any one of
crossovers
2,3,4 and 5 would
break the link
between genes B/b
and C/c
Linked genes 4
• Let R = Red let r = white
• Let S = straight let s = curly
• The genes are linked R and S ,
r and s
•
RrSs x RrSs
• Possible gametes
• RS , rs x RS , rs
• 6400 offspring
• ¾ Red straight = 4800
• ¼ white curly = 1600
• Small numbers of red curly and
white straight by crossing over
RS
rs
RS
RRSS
RrSs
rs
RrSs
rrss
Linked genes test cross
• Let H = hairless, h = hairy
• Let T = tall, t = dwarf
• Genes T and H and t and h are
linked on the same chromosome
HhTt x hhtt
• Possible gametes
• HT , ht x ht
• 1000 offspring
• in 1:1 ratio
• 50% hairless tall = 500 ( 460)
• 50% hairy dwarf = 500 ( 445)
• Small number of recombinants by
crossing over
• Hairless dwarf ( 45)
• Hairy dwarf (50)
X
ht
HT
HhTt
ht
hhtt
Linked genes 5
Gametes
Meiosis 1
H
t
H
Ht
t
X
hT
h
T
h
T
Crossing over
X
HT
ht
Large
numbers
of parental
gametes
Small
numbers of
recombinant
gametes
Linked genes 6
• Let P = purple, p = cut
• Let N = normal, n = twisted
• P and n and p and N are on
the same chromosome
• PpNn x ppnn
• Possible gametes
• Pn , pN x pn
• 1280 offspring
• 50% purple twisted = 610
• 50% cut normal = 600
• Small numbers of recombinant
phenotypes purple normal (33)
and cut twisted (37) by
crossing over
X
pn
Pn
Ppnn
pN
ppNn
Gametes and Ratios
Linked genes
Non linked genes
RrYy
(R+Y) RY, ry
RY,Ry,rY,ry
PpTt
(P+T) PT, pt
PT,Pt,pT,pt
HhSs
(H +s) Hs, hS
HS,Hs,hS.hs
RrYy X RrYy
3:1
9:3:3:1
RrYy X rryy
1:1
1:1:1:1
1:1
1:1:1:1
3:1
9:3:3:1
HhSs X hhss
PpTt X PpTt
Genetics Strategy
•
•
•
•
Monohybrid
One characteristic (two alleles, two phenotypes)
e.g. Eye colour- Red eyes and White eyes
Dominance – Rr x Rr
3 red : 1 white
Rr x rr
1:1 ratio
• Co dominance – two alleles, three phenotypes
RR (red) RW (pink) and WW (white)
• Multiple alleles – more than two alleles
e.g. ABO blood groups
• Sex linkage – show X and Y chromosome
XRY x XRXr
1:1:1:1 ratio
Genetics Strategy
•
•
•
•
•
•
•
•
•
•
Dihybrid
Two characteristics, four phenotypes
Colour and Size
A = red, a = green S = short, s = long
Red short
Non linkage
AaBb x AaBb 4 phenotypes 9:3:3:1 Red long
Green short
AaBb x aabb
4 phenotypes 1:1:1:1
Green long
Linkage
AaBb x AaBb 2 phenotypes 3:1 Red short (A+B)
Green long (a+b)
AaBb x aabb
2 phenotypes 1:1
0R
Red long (A+b)
Green short (a+B)