Presentation Slides

Mendel conducted both monohybrid and dihybrid
crosses
Dihybrid cross – initial parents differ in two traits
Mendel’s breeding scheme for a dihybrid cross
Figure 3-3 step 3
R/R;y/y
R/R;Y/Y
r/r;Y/Y : genes are in trans, as above
r/r;y/y : genes are in cis
Both of these parental combinations = the same F1
Mendel’s dihybrid crosses
round = 315 + 108 = 423
wrinkled = 101 + 32 = 133
Ratio = 3.2 : 1
yellow = 315 + 101 = 416
green = 108 + 32 = 140
Ratio = 3.0 : 1
dihybrid ratio = two
superimposed monohybrid
ratios
gene pairs assort independently
Figure 3-3 step 5
9 : 3 : 3 : 1 ratio
How Two 3:1 Monohybrid Ratios
Produce a 9:3:3:1 Dihybrid Ratio
Monohybrid Ratios:
Seed Shape: Round = 3/4, Wrinkled = 1/4
Seed Color: Yellow = 3/4, Green = 1/4
Resulting Dihybrid Phenotypic Ratios:
Round Yellow
3/4 x 3/4 = 9/16
Round Green
3/4 x 1/4 = 3/16
Wrinkled Yellow
1/4 x 3/4 = 3/16
Wrinkled Green
1/4 x 1/4 = 1/16
Even segregation and
independent assortment of
alleles from two genes
underlies a 9 : 3 : 3 : 1 ratio
From each gene pair, 1/2 of
gametes will receive a
dominant allele and 1/2 a
recessive allele
Figure 3-4 step 4
1/2 will be R (vs. r ) and 1/2
will be Y (vs. y )
1/2 x 1/2 = 1/4 will be R;Y
Other gametic combinations
will be R;y, r;Y and r;y, each
1/4 of the total
A punnett square can
illustrate the genotypes
underlying a 9:3:3:1 ratio
Figure 3-4 step 4
1/4 male R;Y
X
1/4 female R;Y
= 1/16 R/R;Y/Y
Punnett square illustrating
the genotypes underlying a
9 : 3 : 3 : 1 ratio
Note:
the ratio of yellow
to green is 3:1
Figure 3-4 step 10
the ratio of round
to wrinkled is 3:1
Chromosomal Basis for Independent Assortment:
genes on different chromosomes assort independently
Dominant phenotypes are represented
by partial genotypes
(ex. A-, - indicates either dominant or recessive
allele may be present)
• Yellow pea could be either Y/Y or Y/y
• Yellow can be designated as Y/- or Y• This is called a partial genotype
• Yellow Round could be any of four
genotypes
• Y/Y ; R/R Y/y ; R/R Y/Y ; R/r Y/y ; R/r
• For simplicity the partial genotype is
used: Y/- ; R/- or Y- ; R- or Y- R-
The ratio of partial genotypes is equal
to the ratio of phenotypes
A- ; BA- ; bb
aa ; Baa ; bb
3/4 A- x 3/4 B3/4 A- x 1/4 bb
1/4 aa x 3/4 B1/4 aa x 1/4 bb
= 9/16
= 3/16
= 3/16
= 1/16
When genes assort independently
phenotypic and genotypic dihybrid ratios
can be calculated as the product of their
monohybrid ratios
Also see: http://physics.scsu.edu/~dscott/gen/MendelianIntro.html
Use of a Punnett square in dihybrid
crosses is cumbersome
Figure 3-4 step 10
What proportion would be RRyy ?
Figure 3-4 step 10
What proportion would be RRyy ?
1/4 RR x 1/4 yy = 1/16 RRyy
Figure 3-4 step 10
What proportion would be RrYy ?
Figure 3-4 step 10
What proportion would be RrYy ?
1/2 Rr x 1/2 Yy = 1/4 RrYy
Figure 3-4 step 10
What proportion would be R- yy ?
Figure 3-4 step 10
What proportion would be R- yy ?
3/4 R- x 1/4 yy = 3/16 R- yy
Figure 3-4 step 10
Calculation of Phenotypic Ratios
for Crosses with Multiple Genes
Purple, Yellow, Tall, Round
CC
YY
TT
white, green, short, wrinkled
RR
cc
yy
tt
rr
F1 = all Purple, Yellow, Tall, Round
Cc
Yy
Tt
Rr
F1 shows the dominant characters, will be heterozygous for all
Calculation of Phenotypic Ratios
for Crosses with Multiple Genes
Purple, Yellow, Tall, Round
CC
YY
TT
white, green, short, wrinkled
RR
cc
yy
tt
rr
F1 = all Purple, Yellow, Tall, Round
Cc
Yy
Tt
Rr
What fraction of the F2 will be Purple, green, short, Round?
Answer = 3/4 x 1/4 x 1/4 x 3/4 = 9/256
What fraction will be CC Yy Tt rr ?
Answer = 1/4 x 1/2 x 1/2 x 1/4 = 1/64
A dihybrid testcross
= F1 crossed to
double recessive.
It produces a 1:1:1:1
phenotypic ratio
The parental
combinations
• (round, green and
wrinkled, yellow)
equal the
recombinants
• (round, yellow
and wrinkled,
green)
Gametes =
R;Y R;y r;Y r;y
Gametes = r;y
Genetic and Chromosomal Basis for 1:1:1:1 Testcross Ratio
Testcross:
Phenotypes of
testcross progeny
reflect the genotypes
of gametes from
heterozygous parent
Figure 3-13 step 4
Dom/Dom, rec/rec = Parental
Dom/rec, rec/Dom = Recombinant
Independent assortment – cis or trans
Parental genes in cis
Parental genes in trans
p2
F1 produces 4 different gametic genotypes in equal
frequency for cis or trans parental combinations:
AB, ab (cis) = Parental
aB, Ab (trans) = Recombinant
aB, Ab (trans) = Parental
AB, ab (cis) = Recombinant
Linked genes do not assort independently
Linkage:
 genes on same
chromosome
 reduces the
frequency of
recombinants
Figure 4-8 step 2
Linked gene combinations will tend to be
inherited together. F1 will produce 4 different
gametic genotypes in unequal frequency:
mostly parental combinations = AB, ab
fewer recombinants = Ab, aB
Linkage Reduces the Frequency of Recombinants
Parentals in cis
Phenotypes of
testcross progeny
reflect the genotypes
of the F1 gametes
cis combinations
trans combinations
Linkage Reduces the Frequency of Recombinants
Parentals in trans
Phenotypes of
testcross progeny
reflect the genotypes
of the F1 gametes
trans combinations
cis combinations
Introduction to Polygenic Inheritance
Some traits show
intermediate inheritance
in heterozygotes due to
incomplete dominance
This is incomplete
dominance
More than one incompletely
dominant trait may control a trait.
This in combination with
environmental effects on gene
expression can lead to continuous
phenotypic distribution.
Inheritance of incompletely dominant traits follows
Mendelian principles
Petal color in snapdragons:
Red
F1
F2
White
Pink
1/4 Red
1/2 Pink
1/4 White
Pink F1 consistent with blending -
These results are not consistent
with blending
Inheritance of incompletely dominant traits follows
Mendelian principles
Red
RR
White
rr
F1
Pink Rr
F2
1/4 Red RR
1/2 Pink Rr
1/4 White rr
Incomplete dominance: Mendelian F2 genotypic ratios,
Heterozygote phenotype intermediate between homozygotes
A Common Mistake
(I found this on the internet):
Different letters may not
be used for different
alleles of the same gene
Also: RR, RR’, R’R’
Archaic, not commonly
used because:
R and R’ both imply
active pigment alleles,
white results from a null
allele
“Pink” appearance
is the result of less
red pigment on an
unpigmented
background – a
dilution effect
It is not due to
modification of red
pigment to pink
Polygenic Inheritance
• Polygenes – active alleles of two or more
incompletely dominant genes all have the same
effect on phenotype
• This is also called quantitative inheritance
• Each active allele produces one “dose” of
phenotypic determinant (ex. = pigment)
• with a dihybrid cross, gene A and gene B:
• AABB = 4 doses, AaBb or AAbb = 2 doses
• Since both genes have the same effect they are
designated by subscripts: R1 or r1, R2 or r2
• Dihybrid cross =
• R 1R 1 R 2R 2
r1r1 r2r2
R1r1R2r2
• same as: AABB
aabb
AaBb
Polygenes in progeny of a dihybrid self
2 doses
= R 1R 2
Figure 3-15 step 1
1 dose =
R1r2 or r1R2
0 doses
= r 1 r2
Polygenes in progeny of a dihybrid self
Figure 3-15 step 5
Polygenes in progeny of a dihybrid self
Explanation for combinations of R1 or R2
alleles: R1 or R2 = active allele = +
r1 or r2 = inactive allele = -
All R1, R2
=
one r1 or r2
any position
two r1 or r2
any position
++++
+++++-+
+-++
-+++
++---++
+-+-+-+
+--+
-++-
=
1/16 of the total
1/16 each
4 combinations
4/16 of the total
1/16 each
6 combinations
6/16 of the total
For polygenic
traits, the
frequency of a
phenotype
depends on
how many
combinations of
alleles will
result in that
phenotype
16ths with "+" Alleles
Phenotypic Distribution
88
Stronger phenotypic expression
66
44
22
00
0
+
0+
1 + 2+
2+ 3
+ 4+
4+
1+
3+
Number of "+" Alleles
For polygenic
traits, the
frequency of a
phenotype
depends on
how many
combinations of
alleles will
result in that
phenotype
64ths with “+” alleles
20
15
6
1
Additive
polygenic
model for
three genes
with two
alleles each =
6 max active
alleles
15
6
1
0 1 2 3 4 5 6
Number of “+” alleles
For polygenic
traits, genotypic
expression may
be altered by
environmental
effects that
smooth the
phenotypic
distribution into
a continuum.