Dihybrid Cross Experiments
Objectives:
After completing this section, you should:
1. Apply Mendel's Principle of Independent
Assortment to the description of an individuals genotype or the probability of
the inheritance of trait combinations in a specific cross.
2. Recognize the inheritance patterns of two independently inherited genes.
3. Identify the gametes produced by parents for pairs of genes.
4. Design experiments that will test if two traits are controlled by independent
gene pairs.
Key Terms:
Principle of Independent Assortment
dihybrid
gamete
testcross
Online Lesson(s):
The Dihybrid Cross and Independent Assortment
(http://plantandsoil.unl.edu)
Mendel’s Result from Crossing Peas
Trait
Seed form
Cross
Round X Wrinkled
F1
All round
F2 number
5474 Round
1850 Wrinkled
F2 ratio
2.96 to 1
Cotyledon
Color
Yellow X Green
All yellow
6022 Yellow
2001 Green
3.01 to 1
Seed coat
Color*
Gray X White
All gray
705 Gray
224 White
3.15 to 1
Pod form
Inflated X Constricted
All inflated
882 Inflated
299 Constricted
2.95 to 1
Pod color
Green X Yellow
All green
428 Green
152 Yellow
2.82 to 1
Flower
position
Axial X Terminal
All axial
651 Axial
207 Terminal
3.14 to 1
Stem
Length
Tall X Short
All tall
787 Tall
277 Short
2.84 to 1
*Gray seed coat also had purple flowers, White seed coat had white flowers.
Selfing Dominant F2’s to produce F3 rows gave the following results:
F2 type
Round seed
372
Mixed rows
True breeding
193
1.93 to 1
Yellow cotyledon
353
166
2.13 to 1
Gray seed coat
64
36
Inflated Pod
71
29
1.78 to 1
.
2.45 to 1
Green Pod
60
40
1.50 to 1
Axial flower
67
33
2.03 to 1
Tall plant
72
28
2.57 to 1
Average ratio to heterozygote F2 to homozygote F2 was 2.06 to 1
ratio
Mendel’s Explanation
1. “Unit characters” Trait considered is height, tall vs. short. Difference is
assumed to be due to one gene pair.
2. Segregation
a. “Traits are determined by particulate factors.”
Illustrate with symbols
T = gene for tall
t = gene for short
b. “Genes are in pairs in somatic or body cells.”
Tall parent = TT,
short parent = tt
c. “When gametes are formed, members of a pair separate,
member to a gamete.”
TT parent gives T gametes
tt parent gives t gametes
d. “Gametes unite in fertilization to restore the double number of
genes.”
T egg + t sperm = Tt zygote (F1)
3. Dominance. “If T is present, t is not expressed.”
F1, Tt is tall
T is expressed in F1
T is masked F1
The Principal of Segregation: During the formation of gametes, the paired
elements separate and segregate randomly such that each gamete receive one or
the other element.
Mendel’s Dihybrid Cross Experiment
Pattern of inheritance when more than one pair of alleles is considered.
round, yellow seeds
X
wrinkled, green seeds
F1
(self)
F2
Results:
F1 Phenotype: all round, yellow
F2 (as follows)
Phenotype
round, yellow
round, green
wrinkled, yellow
wrinkled, green
Frequency
315
108
101
32
Fraction
9/16
3/16
3/16
1/16
Genotype
R_Y_
R_yy_
rrY_
rryy
(to extent
known)
F2 Ratio of each trait separately:
round
315 + 108 = 423
wrinkled
101 + 32 = 133
yellow
315 + 101 = 416
3 round : 1 wrinkled
3 yellow : 1 green
green
108 + 32 = 140
Mendel’s calculation:
Recognized the fractions obtained in each phenotypic class in the F2 could be
obtained by simple multiplication as follows:
(3/4 round + 1/4 wrinkled) X (3/4 yellow + 1/4 green) =
9/16 round, yellow 3/16 round, green
3/16 wrinkled, yellow 1/16 wrinkled, green
Principle of Independent Assortment: Segregation of the members of a pair
of alleles is independent of the segregation of other pairs during the processes
leading to the formation of the reproductive cells.
Independent Assortment of gene pairs
1)
As many different kinds of gametes as possible gene combinations (see above)
2)
All gametes in equal frequencies
3)
Once made, male and female gametes unite at random
Mendel’s Method
1/4 RY
1/4 Ry
1/4 rY
1/4 ry
1/4 RY
1/4 Ry
1/4 rY
1/4 ry
RRYY 1/16
RRYy 2/16
RrYY 2/16
RrYy 4/16
RRyy 1/16
Rryy 2/16
rrYY 1/16
rrYy 2/16
rryy 1/16
round, yellow
315/556
round, green
108/556
wrinkled, yellow
101/556
wrinkled, green
32/556
Punnett’s Method
1/4 RY
1/4 RY
1/4 Ry
RrYy
1/4 rY
RrYy
RrYy
1/4 ry
RrYy
RRYY
1/4 Ry
1/4 ry
1/4 rY
Testcross
Cross a F1 plant with a wrinkled, green plant:
RrYy X rryy
Results:
progeny phenotypes
round, yellow
round, green
wrinkled, yellow
wrinkled, green
observed freq.
55
51
49
52
207
Do these results make sense with respect to Mendel’s theories?
(Are the observed results similar to the expected results?)
Check with Punnett square:
Gametes expected
from rryy
Yes, the expected results are very similar to the observed results.
Therefore, Mendel’s theories hold in this experiment.
Mendel’s Principles:
What kind of cross?
Segregation
Independent assortment::
Genotype: RrYyTt
How many kinds of gametes?
List them:
Do Not Try This at Home!!!
Problem:
Cross an F2 horned, black, white-faced with a horned, red, colored-face.
What is the probability of a progeny that is horned, red, and colored-face?
Solution:
If the alleles at each locus assort independently (Principle of Independent Assortment),
then consider each locus separately. That is, draw a Punnett square for each locus, with
the genotypes of the F1 gamete along the margins and the F2 genotypes inside.
H
H
h
B
b
h
HH
polled
Hh
polled
Hh
polled
hh
horned
B
b
BB
black
Bb
black
Bb
black
bb
red
H
The genotype of both horned individuals must
be hh; therefore, the probability of a horned
progeny = 1 x 1 = 1
The black F2 could be BB (probability =
1/3, why?) or Bb (probability = 2/3, why?).
The red F2 could only be bb. The progeny
could be Bb (probability = 1/3 x 1 = 1/3) or
bb (probability = 2/3 x 1 = 2/3)
h
H
HH
Wh face
Hh
Wh face
h
Hh
Wh face
hh
Col face
The white-faced F2 could be HH
(probability = 1/3) or Hh (probability = 2/3)
The colored face F2 could only be hh. The
progeny could be Hh (probability = 1/3 x 1 =
1/3 or hh (probability = 2/3 x 1 = 2/3)
Putting this altogether by multiplication (because what happens at each locus is an
independent event),
Probability of a progeny that is horned, red and colored face = 1 x 2/3 x 2/3 = 4/9.
Problem:
Cross an F2
What is the probability of a progeny that is…………………………?
Solution:
???