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 Cross F1 F2 number F2 ratio
Seed form / Round X Wrinkled / All round / 5474 Round1850 Wrinkled / 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 Mixed rows True breeding ratio
Round seed / 372 / 193 / 1.93 to 1Yellow cotyledon / 353 / 166 / 2.13 to 1
Gray seed coat / 64 / 36 / 1.78 to 1
.
Inflated Pod / 71 / 29 / 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
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)
(to extent
Phenotype Frequency Fraction Genotype known)
round, yellow 315 9/16 R_Y_
round, green 108 3/16 R_yy_
wrinkled, yellow 101 3/16 rrY_
wrinkled, green 32 1/16 rryy
F2 Ratio of each trait separately:
round 315 + 108 = 423
3 round : 1 wrinkled
wrinkled 101 + 32 = 133
yellow 315 + 101 = 416
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 round, yellow 315/556
RRYy 2/16
RrYY 2/16
RrYy 4/16
RRyy 1/16 round, green 108/556
Rryy 2/16
rrYY 1/16 wrinkled, yellow 101/556
rrYy 2/16
rryy 1/16 wrinkled, green 32/556
Punnett’s Method
1/4 RY 1/4 Ry 1/4 rY 1/4 ry
RRYY / RrYy
/ RrYy
/ RrYy
RrYy
Testcross
Cross a F1 plant with a wrinkled, green plant:
RrYy X rryy
Results:
progeny phenotypes observed freq.
round, yellow 55
round, green 51
wrinkled, yellow 49
wrinkled, green 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?
SegregationIndependent 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
HHpolled / Hh
polled
Hh
polled / hh
horned
The genotype of both horned individuals must
be hh; therefore, the probability of a horned
progeny = 1 x 1 = 1
B b
BBblack / Bb
black
Bb
black / bb
red
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
HHWh face / Hh
Wh face
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:
???