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 Round
1850 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 1
Yellow 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?

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

HH
polled / 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

BB
black / 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

HH
Wh 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:

???