Module 6 Lesson 4 - 2

Law of Independent Assortment

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Read page 593

Mendel was interested in determining whether the movement of one trait's alleles affected the movement of another trait's alleles. To test this, he conducted crosses between plants that were true breeding for two traits. A cross made to observe two traits at the same time is called a dihybrid cross. If three traits were being analyzed at once, it would be a trihybrid cross, and so on.

Mendel took pure breeding yellow and round seed plants (YYRR) and crossed them with pure breeding green and wrinkled seed plants (yyrr). The resulting offspring were all yellow and round. Then, he self-crossed the F1 generation to obtain the F2 generation. The F2 generation had four phenotypes and resulted in the following ratio: 9 yellow, round : 3 green, round : 3 yellow, wrinkled : 1 green, wrinkled.

Why was this result so different from the 3 : 1 ratios of the monohybrid crosses? Mendel realized that the dihybrid cross ratio was similar to the results from combining two monohybrid crosses at random. When Mendel looked at the seed surface characteristic from the F2 generation, he saw 12 round and 4 wrinkled seeds, which can be reduced to a ratio of 3 : 1. Similarly, when looking at seed colour, he saw 12 yellow and 4 green seeds, which also can be reduced to a ratio of 3 : 1. He realized that the ratio was the same as the results obtained from the monohybrid crosses.

Mendel concluded that the movement of alleles for different genes did not affect each other. This led him to propose his second law, The Law of Independent Assortment, which states that different gene pairs assort independently in gamete formation. 

Read page 593 of your text to learn about this law and about how to complete a large, 16-square Punnett square to track dihybrid crosses.

Dihybrid Punnett Squares (Two Traits)

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Dihybrid crosses involve two different traits (two different genes) located on separate chromosomes. What happens on one chromosome has no effect on the other chromosome. Because of dealing with two different traits, the dihybrid Punnett square must be much larger than a monohybrid Punnett square.

How to Solve Dihybrid Punnett Squares

Example

A pure breeding plant for both inflated pod shape and round seed shape is crossed with a plant that is homozygous for both recessive traits. (Inflated pods are dominant to pinched pods. Round seeds are dominant to wrinkled seeds.) Determine the genotype and phenotype ratios of the F2 generation.

1. From the information about the trait, determine what is dominant and what is recessive.

   Inflated > Pinched  and  Round > Wrinkled

   • Use a capital letter for dominant trait (colour, shape, etc.).
     

     I = inflated and i = pinched

     R = round and r = wrinkled

   • Use the lower case letter of the dominant trait for the recessive trait.
     
   • Write the dominant allele first for a trait. aA is the same as Aa, but the capital letter is always written first.
   • Make the P1 = ______ x  _____ generation.
     
   
   

   P1 Generation = IIRR x iirr

   Keyword	Definition	Genotype
   homozygous	the same alleles, the same letters	RR or rr
   heterozygous	different alleles, different letters	Rr
   pure, or true breeding	homozygous parents	RR or rr
   recessive	homozygous recessive	rr
   x	mating, fertilization, pollination, crossed with

   
   
   
2. Divide the parental genotypes into its gametes (separate alleles).  The Law of Independent Segregation applies here: the allele pair of a gene is separated to form the gametes. Use the FOIL method.
   
   

   IIRR has the gamete IR.

   iirr has the gamete ir.

   
   
   
3. Complete a Punnett square for the P1 cross.
   
   

   	IR	IR	IR	IR
   ir	IiRr	IiRr	IiRr	IiRr
   ir	IiRr	IiRr	IiRr	IiRr
   ir	IiRr	IiRr	IiRr	IiRr
   ir	IiRr	IiRr	IiRr	IiRr

   
   Notice how all the gametes are the same. This means the Punnett square can be simplified as this:
   
   

   	IR
   ir	IiRr

   
   
   

   All 16 squares are identical and resulted in IiRr genotype. (When only one possible gamete comes from each parent, you need to complete only the first square.)

   F1 Generation Genotype = 100% IiRr     F1 Generation Phenotype = 100% Inflated, Round
   



4. Make the F1 generation cross = _____ x _____
   
   

   F1 Generation = IiRr x IiRr
   



5. Form the gametes of the F1 parents using the FOIL method.
   
   

   Steps	Description	Gametes Formed
   	FIRST: Use the first allele of each trait.	IR
   	OUTSIDE: Use the outside alleles.	Ir
   	INSIDE: Use the inside alleles.	iR
   	LAST: Use the last alleles.	ir

   
   
   

   Place these gametes along the top row and left-most column of the Punnett square. 




6. Complete the Punnett square for the F1 generation to determine the genotypes and phenotypes of the F2 generation.
   
   

   	IR	Ir	iR	ir
   IR	IIRR	IIRr	IiRR	IiRr
   Ir	IIRr	IIrr	IiRr	Iirr
   iR	IiRR	IiRr	iiRR	iiRr
   ir	IiRr	Iirr	iiRr	iirr

   
   

   A heterozygous dihybrid cross IiRr x IiRr will result in four phenotypes in a ratio of 9 : 3 : 3 : 1. The F2 generation genotype ratio is 1 IIRR (homozygous dominant) : 2 IIRr : 2 IiRR : 4 IiRr (parental genotype) : 1 IIrr : 2 Iirr : 1 iiRR : 2 iiRr : 1 iirr (homozygous recessive).
   

   
   

   Blue	Inflated Round	9/16
   Green	Inflated Wrinkled	3/16
   Orange	Pinched Round	3/16
   White	Pinched Wrinkled (Homozygous Recessive)	1/16

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Sample Problem 1 

What are the possible phenotypes of the F1 Generation if the parents are heterozygous for brown hair and brown eyes? Brown hair is dominant over blonde hair and brown eyes are dominant over blue eyes.