Module 6 Lesson 2 - 3

Monohybrid Punnett Squares Examples

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Example 1

What are the possible phenotypes and genotypes of the F1 Generation (the first generation of offspring) if the parents are heterozygous for brown hair? Recall that brown hair is dominant over blonde hair.

Solution

1. Let B represent the dominant allele for brown hair. Let b represent the recessive allele for blonde hair.

   Each parent is heterozygous for brown hair; therefore, the allele pair is made of different alleles, Bb.

   P1 Generation = Bb x Bb
   
   

2. Construct a Punnett square for a single trait.
   1. Draw a table that consists of 3 rows and 3 columns.
   2. Place one allele from each parent in the first row and the first column.
   
   

   	First Allele from Parent 1 B	Second Allele from Parent 1 b
   First Allele from Parent 2 B
   Second Allele from Parent 2 b




3. Complete the table.

   1. Considering Parent 1's genotype ( Bb), write one allele in each box in the second and third row.
   2. Considering Parent 2’s genotype ( Bb), write one allele in each box in the second and third row.
   
   




	B	b
B	BB	Bb
b	Bb	bb




4. The phenotypes for this first generation of offspring (F1) are brown hair (BB and Bb) and blonde hair (bb) in a ratio of 3:1, respectively.

   The genotypes for this cross are 1 BB : 2 Bb : 1 bb.
   

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Examine the sample on how to build a Punnett square on the bottom of page 590 of your textbook. Notice how the gametes of one parent are written along the top of the Punnett square and the gametes of the other parent down the side. That way, the square can be completed by carrying the appropriate letter across or down as shown.

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Example 2

True breeding yellow pod plants are crossed with true breeding green pod plants. Write the genotypes and phenotypes of the F1 and F2 generations. Determine the phenotypic ratio of the F2 plants. (Green pods are dominant to yellow pods.)

Solution

1. Let G represent the dominant allele for green pods. Let g represent the recessive allele for yellow pods.

   One parent is true breeding green pod plant; therefore, the allele pair is made of the same alleles, GG. Because the second parent is true breeding yellow pod plant, the allele pair is made of same alleles, gg. 

   P1 Generation = GG x gg



2. Construct a Punnett square for the P1 generation and complete the table.
   

   	First Allele from Parent 1 G	Second Allele from Parent 1 G
   First Allele from Parent 2 g	Gg	Gg
   Second Allele from Parent 2 g	Gg	Gg



3. F1 Generation Genotype = 100% Gg
   F1 Generation Phenotype = 100% Green Pods
   



4. F1 Generation Cross = Gg x Gg
   

   Construct a Punnett square for the F1 generation and complete the table. 

   
   

   	G	g
   G	GG	Gg
   g	Gg	gg




5. F2 Generation Genotypes = 1 GG : 2 Gg : 1 gg = 1 : 2 : 1
   

   F2 Generation Phenotypes = 3 Green Pods : 1 Yellow Pods