1. Genotypes & Phenotypes
Capital letters used for dominant allele
lowercase of the same letter used for recessive
ex. round seeds are dominant over wrinkled.
R = round, r = wrinkled
Genotype
the alleles for an individual, ex. for seeds:
- RR (homozygous round)
- rr (homozygous wrinkled)
- Rr (heterozygous round)
Phenotype
the expressed trait for an individual (what is seen) ex. for seeds:
- round
- wrinkled
Heterozygous genotypes (Rr) always have the phenotype of the dominant allele.

2. Plant Height Experiments
F1 genotype:
- 4 Tt (heterozygous)
F1 phenotype:
- 4 tall
F2 genotypes:
- 1 TT (homozygous dominant)
- 2 Tt (heterozygous)
- 1 tt (homozygous recessive)
- ratio 1:2:1
F2 phenotypes:
- 3 tall
- 1 short
- ratio 3:1 tt
P Generation
F1 Generation
F2 Generation
True-breeding parents
Tall = dominant = T Short = recessive = t TT
Hybrid offspring All heterozygous
all Tt
Monohybrid cross from F1 generation Tt
Tall
all Tall
short
Genotypes 1 TT : 2 Tt : 1 tt
Phenotypes 3 Tall : 1 short

3. Probability & Punnet Squares
Probability – the likelihood of an outcome Punnett square - used to determine the probabilities of allele combinations when the genotypes of the parents are known

4. Making Punnet Squares

5. Practice Problem #1
A geneticist crosses two pea plants. One of the plants is heterozygous for the dominant inflated pea pod trait, and the other plant has constricted pea pods. What would be the expected genotypic and phenotypic proportions of the offspring?
Given
inflated pod (I) - dominant
constricted pod (i) - recessive
II or Ii = Inflated pods
ii = constricted pods
heterozygous inflated x constricted = Ii x ii
Required
Punnet square
gametes from the cross
expected proportions of F1 plants with genotypes II, Ii, and ii
expected proportions of phenotypes (inflated pods and constricted)

6. Solution The Ii parent produces gametes I and i.
Paraphrase
1/2 of the offspring would be heterozygous (Ii) and 1/2 would be ii. Therefore, 1/2 of the offspring would have inflated pods, and 1/2 would have constricted pods.

7. Solutions #2- 4 2. Solution The Aa parent produces gametes A and a.
Paraphrase
½ the offspring would be heterozygous (Aa) and ½ would be aa. Therefore, ½ of the offspring would be brown, and ½ would be albino
3. Solution
The DD parent produces gametes D and D.
The Dd parent produces gametes D and d.
½ of the offspring would be heterozygous (Aa) and ½ would be homozygous (AA). Therefore, all of the offspring would have dimples.
4. Solution
The Tt parents produce gametes T and t.
½ of the offspring would be heterozygous (Tt), ¼ would be homozygous dominant (TT) and ¼ would be homozygous recessive (tt). Therefore, ¾ of the offspring would taste PTC, and ¼ would not. D DD d Dd T t TT Tt tt

8. Test Crosses how genotype can be determined from phenotype
breed an individual with the dominant phenotype with a homozygous recessive
If offspring are all dominant phenotype, the parent was homozygous dominant
If offspring are half dominant phenotype, and half recessive phenotype, the parent was heterozygous dominant.

9. Dihybrid Crosses
a cross between two individuals that differ in two traits
Mendel used to determine whether or not the inheritance of one characteristic influenced the inheritance of another
ex: Does pea shape influence the inheritance of pea colour?
Method:
choose plants homozygous for 2 traits (colour & shape)
crossbred 2 different ones to get dihybrid F1 generation
ex. round yellow seeds x wrinkled green
crossed two dihybrid F1 to get F2 (dihybrid cross)
Results:
all F1 were round yellow
F2 had phenotypic ratio of 9:3:3:1

10. F2 Generation
F1 Generation
P Generation

11. Flower location: A = Axial a = terminal
Dihybrid Crosses
State allele symbols
Follow steps below
Step
1. Draw a square with a 4 by 4 grid.
2. Consider all possible gametes produced by the first parent. (Use FOIL) Write the alleles for these gametes across the top of the square.
3. Consider all possible gametes produced by the second parent. Write the alleles for these gametes down the side of the square.
4. Complete the square by writing all possible allele combinations from the cross.
5. Determine the genotypic and phenotypic proportions of the offspring.
Example
Flower location: A = Axial a = terminal
Height:
T = Tall t = short

12. Practice Problem #1
In mice, the normal long-tail phenotype is dominant to the short-tail trait, and black coat colour is dominant to brown coat colour. If two long-tailed black mice, heterozygous for both traits, are mated, what proportion of their offspring will be brown with short tails?
Given
Long tail (L) is dominant. Short tail (l) is recessive.
LL or Ll - long tails, ll - short tails.
Black coat colour (B) is dominant. Brown coat colour (b) is recessive.
BB or Bb - black coat colour, bb - brown coat colour.
The cross is BbLl X BbLl
Required
Determine the gametes using FOIL.
make a Punnett square
find the proportion of offspring that will be brown with short tails

13. Solution • The BbLl parents produce gametes BL, Bl, bL, bl. Paraphrase
Mice with brown coats and short tails have the genotype bbll. 1/16 of the offspring will be brown with short tails.

14. Law of Independent Assortment
Mendel found that this phenotypic ratio (9:3:3:1) was always found in the offspring of such dihybrid crosses.
law states that the inheritance of alleles for one trait does not affect the inheritance of alleles for another trait.
Different pairs of alleles are passed to the offspring independently of each other
laws apply to all living things, not just plants
many, but not all, traits follow Mendel’s laws