Mendel's Laws of Heredity and Dihybrid Crosses: Understanding Genetic Inheritance

1. Genetics Mendel’s Laws of Heredity and Dihybrid Crosses

2. Meiosis (Review) • Reproductive cells produce haploid gametes (egg or sperm) • During meiosis, the homologous chromosomes separate into different cells • Therefore, each gamete receives only one allele for each type of trait • When the gametes (egg and sperm) combine to form an offspring, 2 alleles are inherited (1 from the egg and 1 from the sperm) for each trait. • ie: The Law of Segregation

3. Mendel’s 1st Law • The Law of Segregation • Organisms inherit two copies of each gene (one from each parent) • Organisms donate only one copy of each gene in their gametes. During meiosis the two copies of each gene separate (“segregate”) during gamete formation.

4. Mendel’s 1st Law – Explained • The Law of Segregation means that if your mom was heterozygous for a particular trait, she could have given you the dominant allele or she could have given you the recessive allele. You only get one from your mom. • (Same thing regarding alleles from your dad.)

5. Mendel’s 2nd Law • The Law of Independent Assortment • Different traits are determined by different genes. • Genes for different traits can be inherited separately from other traits.

6. Mendel’s 2nd Law – Explained • The Law of Independent Assortment means that just because you are tall does not mean you have to have brown eyes… Your alleles for a certain trait are not necessarily linked to your alleles for a different trait. They can get shuffled around during meiosis (gamete formation).

7. Mendel’s 2nd Law – Example In Fruit flies: • Red eyes are dominant to white eyes • Having wings is dominant to no wings • R = red eyes r = white eyes • N = wings n = missing wings

8. Mendel’s 2nd Law – Example • R = red r = white • N = wings n = missing wings • What is the genotype of a heterozygous red eyed fly? • What is the genotype of a heterozygous normal winged fly?

9. Mendel’s 2nd Law – Example • R = red r = white • N = wings n = missing wings • What gametes will a heterozygous red eyed fly produce? • What gametes will a heterozygous normal winged fly produce?

10. Mendel’s 2nd Law – Example • R = red r = white • N = wings n = missing wings • What is the genotype of a fly that is heterozygous for eye color and for wing type?

11. Mendel’s 2nd Law – Example • R = red r = white • N = wings n = missing wings • What gametes will a fly heterozygous for eye color and wing type produce? (Hint: Use FOIL method.)

12. Dihybrid Cross • A dihybrid cross examines the inheritance of two traits at the same time. • We will now examine a dihybrid cross between two flies heterozygous for both eye color and wing type.

13. Dihybrid Cross – Step #1 1) Assign allele letters to represent the traits R = red r = white N = wings n = missing wings

14. Dihybrid Cross – Step #2 & 3 2) Determine genotypes of parents. 3) Determine the gametes they will produce.

15. Dihybrid Cross – Step #4 4) Fill in your Dihybrid Punnett Square. Place gametes on outside and F1s on inside just like before. rN RN Rn rn RN Rn rN rn

16. Dihybrid Cross – Step #4 4) Fill in your Dihybrid Punnett Square. Place gametes on outside and F1s on inside just like before. rN RN Rn rn RN Rn rN rn

17. Dihybrid Cross – Step #4 4) Fill in your Dihybrid Punnett Square. Place gametes on outside and F1s on inside just like before. rN RN Rn rn RN Rn rN rn

18. Dihybrid Cross – Step #4 4) Fill in your Dihybrid Punnett Square. Place gametes on outside and F1s on inside just like before. rN RN Rn rn RN Rn rN rn

19. Dihybrid Cross – Step #4 4) Fill in your Dihybrid Punnett Square. Place gametes on outside and F1s on inside just like before. rN RN Rn rn RN Rn rN rn

20. Dihybrid Cross – Step #4 4) Fill in your Dihybrid Punnett Square. Place gametes on outside and F1s on inside just like before. rN RN Rn rn RN Rn rN rn

21. Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. rN RN Rn rn RN Rn rN rn

22. Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. rN RN Rn rn Red w/wings = Red w/o wings = White w/wings = White w/o wings = RN Rn rN rn

23. Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. rN RN Rn rn Red w/wings = Red w/o wings = White w/wings = White w/o wings = RN Rn rN rn

24. Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. rN RN Rn rn Red w/wings = Red w/o wings = White w/wings = White w/o wings = RN Rn rN rn

25. Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. rN RN Rn rn Red w/wings = Red w/o wings = White w/wings = White w/o wings = RN Rn rN rn

26. Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. rN RN Rn rn Red w/wings = Red w/o wings = White w/wings = White w/o wings = RN Rn rN rn

27. Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. rN RN Rn rn Red w/wings = 9 Red w/o wings = White w/wings = White w/o wings = RN Rn rN rn

28. Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. rN RN Rn rn Red w/wings = 9 Red w/o wings = 3 White w/wings = White w/o wings = RN Rn rN rn

29. Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. rN RN Rn rn Red w/wings = 9 Red w/o wings = 3 White w/wings = 3 White w/o wings = RN Rn rN rn

30. Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. rN RN Rn rn Red w/wings = 9 Red w/o wings = 3 White w/wings = 3 White w/o wings = 1 RN Rn rN rn

31. Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. 9 : 3 : 3 : 1 rN RN Rn rn Red w/wings = 9 Red w/o wings = 3 White w/wings = 3 White w/o wings = 1 RN Rn rN rn