Genetics Mendels Laws of Heredity and Dihybrid Crosses
Genetics Mendel’s Laws of Heredity and Dihybrid Crosses
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
Mendel’s 1 st 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.
Mendel’s 1 st 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. )
Mendel’s 2 nd Law • The Law of Independent Assortment – Different traits are determined by different genes. – Genes for different traits can be inherited separately from other traits.
Mendel’s 2 nd 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).
Mendel’s 2 nd Law – Example In Fruit flies: • Red eyes are dominant to white eyes • Having wings is dominant to no wings • R = red eyes • N = wings r = white eyes n = missing wings
Mendel’s 2 nd Law – Example • R = red • N = wings r = white n = missing wings • What is the genotype of a heterozygous red eyed fly? • What is the genotype of a heterozygous normal winged fly?
Mendel’s 2 nd Law – Example • R = red • N = wings r = white n = missing wings • What gametes will a heterozygous red eyed fly produce? • What gametes will a heterozygous normal winged fly produce?
Mendel’s 2 nd Law – Example • R = red • N = wings r = white n = missing wings • What is the genotype of a fly that is heterozygous for eye color and for wing type?
Mendel’s 2 nd Law – Example • R = red • N = wings r = white n = missing wings • What gametes will a fly heterozygous for eye color and wing type produce? (Hint: Use FOIL method. )
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.
Dihybrid Cross – Step #1 1) Assign allele letters to represent the traits R = red N = wings r = white n = missing wings
Dihybrid Cross – Step #2 & 3 2) Determine genotypes of parents. 3) Determine the gametes they will produce. Parents Gametes Rr. Nn RN Rn r. N rn
Dihybrid Cross – Step #4 4) Fill in your Dihybrid Punnett Square. Place gametes on outside and F 1 s on inside just like before. RN RN Rn r. N rn
Dihybrid Cross – Step #4 4) Fill in your Dihybrid Punnett Square. Place gametes on outside and F 1 s on inside just like before. RN RN RRNN Rn r. N rn
Dihybrid Cross – Step #4 4) Fill in your Dihybrid Punnett Square. Place gametes on outside and F 1 s on inside just like before. RN Rn r. N rn RN RRNn Rr. NN Rr. Nn Rn r. N rn
Dihybrid Cross – Step #4 4) Fill in your Dihybrid Punnett Square. Place gametes on outside and F 1 s on inside just like before. RN Rn r. N rn RN RRNn Rr. NN Rr. Nn Rn r. N rn RRNn RRnn Rr. Nn Rrnn
Dihybrid Cross – Step #4 4) Fill in your Dihybrid Punnett Square. Place gametes on outside and F 1 s on inside just like before. RN Rn r. N rn RN RRNn Rr. NN Rr. Nn Rn r. N rn RRNn RRnn Rr. Nn Rrnn Rr. NN rr. Nn Rr. Nn
Dihybrid Cross – Step #4 4) Fill in your Dihybrid Punnett Square. Place gametes on outside and F 1 s on inside just like before. RN Rn r. N rn RN RRNn Rr. NN Rr. Nn Rn r. N rn RRNn RRnn Rr. Nn Rrnn Rr. NN Rr. Nn rr. NN rr. Nn Rrnn rr. Nn rrnn
Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. RN Rn r. N rn RN RRNn Rr. NN Rr. Nn Rn r. N rn RRNn RRnn Rr. Nn Rrnn Rr. NN Rr. Nn rr. NN rr. Nn Rrnn rr. Nn rrnn
Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. RN Rn r. N rn RN RRNn Rr. NN Rr. Nn Rn r. N rn RRNn RRnn Rr. Nn Rrnn Red w/wings = Red w/o wings = White w/o wings = Rr. NN Rr. Nn rr. NN rr. Nn Rrnn rr. Nn rrnn
Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. RN Rn r. N rn RN RRNn Rr. NN Rr. Nn Rn r. N rn RRNn RRnn Rr. Nn Rrnn Red w/wings = Red w/o wings = White w/o wings = Rr. NN Rr. Nn rr. NN rr. Nn Rrnn rr. Nn rrnn
Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. RN Rn r. N rn RN RRNn Rr. NN Rr. Nn Rn r. N rn RRNn RRnn Rr. Nn Rrnn Red w/wings = Red w/o wings = White w/o wings = Rr. NN Rr. Nn rr. NN rr. Nn Rrnn rr. Nn rrnn
Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. RN Rn r. N rn RN RRNn Rr. NN Rr. Nn Rn r. N rn RRNn RRnn Rr. Nn Rrnn Red w/wings = Red w/o wings = White w/o wings = Rr. NN Rr. Nn rr. NN rr. Nn Rrnn rr. Nn rrnn
Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. RN Rn r. N rn RN RRNn Rr. NN Rr. Nn Rn r. N rn RRNn RRnn Rr. Nn Rrnn Red w/wings = Red w/o wings = White w/o wings = Rr. NN Rr. Nn rr. NN rr. Nn Rrnn rr. Nn rrnn
Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. RN Rn r. N rn RN RRNn Rr. NN Rr. Nn Rn r. N rn RRNn RRnn Rr. Nn Rrnn Red w/wings = 9 Red w/o wings = White w/o wings = Rr. NN Rr. Nn rr. NN rr. Nn Rrnn rr. Nn rrnn
Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. RN Rn r. N rn RN RRNn Rr. NN Rr. Nn Rn r. N rn RRNn RRnn Rr. Nn Rrnn Red w/wings = 9 Red w/o wings = 3 White w/wings = White w/o wings = Rr. NN Rr. Nn rr. NN rr. Nn Rrnn rr. Nn rrnn
Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. RN Rn r. N rn RN RRNn Rr. NN Rr. Nn Rn r. N rn RRNn RRnn Rr. Nn Rrnn Red w/wings = 9 Red w/o wings = 3 White w/o wings = Rr. NN Rr. Nn rr. NN rr. Nn Rrnn rr. Nn rrnn
Dihybrid Cross – Step #5 5)Determine the gametes that give rise to the same phenotypes and write the phenotype ratio. RN Rn r. N rn RN RRNn Rr. NN Rr. Nn Rn r. N rn RRNn RRnn Rr. Nn Rrnn Red w/wings = 9 Red w/o wings = 3 White w/o wings = 1 Rr. NN Rr. Nn rr. NN rr. Nn Rrnn rr. Nn rrnn
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 r. N rn RN RRNn Rr. NN Rr. Nn Rn r. N rn RRNn RRnn Rr. Nn Rrnn Red w/wings = 9 Red w/o wings = 3 White w/o wings = 1 Rr. NN Rr. Nn rr. NN rr. Nn Rrnn rr. Nn rrnn
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