Bell Work • What types of offspring can be made from a mouse who is heterozygous for having whiskers (no whiskers is recessive) and grey (white and black are incompletely dominant) and a white mouse who has no whiskers.
Intro to Biology – Lecture 52 Double Punnett Squares Genotype and Phenotype
Making a Double Punnett Square • Step 1: Decide which possible allele combinations can be passed from each parent to the offspring. • Step 2: Take those 4 allele combinations and set them at the top (parent 1) and sides (parent 2) of the double punnett square. • Step 3: Cross the allele combinations in the punnett square to get the possible genotypes of the children. • Step 4: Determine the phenotype of each gene trait for the children.
TODAY • Step 4: Determine the phenotype of each gene trait for the children.
Additional Info for Step 4 • You must determine the phenotype for EACH letter combination in the double genotype.
Reminder Info for Step 4 • If the gene is dominant/recessive, the presence of 1 capital letter (in that two letter combination) shows the dominant trait. • If the gene is co-dominant or incomplete dominant, both letters must be the same for a gene trait to be present. If the two letters are different, the intermediate phenotype is shown.
Steps for Step 4 • Step 1 – Inventory all of the letter combinations. Make sure that you do not count two of the same letter combinations as different ones. • Step 2 – Count how many of each one is present. • Step 3 – Determine the phenotype for each of the first two letters in each pair of letter combinations. • Step 4 – Determine the phenotype for each of the last two letters in each pair of letter combinations. • Step 5 – Identify, then count the total of the genotypes that result in the same phenotype.
Example 1 • Mr. and Mrs. Mouse both had one copy of each allele for teeth (one normal copy, one vampire fangs copy). In that sense, they were hybrids for the vampire allele. Well, they are also hydrids for the fur allele – one copy of the fluffy fur allele, and one copy of the wiry fur allele. (Let's assume that fluffy fur is a recessive trait, like vampire fangs).
NW Nw nW nw NW Nw nW nw
Step 1 • NNWW • NNWw • NnWW • NnWw • NNWw • NNww • NnWw • Nnww • NnWW • NnWw • nnWW • nnWw • Nnww • nnWw • nnww
Step 2 • NNWW = 1 • NNWw, NNWw = 2 • NnWW, NnWW = 2 • NnWw, NnWw, NnWw, NnWw = 4 • NNww = 1 • Nnww, Nnww = 2 • nnWW = 1 • nnWw, nnWw = 2 • nnww = 1
Step 3 • 1 NNWW = Normal teeth • 2 NNWw = Normal teeth • 2 NnWW = Normal teeth • 4NnWw = Normal teeth • 1 NNww = Normal teeth • 2 Nnww = Normal teeth • 1 nnWW = Vampire teeth • 2 nnWw = Vampire teeth • 1 nnww = Vampire teeth
Step 4 • 1 NNWW = Normal teeth, Wiry • 2 NNWw = Normal teeth, Wiry • 2 NnWW = Normal teeth, Wiry • 4NnWw = Normal teeth, Wiry • 1 NNww = Normal teeth, fluffy • 2 Nnww = Normal teeth, fluffy • 1 nnWW = Vampire teeth, wiry • 2 nnWw = Vampire teeth, wiry • 1 nnww = Vampire teeth, fluffy
Step 5 • 1 NNWW = Normal teeth, Wiry • 2 NNWw = Normal teeth, Wiry • 2 NnWW = Normal teeth, Wiry • 4NnWw = Normal teeth, Wiry • 1 NNww = Normal teeth, fluffy • 2 Nnww = Normal teeth, fluffy • 1 nnWW = Vampire teeth, wiry • 2 nnWw = Vampire teeth, wiry • 1 nnww = Vampire teeth, fluffy
Step 5 Totals • 9 Normal teeth, Wiry = 1 NNWW, 2 NNWw, 4 NnWW, and 2 NnWw • 3 Normal teeth, fluffy = 1 NNww and 2 Nnww • 3 Vampire teeth, wiry = 1 nnWW and 2 nnWw • 1 Vampire teeth, fluffy = nnww
Practice 1 • Cross a black, long haired mouse with a white, short haired mouse. Black coat color and long hair are dominant. Both animals are homozygous for all their traits.
Phenotypes
Practice 2 • Cross a blue, fat aardvark with a white, skinny aardvark. Blue and fat are dominant. The blue, fat aardvark is heterozygous for its traits.
Phenotypes
Practice 3 • Cross a heterozygous white homozygous spotted dog with a black, heterozygous non-spotted dog. White and non-spotted are dominant.
Phenotypes
