Scenario: In dogs, coat color and tail length are determined by two separate genes. Coat color is controlled by the B gene, where B represents the dominant allele for black coat color, and b represents the recessive allele for brown coat color. Tail length is controlled by the T gene, where T represents the dominant allele for long tails, and t represents the recessive allele for short tails. Parent 1 is heterozygous for both coat color (Bb) and tail length (Tt), exhibiting black coat color and long tail. Parent 2 is homozygous recessive for both traits (bb tt), displaying brown coat color and short tail. Task 1. Complete Punnett Square Task 2. What is the expected phenotypic ratio of the offspring? Task 3. What is the probability of producing an offspring with the genotype BbTt?
Answer (1)
GivenParent 1: BbTt (heterozygous for both traits)Parent 2: bbtt (homozygous recessive for both)Task 1: Complete the Punnett SquareWe cross: BbTt × bbttUse the FOIL method to get the gametes.Parent 1 (BbTt):Gametes: BT, Bt, bT, btParent 2 (bbtt):Gametes: bt, bt, bt, bt (only one type, all recessive) bt bt bt btBT BbTt BbTt BbTt BbTtBt Bbtt Bbtt Bbtt BbttbT bbTt bbTt bbTt bbTtbt bbtt bbtt bbtt bbttTask 2: Expected Phenotypic RatioLet’s break it down:Genotype PhenotypeBbTt Black coat, Long tailBbtt Black coat, Short tailbbTt Brown coat, Long tailbbtt Brown coat, Short tailCount:BbTt = 4Bbtt = 4bbTt = 4bbtt = 4Total = 16Phenotypic ratio:Black, Long tail = 4Black, Short tail = 4Brown, Long tail = 4Brown, Short tail = 4So the ratio is: 1 : 1 : 1 : 1Task 3: Probability of BbTtThere are 4 out of 16 offspring that are BbTt.
