Unordered events puppies black-white In a breed of dogs, black coat color is dom

Question

Unordered events puppies black-white In a breed of dogs, black coat color is dominant and white coat is recessive. Presume for this question there is only 1 gene controlling the coat color. If you have two heterozygous black dogs (i.e. Bb) and they mate what is the probability that you will have 4 out of 7 puppies that are black, thus 3 out of 7 pupples that are white? Note this is an unordered set of independent events, each puppy was independently fertilized. Give answer as a percentage out of OO not a decimal value. Round to the nearest whole percentage. Example if your calculated an er s 06 7 e ter67 Do not use apercentsign Arange ofea esvil e accepted so you do not need to worry about minor differences due to rounding and significant figures. Numeric Response

Explanation / Answer

Bb x Bb

The offspring produced would be 1 homozygous dominant black (BB), 2Heterozygous black (Bb) and 1 homozygous white (bb).

The probability of 4 out of 7 puppies black is = 4/7 = 0.57 = 57

The probability of 3 out of 7 puppies white is = 3/7 = 0.428 = 43

male/ female B b B BB Bb b Bb bb

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