Question

a bobtail is dominant over a normal tail. based on the punnett square shown, how many puppies can we expect to have a bobtail? one four two zero

Explanation:

Step1: Analyze the Punnett square

The Punnett square shows the cross between two parents. Let's assume the dominant allele for bobtail is \( B \) and the recessive allele for normal tail is \( b \). The parents' genotypes are \( Bb\times Bb \).

Step2: Determine the genotypes of offspring

Using the Punnett square method:

• \( B\times B = BB \) (bobtail)
• \( B\times b=Bb \) (bobtail)
• \( b\times B = Bb \) (bobtail)
• \( b\times b=bb \) (normal tail)

Out of 4 possible offspring genotypes (\( BB, Bb, Bb, bb \)), 3 have the dominant allele (\( B \)) which expresses the bobtail phenotype.

Answer:

Three - fourths of the offspring (i.e., 3 out of 4) are expected to have a bobtail. But since the options are Zero, Two, Four, One, and assuming the Punnett square has 4 boxes (representing 4 offspring in a typical simple cross analogy here), 3 out of 4 is equivalent to three (if we consider the count in a simple whole - number per - box sense, as each box represents one offspring). But if we strictly go by the proportion, in terms of the count of dominant - expressing boxes (where \( B \) is present), there are 3. But if we assume a mis - phrasing and count the number of non - \( bb \) (normal tail) boxes (since \( bb \) is normal and others are bobtail), there are 3. However, if we consider the options given and a wrong - but - common - student - error approach of just counting the number of \( B \) - containing genotypes ( \( BB, Bb, Bb \)) as 3, but if the question is a multiple - choice with options as given (Zero, Two, Four, One) and there is a mistake in the problem setup (maybe a 2x2 Punnett square mis - interpreted), if we assume each box is one puppy and 3 out of 4 have bobtail. But if we consider a wrong keying (maybe a typo in the problem), if we assume that in a \( Bb\times Bb \) cross, the ratio is 3:1 (bobtail:normal), and if we consider the number of bobtail puppies in 4 (as per Punnett square boxes), it's 3. But since the options are Zero, Two, Four, One, there is an error. But if we assume a mis - drawn Punnett square (maybe \( Bb\times bb \) which would be 2 bobtail ( \( Bb \)) and 2 normal ( \( bb \))). But given the Punnett square in the image (with \( b,b \) on top and \( B,b \) on the side, assuming it's \( Bb\times Bb \)), the correct number is 3. But since the options don't have three, and if we assume a mis - take in the problem (maybe a \( Bb\times bb \) cross which would give 2 \( Bb \) (bobtail) and 2 \( bb \) (normal)), then the answer is Two.

So, the answer is Two.