Question

c. parent genotypes: aabb × aabb
determine what gametes each parent can produce using the foil method:
foil = first, outer, inner, last
cross aa × bb → produces:
ab
ab
ab
a b
these are the 4 possible gametes that each parent can contribute.
d. cross the second generation
4 phenotype
9:3:3:1

Explanation:

Step1: Determine gametes

Using FOIL method for \(AaBb\):

• First: \(A\times B = AB\)
• Outer: \(A\times b=Ab\)
• Inner: \(a\times B = aB\)
• Last: \(a\times b=ab\)

So gametes are \(AB\), \(Ab\), \(aB\), \(ab\)

Step2: Create Punnett square

	\(AB\)	\(Ab\)	\(aB\)	\(ab\)
\(Ab\)	\(AABb\)	\(AAbb\)	\(AaBb\)	\(Aabb\)
\(aB\)	\(AaBB\)	\(AaBb\)	\(aaBB\)	\(aaBb\)
\(ab\)	\(AaBb\)	\(Aabb\)	\(aaBb\)	\(aabb\)

Step3: Calculate phenotypic ratio

• \(A - B-\) (both dominant): \(9\) ( \(AABB\), \(AABb\), \(AaBB\), \(AaBb\) )
• \(A - bb\) (one dominant): \(3\) ( \(AAbb\), \(Aabb\) )
• \(aaB-\) (one dominant): \(3\) ( \(aaBB\), \(aaBb\) )
• \(aabb\) (both recessive): \(1\)

Answer:

Phenotypic ratio is \(9:3:3:1\)