Question

in boxes a and b, the genotypes are shown for you. you need to determine the phenotypes. in boxes c and d, you must figure out the genotypes yourself first. then, determine the phenotypes.

Explanation:

(Step-by-Step for Each Box):

Box A:

Step 1: Analyze Genotype \( BBEe \)

• Fur: \( BB \) (dominant homozygous) → black fur.
• Eyes: \( Ee \) (dominant allele \( E \)) → black eyes.

Step 2: Determine Phenotype

Combine traits: black fur, black eyes.

Box B:

Step 1: Analyze Genotype \( BBee \)

• Fur: \( BB \) → black fur.
• Eyes: \( ee \) (recessive homozygous) → red eyes.

Step 2: Determine Phenotype

Combine traits: black fur, red eyes.

Box C:

Step 1: Find Genotype (Parent 1: \( bE \); Parent 2: \( be \))

• Fur alleles: \( b \) (Parent 1) + \( b \) (Parent 2) → \( bb \) (white fur).
• Eye alleles: \( E \) (Parent 1) + \( e \) (Parent 2) → \( Ee \) (black eyes).
• Genotype: \( bbEe \).

Step 2: Determine Phenotype

Combine traits: white fur, black eyes.

Box D:

Step 1: Find Genotype (Parent 1: \( be \); Parent 2: \( be \))

• Fur alleles: \( b \) + \( b \) → \( bb \) (white fur).
• Eye alleles: \( e \) + \( e \) → \( ee \) (red eyes).
• Genotype: \( bbee \).

Step 2: Determine Phenotype

Combine traits: white fur, red eyes.

(Note: Assumptions about dominance (B = black, b = white; E = black, e = red) are standard for Mendelian genetics. Adjust if the problem’s dominance rules differ, but this follows typical patterns.)

Answer:

To solve for the genotypes and phenotypes in the Punnett square (assuming dominant/recessive relationships: let's assume \( B \) = black fur (dominant), \( b \) = white fur (recessive); \( E \) = black eyes (dominant), \( e \) = red eyes (recessive)):

Box A: Genotype \( BBEe \)

• For fur: \( BB \) → black fur (dominant homozygous).
• For eyes: \( Ee \) → black eyes (dominant allele present).
• Phenotype: black fur, black eyes.

Box B: Genotype \( BBee \)

• For fur: \( BB \) → black fur.
• For eyes: \( ee \) → red eyes (recessive homozygous).
• Phenotype: black fur, red eyes.

Box C: Determine Genotype (Parent 1: \( BbEe \) (gamete \( bE \)); Parent 2: \( Bbee \) (gamete \( be \)))

• Genotype: \( bE \times be = BbEe \)? Wait, no—wait, Parent 1 (top mouse: \( BbEe \)) gametes: \( BE, Be, bE, be \). Parent 2 (left mouse: \( Bbee \)) gametes: \( Be, Be, be, be \).
• Box C is at row \( be \) (Parent 2 gamete) and column \( bE \) (Parent 1 gamete). So genotype: \( bE \times be = BbEe \)? Wait, no—alleles: \( b \) (from Parent 2) and \( B \)? Wait, Parent 2 is \( Bbee \), so gametes are \( Be \) (B and e) or \( be \) (b and e). Parent 1 gametes: \( BE, Be, bE, be \).

Wait, correction: Parent 2 (left mouse) is \( Bbee \), so its gametes are formed by independent assortment: \( B \) or \( b \) (from \( Bb \)) and \( e \) or \( e \) (from \( ee \)) → so gametes are \( Be \) (B + e) and \( be \) (b + e) (each repeated twice, since \( ee \) is homozygous).

Parent 1 (top mouse) is \( BbEe \), so gametes: \( BE, Be, bE, be \) (each once, assuming equal segregation).

Box C: row is \( be \) (Parent 2 gamete: \( be \)) and column is \( bE \) (Parent 1 gamete: \( bE \)). So genotype: \( bE \times be = BbEe \)? Wait, no—alleles: \( b \) (from Parent 2’s \( be \)) and \( B \)? No, Parent 2’s \( be \) is \( b \) (fur) and \( e \) (eyes); Parent 1’s \( bE \) is \( b \) (fur) and \( E \) (eyes). Wait, no—Parent 2 is \( Bbee \), so \( Bb \) for fur: so \( B \) or \( b \); \( ee \) for eyes: only \( e \). So Parent 2’s gametes: \( Be \) (B + e) and \( be \) (b + e).

Parent 1’s gametes: \( BE \) (B + E), \( Be \) (B + e), \( bE \) (b + E), \( be \) (b + e).

So Box C: row \( be \) (Parent 2: \( be \)) and column \( bE \) (Parent 1: \( bE \)) → genotype: \( bE \) (Parent 1) × \( be \) (Parent 2) = \( b \) (from Parent 1) and \( b \) (from Parent 2) for fur? Wait, no—Parent 1’s \( bE \) is \( b \) (fur allele) and \( E \) (eye allele). Parent 2’s \( be \) is \( b \) (fur allele) and \( e \) (eye allele). So combining: fur alleles: \( b \) (Parent 1) and \( b \) (Parent 2) → \( bb \)? Wait, no—Parent 2 is \( Bbee \), so \( Bb \) for fur: so \( B \) or \( b \). Wait, I made a mistake: Parent 2 is \( Bbee \), so fur genotype is \( Bb \) (not \( bb \))! Oh right—\( Bbee \) means \( Bb \) (fur) and \( ee \) (eyes). So Parent 2’s gametes: \( Be \) (B + e) and \( be \) (b + e) (because \( Bb \) segregates into \( B \) or \( b \), and \( ee \) into \( e \) or \( e \)).

So Parent 2’s \( be \) gamete is \( b \) (fur) + \( e \) (eyes). Parent 1’s \( bE \) gamete is \( b \) (fur) + \( E \) (eyes).

Thus, Box C genotype: \( b \) (fur, Parent 1) + \( b \) (fur, Parent 2) = \( bb \)? No—wait, Parent 1’s \( bE \) is \( b \) (fur) and \( E \) (eyes); Parent 2’s \( be \) is \( b \) (fur) and \( e \) (eyes). So fur alleles: \( b \) (Parent 1) and \( b \) (Parent 2) → \( bb \) (white fur, recessive). Eye alleles: \( E \) (Parent 1) and \( e \) (Parent 2) → \( Ee \) (black eyes, dominant). So genotype: \( bbEe \).

• Phenotype: \( bb \) → white fur; \( Ee \) → black eyes. So white fur, black eyes.

Box D: Determine Genotype (row \( be \) (Parent 2 gamete) and column \( be \) (Parent 1 gamete))

• Parent 1 gamete: \( be \) (b + e); Parent 2 gamete: \( be \) (b + e).
• Genotype: \( be \times be = bbee \).
• Phenotype: \( bb \) → white fur; \( ee \) → red eyes. So white fur, red eyes.