Question

two fruit flies that are heterozygous for body color and eye color are crossed. brown body color is dominant to black body color. red eye color is dominant to brown eye color. determine the genotypes of offspring 1-8. 1 2 3 4 5 6 7 8 be be be be be 1 bbee bbee 2 be bbee 3 bbee bbee be 4 bbee 5 6 be bbee 7 bbee 8

Explanation:

Step1: Analyze Parent Genotypes

The parents are heterozygous for body (B/b) and eye (E/e) color, so genotype is \( BbEe \). Gametes: \( BE, Be, bE, be \) (both parents, since heterozygous).

Step2: Offspring 1 (BE × BE)

Combine alleles: \( B \) from first \( BE \), \( B \) from second \( BE \); \( E \) from first, \( E \) from second. Genotype: \( BBEE \).

Step3: Offspring 2 (BE × be)

\( B \) (BE) + \( b \) (be) = \( Bb \); \( E \) (BE) + \( e \) (be) = \( Ee \). Genotype: \( BbEe \).

Step4: Offspring 3 (Be × Be)

\( B \) (Be) + \( B \) (Be) = \( BB \); \( e \) (Be) + \( e \) (Be) = \( ee \)? Wait, no: \( Be \) (parent1) and \( Be \) (parent2): \( B \times B = BB \), \( e \times e = ee \)? Wait, no, eye color: \( E \)? Wait, no, parent gametes: \( Be \) is \( B \) (body) and \( e \) (eye)? Wait, original: brown body (B) dominant to black (b); red eye (E) dominant to brown (e). So heterozygous is \( BbEe \), so gametes: \( BE, Be, bE, be \) (each gamete has one body and one eye allele). So \( Be \) is \( B \) (body) and \( e \) (eye); \( BE \) is \( B \) and \( E \); \( bE \) is \( b \) and \( E \); \( be \) is \( b \) and \( e \).

So Offspring 3: \( Be \times Be \): \( B \times B = BB \), \( e \times e = ee \)? No, wait: \( Be \) (gamete1: B, e) and \( Be \) (gamete2: B, e). So body: \( B \times B = BB \); eye: \( e \times e = ee \)? Wait, no, the table shows Offspring 3 is in row \( Be \), column \( Be \). Wait, row is first parent's gamete, column is second parent's gamete. So row: Be (parent1 gamete: B, e), column: Be (parent2 gamete: B, e). So genotype: \( B \) (from row) + \( B \) (from column) for body; \( e \) (row) + \( e \) (column) for eye? Wait, no, each gamete has one body and one eye allele. So combining \( Be \) (B, e) and \( Be \) (B, e): body alleles \( B \) and \( B \) → \( BB \); eye alleles \( e \) and \( e \) → \( ee \)? Wait, but the table's Offspring 3 is in row Be, column Be. Wait, maybe I messed up. Wait, original table:

Row 1: BE (gamete1: B, E)
Column 1: BE (gamete2: B, E) → Offspring 1: BBEE (correct, Step2)

Row 1, Column 4: BE × be → B (BE) + b (be) = Bb; E (BE) + e (be) = Ee → BbEe (Offspring 2: BbEe)

Row 2 (Be: B, e), Column 2 (Be: B, e): Offspring 3. So \( B \) (Be) + \( B \) (Be) = BB; \( e \) (Be) + \( e \) (Be) = ee? Wait, no, eye color: red (E) dominant to brown (e). Wait, maybe I mixed up eye alleles. Wait, red is E, brown is e. So heterozygous is Ee (red eyes). So \( Be \) is B (body) and e (eye: brown)? Wait, no, heterozygous for eye color: Ee (red eyes, since E is dominant). So gametes: BE (B, E), Be (B, e), bE (b, E), be (b, e). So when two Be gametes (B, e) combine: body: B + B = BB; eye: e + e = ee. So genotype: BBee? Wait, but the table's Offspring 3 is in row Be, column Be. Wait, maybe the table has a typo? No, let's check Offspring 3: row Be (B, e), column Be (B, e). So \( B \times B = BB \), \( e \times e = ee \) → BBee.

Step5: Offspring 4 (bE × BE)

Row: bE (b, E), Column: BE (B, E). Body: \( b + B = Bb \); Eye: \( E + E = EE \). Genotype: \( BbEE \).

Step6: Offspring 5 (bE × bE)

Row: bE (b, E), Column: bE (b, E). Body: \( b + b = bb \); Eye: \( E + E = EE \). Genotype: \( bbEE \).

Step7: Offspring 6 (bE × be)

Row: bE (b, E), Column: be (b, e). Body: \( b + b = bb \); Eye: \( E + e = Ee \). Genotype: \( bbEe \).

Step8: Offspring 7 (be × Be)

Row: be (b, e), Column: Be (B, e). Body: \( b + B = Bb \); Eye: \( e + e = ee \). Genotype: \( Bbee \).

Step9: Offspring 8 (be × be)

Row: be (b, e), Column: be (b, e). Body: \( b + b = bb \); Eye: \( e + e = ee \). Genotype: \( bbee \).

Wait, let's recheck each:

1. BE × BE: \( B \times B = BB \), \( E \times E = EE \) → BBEE
2. BE × be: \( B \times b = Bb \), \( E \times e = Ee \) → BbEe
3. Be × Be: \( B \times B = BB \), \( e \times e = ee \) → BBee
4. bE × BE: \( b \times B = Bb \), \( E \times E = EE \) → BbEE
5. bE × bE: \( b \times b = bb \), \( E \times E = EE \) → bbEE
6. bE × be: \( b \times b = bb \), \( E \times e = Ee \) → bbEe
7. be × Be: \( b \times B = Bb \), \( e \times e = ee \) → Bbee
8. be × be: \( b \times b = bb \), \( e \times e = ee \) → bbee

Answer:

1. \( BBEE \)
2. \( BbEe \)
3. \( BBee \)
4. \( BbEE \)
5. \( bbEE \)
6. \( bbEe \)
7. \( Bbee \)
8. \( bbee \)