A girl crossed a pure short, black haired animal with a pure long, brown haired individual (cross A). The F1 consisted of 10 individuals who had all the same phenotype. She interbred this F1 generation (cross B) and counted the offspring, which consisted of 160 guinea pigs with the following phenotypes:
91 with short, black hair
29 with short, brown hair
28 with long, black hair
12 with long, brown hair
(This is called a dihybrid cross).
1)Show cross A and give the results.
2)show cross B and in doing so give the phenotypes and genotypes the breeder obtained.
PLEASE HELP!
To solve this problem, we need to use Punnett squares to determine the genotypes and phenotypes of the offspring in each cross. Let's start with cross A:
1) Cross A: Pure short, black-haired individual (SSBB) x Pure long, brown-haired individual (ssbb)
To set up the Punnett square, we'll separate the alleles for hair length and hair color. The alleles for hair length are S (short) and s (long), and the alleles for hair color are B (black) and b (brown).
The genotype of the pure short, black-haired individual is SSBB (S from short hair and B from black hair), and the genotype of the pure long, brown-haired individual is ssbb (s from long hair and b from brown hair).
The Punnett square for cross A would look like this:
B b
------------------
S | SB SB
| SB SB
s | sb sb
| sb sb
The results of cross A would be all heterozygous (SsBb) offspring, which means they all have the same phenotype: short, black hair.
2) Cross B: Interbreeding the F1 generation (SsBb) from cross A
To determine the potential genotypes and phenotypes of the offspring in cross B, we'll use another Punnett square:
The genotype of the F1 generation is SsBb (derived from cross A).
The Punnett square for cross B would look like this:
B b
------------------
S | SB Sb
| SB Sb
s | sB sb
| sB sb
Now, let's fill in the Punnett square to determine the genotypes and phenotypes of the offspring:
- Short, black hair (SB) - 2 SsBb
- Short, brown hair (Sb) - 2 Ssbb
- Long, black hair (sB) - 2 ssBb
- Long, brown hair (sb) - 2 ssbb
Counting the outcomes, we have:
- 2 + 2 = 4 individuals with short, black hair (SB)
- 2 individuals with short, brown hair (Sb)
- 2 individuals with long, black hair (sB)
- 2 individuals with long, brown hair (sb)
Therefore, the breeder obtained the following phenotypes and genotypes in cross B:
- 4 individuals with short, black hair (SsBb)
- 2 individuals with short, brown hair (Ssbb)
- 2 individuals with long, black hair (ssBb)
- 2 individuals with long, brown hair (ssbb)
That's how you can determine the phenotypes and genotypes of the offspring in both cross A and cross B.