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 phenotypes. 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 and 12 with long, brown hair.
Show cross A and give the results.
Show cross B and in doing so give the phenotypes and genotypes the breeder obtained.
How do you do this???
To solve this problem, we need to understand the principles of Mendelian genetics and apply them to the given information. Let's break it down step by step to find the answers to Cross A and Cross B.
Cross A:
The girl crossed a pure short, black-haired animal (let's call it SB) with a pure long, brown-haired individual (let's call it LB).
From this information, we can infer the genotypes of the parents:
- SB = homozygous dominant for short hair and black color (SSBB)
- LB = homozygous recessive for long hair and brown color (ssbb)
Now, let's find the phenotypes of the F1 generation:
When we cross SB and LB, the resulting offspring will have genotypes of SsBb, with a phenotype of short, black hair (dominant traits always determine the phenotype).
Therefore, all the F1 individuals will have the same phenotype (short, black hair).
Cross B:
Now, the breeder interbreeds the F1 generation (SsBb x SsBb).
To determine the phenotypes and genotypes of the offspring in Cross B, we need to use Punnett squares.
Let's fill out the Punnett square for the cross:
| Ss | Ss |
------------------------
Ss | SSbb | SSbb |
-------------------------
Ss | SSbb | SSbb |
Phenotypes:
- SSbb: short, brown hair
- SSbb: short, brown hair
- SSbb: short, brown hair
- SSbb: short, brown hair
- SSbb: short, brown hair
- SSbb: short, brown hair
- SSbb: short, brown hair
- SSbb: short, brown hair
- SSBB: short, black hair
- SSBB: short, black hair
- SSBB: short, black hair
- SSBB: short, black hair
- SSBB: short, black hair
- SSBB: short, black hair
- SSBB: short, black hair
- SSBB: short, black hair
Counting the phenotypes in the offspring, we get:
- 91 with short, black hair
- 29 with short, brown hair
- 28 with long, black hair
- 12 with long, brown hair
Therefore, the breeder obtained these phenotypes in the offspring of Cross B.
In terms of genotypes, we have:
- 91 individuals with the genotype SsBb
- 29 individuals with the genotype Ssbb
- 28 individuals with the genotype SSBb
- 12 individuals with the genotype SSbb
I hope this explanation helps you understand how to solve the problem.