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.
from the phenotypes, it is apparent that short is dominant to long, and black is dominant to brown.
ss BlBl x ll Br Br
sl Bl Br
Sl Bl Br
these are the two genotypes in Cross A, all have the same phenotype (short, Black)
Cross B.
Sl Bl Br x Sl Bl Br
ss Bl Bl
ss Br Br
ss Br Bl
Sl Bl Bl
Sl Bl Br
Sl Br Br
ll Bl Br
ll Bl Bl
ll Br Br
Can you figure it now? It is making each possible combination of
each gene
To understand the results of cross A and cross B, we need to determine the genotypes and phenotypes of the parents and offspring.
In cross A, a pure short-haired, black animal was crossed with a pure long-haired, brown animal. Let's use the letters "B" and "b" to represent the alleles for black and brown hair respectively, and "S" and "s" for short and long hair respectively.
The genotype of the pure short-haired, black individual can be represented as "BBss" (since it is pure for both black hair and short hair), and the genotype of the pure long-haired, brown individual can be represented as "bbSS" (since it is pure for both brown hair and long hair).
When these two individuals are crossed, we can use Punnett square to determine the possible genotypes of the offspring:
b b
----------------
S | BbSs | BbSs |
s | Bbss | Bbss |
----------------
From this Punnett square, we can see that all the F1 individuals will have the genotype "BbSs" (hybrids) since they received one copy of the black hair allele (B) and one copy of the short hair allele (s) from each parent. This explains why all the F1 individuals in cross A have the same phenotypes.
Now let's move on to cross B. In this generation, the F1 individuals with genotype "BbSs" are interbred. Again, let's use Punnett square to examine the possible genotypes of the offspring:
B b S s
-----------------------------
B| BBSS | BBss | BBSS | BBss |
b| BbSS | Bbss | BbSS | Bbss |
S| BBSS | BBss | BBSS | BBss |
s | BbSS | Bbss | BbSS | Bbss |
-----------------------------
From this Punnett square, we can now determine the phenotypes and genotypes of the offspring:
- 91 guinea pigs have short, black hair (phenotype: Bbss)
- 29 guinea pigs have short, brown hair (phenotype: Bbss)
- 28 guinea pigs have long, black hair (phenotype: BBSS)
- 12 guinea pigs have long, brown hair (phenotype: BBss)
This information gives us a breakdown of the phenotypes obtained from cross B.
So, in summary:
- Cross A (pure short-haired, black individual x pure long-haired, brown individual) resulted in F1 generation with all individuals having the same phenotypes.
- Cross B (interbreeding of F1 individuals) produced 160 offspring with a distribution of different phenotypes, indicating a variety of genotypes resulting from the crossing of the F1 generation.