Genetics as it applies to evolution, molecular biology, and medical aspects.
5 posts • Page 1 of 1
I have a question and I was hoping someone will answer me A brown mink is crossed with a silver mink and produce all brown offspring. When these brown F1 offspring were crossed among themselves they produced 47 brown and 15 silver F2 offspring. Which is the dominant allele? How many of the brown F2 animals would be expected to be homozygous? How many silver F2 animals would be homozygous?
I was hoping someone could answer me because its really giving me a headache...
gellybelly08, don't be worried and have a rest if you have headache (with headache only you'll be more angry with the problem, he he). Here you have the solution:
1. If we do the first cross, we obtain this result:
P: Brown x Silver ---> F1: Brown.
2. If we do the second cross, we obtain this result:
Brown (F1) x Brown (F1) ---> F2: 47 Brown and 15 Silver.
3. We can expect that it is a 3:1 segregation (but only it is a hypothesis that we have to prove), because if we do 47/15 and 15/15, we obtain 3'13 (that is aproximate to 3) and 1 respectively.
4. This 3:1 segregation means that it is a monohybrid cross and this:
- In the first cross: P: Brown (AA) x Silver (aa) ---> F1: Brown (Aa).
- In the second cross: Brown [F1] (Aa) x Brown [F1] (Aa) ---> AA, Aa and aa. A_ has a probability of 3/4 (1/4 AA + 2/4 Aa) and aa has a probability of 1/4. This accepts our hypothesis of 3:1 segregation that we expected before.
NOTE: To accept completely the hypothesis that we expected about 3:1 segregation, you should do the Chi squared test.
5. So, as we have called (A,a) to this gen which controls colour, we say that:
- Brown minks have this genotype: A_ (AA or Aa) [Dominant homozygous or heterozygous].
- Silver minks have this genotype: aa [Recesive homozygous].
Did I answered your doubt? Take a look to your notes (about monohybrid crosses) and tell me if you don't understand something.
from the fact that the F1 is only brown directly indicates that brown is dominant and silver is recessive and also that the only possibility is that the P-generation is BB x bb where B = brown and b = silver.
--> otherwise you'll never get 100% brown in F1 and these are all Bb (heterozygous)
--> then we get the next crossbreeding of F1: Bb x Bb. This leads to 25% BB, 50% Bb and 25% bb
--> thus now you know that of the brown minks (25/75) * 100% = 33,33% = BB and 66,67% = Bb
--> we have 47* 0,3333 = 15 homozygous brown minks and thus 32 heterozygous brown minks.
--> all the silver minks are homozygous, since we determined bb to be silver: homozygous recessive.
5 posts • Page 1 of 1
Who is online
Users browsing this forum: No registered users and 3 guests