The best way to know the probabilities here is to do each cross and see the results.
a. Aa Bb Cc Ddx aa bb cc dd
Aa Bb Cc Dd
Aa Bb Cc Dd
aa bb cc dd
aa bb cc dd
You have 50% chances of Aa Bb Cc Dd which is phenotypically the same as one of the parents, and 50% chances of aa bb cc dd which are phenotypically the same as one of the parents.
b. aa bb cc dd x AA BB CC DD
aA bB cC dD
aA bB cC dD
aA bB cC dD
aA bB cC dD
You have 100% chances of having aA bB cC dD offspring, and it will be phenotypically the same as the AA BB CC DD because of the dominant allele.
c. Aa Bb cc Ddx Aa Bb cc Dd
AA BB cc DD
Aa Bb cc Dd
aA bB cc dD
aa bb cc dd
You have 25% chances of AA BB cc DD, 50% chances of Aa Bb cc Dd which are phenotypically the same as both the parents and 25% chances of aa bb cc dd.
d. aa bb cc dd x aa bb cc dd
aa bb cc dd
aa bb cc dd
aa bb cc dd
aa bb cc dd
You have 100% chances of having aa bb cc dd which are phenotypically the same as both the parents.
Answer:
Active transport is what i'm going to go with because it makes the most sense and matches the statements
Explanation:
So therefore Active transport is your answer. (tell me if i'm wrong)
