Answer:
a)BB*Bb
b)0:2:2(short: intermediate:long)
c)0
Explanation:
A naturalist visiting an island in the middle of a large lake observes a species of small bird with three distinct types of beaks. Those with short, crushing beaks (BB) consume hard shelled nuts, those with long, delicate beaks (bb) pick the seeds from pine cones, and those with intermediate beaks (Bb), consume both types of seeds though they are not as good at either. Assume that this difference in beak morphology is the result of incomplete dominance in a single locus gene. Which of the mated pairs below will have the best adapted offspring in a year in which most of the food available is in the form of hard shelled nuts? What would be the phenotypic ratio of the F1 generation resulting from a cross of Bb x bb(Short:Intermediate:Long)? How many offspring of an intermediate x short beak cross will have long beaks (assume 4)?
A) since the feed is put into consideration, then the offspring that is best adapted to feed on hard shelled but is to be considered. Cross-linking a gene BB with Bb is the best. If BB is said to be dominant then the offspring are produced will be short beak and intermediate beak in the ratio of 2:2 where both are suitable for feeding on hard shelled but.
B) The phenotypic ratio of the F1 generation (first generation or offspring) between Bb and bb is that there won't be a pure short beak(BB). Therefore 0:2:2 for short: intermediate: long.
C) for the intermediate and short beak crosslinking there won't be a long beak in the first generation when we limit the number offspring to four.
