Answer:
1.31 cM
Explanation:
Total offspring = 2205
Since two genes are involved, F1 progeny should have four types of combination. Out of them two are 17 and 12 which definitely means they are in lesser number. Since recombinants are always less than parental progeny in linkage, the given two types are recombinants.
Recombination frequency = (Number of recombinants / Total progeny) * 100
= [ ( 17 + 12 ) / 2205 ] * 100
= ( 29 / 2205 ) * 100
= 1.31 %
Map distance = Recombination frequency
Hence, distance between two genes = 1.31 cM
Thornscrub has well-illustrated dry and wet seasons, with the majority of the wood plants showing drought deciduousness. Thornscrub is a transitional biome between the tropical and desert forest. The thornscrub ecosystem belongs to shrubland. The scrubland, shrubland, brush, scrub, or bush refers to a plant community featured by the vegetation subjugated by shrubs, generally also comprising herbs, grasses, and geophytes. The shrubland may take place either naturally or due to human activity.
Water's ionization property creates positive and negative ions, which are essential for chemical reactions.
Answer:
B
Explanation:
It would not be D because D is the location of a common ancestor, AKA an animal that does not haver jaws or bones. It would not be C because the question states that the fish has jaws, and C does not have jaws since the arrow does not point to that. It would not be A because the question states the fish has "NO TRUE BONES", meaning that there are no bones present. And in A, there is the bones arrow pointing to it, meaning that A has bones since that trait is on the same horizontal line as A. Only leaving us with option B left, since it came from an ancestor with jaws but not from an ancestor with bones. Sorry, I'm not the best with explaining but hopefully that makes sense why it is B, it's like a family tree, you just have to follow it carefully :) I hope that helps
Answer:
it's considered autotrophic ;)
