Answer:
D. Fruit trees, birds/monkeys, tigers
Explanation:
Fruit trees are the primary producers and produce fruits and other plant parts as food for birds. Monkeys are omnivorous and derive nutrition from eggs of birds or fruits, seeds and other parts of the plant. Tigers are top consumers and are carnivores as they feed on monkeys and even birds. Therefore, the given ecosystem may have the following food chain:
Fruit trees (first trophic level and producers)----> birds /monkeys (second trophic level and first consumers)---> tigers (third trophic level and top consumers).
Ground Substance is the correct answer
<span>1) The probability of F2 seed chosen at random will be yellow is the same as asking about the probability of forming a yellow seed in that generation. The answer is 3/4 (75%) because 1/4 = 25% GG is homozygous dominant and the rest 2/4 = 50% Gg are heterozygous.
2) In the F2 generation there are
two genotypes that will breed true, which is homozygotic genotypes: GG and gg. But of these two, only one of
them is yellow. The answer is 1/3 because from all the yellow seeds that resulted (GG, Gg and gG) only one has the genotype for true breeding.
3) The probability of taking out three seeds in which a least one is yellow can be calculated by subtracting the only
probability that doesn't fit the criterion, which is taking out a green seed 3 times:
= 1/64.
Then substract that to 1
=
4) there are three possible groupings in which the green is taken once and the yellow seeds taken twice. The first possibility (follow the example) is the green being the first to come out, another possibility is being the second and the last possibility is being taken in the third time.
For example, Green yellow yellow- probability : </span>
<span><span><span>
</span> = 9/64
This is one possibily, but since there are 3, multiply it by 3 and you obtain the final answer:2</span>7/64. </span>