Answer:
Mechanical Isolation
A snail with a flat disc-like shell
will not be able to mate with a
snail having a conical shell.
The reproductive organs of
male bush babies do not
match with the reproductive
organs of females of other
bush baby species
Behavioral Isolation
The mating call of a cricket is not
recognized by a cricket of other
species.
The signals sent by a male firefly
are not recognized by the female
firefly of other species.
Explanation:
Mechanical isolation occurs in the scenario where two organisms are willing to mate but cannot do so due to the physical challenges. As in the case of snail a flat disk and a conical disk snail are willing to mate but physical barriers restrict them in doing so.
The same case is with the male and female bush babies where the difference in reproductive organs does not allow them to mate.
Behavioral isolation represents a scenario where two organisms are unaware of mating rituals, season, pattern or signals of each other and hence cannot mate.
For example – male and female cricket do not recognize the mating call of each other. Likewise male and female firefly is also not able to understand each other’s mating signal.
Answer: They would not survive because they have already adapted to the humid environment of the rain forest and could not adapt fast enough to survive the frigid climate of the artic.
Explanation:
Answer:
Option (A) is the correct answer to this question.
Explanation:
If the hypothesis is correct then we shall be higher than Fst in the rest of the genome in the area environment the cold-shock protein in the genomic region surrounding the cold-shock gene, except for the Fst.
- Cold shock components are those product uses that are produced in cold and defend against cold stress. Fst is a gene of ice, present offshore of proteins from cold shock. During cold stress, the fist gene expression increase by several folds to increase cold shock protein production.
- Other options are incorrect because they are not related to the given scenario.
Answer/ Explanation:
a. The genotype of a homozygous white eyed long winged female would be Vg+Vg+XrXr. We denote the white allele as recessive (r) because the XY male only has one copy and yet has red eyes, so the red eye trait (R) must be dominant. A homozygous red eyed vestigial winged male would have be VgVgXRY. The possible gametes for the female are Vg+Xr only. For the male, the possible gametes are VgXR or VgY
The attached punnett square shows the results of the cross. The females will all be Vg+VgXRXr. The males will all be Vg+VgXRY (must inherit Y from father). That means they will all have normal length wings, the males will have white eyes and the females will have red eyes.
b. The F2 flies arise from intercrossing the F1, so the cross will be Vg+VgXRXr x Vg+VgXRY. The possible gametes for the mother are: Vg+XR, Vg+Xr, VgXR or VgXr. The possible gametes for the father are Vg+Xr
, Vg+Y
, VgXr
, VgY
. The attached punnet square shows this cross. The ratio of the phenotypes will be 6:6:2:2, or 3:3:1:1 (long-winged red eye: long-winged white eye: vestigial wing red eye: vestigial wing white eye), genotypes shown in the attachment.
c. F1 cross back to the mother would be Vg+VgXRY x Vg+Vg+XrXr. The genotypes are shown in the attached punnet square. The offspring will all be long-winged with white eyes. The F1 to the father would be Vg+VgXRXr x VgVgXRY. The ratio would be 3:3:1:1 long-winged red eye: long-winged white eye: vestigial wing red eye: vestigial wing white eye
The first trophic level
they would want to eat the primary producers, which are located at the first trophic level. due to the 10% rule only 10% of the energy from the previous trophic level is passed on.
