Answer:
the second one is 1, the first one is 3, the last one is 5, the fourth one is 2 and the third one is 4
Answer:
The importance of mitosis lies in its condition of cellular reproduction par excellence, characteristic of the vast majority of life forms that currently populate the Earth.
Explanation:
Mitosis is the cellular process by which two identical nuclei are produced in preparation for cell division. In general, mitosis is immediately followed by the equitable distribution of the cell nucleus as well as the rest of the cellular content in two daughter cells.
We all start being a cell. That is quite difficult to imagine, but it is true. And now there are about a hundred billion cells in each of our bodies, and each one has to carry the complete set of DNA instructions. That means that when cells divide, all that has to be copied by some process. And that is mitosis, where each chromosome has to be converted into a pair, and then they have to separate properly so that each of the daughter cells receives a complete set.
Well a small baby whale can eat about 200 pounds in a day
An adult whale can eat roughly about 6 TONS of food per day-
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
