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insens350 [35]
3 years ago
15

Three cells are placed in three solutions of different concentrations. Determine the concentration of the solutions based on the

predominant
movement of water molecules through the cells.

isotonic

hypertonic

hypotonic

Solution 1

Solution 2

Solution 3
Biology
2 answers:
Pie3 years ago
3 0

Answer: i also need this so please anyone

Explanation:

Readme [11.4K]3 years ago
3 0
The solutions around the cells can be:
1. Isotonic-when the concentration of solutions is equal inside and out the cell,
2. Hypotonic-when the concentration of solutions outside the cell is lower then inside the cell. The water will get into the cell via osmosis and as a consequence the cell will swell.
3. Hypertonic-the concentration of solutions outside the cell is higher then inside the cell, so the water will move outside. The cell will shrink.
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In Drosophila, vestigial (partially formed) wings (vg) are recessive to normal long wings (vg+) and the gene for this trait is a
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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

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. 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

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