Answer:
a. fatty acids is the correct answer.
Explanation:
fatty acids is NON-polar molecule.
Nonpolar molecules are that molecules which are not soluble in water and fatty acids are not soluble in water.
fatty acids are hydrophobic and the hydrocarbon chain present in fatty acids are nonpolar, thus that's the reason fatty acid is nonpolar.
examples of fatty acids
Answer:
<u>Chloroplast</u>
The chloroplast is where photosynthesis occurs in plant cells. During photosynthesis, sunlight energy, carbon dioxide, and water are converted into simple sugars (glucose). These simple sugars are cell's version of food.
<u>Mitochondria</u>
The mitochondria is where cellular respiration occurs. During the process of cellular respiration, glucose, which is produced through photosynthesis in the chloroplasts, is broken down into cellular energy that can be used by the cell.
- Both chloroplasts and mitochondria are organelles that generate metabolic energy.
~Hope this Helps!~
Lafayette
the ansewer is lafayette its the biggest ship in the world for fish processing
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
