Answer:
1.- Arachnoid mater: Middle meninx; web-like.
2.- Dura mater<u>:</u> Tough, outer meninx.
3.- Epidural space: Space filled with adipose tissue.
4.- Pia mater: Thin meninx intimate with spinal cord.
5.- Subarachnoid mater<u>:</u> Contains cerebrospinal fluid.
6.- Denticulate ligaments
: Extension of pia mater attaching to dura.
The solution of nutrients and organelles is called CYTOPLASM.
The cytoplasm of a cell is made up of all the organelles that are located outside of the nucleus which are enclosed within the cell membrane of a cell. The cytoplasm is a colorless, clear, jelly like substance whose constituent is majorly water [80%]. It also contains enzymes, salts, organelles and various organic molecules.
The answer is C, because if something contagious we’re to threaten the population of the species then not having any variant means they’re all the same and it would kill them all and cause them to go extinct
Answer:
Hurricanes are not common in the Midwest. However, if a hurricane is particularly strong, it can move far enough northward and inland to cause a significant rain event for areas in the Midwest.
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
