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
Answer:
Runners
Explanation:
stems are usually upright and prop up the plant to gather sunlight, however, some stems grow along the ground. These stems are called runners and the plant itself is called a stolon. Runners are involved in asexual reproduction of plants. They have nodes along them which can grow into another plant.
Answer:
C low competition for glucose
Explanation:
From the given answer choices and information,
It cannot be A since there is no visual disease or any indication of disease in the experiment
Cannot be B, since space would not be an issue since it clearly hit 111 on day 3
and cannot be D since there is no indication of a change in temperature.
However, we know Petri Dishes have nutrients to stimulate cell growth, and those resources are not unlimited therefore we can only attest that a large portion of the nutrients have been consumed and starved some of the cells thus causing a population decrease
Explanation:
At the end of mitosis, the two daughter cells will be exact copies of the original cell. Each daughter cell will have 30 chromosomes
