Complete question:
A female is born with attached earlobes, which is a recessive phenotype. Which of the following statements about her parents must be true?
a. neither has the codominant allele.
b. Her father has an inactivated allele.
c. Both parents have the recessive allele.
d. Her mother carries the dominant allele.
Answer:
c. Both parents have the recessive allele.
Explanation:
A recessive trait is expressed in homozygous recessive genotypes only. If a female expresses the attached earlobe phenotype, she should be homozygous for the allele since the attached earlobe is a recessive trait. If allele "a" gives attached earlobe phenotype in homozygous genotype, the genotype of the female would be "aa". This means that the female has got one copy of the allele "a" from her either parents. Therefore, both her parents should have at least one copy of the recessive allele "a".
Answer:
In contrast, applied science or “technology,” aims to use science to solve real-world problems, making it possible, for example, to improve a crop yield, find a cure for a particular disease, or save animals threatened by a natural disaster. In applied science, the problem is usually defined for the researcher.
Explanation:
I hope this helps, but when you use this make sure you copy and paste this to paraphrasing tool.
Answer:
I believe you just answered it yourself. they development an immunity to insecticides the more they are exposed to it.
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:
movement
Explanation:
Both of them can move and can be move
