A viral hepatitis screen with positive hepatitis B surface antigen (HBsAg) should be interpreted as ________ hepatitis B
1 answer:
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Answer:
Pollen grain lands on stigma
Explanation:
A) Pollen grain lands on stigma
- pollen tube grows down the style
- generative cell divides, forming two sperm
- two sperm are discharged to them female gametophyte
- sperm fuse with the egg and two polar nuclei
- zygote forms and divides into a terminal cell and a basal cell
- cells of embryo differentiate into three tissue types---- seed dries out and becomes dormant.
B)
- During pollination, a pollen grain is transferred from an anther to a stigma. Once the pollen grain lands on a suitable stigma, it germinates and forms a pollen tube, a structure that grows down through the style to the ovary.
- Once in the pollen tube, the generative cell from the pollen grain divides by mitosis, forming two sperm. The sperm travel down the pollen tube and are discharged into the female gametophyte. In a process called double fertilization, one sperm fertilizes the egg, forming the zygote; the other sperm fuses with two polar nuclei in the female gametophyte, forming a triploid (3n) nucleus. The zygote develops into the plant embryo, and the triploid nucleus divides and gives rise to the endosperm. Double fertilization prevents the waste of the plant's resources by ensuring that the nutrient-rich endosperm only develops if the egg is fertilized.
- After double fertilization, the ovule starts to develop into a seed containing the plant embryo. As the embryo develops, the three tissue systems are established, and the cotyledons (seed leaves) form. The seeds of many species dry out as they mature. These dry seeds lie dormant until suitable germination conditions occur.
Answer:
The correct answer would be 48 percent.
It can be solved by using the Hardy-Weinberg equation.
p² + 2pq+ q² = 1
p + q = 1
where, p² represents homozygous dominant's frequency that is, SS in this case,
q² represents homozygous recessive's frequency that is, ss in this case,
2pq represents heterozygous dominant's frequency that is, Ss in this case.
p represents dominant allele's frequency and q represents the recessive allele's frequency.
Now, the percentage of homozygous recessive or short stems is 35%. Therefore, the frequency of homozygous recessive would be 0.35.
Thus, the value of q² = 0.35
So, q = = 0.59 which is approximately equals to 0.6.
As p + q = 1 so, q = 1 - 0.6 = 0.4
The frequency of heterozygous (Ss) = 2pq = 2(0.6)(0.4) = 0.48
Hence, the frequency of heterozygous would be 48 percent.