Estrogens from the vesicular follicle trigger an LH surge from the anterior pituitary. This is an example of: Estrogens from the vesicular follicle trigger an LH surge from the anterior pituitary. This is an example of: negative feedback first tier of control positive feedback fourth tier of control

Answers

Answer 1

Answer:

The correct answer is: positive feedback.

Explanation:

Positive and negative feedback are terms that are used to define the way a system is regulated by its products. For example, when the production of a certain substance inhibits its producer that is called negative feedback. On the other hand, when a substance stimulates its producer to keep secreting said substance, this is called positive feedback.

Estrogen is a hormone of great importance in the female reproductive cycle and is secreted by the follicle cells in response to the presence of FSH (follicle-stimulating hormone), released by the anterior pituitary when stimulated by the GnRH (gonadotropin-releasing hormone).

Estrogen appears at the beginning of the cycle to stimulate the development of the ovarian follicles, and the hormone levels continue to rise as the cycle advances. Up until the 14th day of the cycle, estrogen has a negative feedback on GnRH and FSH, but then it makes an abrupt change to now exert positive feedback on GnRH to stimulate the release of LH (luteinizing hormone), which will be responsible for ovulation.


Related Questions

Antimicrobial peptides can do all of the following except:_________A) regulate body temperature.B) target intracellular components.C) insert themselves into target cell membranes.D) disrupt plasma membrane and/or cell wall.E) stimulate leukocytes.

Answers

Answer:

Option A

Explanation:

Antimicrobial peptides cannot regulate body temperature. They can do all of the rest.

=> target intra cellular components.

=> insert themselves into target cell membranes.

=> disrupt plasma membrane and/or cell wall.

=> stimulate leukocytes.

Which combination forms a covalent bond?
A. Two carbon atoms
B. Two oxygen molecules
C. Two water molecules
D. Two sodium atoms
SUSMIT
!!!

Answers

Answer:

A. Two carbon atoms

Explanation:

We know that covalent bond occurs when there are two non - metallic atoms and carbon atoms are non-metallic so it is the answer.

Hope it helps and pls mark as brainliest :)

what is cellular affinity in biology?

Answers

Answer:

affinity. Attraction between particles. Biology: A measure of the attraction of one biological molecule toward another molecule, either to modify it, destroy it, or form a compound with it. Examples are enzymes and their substrates, or antibodies and their antigens.

Fill in the blanks with appropriate wods.


1. A process called__________ helps to propel first the bolus and then chyme forward through the esophagus, stomach, and small intestine.
2. Nutrients are absorbed through the lining of the small intestine via a variety of mechanisms. Most nutrients require______________ , where a specific carrier protein is utilized, but no energy is required.
3. Some smaller molecules can cross via_____________ , Like with facilitated diffusion, no energy is required because the concentration of a nutrient is higher inside the GI tract than in the enterocyte. Unlike with facilitated diffusion, however, no ____________ is needed.
4. Nutrients like glucose and amino acids are absorbed through ______________ , which requires both a carrier protein and energy in the form of ATP.
5. A fourth method is called _____________ , which involves a vesicle that forms from the enterocyte's cell membrane, engulfing the nutrient and then drawing the nutrient into the cell.
6. Once absorbed, water-soluble nutrients like carbohydrates and amino acids enter capillaries and travel to the liver via the ___________ .
7. Fat-soluble nutrients initially travel via the _____________ before entering the bloodstream.
8. Most water is absorbed in the _____________ , and solid waste is then eliminated.


a. Large intestine
b. Peristalsis
c. Passive diffusion
d. Carrier protein
e. Active transport
f. Facilitated diffusion
g. Endocytosis
h. Lymphatic system
i. Hepatic portal vein

Answers

Answer:

1. b. Peristalsis

2. c. Passive diffusion

3.  Facilitated diffusion, d. Carrier protein

4. e. Active transport

5. g. Endocytosis

6. i. Hepatic portal vein

7. h. Lymphatic system

8. a. Large intestine

Explanation:

1. peristalsis helps to propel first the bolus and then chyme forward through the esophagus, stomach, and small intestine.

2. Nutrients are absorbed through the lining of the small intestine via a variety of mechanisms. Most nutrients require passive diffusion, where a specific carrier protein is utilized, but no energy is required.

3. Some smaller molecules can cross via facilitated diffusion Like with facilitated diffusion, no energy is required because the concentration of a nutrient is higher inside the GI tract than in the enterocyte. Unlike with facilitated diffusion, however, no carrier protein is needed.

4. Nutrients like glucose and amino acids are absorbed through active transport which requires both a carrier protein and energy in the form of ATP.

5. A fourth method is called endocytosis which involves a vesicle that forms from the enterocyte's cell membrane, engulfing the nutrient and then drawing the nutrient into the cell.

6. Once absorbed, water-soluble nutrients like carbohydrates and amino acids enter capillaries and travel to the liver via the hepatic system

7. The lymphatic system helps in traveling of the Fat-soluble nutrients initially, before entering the bloodstream.

8. Most water is absorbed and solid waste is then eliminated. in the large intestine

DNA damage checkpoints are crucial for preventing cell proliferation when DNA damage has occurred. The role of DNA checkpoint proteins in the cell cycle is closely tied to the roles of DNA repair enzymes, although these proteins perform different functions.1. Which of the following statements is true for DNA checkpoint proteins and not for DNA repair enzymes? a. These proteins remove a single nucleotide of damaged DNA. b. These proteins nick the phosphodiester backbone of the damaged DNA. c. These proteins prevent the cell cycle from progressing if damaged DNA is detected. d. These proteins remove the region of damaged DNA. 2. Which cell cycle phases contain critical DNA damage checkpoints?a. Gob. G1c. G2d. M

Answers

Answer:

1. C. These proteins prevent the cell cycle from progressing if damaged DNA is detected.

2. C. G2

Explanation:

1. All other choices apply uniquely to DNA repair enzymes (i.e. not to DNA checkpoint proteins).

2. Go doesn't contain any critical DNA damage checkpts. because it's not even an official part of the cell cycle; it's simply a phase in which cells are arrested. G1 doesn't contain any critical DNA damage checkpts. because it's before DNA is replicated. G2 is right after S phase when the DNA is replicated; thus, it contains critical DNA damage checkpts.

About 16% of the world's total oil output:
A) comes from offshore drilling
B) is made by the United States
C) is mined deep beyond the continental shelf
D) comes from oil rigs in the Atlantic Ocean

Answers

The answer is D


Have a nice day

Answer:

The answer is A,  It comes from offshore drilling.

Explanation:

Cause i got it right.

1. You are given a 1 gram soil sample of unknown bacterial load. After doing 10-fold serial dilutions of the soil in sterile water, 100 uL volumes are taken from each dilution for preparation of pour plates. Following incubation, each half of the 10-8 plate has 46 colonies.
a) What was the dilution factor?
b) How many bacteria were present in the soil?

2. Staphylococcus aureus divides every 20 minutes. A culture begins with 10 bacterial cells.
a) After 5 hours, how many generations have occurred
b) After 5 hours, how many bacteria are present?

3. How many milliliters would you need to prepare a 10-2 dilution from a 10ml starting culture?

Answers

Answer:

1a. 1:10

1b. 4.6×10¹³

2a. 15generations

2b. 327,680

Explanation:

10 fold serial dilution means 1ml of stock is mixed in 9ml distilled water. This is called as 10⁻¹ solution.

In this way, take 1ml each of every dilution and mix in 9ml distilled water to make 10⁻², 10⁻²-------10⁻⁸ dilutions.

Now, CFU/ml = no. of colonies × dilution factor/ volume of sample plated.

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