Answer:
DNA polymerase III only synthesizes DNA in the 5'-3' direction.
Explanation:
DNA replication is an important phenomenon for every living cell. It is the process whereby the double-stranded DNA is doubled to form two new separate double strands. In order for DNA replication to occur, the double strand of the DNA molecule must first be unwound by an enzyme called DNA HELICASE. This gives two separate single strands, which individually acts as a template for the newly synthesized strands.
DNA polymerase III is the enzyme responsible for synthesizing new DNA strand by pairing complementary nucleotides to the old strands it attaches to. However, one of the old strands called LEADING STRAND runs in the 3'-5' direction while the other strand called LAGGING STRAND runs in the 5'-3' direction.
DNA polymerase III only attaches to the 3' hydroxyll end of the DNA and synthesizes new strand of DNA in the 5'-3' direction. Since the lagging strand runs in 5'-3', it is synthesized in small separate fragments called OKAZAKI FRAGMENTS which are later joined together by an enzyme called LIGASE.
N.B: As DNA polymerase synthesizes DNA strand on the leading strand(5'-3'), it is moving in an opposite direction of the lagging strand. Hence, it has to detach and come back to synthesize on the lagging strand. This causes the lagging strand to be synthesized discontinuously.
Which is a characteristic of all waves?
Answer:
All kinds of waves have the same fundamental properties of reflection, refraction, diffraction and interference, and all waves have a wavelength, frequency, speed and amplitude. A wave can be described by its length, height (amplitude) and frequency. All waves can be thought of as a disturbance that transfers energy.
Explanation:
Answer: there's three characteristics
Explanation:
Three characteristics of waves can be measured: amplitude, wave-. length and frequency.
Most gastrointestinal infections are treated with
A. Antitoxin
B. Penicillin
C. Water and electrolytes
D. Quinacrine
E. Thorough cooking
Explanation:
Through water and electrolytes.
Why are G protein only found in Eukaryote cell ?
Answer:
they bind to protein-coupled transmembrane receptors with higher complexity than those found in prokaryotes
Explanation:
G-proteins are proteins found inside the cells that function as molecular switches which are activated by binding to guanosine triphosphate (GTP), while they are inactive by binding to guanosine diphosphate (GDP). The G-proteins bind to G-protein-coupled transmembrane receptors (GPCRs) in the cytoplasmic region. The GPCRs are a very diverse group of proteins that are activated by extracellular molecules ranging from small peptides to large proteins, including pheromones, neurotransmitters, light-sensitive compounds, etc, thereby allowing them to respond to diverse stimuli from the extracellular environment. In consequence, it is reasonable to suppose that the signaling pathways in which G proteins are involved have a higher complexity level than those observed in primitive prokaryotic organisms.
Complete each sentence by putting the labels to the appropriate blanks. Then place each sentence in a logical order beginning with blood entering the right side of the heart.
a. Blood enters the right ventricle from the right atrium through the right atrioventricular valve which is also known as the_________.
b. From the left ventricle, the blood is pumped up to the___________ through the left semilunar valve which is sometimes referred to as the___________.
c. Beginning with the return from the systemic circulation, blood enters the________.
d. The right ventricle contracts next, which forces blood up through the right semilunar valve which is otherwise known as the__________.
e. After circulating through the blood returns to the_______.
f. From the left atrium, blood moves through the left________ which is also referred to as the________ or mitral valve.
1. bicuspid
2. tricuspid valve
3. right atrioventricular valve
4. right atrium
5. aortic valve
6. left atrioventricular valve
7. atrioventricular valve
8. pulmonary valve
9. left atrium
10. aorta
11. pulmonary capillaries
12. systemic capillaries
Answer:
c. Beginning with the return from the systemic circulation, blood enters the right atrium.
a. Blood enters the right ventricle from the right atrium through the right atrioventricular valve which is also known as the tricuspid valve
d. The right ventricle contracts next, which forces blood up through the right semilunar valve which is otherwise known as the pulmonary valve .
e. After circulating through the pulmonary capillaries, the blood returns to the left atrium.
f. From the left atrium, blood moves through the left atrioventricular valve which is also referred to as the bicuspid or mitral valve.
b. From the left ventricle, the blood is pumped up to aorta the through the left semilunar valve which is sometimes referred to as the aortic valve.
Explanation:
c. Beginning with the return from the systemic circulation, blood enters the right atrium.
a. Blood enters the right ventricle from the right atrium through the right atrioventricular valve which is also known as the tricuspid valve
d. The right ventricle contracts next, which forces blood up through the right semilunar valve which is otherwise known as the pulmonary valve .
e. After circulating through the pulmonary capillaries, the blood returns to the left atrium.
f. From the left atrium, blood moves through the left atrioventricular valve which is also referred to as the bicuspid or mitral valve.
b. From the left ventricle, the blood is pumped up to aorta the through the left semilunar valve which is sometimes referred to as the aortic valve.
Alex, his mother, his father, and his older sister have blood type B. His younger sister and his brothers have type O. Anna, her mother, her father, and her younger brother have type A. Her sister and her older brother have type O. If Alex and Anna have three children, what is the probability at least two of them will have type O
Answer:
25%
Explanation:
the probability at least two of them will have type O can be solved as follows
Let the chances be
(Chance of Alex is I^B I ) x (1-none have type O) X (Chance of Anna is I^A i)
=>>(2/3) x (1-.75 3 ) x (2/3)= 0.25
So 0.25x 100 = 25%
The blood type of the offspring is dependent on the probable blood type
their parents.
The probability that at least 2 of them will have type O is [tex]\underline{\frac{5}{144}}[/tex]
Reasons:
Alex; Both parents are [tex]I^Bi[/tex]
Anna; Both parents are [tex]I^Ai[/tex]
Alex
[tex]\left[\begin{array}{lll}&I^B&i\\I^B&I^BI^B&I^Bi\\i&I^Bi&ii\end{array}\right][/tex]
Anna
[tex]\left[\begin{array}{lll}&I^A&i\\I^A&I^AI^A&I^Ai\\i&I^Ai&ii\end{array}\right][/tex]
Probability that [tex]I^Bi[/tex] from Alex = [tex]\frac{2}{3}[/tex]
Probability that [tex]I^Ai[/tex] from Anna = [tex]\frac{2}{3}[/tex]
Probability that two have type O = [tex]\frac{2}{3}[/tex] × [tex]\frac{2}{3}[/tex] × [tex]\left(\frac{1}{4}\right)^2[/tex] = [tex]\frac{1}{36}[/tex]
Probability that three have type O = [tex]\frac{2}{3}[/tex] × [tex]\frac{2}{3}[/tex] × [tex]\left(\frac{1}{4}\right)^3[/tex] = [tex]\frac{1}{144}[/tex]
Probability that at least two have type O is
[tex]P(At \ least \ 2)=\dfrac{1}{144} + \dfrac{1}{36} = \dfrac{5}{144}[/tex]
The probability that at least 2 of them will have type O is [tex]\underline{\frac{5}{144}}[/tex]
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the absence of the sry gene will result in: a) external genitalia of a female; internal genitalia of a male b) A phenotypic female c) and intersex individual d) A phenotypic male e) external genitalia of a male; internal genitalia of a female
Answer:
The correct answer is option b.
Explanation:
Sex determining Region of the Y chromosome gene or sry gene is a important gene found on the Y chromosome which is responsible for testis development or so the male characteristic. An XX individual with the SRY gene of the Y chromosome, will develop as male.
Hence, Absence of SRY gene will not result in phenotypic male So,option d is not correct. intersex have male characteristic so option c is not correct.
Absence of SRY gene will lead to development of phenotypic female due to form ovary and Fallopian tube in absence of this gene
Hence, option B is correct
Which of the following structures would you expect to see in meiosis of a cell with a previously existing reciprocal balanced translocation?
a. etravalent complex in meiosis I
b. tetravalent complex in meiosis II
c. unpaired loop in meiosis I
d. inversion loop in meiosis I
Answer:
A
Explanation:
my answer is A because it is correct
During genetic engineering, how do Restriction enzymes know what base pairs to act on? A. Ligase bonds it to the correct spot. B. They don’t, the scientist must target it manually. C. They bond to the Recognition site. D. It recognizes the sticky ends.
Answer:
C
Explanation:
Restriction enzymes, also known as restriction endonucleases or DNA-cutter identify the base pairs to act on by targeting and binding to the recognition site in a gene.
The recognition site in a gene/DNA consists of four to eight palindromic nucleotide sequence. It is a region that the restriction enzyme cuts during the genetic engineering process in order to produce sticky or blunt ends. A gene might contain more than one recognition site.
Restriction enzymes are usually found in prokaryotic organisms, especially bacteria where they function primarily as defense molecules against bacteriophages.
Correct option: C
A purebred tall pea plant is cross-pollinated with a tall, heterozygous pea plant. Use a Punnett square to determine the probability the offspring inherita
recessive short allele. (I point)
75%
25%
0%
50%
Answer:
The correct answer is - 25%
Explanation:
A cross between true tall pea plant and heterozygous tall pea plant is the cross of tall allele whic is TT and heterozygous which is Tt, so the gametes will be formed would be - T and T by true tall plant and T, and t allele by heterozygous plant.
The Punnett square of this cross is attached with the answer, where 2 heterozygous offspring and two tall offspring produced. In which there is only two recessive short allele formed in this generation out of 8 alleles.
So the probability of short allele would be:
= (2/8) *100
= (1/4) *100
= 25%
Which is/are possible genotypes for a person with straight little finger and brown eyes?
a. bbrr
b. bbRR
c. BbRr
d. bbRr
Answer:
The correct answer is b. bbRR.
Explanation:
Brown eyes color is dominant over blue eyes color. So, we can assign the letter R to the dominant allele that expresses brown eyes and r to the recessive allele that expresses the blue eyes. The phenotype bent little finger is dominant over the phenotype straight little finger. We can assign the letter B to the dominant allele, and b to the recessive allele.a. bbrr : This genotype belongs to a person with blue eyes and straight little finger
b. bbRR : This genotype belongs to a person with brown eyes and straight little finger
c. BbRr : This genotype belongs to a person with brown eyes and bent little finger
d. bbRr: This genotype belongs to a person with brown eyes and straight little finger.
For the partial diploid you've created, F' I+ P+ Oc Z− Y+ / I− P+ O+ Z+ Y− , you already know that the promoter sequences for copies are functional, so you can focus on the repressor proteins and operator regions that control expression.Indicate the source of repressor protein and the operator region(s) to which it binds. Then indicate how this affects expression of lac genes from the F' plasmid and from the bacterial chromosome.
MATCH
1. Repressor protein is made from__________.
2. Repressor protein is able to bind to the operator region on_________.
3. Functional beta-galactosidase protein could be made from__________.
4. Functional permease protein could be made from___________.
a. the F' plamid.
b. the bacterial chromosome.
c. both the F' plasmid and the chromosome.
d. neither the F' plasmid nor the chromosome.
Answer:
1. the F' plasmid.
2. the bacterial chromosome
3. the bacterial chromosome
4. the F' plasmid.
Explanation:
The F' plasmid is an extrachromosomal DNA molecule composed of genes that act during the transference of genetic material. This plasmid can also be defined as an episome capable of transferring itself into another bacterial chromosome. The bacteria that contain F factor are named F+, while bacteria without this factor are called F-. During conjugation, the F+ factor interacts with F- cells to enable DNA transference among bacteria.
A research team discovered that a novel hormone X stimulates an enzyme that hydrolyzes proteins in the small intestine. However, when hormone X was added to a test-tube mixture containing the enzyme and proteins, no breakdown occurred.
Required:
a. Why is this the case?
b. Name ONE signal transduction pathway that hormone X is using to transfuse its signal. (write all the steps and intermediates)
Answer:
a. It is a hormone that acts upstream in the signaling pathway (i.e., acting as a ligand by binding to cell receptors)
b. Dopamine-mediated pathways
Explanation:
It is well known that many cellular transduction pathways are activated via binding of signaling molecules (ligands) to receptors capable of triggering signaling cascades into the cell. For example, dopamine is a signaling hormone (neurotransmitter) involved in diverse brain processes (i.e., motor control, emotional response, etc). It is a neurotransmitter that binds to dopamine receptors and then stimulates the enzyme adenylate cyclase in a process that subsequently activates the production of the second messenger cyclic adenosine monophosphate (cAMP). Finally, cAMP acts as an intracellular signal transductor that play roles in many biological processes by transferring into the cell the effects of dopamine.
Which is NOT a function of magnesium in the body?
znutritio acting as a cofactor for enzyme systems in the Irody
regulating bone and mineral status
protecting teeth from acid-forming bacteria
making up one component in bone mineral
Answer: I think it is the third one which states, "Protecting teeth from acid-forming bacteria".
Explanation:
Which of the following describes the most likely impact that exposure to pollutants in the atmosphere would have on one’s personal health?
Answer:
It will lead to upper respiratory infections and pneumonia.
Explanation:
The backbone vertebrae, skull, and rib cage make up the _______ skeleton. Among other types of cells, bone marrow produces _______ blood cells, which carry oxygen in the blood. _______ can eventually become osteocytes. If bones rapidly deconstruct faster than new bone tissue grows, this can lead to less dense and more fragile bones. When severe, this condition is called _______. _______ such as the MCL and ACL connect bone to other bone at joints.
ANSWERS :
1.axial
2. red
3. Osteoblasts
4. osteoporosis
5. Ligaments
6. Osteoblasts are cells that build bone tissue, and are called osteocytes when they become surrounded by the bone that they have built. Osteoclasts are cells that break down bone tissue. The building and breaking down of bone tissue continues throughout a person’s lifetime and makes bones strong. If osteoclasts break down more bone than osteoblasts build, this can lead to osteopenia and osteoporosis, causing bones to be fragile and break more easily.
Answer:
1. axial
2. red
3. Osteoblasts
4. osteoporosis
5. Ligaments
Explanation:
The axial skeleton consists of the bones of the head and trunk of a vertebrate including skull, and rib cage.
Bone marrow produces red blood cells, which carry oxygen in the blood.
Osteoblasts can become osteocytes, which are the third type of bone cells.
Osteoporosis is a bone resorption disease in which bones rapidly deconstruct faster than new bone tissue grows, and decreases the mechanical strength of bones.
Ligaments connects one bone to other at joint such as medial collateral ligament (MCL) and anterior cruciate ligament (ACL) that joins knee.
Hence, the correct answer for the question is as follows:
1.axial
2. red
3. Osteoblasts
4. osteoporosis
5. Ligaments
Answer:
1. axial
2. red
3. Osteoblasts
4. osteoporosis
5. Ligaments
6. Osteoblasts are cells that build bone tissue, and are called osteocytes when they become surrounded by the bone that they have built. Osteoclasts are cells that break down bone tissue. The building and breaking down of bone tissue continues throughout a person’s lifetime and makes bones strong. If osteoclasts break down more bone than osteoblasts build, this can lead to osteopenia and osteoporosis, causing bones to be fragile and break more easily.
Explanation:
Penn Foster
Which of the following small GTPases are NOT involved in vesicle budding or docking? A. ARF B. Rab1 C. Ras D. Sar1p
Answer:
option c is correct that is Ras
Explanation:
Apoptosis differs from necrosis in that necrosis:_______ a. requires the reception of an extracellular signal. b. causes DNA to fragment. c. causes cells to swell and burst, whereas apoptotic cells shrink and condense.d. involves a caspase cascade
Answer:
c. causes cells to swell and burst, whereas apoptotic cells shrink and condense
Explanation:
These two mechanism are responsible for death of cells in multi cellular organisms.Apoptosis is naturally occurring process which is well organised and programmed.It is ensures the automatic death of old and worn out cells, needed for certain cellular developments. This is demonstrated by DNA fragmentation,cell shrinkage,and the break down of the nuclear walls.It can be initiated through two pathways of intrinsic and extrinsic,The latter leads to death from internal factors,while in the former is when external factor causes the death .
However,when cell death is influenced by external factors(toxins) in the cell environments,which lead to unpredictable or unregulated process leading to cell death ,the process is called Necrosis.it is demonstrated with the swelling and busting of the cell.and characterised with inflammation. It is an accidental death which causes injuries to the cells and tissues.
21. When looking at the effect of pH on sucrase activity, what was the wavelength of light
used in the spectrophotometer?
Answer:
540nm
Explanation:
Sucrase is one of the digestive enzymes responsible for the digestion of sucrose containing foods. Its major function is the catalysis of the hydrolysis of sucrose into its subunits; fructose and glucose.
Enzymes, as biochemical catalysts always function maximally at an optimum pH. As the pH of the system departs from the ideal pH for any enzyme, the activity of that enzyme decreases accordingly.
The activity of sucrose is studied using a spectrophotometer of wavelength about 540nm. As the pH of the solution departs from the specific ideal pH of sucrase, its activity decreases as stated above.
A.While the traits studied in Exercise 1 were hypothetical genetic traits, what type of genetic traits do you think are important to study and predict
Answer:
Genetic disturbances harmful to the organism are genetic characteristics that must be studied and predicted.
Explanation:
Genetic disorders that harm an organism such as sickle cell anemia, hemophilia, Turner syndrome, among others, are important to be studied and predicted, because it will provide better maintenance of that organism. Allowing people who inherit these characteristics to have more efficient treatments and a better life, since the prediction of these characteristics allows the family to establish better adapapitativo methods, that will make the life of this individual easier and with more quality.
What are two theories proposed to explain Primate Origins and how do they relate to traits we see in modern primates
Answer:
The level of specialization is given by the function.
The function of the organ makes the cell as it becomes mature, its more specific function becomes.
First, a stem cell differentiates itself into a hepatic or neuronal cell, thus generating that its functions are limited compared to the rest of the stem cells.
After this process, throughout the cell development and maturation of the organ, its cells become more specific in a single activity.
Explanation:
The differentiation of the stem cells is regulated by the genetic code, the migration of these during the embryonic stage of the mouse and the external stimulation that these stem cells perceive.
Draw concentric circles and label the four layers of the gut. Also label the hole at the inner most circle. List what is in these four layers.
Explanation:
here is your answer hopes it's helps u
We can find four principal layers composing the digestive tube walls. From the lumen to the exterior: Mucosa, Submucosa, Muscle, and Serosa Layers.
-----------------------------
There are four principal layers composing the digestive tube walls.
From the lumen to the exterior of the tube, we can find,
Mucosa Layer
Composed of lining epithelium, a propria lamina of connective tissue, and a smooth muscle layer.
The epithelium works as a barrier that separates the lumen from the organ.
Propria lamina has glands, vessels that get the absorbed substances, and immune system elements.
The smooth muscle delimitates the mucose from the submucosa layer. It produces movements that are independent of the rest of the organ's movements.
Submucosa Layer
Dense connective tissue is placed under the mucosa.
It has bigger vessels that ramification to the mucose, muscle layer, and serosa layer.
In certain areas, this layer also has glands.
Muscle Layer
Composed of two or three concentric smooth muscle layers that differ in their orientation.
The contraction of these layers mixes and propels the content of the lumen of the digestive tube.
Serosa layer
Connective tissue membrane covered by simple plane epithelium.
High caliber vessels are placed in this layer and limphatic vessels.
There are regions of the digestive tube where there is no serosa layer. Instead, there is a connective tissue named the adventitious layer.
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g UV radiation causes covalent dimerization of two subsequent ______ bases in DNA. This dimer is recognized and repaired by ______.
Answer: The options are not given.
Here are the options.
a. C and C ;;;; DNA photolyase
b. T and T ;;;;; DNA photolyase
c. C and C ;;;; DNA photoisomerase
d. T and T ;;;;; DNA photoisomerase
e. None of the above
Explanation:
UV light damages the DNA of cells that are exposed by making bonds to be formed between adjacent pyrimidine bases, usually thymines, in the DNA chains. The thymine dimers inhibit or hinder the DNA correct replication during reproduction of the cell.
UV radiation causes covalent dimerization of two subsequent T and T because Thymine bases of DNA directly absorbs a UVB photon . UVB light causes thymine base pairs close to each other in genetic sequences to bond together into pyrimidine dimers, thereby causing a disruption in the strand, which reproductive enzymes cannot copy.
UV-induced thymine dimers can be repaired by photoreactivation, in a process where energy from visible light is used to split the bonds forming the cyclobutane ring through the action of DNA photolyase, an enzyme that repaired damaged cause by uv radiation to dna.
Which organisms transform nitrogen to a form that is useful to plants?
bacteria
animals
fungi
humans
Answer:
Bacteria
Explanation:
Nitrogen fixation is the step of the nitrogen cycle that occurs with the help of microbes. Bacteria transform the nitrogen into a usable form for the plants. Thus, option a is correct.
What are nitrogen-fixing bacteria?The nitrogen cycle is the part of the biogeochemical cycle that converts nitrogen into other forms that can be used by plants. The nitrogen is converted through the nitrogen-fixing bacterias.
The nitrogen-fixing bacterias are heterotrophic in nature that lives freely in the soil. A few examples are Bacillus, Azotobacter, Clostridium, etc. These bacterias transform the atmospheric nitrogen into usable compounds called ammonia which is the fixed nitrogen and can be used by the plants.
Therefore, nitrogen-fixing bacterias transform nitrogen.
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While cooking, Lisa spills boiling water on her hand. Her skin turns blotchy with blisters. Which type of burn does she have? A. first-degree burn B. second-degree burn C. third-degree burn
Answer:
B
Explanation: First degree only makes the skin turn red
. Buffer solution resists change to its pH when small amounts of an acid or an alkali are added to it. Buffer solutions can be used to keep the pH of a substance constant during an experiment. For example, if pH 5.5 buffer solution is added to a mixture of amylase and starch solution, the pH of the mixture will remain constant at 5.5. The student has the following buffer solutions available: 5.5, 6.0, 6.5, 7.0, 7.5 and 8.0. Describe how the student can adapt the tube experiment to investigate the effect of pH on the action of amylase.
Answer:
hey
Explanation:
In pea plants there is a dominant allele (A) for green pods and a recessive allele (a) for yellow pods. Suppose a heterozygous plant is crossed with a plant that has yellow pods. Complete the sentences about this cross with the correct terms.
1. The phenotype of the heterozygous plant is_________.
2. The genotype of the heterozygous plant is_________.
3. The genotype of the plant with yellow pods is________.
4. The genotypes of gametes produced by a heterozigous plant is______.
5. The genotypes of gametes produced by a yellow plant is_______.
A. Green pods
B. 75% AA and 25% Aa
C. AA
D. 100% A
E. Yellow pods
F. 50%
G. 25%
H. 100% a
I. 50% A and 50% a
J. Aa
K. 0%
L. 75%
M. 75% A and 25% a
Answer:
1. Green
2. Aa
3. aa
4. A and a
5. a and a
Explanation:
1. The phenotype of the heterozygous plant (Aa) would be green since the allele responsible for green color (A) is dominant over the allele responsible for yellow color (a).
2. The genotype of a heterozygous plant would be Aa. Heterozygous individuals have different alleles for the same gene.
3. The genotype of the plant with a yellow pod would be aa. Since the allele for green color (A) is dominant over the allele for yellow color (a), the yellow color can be expressed only in the absence of the (A) allele. Hence, the genotype fo yellow color has to be aa.
4. The genotypes of gametes produced by a heterozygous plant would be (A) and (a) because heterozygous individuals have two different alleles for a gene.
5. The genotypes of gametes produced by a yellow plant would be (a) and (a) because a yellow plant has the genotype (aa).
What causes potential energy that accumulates in a fault to transfer to the surface during an earthquake? A. refraction B. Triangulation C. Seismic waves D. Lithostatic pressure
Answer: C.
Explanation: Seismic waves causes potential energy that accumulates in a fault to transfer to the surface during an earthquake.
Seismic waves cause potential energy that accumulates in a fault to transfer to the surface during an earthquake. So, the correct option is C.
What are Seismic waves?A seismic wave is defined as a wave of acoustic energy that travels through the Earth or any other planetary body that can be generated by an earthquake, volcanic eruption, magma movement, a large landslide, and a large man-made explosion that produces low frequency acoustics energy.
There are three basic types of seismic waves which are as follows:
P-wavesS-waves Surface waves.P-waves and S-waves are sometimes collectively referred to as body waves.
Thus, Seismic waves cause potential energy that accumulates in a fault to transfer to the surface during an earthquake. So, the correct option is C.
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Which of the following reflects the relationship between the posterior and the anterior pituitary glands?A. Unlike the posterior pituitary, the anterior pituitary makes and releases its own hormones without input from the hypothalamus.B. One is always a little behind.C. The anterior pituitary is under the control of the hypothalamus but the posterior pituitary is not.D. Unlike the posterior pituitary, the anterior pituitary releases "releasing or inhibiting" hormones.E. The posterior pituitary releases neurohormones and the anterior pituitary releases hormones.F. They are both extensions of the hypothalamus.
Answer:
The correct answer is - option E.
Explanation:
The anterior and posterior pituitary glands are he two distinct lobes of the pituitary gland which is controlled by the hypothalamus. The anterior lobe of the master gland is receives its signals from the specific neurons of the hypothalamus known as parvocellular neurons and produces six hormones that release in to circulation by itself.
On other hand the extension of the hypothalamus or the posterior pituitary gland receives the neurohormones from the hypothalamus such as oxytocin and ADH. Neurohormones are the hormones that are released by neurons of the posterior pituitary gland.
Thus, the correct answer is option E.
g dGDP is made from ________ by the ribonucleotide reductase. This enzyme is inactive when ______ is bound to its master regulatory pocket.
Answer:
1. GTP dephosphorylation
2. hydrolyzed or removed
Explanation:
GDP, (Guanosine diphosphate) is a biological term, that is made of composition including pyrophosphate group, a pentose sugar ribose, and the nucleobase guanine and it is made from GTP ( Guanosine triphosphate ) dephosphorylation by the ribonucleotide reductase. This enzyme is inactive when hydrolyzed or removed, and then eventually bound to its master regulatory pocket.
Vaccination is a process of injecting a dead or weakened form of a pathogen into the body. How does this help strengthen the immune system? It allows the body to increase its internal temperature. It allows the body to regulate the inflammatory response. It allows the body to keep pathogen information for future infections. It allows the body to increase its desire to eat healthy food.
Answer:
It allows the body to keep pathogen information for future infections.
Explanation:
Vaccination is the process whereby a substance called VACCINE is injected into body in order to prepare the immune system for resistance against infections. A vaccine contains the harmless or dead version of a pathogen called ANTIGEN, which functions to give the immune system defense mechanism against that form of pathogen in future.
The vaccine works by stimulating the immune system of the body to fight against the antigens in the same way it will fight against an actual pathogen in the future. Hence, the body keeps the information on how to combat such pathogens when there is an actual infection.
Answer
The answer is C edge 2020
Explanation:
