Answer:
Epidermis layer is responsible for the protection of meristematic region.
Explanation:
Meristematic region is protected in the stem tip by epidermis which consist of dead layer of cells. Epidermis is the outer layer of stems, leaves, flowers and fruits which is responsible for the protection of inner part from damage. Vascular bundle such as phloem present near the boundary of the stem while the xylem is present in the inside of the stem. In the inside layer of the stem, all the xylem cells are genetically identical while the layer that is present at the edge of the stem is phloem in which all the cells are genetically identical to each other.
Describe the structure of G protein ?
Answer:
G protein can refer to two distinct families of proteins.
Explanation:
heterotrimric G proteins are large in shape, G proteins are activated by G protein coupled receptors and are made up of alpha (a) beta(b) and gamma(y) subunits.
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.
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).
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:
While not an official step in the process, Acetyl CoA formation is a transition between glycolysis and krebs, and occurs only in the presence of oxygen.
A. True
B. False
Answer:
A. True
Explanation:
Acetyl coA is a compound which helps in the transport and availability of the acetyl group to the citric acid cycle (Krebs cycle).
Though it is not an official step in the process, Acetyl CoA formation is a transition between glycolysis and krebs, and occurs only in the presence of oxygen.
The Acetyl group helps in the oxidation for energy production.
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 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.
What are the similarities and differences between prokaryotic cells and eukaryotic cells?
Which two Neolithic activities came about because of climate change?
Answer:
1. Domesticating animals
2. Farming grains
Explanation:
Poeple started doing these without machines because of climate change.
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:
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:
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 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.
____ are thought to be present before vertabrates
Answer:
Jawless fish was present before vertebrates.
Explanation:
Jawless fish has some characteristics of vertebrates and also considered as the ancestor of vertebrates on earth which is similar in appearance with the hagfish. It was present about 500 to 600 million years ago. They consist of cranium but vertebral column is absent. but with the passage of time, evolution occurs and many animals which was the descendant of this jawless fish having vertebral column or backbone in their body.
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.
Circle
1
2
Circle the most reactive metal in each set.
1) Magnesium / Potassium
2) Aluminum / Gold
3) Cobalt / Cesium / Calcium
4) Iron / Titanium / Potassium
5) Francium / Lithium / Beryllium
لی
Answer:
Explanation:
1) potassium
2)Aluminium
3) Cesium
4) potassium
5) Beryllium
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:
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%
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 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.
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
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.
Name two structural characteristics that triglycerides and phospholipids have in common.
Answer:
They both have fatty acids.
Explanation:
which vary in number whereby triglycerides has three fatty acids while phospholipids have two fatty acids.
Answer:
Explanation:
which vary in number whereby triglycerides has three fatty acids while phospholipids have two fatty acids.
make sure to add
two fatty acids
glycerol
carboxyl group
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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. 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:
Most gastrointestinal infections are treated with
A. Antitoxin
B. Penicillin
C. Water and electrolytes
D. Quinacrine
E. Thorough cooking
Explanation:
Through water and electrolytes.
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.
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.
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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