Answer:
D) Pseudomonas
Explanation:
Pseudomonas is the group of gram negative bacteria which are considered opportunistic pathogens. This means that they infect in presence of an already existing ailment like cystic fibrosis. They usually target immuno compromised people.
Normally they are present on soil, water and skin. They are a major cause of hospital acquired infections like sepsis syndrome and ventilator-associated pneumonia. They thrive in moist conditions hence can be found on medical equipment causing cross infections in hospitals. They are resistant to many antibiotics hence the treatment is difficult.
Which of the following explains what happens to oxygen produced by the light-dependent reactions?
A) It is used in the Calvin cycle
B) It is released into the atmosphere.
C) It combines with NADPH to produce water.
D) It is recycled as a reactant in another light-dependent reaction
Answer:
The correct answer is option B.
Explanation:
The light-dependent reactions in the process of photosynthesis utilize Sun's light energy to dissociate water, known as photolysis. Water after getting dissociated produces hydrogen, oxygen, and electrons. The electrons move through the compositions in the chloroplasts and by the process of chemiosmosis, produce ATP.
The hydrogen gets transformed into NADPH, which is further utilized in the light-independent reactions. While oxygen diffuses out of the plant as a waste component of photosynthesis into the atmosphere. All this takes place in the grana thylakoids of the chloroplasts.
What enzyme complex is responsible for transcription?
a. DNA polymerase
b. RNA polymerase
c. hexokinase
d. amylase
Answer:
The correct answer will be option-B.
Explanation:
Transcription is a process which produces a complementary transcript molecule of DNA called RNA. The RNA formation takes place in three steps: initiation, elongation and termination.
The initiation step involves a complex enzyme called RNA polymerase which synthesize a complementary strand using one strand of DNA. the enzyme adds a new nucleotide at 3' end of the strand thereby proceeding the reaction in 5' to 3' direction.
Thus, Option-B is the correct answer.
Answer:
RNA polymerase.
Explanation:
Transcription may be defined as the process of formation of RNA molecules from the DNA template. The process of translation occurs in the 5' to 3' direction.
The main enzyme responsible for the transcription is DNA dependent RNA polymerase or RNA polymerase. This enzyme is important for the synthesis of the RNA molecule complementary to the DNA template. RNA pol II is the main transcription enzymes of the mRNA molecule in case of eukaryotes.
Thus, the correct answer is option (b).
is the transport of potassiun ions into the cell endergonic
orexergonic and why?
Answer: Endergonic
Explanation:
The sodium-potassium pump (Na+/K+ pump) moves sodium and potassium ions against its concentration gradient, which means in opposite directions through the membrane. Two potassium ions are imported into the cell while three sodium ions are exported.
This pump uses the energy derived from exergonic ATP hydrolysis. This means, ATP releases energy that will be used by the pump. Thereby the use of energy by the pump is an endergonic reaction, because it requires the absorption of energy to function.
So, when ATP is hydrolyzed, it donates free energy to the pump which goes under a conformational change, allowing it to move the ions.
A column on the periodic table is called a group. Which statement would not support the conclusion that two elements are in the same group?? A) The two elements both have similar atomic radii B) The two elements both have very high melting points C) The two elements both are highly reactive in oxygen D) The two elements both bonds with chlorine to form a salt in a 1 to one ratio
Answer: Option (A) is the correct answer.
Explanation:
A group in a periodic table is represented by a column. And, when we move down a group then there occurs an increase in atomic size or atomic radii of the elements.
This is because number of electrons add into new shells which leads to increase in atomic radii of the element.
Therefore, if we say that two elements in a group have similar atomic radii then it is false.
Whereas the two elements of a group might have a high melting point.
Also the statements, the two elements both are highly reactive in oxygen and the two elements both bonds with chlorine to form a salt in a 1 to one ratio can be true for elements of the same group.
Thus, we can conclude that the statement two elements both have similar atomic radii would not support the conclusion that two elements are in the same group.
The statement that would not support the conclusion that two elements are in the same group is B) The two elements both have very high melting points.
Explanation:The statement that would not support the conclusion that two elements are in the same group is B) The two elements both have very high melting points. Grouping elements in the same group is based on their similar chemical properties, not on their physical properties such as melting points. For example, elements in the same group may have different melting points due to varying atomic structures and bonding.
How might a protein kinase select the correct residues
tophosphorylate given the availability of serine and
threonineresidues on the surface of globular proteins?
Answer:
The flanking residues facilitate recognition and specificity.
Explanation:
The kinase is able to to specifically phosphorylate serine and threonine due to the recognition of the flanking residues. Altogether, these form a consensus sequence across the substrate.
Where would you find tubulin in a cell? List 2-3 structures.
Answer:
Tubulin are the monomers of the microtubules that plays an important role in the formation of cytoskeleton. The cytoskeleton helps in providing structural framework to the cytoplasm and maintains the cell mobility.
Tubulin is numerous inside the cell. The centrosomes is the main organizing center of the microtubules that plays an important role in the cell division. The golgi bodies also constitute of the microtubules that maintain its structure. The flagella and cilia are made of tubulin that helps in the cell mobility.
How many membranes does the chlorplast have?
a. 1
b. 2
c. 3
d. 4
Answer:
B
Explanation:
Chloroplasts have two membranes: external and internal membrane. These membranes contain 60% of lipids and 40% of proteins, just like the typical cell membranes. The inner or internal membrane is virtually non-permeable to any substances but it has transport proteins, on the other hand, the outer or external membrane is permeable to most of the small molecules
Bacteria reproduce by simple cell division. What is the official name for this class of cell division?
a. Sexual reproduction
b. Budding reproduction
c. Binary fusion
d. Binary fission
e. None of the above
Which of the following statements concerning photosynthesis is NOT true?
a. Photosynthesis occurs in the chloroplast
b. In photosynthesis, CO2 is reduced
c. During photosynthesis, the hydrogen carrier is NAD
d. During photosynthesis, O2 is released
Answer: c. During photosynthesis, the hydrogen carrier is NAD
Explanation:
The photosynthesis is a process that occurs in the chloroplast organelle of the plant cells. The typical light reaction of the photosynthesis takes place inside the thylakoid membrane of the chloroplast. The chief reactants such as water and carbon dioxide participate in the reaction. The water gets oxidized in the reaction to release the oxygen as a product. The carbon dioxide get reduced.
The light accepted by the pigment chlorophyll in the chloroplast basically transfers the electron and hydrogen from water to the acceptor that is called as NADP⁺.
Answer:
c. During photosynthesis, the hydrogen carrier is NAD
Explanation:
The co-factor during photosynthesis is Nicotinamide Adenine Dinucleotide Phospahte .(NADP), and not NAD.The latter is a coenzyme in cellular respiration for ferrying H+(as NADH+) together with FADH+ from Kreb's Cycle into the matrix of mitochondria for oxidative phospshorylation, needed for ATPs synthesis,
NADPH is a co-factor, needed as reducing agent in photosynthesis's cyclic and non cyclic photophosphorylation reactions.Therefore option C is not true.
Which of the following combinations is most likely to result in digestion?
a. pepsin, protein, water, body temperature
b. pepsi, protein, water, body temperature, HCl (hydrochloric acid)
Explain.
Answer:
Option (b).
Explanation:
Digestion may be defined as the process of break down of large food particles into simpler substances that be absorbed by the body. Specific enzymes are required for the breakdown of specific chemicals.
The combination that contains protein, water, HCl (hydrochloric acid), temperature and pepsin result in digestion. HCl (hydrochloric acid) activates the enzyme pepsinogen into pepsin. The pepsin than digest the proteinsat the body temperature in presence of water. Proteins are broken down into simple amino acids that can be seen as a result of digestion.
Thus, the correct answer is option (b).
5) Strawberry plant clones are given different amounts of water for a 3-week period. First strawberry plant
receives 400 millilitres (ml) a day. The second strawberry plant receives 200ml a day. The third strawberry
plant receives 100ml a day. The fourth strawberry plant does not receive any extra water; this plant only
receives natural ways of receiving water. The height of the strawberry plants is recorded daily.
IV :
DV :
CG :
Con :
Answer:
IV: Amount of water
DV: Height of the strawberry plant
CG: The fourth strawberry plant
Explanation:
The Independent Variable is the amount of water given to the strawberry plants; the Dependent Variable is the height of the strawberry plants; the Control Group is the fourth strawberry plant that only receives natural ways of water, without any extra water treatment.
What information does this experiment include?In an experiment with strawberry plant clones, different amounts of water are provided to observe their effect on plant height. The amount of water is the independent variable (IV), while the height of the plants is the dependent variable (DV).
One plant receiving no extra water serves as the control group (CG) for comparison. Constants include the experiment's duration, environmental conditions, plant type, measurement method, and initial plant height. This investigation aims to understand the impact of water availability on plant growth.
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Describe the substrate specificity of chymotrypsin and the structural feature that determines this specificity.
Answer:
The main substrate of chymotrypsin includes tryptophan, tyrosine, phenylalanine and methionine
Explanation:
1.
Histidine yields a proton to aspartate and recovers it from serine.
Seen in another way: aspartate captures a proton from the serine through histidine.
2.
a) The serine (deprotonated) is thus capable of attacking the peptide bond (nucleophilic carbonyl attack) and forms a tetrahedral intermediate; The substrate is thus covalently bound to the enzyme (now it is a transition state).
b) The peptide bond is broken and the released amino terminus (R) recovers a proton from histidine.
c) Histidine, in turn, recovers it from aspartic.
3.
a) Aspartate captures a proton of histidine again, so that it can capture it in turn from water.
b) This generates a hydroxide anion that attacks the ester intermediate between the serine and the carboxyl part (R ′) of the substrate peptide.
c) A new tetrahedral intermediate bound to the enzyme is formed (via serine residue).
4.
a) The carboxyl group of the peptide is regenerated, the serine being separated and the other peptide fragment being free (the R ′ part with a free carboxyl end)
b) The serine recovers the proton at the expense of histidine, which in turn captures it from aspartic acid.
c) The catalytic triad (Asp, His, Ser) has been regenerated in its original state.
The net reaction is:
R–NH – CO –R ′ + H2O ⟶ R– NH2 + HOOC –R ′ ⟶ R – NH3 + + −OOC –R ′
The active site or catalytic center of chymotrypsin is formed by several amino acid residues, among which the essential role corresponds to the "catalytic triad".
What do B cells do when they identify a pathogen?
A. Engulf and eat it
B. Kill it
C. Produce antigens to bind to the pathogen and memory cells to guard against future infection.
D. Produce antibodies to bind to the pathogen and memory cells to guard against future infection.
E. Engage helper T cells to fight the infection
D
Explanation:
With the help T cells , B cells make special proteins called antibodies which stick to antigens on surface of germs stopping them in their track.
Humerus bones from the same species of animal have approximately the same length-to-width ratios. It is known that Species A has a mean ratio of 8.5. Suppose that 41 fossil humerous bones were unearthed at a site where Species A is known to have flourished. (We assume that all bones are from the same species.) The length-to-width ratios of these bones has sample mean 9.26 and sample standard deviation 1.20. Can we conclude that these bones belong to Species A? Perform a level 0.05 z-test to check.
Answer:
The unearthed Humerous bones don't belong to the species A.
Explanation:
Hello!
You are studying the Humerus bones species A, who is known to have a mean ratio of 8,5. This value corresponds to the population mean of the length-to-with ratio of bones of species A, symbolized as μ.
The hypothesis you want to study is "The Humorous bones unearthed belong to the species A" if you assume this to be true, then the mean of the length-to-with ratio should be equal to the known population mean of the length-to-with ratio.
Symbolized:
H₀: μ = 8,5
H₁: μ ≠ 8,5
Significance level: α: 0,05
You are asked to use a Z-test, since you don't know the value of the population variance, but have the sample values, the sample size is big enough (more than n=30). Assuming that the sample values are independent, the statistic test of choice is the approximation:
[tex]Z= \frac{x_bar-μ}{S\sqrt[]{n} }[/tex]≅N(0;1)
The critical region, in this case, it's a two-tailed test (remember the type is determined by the null hypothesis) so you'll have two critical values.
Left value [/tex][tex]Z_{\alpha/2} = Z_{0,025} = -1,96[/tex]
Right value [tex]Z_{1-\alpha/2} = Z_{0,975} = 1,96
So you'll reject the null hyphotesis if the calculated [tex]Z_{obs}[/tex] value is ≤-1,96 or ≥1,96 and you'll support it if -1,96<[tex]Z_{obs}[/tex]<1,96
Now we calculate the statistic by replacing the formula with the data:
x_bar = 9,26
S = 1,20
n = 41
[tex]Z_{obs}= \frac{x_bar-μ}{S\sqrt[]{n} }[/tex]
[tex]Z_{obs}= \frac{9,26-8,5}{1,20\sqrt[]{41} }[/tex]
[tex]Z_{obs}[/tex]= 4,0553
Since the calculated value falls in the rejection region, this means, you have statistically significant results. In other words you can reject the null hipothesis (H₀: μ = 8,5) and asume that the unearthed Humerous bones don't belong to the species A.
I hope you have a SUPER day!
Reject null hypothesis; mean ratio of fossil bones significantly greater than Species A mean, based on z-test at 0.05 level.
To check if the fossil humerus bones belong to Species A, we can perform a z-test for a population mean.
Given:
- Population mean [tex](\( \mu \))[/tex] of Species A = 8.5
- Sample mean [tex](\( \bar{x} \))[/tex] of the fossil humerus bones = 9.26
- Sample standard deviation [tex](\( s \))[/tex] of the fossil humerus bones = 1.20
- Sample size [tex](\( n \))[/tex] = 41
- Significance level [tex](\( \alpha \))[/tex] = 0.05
The null hypothesis ([tex]\( H_0 \)[/tex]) is that the mean length-to-width ratio of the fossil humerus bones is equal to the mean ratio of Species A [tex](\( \mu = 8.5 \))[/tex].
The alternative hypothesis ([tex]\( H_1 \)[/tex]) is that the mean length-to-width ratio of the fossil humerus bones is greater than the mean ratio of Species A [tex](\( \mu > 8.5 \))[/tex].
We'll use the z-test formula:
[tex]\[ z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \][/tex]
Substituting the given values:
[tex]\[ z = \frac{9.26 - 8.5}{\frac{1.20}{\sqrt{41}}} \][/tex]
[tex]\[ z \approx \frac{0.76}{\frac{1.20}{\sqrt{41}}} \][/tex]
[tex]\[ z \approx \frac{0.76}{\frac{1.20}{\sqrt{41}}} \][/tex]
[tex]\[ z \approx \frac{0.76}{0.187} \][/tex]
[tex]\[ z \approx 4.07 \][/tex]
Now, we compare the calculated z-value with the critical z-value at [tex]\( \alpha = 0.05 \)[/tex] for a one-tailed test.
From the z-table, [tex]\( z_{\alpha} = 1.645 \)[/tex] (approximately).
Since the calculated z-value (4.07) is greater than the critical z-value (1.645), we reject the null hypothesis.
Therefore, we can conclude that the mean length-to-width ratio of the fossil humerus bones is significantly greater than the mean ratio of Species A, at the 0.05 level of significance. Thus, we cannot conclude that these bones belong to Species A.
Cystic fibrosis is a genetic disorder caused by a recessive allele. If two parents each carry an allele for the disorder but have normal phenotypes, indicate what percentage of their children will be phenotypically normal and what percentage will have cystic fibrosis.
Answer: 75% normal phenotype, 25% cystic fibrosis.
Explanation:
Since cystic fibrosis is caused by a recessive allele, then it needs both affected alleles to develop. If the parents have normal phenotyps but they carry one allele for the disorder, then their genotypes will be Cc for both (they are heterozygous).
C is the dominant allele that codes for the trait, and c is the affected allele.
They only have one copy of the affected allele, thereby they will not develop cystic fibrosis.
The next step is to find out the genotypes of the gametes.
Gametes are sex cells, sperm or eggs produced by the parents. Those cells only have one allele of the gene, that means the gametes produced by them can be either C or c. Those gametes are used in the punnett square as shown in the picture.
There we can see 50% of the offspring is Cc, 25% is cc and 25% is CC.
As it was stated before, individuals who are cc will develop cystic fribrosis. Then, 25% of them will have it. And 75% of them (25% of CC + 50% of Cc) will not.
Males have hemophilia when they are hemizy-gous for a nonfunctional recessive mutant allele of the X-Iinked gene for clotting factor Vlll. Factor Vlll is normally secreted into the blood serum by cells in the bone marrow that produce it. a. Do you think that females heterozygous for the hemophilia disease allele could have hemophilia in some parts of their bodies and not others? b. If such a female "carrier" of hemophilia suffered a cut, would her blood coagulate (form clots) faster, slower, or in about the same time as that of an individual homozygous for a normal allele of the factor Vlll gene? Would the rate of clotting vary significantly among heterozygous females?
Answer:
a. No, it is not possible.
A heterozygous female carries one copy of functional gene which is enough for the production of clotting factor. Hemophilia does not show continuous variation or polygenic inheritance and thus, its level does not depend on the number of normal alleles.
It that was the case, then all males would show hemophilia in some parts of the body as they only carry one X chromosome and thus, only one functional gene.
Thus, heterozygotes are only the carriers of the disease, they do not show any symptom of the disease.
b. In perspective of homozygosity or heterozygosity, the rate of blood clotting should be the same as both of them have functional gene. As mentioned above, it does not show continuous variation so, it will not show any increased or decreased rate of clotting in homozygotes or heterozygotes.
However, in reality, the rate of clotting depends on the concentration of clotting factor present in blood plasma. This percentage depends on the physiology of a person but not on the number of alleles present. For example, proteins or enzymes required for gene expression, et cetera.
Which of the following statements about the basic structural features of DNA are true? Select all true statements. Select all true statements. The major and minor grooves prevent DNA binding proteins from making contact with nucleotides. In a DNA macromolecule, the two strands are complementary and antiparallel. The twisting of the DNA double helix is attributed to the tight packing of DNA bases and base-stacking. The major and minor grooves form in the DNA helix because the DNA strands are antiparallel.
Answer:
Option (2) and (3).
Explanation:
DNA is the genetic material of all organisms except some viruses. DNA is composed of the nitrogenous base, pentose sugar and the phosphate group. DNA strands are complimentary to each other.
The DNA strands run in the opposite direction, one strand in 5' to 3' direction and other strand in 3' to 5' direction thus they are anti parallel with each other. The DNA strand can twist with each other which result in the tight packing of DNA base and base stacking. DNA consists of major and minor grooves.
Thus, the correct answer is option (2) and (3).
The true statements about the basic structural features of DNA include the complementary and antiparallel nature of the two strands, the twisting of the double helix due to base packing, and the formation of major and minor grooves.
Explanation:The true statements about the basic structural features of DNA are:
1. In a DNA macromolecule, the two strands are complementary and antiparallel. This means that one strand of DNA runs in the 3' to 5' direction, while the other strand runs in the 5' to 3' direction.
2. The twisting of the DNA double helix is attributed to the tight packing of DNA bases and base-stacking. The hydrogen bonds between the nitrogenous bases hold the two strands together and contribute to the helical structure of DNA.
3. The major and minor grooves form in the DNA helix because the DNA strands are antiparallel. These grooves are binding sites for DNA binding proteins during processes such as transcription and replication.
Where would you find a Riboswitch?
Answer:
The correct answer will be- in the 5' untranslated region of prokaryotic mRNA.
Explanation:
Riboswitches are the cis-regulatory RNA elements present in the non-coding segment of the mRNA.The riboswitches are common in prokaryotes but research showed that they are also present in the few eukaryotes. They are located in the non-coding 5' untranslated region of prokaryotic mRNA.
The riboswitches are usually made of 35 to 200 nucleotides which can bind to co-enzymes, small molecules and metabolites and change their conformation to regulated gene expression.
Thus, in the 5' untranslated region of prokaryotic mRNA is the correct answer.
Identify and describe the portals through which pathogens invade the human body.
Answer:
Contamination signifies the possible existence of microbes in or on the body. Infection is considered as the successful invasion of the pathogen in the body. The pathogens can acquire access or can invade the body of a human being with the help of portals, like via the skin, placenta, and mucous membranes.
The parenteral route is not precisely considered as a portal, however, it is a way of circumventing the general portals.
Describe the "blender experiments" of Hershey and Chase and what they revealed about DNA versus protein.
Answer:
Explanation:
Alfred Hershey and Martha Chase showed that DNA is the genetic material by their blending experiment. Earlier it was believed that protein is the genetic material. As it is found in a larger amount in the cell.
Hershey and Chase infected the bacteria E coli with the T2 phage virus. In the bacteria, the viral DNA is replicated as the bacterial DNA. They prepare 2 culture media one contains radiolabeled phosphorus (32P) and another has radiolabeled (35 S)sulfur. Then he cultures the bacteria in separate culture medium and infected with many T2 phages.
After the infection, they isolated the infected cell and centrifuge the cells. When they tested separate supernatant of 2 culture mediums, they found the virus of radiolabeled phosphorus doesn't show any radioactivity. But the virus in the radiolabeled sulfur contains the radiolabeled.
The viruses have 2 biomolecules one is protein and the other is DNA/RNA. When they have centrifuged the cells which are the palette, the phosphorus is found there and it also degrades. But the sulfur which is present in the supernatant is present in the newly replicated viruses.
He also treated the virus with both the culture medium but newly infected cells show radiolabeled sulfur, not the phosphorus. Thus they concluded DNA is the genetic material.
According to the punctuated equilibria model,
a. given enough time, most existing species will branch gradually into new species.
b. most new species accumulate their unique features relatively rapidly as they come into existence, then change little for the rest of their duration as a species.
c. most evolution occurs in sympatric populations.
d. speciation is usually due to a single mutation.
Answer:
b. most new species accumulate their unique features relatively rapidly as they come into existence, then change little for the rest of their duration as a species
Explanation:
Species can evolve fast or slowly. When species accumulate changes at a slow rate this is called gradualism.
If a species suddenly has a an enormous change for example through mutations in the genes of a few individuals or a single individual the species can has evolve and may become suddenly a new one. A good example can be found in the following link: https://blog.education.nationalgeographic.org/2018/02/07/cloned-crayfish-are-taking-over-the-world/
the article provides evidence of sudden changes that may lead to speciation in at a really fast rate.
The most accurate statement is: 'most new species accumulate their unique features relatively rapidly as they come into existence, then change little for the rest of their duration as a species'. So, option b is true.
Explanation:According to the punctuated equilibria model, the most accurate statement is: 'most new species accumulate their unique features relatively rapidly as they come into existence, then change little for the rest of their duration as a species'. This model suggests that evolution happens in rapid bursts, interspersed with long periods of stability, often called stasis. Essentially, a species experiences little evolutionary change for most of its existence until a sudden shift in its environment or population pressures spur a rapid period of adaptation and evolutionary change, leading to the formation of a new species. Hence, this model disputes the idea that evolution is a slow, gradual process happening over a long period, a theory known as gradualism.
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Griffith, in his 1928 experiments, demonstrated that bacterial strains could be genetically transformed. The evidence that DNA was the "transforming principle" responsible for this phenomenon came later. What was the key experiment that Avery, MacCleod, and McCarty performed to prove that DNA was responsible for the genetic change from rough cells into smooth cells?
Griffith's experiment worked with two types of pneumococcal bacteria (a rough type and a smooth type) and identified that a "transforming principle" could transform them from one type to another.
At first, bacteriologists suspected the transforming factor was a protein. The "transforming principle" could be precipitated with alcohol, which showed that it was not a carbohydrate. But Avery and McCarty observed that proteases (enzymes that degrade proteins) did not destroy the transforming principle. Neither did lipases (enzymes that digest lipids). Later they found that the transforming substance was made of nucleic acids but ribonuclease (which digests RNA) did not inactivate the substance. By this method, they were able to obtain small amounts of highly purified transforming principle, which they could then analyze through other tests to determine its identity, which corresponded to DNA.
Final answer:
The key experiment by Avery, MacLeod, and McCarty that demonstrated DNA as the transforming principle involved degrading specific cellular components and observing that only when DNA was degraded did the transformation of R strain to S strain fail to occur.
Explanation:
Avery-MacLeod-McCarty Experiment
The pivotal experiment that proved that DNA was responsible for the genetic change from rough cells into smooth cells was performed by scientists Oswald Avery, Colin MacLeod, and Maclyn McCarty in 1944. Their experiment built upon Frederick Griffith's transformation experiments with Streptococcus pneumoniae, where he observed the transformation of nonpathogenic rough (R) strain into a pathogenic smooth (S) strain.
Avery, MacLeod, and McCarty systematically used enzymes to degrade specific components (proteins, RNA, and DNA) and observed the ability to transform the R strain. When DNA was destroyed, the transformation did not occur, leading them to the conclusion that DNA was the transforming principle. Thus, it was DNA that carried the hereditary information necessary for the transformation.
What are the advantages of being a respiratory parasite (the tongue worms) over one in the digestive system (if any)?
Answer:
Explanation:
The parasites living at the respiratory system or tongue worms will hide under the soft tissues of the oral cavity, beneath the tongue and even beneath the throat or esophagus. These worms remain undetected during diagnosis. These worms are not subjected to the treatment of acids which worms in the digestive system are exposed to. Thus these worm parasites in the respiratory system or tongue survive comparatively for long as compared to the worms in the digestive system.
Luteinizing hormone is bound to transport proteins in the plasma.
a. True
b. False
Answer:
FALSE
Explanation:
Luteinizing hormone, also known as the lutropin, is a heterodimeric glycoprotein produced by the gonadotropic cells of the anterior pituitary gland.
The function of the luteinizing hormone in males is the secretion of the progesterone hormone. Whereas, in females, the acute rise of this hormone triggers ovulation, maintains the corpus luteum and is also responsible for the secretion of progesterone hormone.
Explain how genetic information is stored in DNA and how this information is used to make functional proteins.
Answer:
The correct answer will be- codons and each codon specific for amino acids.
Explanation:
Deoxyribose nucleic acid is the genetic material of the organism which provides instructions for the organism. DNA is made up of nucleotide monomers which are composed of five-carbon sugar deoxyribose, a phosphate group and four types of nitrogenous bases (adenine, guanine, cytosine and thymine).
It is the arrangement of these nitrogenous bases which provide codes to the organism as it forms mRNA molecule through transcription. The sequence of the nitrogenous bases in mRNA is read by the ribosome during translation.
The ribosome reads the bases in triplets called "codon" which code for a specific amino acid. If the sequence of the base changes, therefore, the amino acid also changes. These amino acids bond to each other by peptide bond and form a protein molecule.
A nucleosome forms hydrogen bonds with what part of the DNA? A nucleosome forms hydrogen bonds with what part of the DNA? only with bases via the major groove with the phosphodiester backbone and with bases via the minor groove only with bases via the minor groove with the phosphodiester backbone and with bases via the major groove
Answer:
A nucleosome forms hydrogen bonds with the phosphodiester backbone and with the bases through the minor groove. The histone-fold hydrogen bonds with both the A: T enriched bases and the phosphodiester backbone. The histone-fold domains' association with the minor groove is responsible for the majority of the associations in the nucleosome.
Final answer:
A nucleosome forms hydrogen bonds primarily with the phosphodiester backbone of DNA, facilitated by interactions at the L1-L2 loops and αl helices, including minor groove entry by arginine side chains from histone folds. These arrangements enable efficient DNA packaging and sequence-independent binding necessary for nucleosome function.
Explanation:
A nucleosome forms hydrogen bonds with the phosphodiester backbone of DNA. These interactions are fundamental for DNA packaging within the nucleus, ensuring DNA is wrapped around histone proteins to form nucleosomes. The L1-L2 loops at the ends of each histone dimer and the αl helices at the center of the DNA binding site are primary contact points, where hydrogen bonds are formed between amino acids of the histones and the phosphate backbone of DNA. Additionally, arginine side chains from histone folds enter the minor groove of DNA, influencing the DNA's structural dynamics within the nucleosome.
These interactions allow nucleosomes to bind to DNA in a non-sequence-specific manner, which is crucial for their role in DNA packaging and regulation. The presence of water-mediated interactions and the flexibility in binding accommodate the DNA's varying sequences and structures, enabling nucleosomes to organize DNA efficiently within the cell nucleus.
Which of the following BEST describes transcriptional regulation of the lac operon in E. coli?
a. On in the presence of lactose
b. On in the presence of lactose and presence of glucose
c. Off in the presence of glucose
d. On in the absence of lactose and presence of glucose
e. On in the presence of lactose and absence of glucose
Answer:
E. On in the presence of lactose and absence of glucose
Explanation:
Expression of lac operon synthesizes the enzymes required for catabolism of lactose sugar. When both glucose and lactose are available, glucose is preferred as a nutrient and the lac operon is not expressed.
Lac operon is expressed only when glucose is absent in the medium and lactose is present. If any of the two conditions deviate, the operon is not expressed.
In the absence of glucose and the presence of lactose, the repressor is rendered inactive to bind to the operator. RNA polymerase enzyme is free to bind to the promoter and continue the process of transcription.
The reduced levels of glucose increase the cAMP levels which in turn bind to the Catabolite activator protein (CAP). CAP is a positive regulator that binds to the promoter to facilitate the transcription of the operon by RNA polymerase.
The lac operon in E. coli is 'on' in the presence of lactose and the absence of glucose, allowing the bacteria to respond efficiently to changes in the nutritional environment.
Explanation:The transcriptional regulation of the lac operon in E. coli is a complex process. 'e. On in the presence of lactose and absence of glucose' best describes this regulation. The lac operon is activated, or 'turned on', in the presence of lactose when glucose is not present. If glucose is available, the bacteria preferentially use glucose, and the lac operon is turned off. The lac operon system thus helps the bacteria efficiently respond to changes in the nutritional environment.
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Explain how human cells compensate for the X-Iinked gene dosage difference in XX and XY nuclei.
Answer:
Dosage compensation in human includes inactivation of one X chromosome in all the cells of human females.
Explanation:
Human females have two copies of X chromosomes while the human males have only one X chromosome. To balance the X linked genes among human males and females, one of the two X chromosomes in the human females is inactivated randomly during early embryonic development.
The inactivated X chromosome is present in the form of a dark spot of chromatin near the nuclear envelope in each cell of human females. This inactivated X chromosome is called the Barr body.
Inactivation of one X chromosome in human females balances the total number of X linked genes between human males and females.
1. Plants carry out chemical reactions that make sugars, which contain carbon, hydrogen, and oxygen, from carbon dioxide (CO2) and water (H2O). Researchers want to know whether the oxygen atoms in sugar come from the carbon dioxide or the water. How could they use radioactive tracers to find out? Design an experiment that investigates this question. Be sure to include the following elements in your experimental design: hypothesis, procedure, control group, experimental group.
Answer:
Radioactive labeling is a procedure used to monitor the path followed by a chemical element within a biological system to demonstrate the source.
Hypothesis:
The free oxygen produced during photosynthesis comes from water.
Procedure.
Single-celled algae were placed in four petri dishes containing carbon dioxide, glucose dissolved in Water.
In each box the oxygen was radioactively marked.
1 petri dish with CO2 marked. Experimental groups
2 petri dish with H2O marked. Experimental groups
3 petri dish with Glucose marked. Experimental groups (control)
4 petri dish with CO2 + H2O + Glucose, all marked. (control).
How many different types of genotypes are possible when working various pea seed color problems?
a. 3
b. 4
c. 5
d. 6
e. 2
Answer:
Option A, 3
Explanation:
Primarily all pea seeds have two colors one is yellow and other is green.
Let green color allele be represented by "G" and yellow color allele be represented by "g".
When we start working on pea seed color problem, there can be following possible genotypes -
a) Heterozygous green color - Gg
b) Homozygous yellow color - gg
c) Homozygous green color - GG
Hence, option A is correct.