Bioinformatics involves using computer programs to align DNA sequences, developing computer-based tools for genome analysis, and using math tools to interpret biological systems. It does not directly involve using DNA technology to combine DNA from different sources.
Explanation:The question at hand is asking which of the given options is not included in the field of bioinformatics. Bioinformatics is a multidisciplinary field involving biology, computer science, and statistics. It utilizes computer programming for aligning DNA sequences, developing computer-based tools for genome analysis, and mathematical tools to interpret biological systems. So, options a, c, and d are within its scope. However, option b describes a process known as recombinant DNA technology, not directly related to bioinformatics. Thus, the answer must be option b: Using DNA technology to combine DNA from two different sources in a test tube.
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Option b, which refers to the use of DNA technology to combine DNA from different sources, is not a part of bioinformatics. This task is more closely associated with the field of molecular biology.
Explanation:Bioinformatics is a rapidly developing branch of biology that is mainly involved in the development and application of computational tools to analyze, integrate, and interpret large sets of biological data. Among the four options presented, the only task that doesn't fall under bioinformatics is b. using DNA technology to combine DNA from two different sources in a test tube. This is more closely related to the field of molecular biology, specifically recombinant DNA technology or genetic engineering. The tasks described in a, c, and d are all common in bioinformatics. They involve the use of computer programs, development of computational tools for genome analysis, and the application of mathematical tools to interpret complex biological systems.
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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 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.
Which of the following can contribute to infertility in a young male body builder taking anabolic steroids?
a. suppression of LH release from the pituitary
b. decreased circulating levels of dihydrotestosterone (DHT)
c. increased muscle mass
d. decreased testicular testosterone concentrations
e. B and D
Answer:
The correct answer is a. suppression of LH release from the pituitary.
Explanation:
Anabolic steroids are synthetic steroids made up of mainly androgenic hormone like testosterone. These anabolic steroids are usually taken by young male bodybuilders because it helps in gaining muscles.
These anabolic steroids have many side effects and infertility in males is one of its side effects. Excessive use of these steroids stops the secretion of gonadotropic hormones such as LH in males from pituitary glands by inducing negative feedback to the pituitary.
LH is important in the production of testosterone from Leydig cells and testosterone is important for sperm production. Therefore if LH is stopped no sperm production will occur and it leads to infertility in males.
Briefly, in two to three sentences, contract the functions of innate, humoral, and cell-mediated immunity.
Answer:
Innate immunity includes non-specific immune responses to prevent the entry of pathogens and to kill the entered pathogen by first line of defenses.
Humoral immunity kills the entered pathogens by production of antibodies specific to the particular type of antigen. B lymphocytes are central to the humoral immunity.
Cell mediated immunity is another type of adaptive immune response wherein the killer T cells directly kill the infected cells.
Explanation:
Cell-mediated immunity is provided by killer T cells against virus-infected cells, foreign cells, and cancer cells. The cytotoxic T cells directly kill the entered pathogens.
Antibody-mediated immunity refers to the specific resistance to disease-causing agents. It includes the production of specific antibodies by B lymphocytes. it is also called humoral immunity.
Innate immune responses are the nonspecific immune responses that prevent the entry of all the diseases causing pathogens and antigens into the body. It includes the first line of defenses such as skin and mucous membranes as well as phagocytes and natural killer cells that kill all the pathogens and antigens non-specifically.
Explain how the independent alignment of homologs, and also crossing-over during the first meiotic division, each contribute to the genetic diversity of gametes.
Answer: Homologous chromosomes are randomly distributed to daughter cells, this means different chromosomes segregate independently of each other. And they exchange segments of DNA during crossing over. This recombination creates genetic diversity because genes from each parent are exchanged.
Explanation:
Meiosis is a type of cell division that produces gamete cells, which are sex cells (egg and sperm)
Chromosomes that form a pair and are found together are called homologous chromosomes, and they are inherited from each parent. During prophase of meiosis I, the homologous chromosomes exchange segments of DNA in a process called crossing over. This recombination creates genetic diversity because genes from each parent are exchanged. It results in new combinations of genes on each chromosome.
After that, during the anaphase of meiosis I, the two chromosomes line up on the equatorial plane of the cell. Then, they are separated and each will go to a new daughter cell. So homologous chromosomes are randomly distributed to daughter cells, this means different chromosomes segregate independently of each other.
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.
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.
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.
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.
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.
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
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.
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.
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).
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".
How does eating influence high density lipoproteins (HDL) in the body?
Answer: The element with the greatest influence on cholesterol levels is the fat content of food. Not only the amount of cholesterol, but also the type of fat.
Explanation:
Cholesterol is a type of fat found in the body, which is used to make hormones and vitamin D. The liver makes cholesterol to handle these tasks, but cholesterol can also be incorporated into the body through foods such as meat, dairy, and poultry. If you consume a lot of these, the cholesterol levels could become too high.
There are two types of cholesterol:
High-density lipoprotein (HDL) or “good cholesterol”. It transports the excess of cholesterol to be expelled so it wont end up in the arteries. Low-density lipoprotein (LDL) or “bad cholesterol”. It transports cholesterol to the arteries, where it may collect in artery walls. This could lead to a buildup of plaque known as atherosclerosis which can increase the risk of blood clots and if one of them breaks away and blocks an artery, it could cause a stroke or heart attack.If there is a predominance of saturated fatty acids in the fat of the diet, a rise in serum cholesterol and LDL-cholesterol is caused. To lower the bad cholesterol, you should limit foods such as milk fats, fatty meats or sausage and include more fibre in your diet. On the other hand, polyunsaturated fatty acids - for example, those present in seed oils or fish - produce their reduction. Monounsaturated fatty acids, such as those provided by olive oil, act like polyunsaturated ones and, in addition, tend to raise HDL-cholesterol, so their use entails a greater benefit.
Banana plants, which are triploid, are seedless and therefore sterile. Propose a possible explanation.
Answer:
The correct answer will be- homologous chromosomes fail to pair.
Explanation:
The seedless variety of the banana are triploid that is each cell contains three copies of the chromosomes. The triploid variety of the organism is considered sterile or not able to produce offspring as they are unable to generate gametes for fertilization.
The formation of gametes relies upon a type of cell division called meiosis which proceeds in four phases of karyokinesis stage.
At metaphase of the meiosis, chromosomes fail to pair at the equatorial plate as two chromosomes pair leaving one extra chromosome which is not paired. The gametes are produced with unequal chromosomes and on fertilization zygote is not formed.
Thus, homologous chromosomes fail to pair is the correct answer.
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.
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.
Starch is broken down into ____________ by the enzyme ___________
Starch is broken down into maltose by the enzyme amylase. Maltose is then further hydrolyzed into glucose by the enzyme maltase, primarily in the small intestine.
Starch is broken down into maltose by the enzyme amylase. Initially, starch is hydrolyzed by salivary amylase in the mouth and later by pancreatic amylase in the small intestine, producing shorter polysaccharides, dextrins, and disaccharides like maltose. These smaller sugars are easier for the body to absorb. Following this, the enzyme maltase, which is produced by the cells in the small intestine, further breaks down maltose into individual glucose molecules, which are used by the body for energy.
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.
What kind of bonds connect the bases of DNA to each other?
a. ionic
b. hydrogen
c. covalent
d. van der Waals
e. protonic
Answer:
The correct answer will be option-B.
Explanation:
Deoxyribose nucleic acid or DNA is the genetic material of the organism which is made up of nucleotide monomer. Each monomer is made up of a five-carbon deoxyribose sugar, a phosphate group and four different nitrogenous bases.
The nitrogenous bases- purines (adenine and guanine) bond with pyrimidines (cytosine and thymine) via hydrogen bond between them. The hydrogen bonds are weak bonds but establish the DNA structure.
Thus, Option-B is the correct answer.
The bases of DNA are connected by hydrogen bonds.
Explanation:The bases of DNA are connected to each other by hydrogen bonds. These bonds form between specific complementary base pairs: adenine (A) with thymine (T) and guanine (G) with cytosine (C). Hydrogen bonds are relatively weak compared to covalent bonds, but they are crucial for maintaining the structure of the DNA helix.
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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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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).
When a large number of species, sometimes entire major taxa, go extinct in a short period of time on the geological scale, the cause is usually
A. Competition with other species
B. A large forest fire
C. Radical global climate change
D. Predation by a new type of hunter
Answer: C. Radical global climate change.
Explanation:
The entire taxa of the species can be wiped out due to an large catastrophic event associated with the change in the climate. As the climate becomes warm the chances of undergoing the global precipitation rate will change. The climate being an important abiotic factor under unstability is likely to affect the amount of rainfall, snow fall, drought and conditions of desertification. All these factors can be responsible for the removal of the entire taxa of the species in short time.
The enzyme helicase unwinds the DNA duplex. Which of the following is likely to be true?
A) A pre-existing 3'OH is required for helicase to work
B) Helicase uses energy from ATP Hydrolysis
C) Helicase is itself made of RNA
D) Energy is released when the DNA helix is unwound.
Answer:
B) Helicase uses energy from ATP Hydrolysis
Explanation:
Helicase's primary function is to separate the annealed nucleic acid strands. It is a motor protein and moves directionally along the phosphodiester backbone. It usually separates strands of double helix DNA or self annealed RNA. It used the energy from ATP hydrolysis and breaks hydrogen bonds between nucleotide bases.
In human body 95 types of helicases are found. They have sequence motifs required for ATP binding, ATP hydrolysis and translocation along nucleic acid phosphodiester backbone. The variable portion in their amino acid sequence imparts specific feature to each helicase.
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.
In October of 2002, General Mills Inc. recalled over 4,000 cans of refried beans suspected to be contaminated with Clostridium botulinum. Which of the following statements regarding adult botulism is correct?
a. the anaerobic environment of the sealed can provides a suitable environment for the spores to germinate.
b. the spores produce a potent neurotoxin
c. symptoms of the disease are stiff, contracted muscles, such as "lockjaw", and eventual paralysis of the respiratory muscles, which results in death
d. reheating the beans to kill the bacteria will prevent the disease.
e. all of the above statements are correct
Answer:
A and B
Explanation:
A and B are correct answers.
C is incorrect because Botulism does not produce stiff muscle, in fact it produces flaccid paralysis. Stiff muscles and lockjaw is produced by C. Tetani.
D is somewhat incorrect because, although you can kill the toxins via heat, you would need to heat your food to over 100°C for around 10 minutes, which no one does. Thus, reheating beans would not protect you from the bacteria and/or its toxins.
How is a spider web like a food chain?
Answer:
food webs are drawn in a diagram similar to a spider web
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.
Learn more about experiments in: https://brainly.com/question/17314369
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