The calculated chi-square value for the analysis is approximately 0.244.
To perform a chi-square analysis, we need to compare the observed frequencies in the F2 generation with the expected frequencies based on the Mendelian inheritance hypothesis. Let's proceed with the calculations:
Observed Frequencies (O):
Normal River Birch (observed) = 811
Dwarf River Birch (observed) = 261
Expected Frequencies (E):
Based on Mendelian inheritance, if we cross a pure breeding normal plant (NN) with a pure breeding dwarf plant (nn), all F1 plants would be heterozygous (Nn), and in the F2 generation, we would expect a phenotypic ratio of 3 normal (N-) : 1 dwarf (nn).
Total F2 plants = 811 + 261 = 1072
Expected normal = (3/4) * 1072 ≈ 804
Expected dwarf = (1/4) * 1072 ≈ 268
Calculating the Chi-Square Value:
The chi-square test formula is: χ² = Σ((O - E)² / E)
Calculating for normal:
χ²[tex]_{normal}[/tex] = ((811 - 804)² / 804) = 0.061
Calculating for dwarf:
χ²[tex]_{dwarf}[/tex]= ((261 - 268)² / 268) = 0.183
Adding both chi-square values:
χ²[tex]_{total}[/tex] = χ²[tex]_{normal}[/tex] + χ²[tex]_{dwarf}[/tex] = 0.061 + 0.183 = 0.244
Rounding the calculated chi-square value to three decimal places:
χ²[tex]_{total}[/tex] ≈ 0.244
The calculated chi-square value for the analysis is approximately 0.244.
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The calculated chi-square value for the inheritance of height in river birch, based on the given F2 data, is 0.245. This value supports the hypothesis that height follows a simple Mendelian inheritance pattern.
Explanation:To determine if the mode of inheritance for height in river birch is a simple Mendelian trait, we perform a chi-square analysis on the F2 generation data. Given that all F1 plants are of normal height, the trait for dwarfism is likely recessive.
If dwarfism follows Mendelian inheritance, we would expect a 3:1 ratio of normal to dwarf plants in the F2 generation. Using the observed numbers of normal (811) and dwarf (261) river birch plants, we calculate the expected values based on a total of 1072 F2 plants (811+261). Expected normal plants would be 3/4 of 1072, and expected dwarf plants would be 1/4 of 1072.
Expected normal = 1072 * 3/4 = 804
Expected dwarf = 1072 * 1/4 = 268
Now we apply the chi-square formula:
χ² = Σ[(Observed - Expected)² / Expected]
For normal = (811 - 804)^2 / 804 = 0.061
For dwarf = (261 - 268)^2 / 268 = 0.184
Calculated chi-square value = 0.061 + 0.184 = 0.245 (rounded to three decimal places).
The following enzymes or processes are known to regulate eukaryotic gene expression. In the blank provided by each enzyme or process, enter the number 1 if the enzyme or process increases gene expression or the number 2 if the enzyme or process decreases gene expression.
(a) histone acetyltransferases (HATs) ____
(b) histone deacetylases (HDACs) ____
(c) acetylation ____
(d) deacetylation ____
(e) decreased attaraction between DNA and histones ____
(f) increased attraction between DNA and histones ____
Answer:
a) 1 b) 2 c) 1 d) 2 e) 1 f) 2
Explanation:
a) A coactivator CBP contains a subunit that has histone acetyltransferase (HAT) activity. These enzymes transfer acetyl groups from an acetyl CoA donor to the amino groups of specific lysine residues on histone proteins.
-Gradually this leads to activation of several other coactivators and as a result initiation of transcription takes place.
b) HDACs are associated with transcriptional repression. HDACs are present as subunits of larger complexes described as corepressors.
-Corepressors are similar to coactivators, except that they are recruited to specific genetic loci by transcriptional factors (repressors) that cause the targeted gene to be silenced rather than activated.
c) Acetylation of histone residues is thought to prevent chromatin fibers from folding into compact structures, which helps to maintain active, euchromatic regions.
-On a finer scale, histone acetylation increases access of specific regions of the DNA template to interacting proteins, which promotes transcriptional activation.
d) Deacetylation of the the histone tails makes the DNA more tightly wrapped around the histone cores, making it difficult for transcription factors to bind to the DNA. This leads to decreased levels of gene expression.
e) Decreased attraction is directly proportional to loosening of the DNA from histone core hence accessible to the transcriptional factors which further leads to transcription and gene expression
f) Increased attraction means DNA gets tightly wound around the histone core and inaccessible to the transcription factor hence decreased gene expression.
Between glycolysis and the citric acid cycle (also known as
theKrebs cycle), there are two molecules of ATP and two molecule
ofGTP generated directly. Most of the energy derived from
theoxidation of glucose comes from the oxidation of
FADH2generated from the citric acid cycle.
NADPH generatedfrom the citric acid cycle.
NADH generatedfrom the citric acid cycle.
NADH generatedfrom glycolysis.
Answer:
The correct answer will be option-NADH generated from the citric acid cycle.
Explanation:
During cellular respiration, few ATP molecules are synthesized by substrate-level phosphorylation in glycolysis and citric acid cycle.
The process also produces high energy equivalents which are NADH and FADH₂ during the citric acid cycle. Two pyruvate molecules generate 6 NADH and FADH₂ molecules. These molecules release electron and hydrogen during the electron transport chain which is utilized to generate the ATP.
Each NADH produces 3 ATP molecule whereas FADH₂ molecule produces 2 ATP molecules. Since 6 NADH (6x3=18 ATP) produced and 2 FADH₂ (2x2).
Thus, NADH provides the highest amount of ATP molecules.
Cellular Respiration (there is more than one correct answer)
a. produces oxygen
b. uses carbon dioxide
c. uses glucose
d. produces carbon dioxide
Answer: Uses glucose ;produces carbon dioxide
Explanation:
Cellular respiration can be defined as the process by which the chemical energy in the body is transferred into ATP and some of it is released in the form of heat.
During this process the glucose that is broken down and the energy released is converted into ATP.
Carbon dioxide is released as a by-product of this process.
Hence, the correct answer is option C and D
Regarding fiber, which one is not correct?
a. If you want to add fiber in your diet also eat the seeds and skins in fruits and vegetables.
b. The removal of the outer husk or bran from our wheat diminished the integrity of our grains.
c. The lack of fiber in the Western diet (that's our diet) was causing gastrointestinal problems that many countries and continents had not seen: diverticulitis, diverticulosis, constipation, increase in colon cancers, etc.
d. You cannot eat too much fiber.
Answer:
d. You cannot eat too much fiber.
Explanation:
Fiber also called as roughage forms the part of plant based foods. Body is not able to digest it so it passes through the digestive system and makes the bulk of stool. It keeps the digestive system clean and eases bowel movements.
The daily required intake of fiber is 25-30 grams. Many people do not take this much fiber hence suffer from problems like constipation, upset stomach etc. But crossing the threshold of 70 grams of fiber can also lead to another set of problems like:
bloatinggasdehydrationpoor absorption of some nutrients like calcium, zinc, iron and magnesium because they bind with the fiber which hinders the absorption process.From the bacteria's perspective, why is it helpful that it produce diarrhea in people?
A. Because it gets the bacteria out of the person and, likely, into the next one
B. It's not helpful really. That's just what that toxin causes.
C. Because that quickly kills the person
D. Because it makes the patient too unpleasant to be around
E. Because there is no real treatment for that
Answer:
A
Explanation:
The vast majority of bacteria that produce gastrointestinal symptoms are transmitted via fecal-oral transmission. This means that when sub-optimal hygiene is present, bacteria from the hands of the infected person that got there from contact with its own feces (after going to the bathroom, for example) is passed to the next person. This happens a lot in the food business, and it's the mode of transmission of the much famous Salmonella.
What is osmolarity? Hemopoiesis?
Answer:
Osmolarity:
Osmolarity may be defined as the number of solutes dissolved in per solution of liters. The osmolarity plays an important role in the human body as it maintains the water electrolyte balance in the body. The osmolarity of the human plasma is about 275-299 milli-osmoles/kilogram.
Hemopoiesis:
Hemopoiesis may be defined as the process of formation of blood cells in the body. The main types of blood cells formed during this process are red blood cells, white blood cells and platelets. The hemopoiesis majorly occurs in the bone marrow cells.
Which DNA segment is not required specifically for an expression vector (for use in E. coli)?
a. origin of replication
b. operator
c. ribosome binding site
d. promoter
e. transcription termination sequence
Answer:
b. operator
Explanation:
An expression vector is usually a plasmid or a virus designed for gene expression in cells, it's designed to act as an enhancer and promoter to efficient transcription of the gene that it carries, meaning it is used for protein production.
As any other vector expression vector may have an origin of replication, a selectable marker, and multiple cloning sites, it also needs elements that are necessary for gene expression, a promoter, a ribosomal binding site, and a start, terminal codon and a transcription termination sequence.
An operator contains the code to begin the process of transcribing the DNA message of one or more structural genes into mRNA, it works as a repressor to regulate the gene expression, since the function of an expression vector is the protein production an operator is not specifically required.
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Compare the chemical structures of DNA and proteins.
Answer:
DNA:
Dexoyribonucleic acid is present as a genetic material in all organism except some virus. DNA is made up of polymers of nucleotides. They are made up of nitrogenous base (adenine, guanine, thymine and cytosine) , deoxyribose pentose sugar and phosphate group. The nitrogenous bases are linked with each other with hydrogen bonds and nucleotides are linked together with phosphodiester bond.
Protein:
Proteins are the main building block of the body. Proteins are made of the polymers of amino acids. The amino acid consist of amine group, carboxylic group, hydrogen and R group attached with the carbon. The amino acids are linked together through the peptide bond.
List a few terms that are unique for regulation of transcription in eukaryotic cells?
Answer:
Eukaryotic transcription refers to an elaborate procedure, which is used by the eukaryotic cells to duplicate the genetic information present in DNA into the units of RNA replica. The transcription of the gene takes place in both prokaryotic and eukaryotic cells. The expression of eukaryotic genes is mainly monitored at the level of the beginning of transcription, though in certain cases, transcription may be monitored and attenuated at subsequent steps.
Like in bacteria, the transcription in eukaryotic cells is monitored by the proteins, which combines with the unique regulatory sequences and modulate the activity of RNA polymerase. The intricate task of monitoring the expression of gene in various kinds of cells of multicellular species is achieved mainly by the combined activities of various different transcriptional regulatory proteins.
In supplementation, the packaging of DNA into chromatin and its variation by methylation produce further levels of complexity to the control of eukaryotic gene expression. The transcription in bacteria is monitored by the combination of proteins to the cis-acting sequences, which regulate the transcription of adjacent genes.
The same kind of cis-acting sequence monitors the expression of eukaryotic genes. These sequences have been determined in mammalian cells mainly by the application of gene transfer assays to examine the activity of suspected regulatory regions of cloned genes.
Cell differentiation always involves
a. transcription of the myoD gene.
b. the movement of cells.
c. the production of tissue-specific proteins.
d. the selective loss of certain genes from the genome.
Answer: (C) The production of tissue-specific proteins.
Explanation:
The cell differentiation is the process in which the cell change from one cell place to another place. It basically occur due to the process of gene expression.
The cell differentiation involve the production of the specific tissue protein known as muscle actin. In cell differentiation, the pluripotent stem cell basically go in the specific differentiation level and then reach in the state of fully differentiation.
The fully differentiation produced a specific function that the production of the protein. Therefore, option (C) is correct.
Cell differentiation involves the production of tissue-specific proteins. The correct option is C.
Thus, the process through which cells develop specialized properties and functions is referred to as cell differentiation. Cells go through a variety of alterations during this process that equip them to carry out certain tasks in a particular tissue or organ.
The creation of tissue-specific proteins is one of the crucial components of cell differentiation. These proteins, which are frequently unique to a given cell type or tissue, carry out the specialised duties of differentiated cells. They are essential in determining the differentiated cell's identity and purpose.
Thus, the ideal selection is option C.
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A fruit fly population has a gene with two alleles, A1 and A2. Tests show that 70% of the gametes produced in the population contain the A1 allele. If the population is in Hardy-Weinberg equilibrium, what proportion of the flies carry both A1 and A2?
a. 0.7 b. 0.49 c. 0.42 d. 0.21
Answer:
Option C
Explanation:
Given ,
A1 allele is carried by [tex]70[/tex] % people
Let us assume A1 s dominant genotype
This means [tex]p= \frac{70}{100} = 0.7[/tex]
Thus, frequency of allele in the given population is [tex]0.7[/tex]
It is also given that the population is in Hardy-Weinberg equilibrium thus
[tex]p+q=1\\0.7+q=1\\q = 1-0.7\\q= 0.3[/tex]
Frequency of fruit fly with genotype A2A2 will be
[tex]q^2\\= (0.3)^2\\= 0.09[/tex]
As per Hardy-Weinberg's second equation of equilibrium
[tex]p^{2} + q^{2} + 2pq = 1\\0.49+0.09+2pq = 1\\2pq = 1-0.49-0.09\\2pq= 0.42[/tex]
Hence, option C is correct
Which of the following intermediary metabolites enters the Krebs cycle and is formed, in part, by the removal of CO2 from a molecule of pyruvate?
A) lactate
B) glyceraldehyde phosphate
C) oxaloacetic acid
D) acetyl CoA
E) citric acid
Answer:
The correct answer will be option-D.
Explanation:
Before Citric acid cycle or Krebs cycle, an intermediate reaction takes place which converts the pyruvate into Acetyl CoA. This reaction is known as pyruvate decarboxylation as it produces Carbon dioxide.
Coenzyme A reacts with pyruvate which causes the release of two oxygen atoms and one carbon to form CO₂ along with the reduction of NAD+ to NADH and produce "Acetyl CoA."
Thus, option-D is the correct answer.
The intermediary metabolite formed by the removal of CO2 from pyruvate and that enters the Krebs cycle is acetyl CoA. This occurs in the phase of cellular respiration known as decarboxylation.
Explanation:The intermediary metabolite that enters the Krebs cycle and is formed, in part, by the removal of a CO2 molecule from a molecule of pyruvate is acetyl CoA (option D).
The process occurs during cellular respiration. Specifically, one carbon atom from pyruvate is released as one molecule of CO2 in a step called decarboxylation, facilitated by an enzyme complex in the mitochondrial matrix. This remaining two-carbon molecule then combines with Coenzyme A to form Acetyl CoA, which enters the Krebs cycle.
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In Drosophila, males from a true-breeding stock with raspberry-colored eyes were mated to females from a true-breeding stock with sable-colored bodies. In the F1 generation, all the females had wild-type eye and body color, while all the males had wild-type eye color but sable-colored bodies. When F1 males and females were mated, the F2 generation was composed of 216 females with wild-type eyes and bodies, 223 females with wild-type eyes and sable bodies, 191 males with wild-type eyes and sable bodies, 188 males with raspberry eyes and wild-type bodies, 23 males with wild-type eyes and bodies, and 27 males with raspberry eyes and sable bodies. Explain these results by diagramming the crosses, and calculate any relevant map distances.
Answer:
This results are explained by the fact that these two genes are in the X chromosome and both mutations are recessive.
There is a estimated map distance of 5.76 cM.
Explanation:
First step is to know what are the initial phenotypes, they say "males from a true-breeding stock" that means that they are homozigous for both genes. Also, they say that in F1 generation, all females had the same phenotype but all males had a different phenotype than females, so this suggest a sex-linked inheritance. In the F1 generation, females had a wild type phenotype, but we know these females had to carry the mutant copies of the two genes, so mutant alleles must be recessive.
Therefore, if r is the allele for raspberry-colored eyes and s for sable-colored bodies, the genotypes of parents should be:
- Males with raspberry-colored eyes (wild type for body color):
[tex]X^{rS}Y^{}[/tex]
- Females with sable-colored bodies (wild type for eyes color):
[tex]X^{Rs}X^{Rs}[/tex]
If they are crossed, we obtain the results on the Punnet Square.
If there is not recombination, we expect F2 as in Punnter Square, but in the results, there are two phenotypes that were not expected: wild type males (23) and males with raspberry-colored eyes with sable-colored bodies (27). These two should be recombinants. To calculate the distantance between the two genes, we use:
[tex]Distance=\frac{recombinants}{Total }*100[/tex]
[tex]Distance=\frac{23+27}{868}*100=\frac{50}{868}*100=5.76 cM[/tex]
Therefore, distance between s and r is 5.76 centimorgans.
Cocaine acts in the nervous system by blocking
a. the dopamine transporter.
b. the serotonin transporter.
c. the norepinephrine transporter.
d. voltage-gated sodium channels in axons.
e. All of the above
Answer:
The correct option is: e. All of the above
Explanation:
Cocaine is a habit-forming illegal recreational drug and a very strong nervous system stimulant. The effects of this drug include euphoria, hallucinations, paranoid delusions, increased heart rate, large pupils, perspiration, itching and high body temperature.
Cocaine acts on the nervous systems of the human beings by blocking the reuptake of dopamine, serotonin, and norepinephrine neurotransmitters in the brain.
It also interferes with the action potential propagation by blocking the voltage-gated sodium channels.
During the light dependent reaction, the hydrogen ions go down their concentration from __________ producing ATP in the process.
a. The stroma into the thylakoid lumen
b. The thylakoid lumen into the stroma
c. Across the inner membrane into the stroma
d. The stroma across the inner membrane
Answer:
b. The thylakoid lumen into the stroma
Explanation:
During light reaction of photosynthesis, when electrons which have high energy travel across electron transport chain some energy is released by them. This energy causes some H+ from stroma to move into thylakoid lumen. Due to this, H+ concentration becomes more in lumen as compared to stroma. Soon after that because of concentration gradient, these H+ then start moving from high concentration to low concentration i.e. from thylakoid lumen to stroma. There are some reverse pumps i.e. ATP synthase located in the thylakoid membrane which are responsible for generating energy in the form of ATP from the movement of H+. From the movement of 3 H+ , 1 ATP is generated.
If an enzyme in solution is saturated with substrate, the most effective way to obtain a faster yield of products is to
a. add more of the enzyme.
b. heat the solution to 90°C.
c. add more substrate.
d. add a noncompetitive inhibitor.
Answer:
a. add more of the enzyme.
Explanation:
Enzymes speed up the rate of reaction by lowering the activation energy for the reaction. They are not used themselves in the reaction. They are specific to substrate molecule. Substrate molecule binds to enzyme's active site and they undergo the reaction to form the product and release back the enzyme.
Rate of reaction depends on both substrate and enzyme concentration. Maximum rate of reaction is reached when all the active sites of enzyme molecules have been occupied by the substrate molecules which means that they are saturated. If more substrate is added at this point it wont have any effect on rate of reaction since there are no free active sites. Hence more enzyme is required to be added so that extra substrate can be utilized and rate of reaction can further be increased.
Which of the following cells or structures are associated with asexual reproduction in fungi?
a.ascospores c.zygosporangia
b.basidiospores d.condiophores
Answer:
d.condiophores is the correct answer.
Explanation:
condiophores is cells or structures are associated with asexual reproduction in fungi
Conidium, a kind of asexual spore-bearing bodies of fungi normally produced at the top of hyphae on specific spore-producing stalks called conidiophores.
The spores get released and detach when it is mature.Then spores germinate into a new hypha and hypha develops into a mycelium.
Thus the Axseual reproduction in Fungi(ascomycetes) occurs by the generation of conidia which are produced on a specialized structure called conidiophores.
Answer:
d. condiophores
Explanation:
As you may know, fungal spores are the main structure for the asexual reproduction of these organisms. These spores are formed in hyphae, but some fungi have special hyphae called conidiophores and their spores called conidia, which are structures for the asexual reproduction of these fungi. For this reason, we can say that, among the options given in the question above, "conidiophores" is the structure that is associated with asexual reproduction in fungi.
Which of the following is a hydrophobic material?
a. paper c. wax
b. table salt d. sugar
Answer:
Wax.
Explanation:
Hydrophobic material may be defined as the material that contains the non polar molecule and hydrophobic interaction in their structure. The hydrophobic molecule are soluble in non polar solvents.
Wax is made of lipids. The lipids are non polar molecule and are hydrophobic in nature. Wax are made of large hydrocarbon part that are insoluble in water and can be easily soluble in non polar solvents.
Thus, the correct answer is option (c).
The term hydrophobic refers to substances that repel water. Out of the mentioned options, wax best fits this description as it doesn't mix with water. Paper, table salt, and sugar dissolve in or absorb water and are thus hydrophilic, or attracted to water.
Explanation:In the context of the question, which asks for the material that is hydrophobic, the correct answer is c. wax. Hydrophobic substances are those that don't mix with or repel water. For instance, wax is a hydrophobic substance you commonly encounter. When you pour water on wax, it simply rolls off or forms beads and doesn't mix, this is due to its hydrophobic characteristic. Materials like table salt, paper, and sugar, are all hydrophilic, they tend to dissolve in and interact with water.
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What are two types of membrane diffusion? Describe the enzymes that promote membrane diffusion.
Answer:
Simple diffusion and Facilitated diffusion
Explanation:
The simple diffusion process involves movement of molecules. This movement occurs from high concentration to low concentration. The facilitated diffusion process is a passive diffusion process. This process requires proteins for movement of molecules, the proteins are carrier and channel proteins.
The enzymes that promote membrane diffusion are carrier proteins. These proteins act as enzymes and allows transport by changing the conformation and opening of channels.
The two types of membrane diffusion are simple diffusion and facilitated diffusion. Simple diffusion does not involve any enzymes, as it's the movement of small and nonpolar molecules down their concentration gradient through the phospholipid bilayer directly. Facilitated diffusion may involve carrier proteins or channel proteins but does not involve classical enzymes that catalyze reactions; instead, these proteins facilitate the movement of substances across cell membranes that cannot easily pass through the lipid bilayer directly.
Diffusion is critical for the transport of materials across biological membranes, helping to maintain homeostasis within cells and organisms. Simple diffusion allows for the passive movement of substances like oxygen and carbon dioxide, relying on the concentration gradient to drive the process without energy input. Facilitated diffusion also moves substances along their concentration gradient but through specific carrier proteins or channels. These proteins provide a pathway for larger or polar molecules that cannot diffuse through the lipid bilayer unaided. Temperature is one of the factors that can affect the rate of diffusion; higher temperatures increase molecular kinetic energy, which generally speeds up diffusion rates.
Testosterone is produced in the
A. Sperm cells.
B. Seminiferous tubules of the testes.
C. Interstitial cells of the testis.
D. Epididymis.
E. Anterior lobe of the pituitary.
The correct answer is C. Interstitial cells of the testis.
Explanation:
The testosterone is one of the main sex hormones in males, due to this, this hormone is linked to sexual characteristics as well as behavior, mass muscle development, among others. In terms of testosterone production, the production of testosterone occurs mainly on the testis (reproductive glands or gonads) and more specifically in the interstitial cells of these called Leydig cells which are specialized cells that besides producing testosterone produce androstenedione and dehydroepiandrosterone. Additionally, the production of any hormone in this cell is stimulated by the luteinizing hormone. According to tho this, it can be concluded testosterone is produced in the interstitial cells of the testis.
Testosterone is produced in the interstitial cells of the testis.
Explanation:Testosterone is produced in the interstitial cells of the testis. These cells, also known as Leydig cells, are located in the testes and are responsible for the production of testosterone. The hormone is then released into the bloodstream and has various effects on the body, including the development of male reproductive organs and secondary sexual characteristics.
The hormone Testosterone is primarily produced in the Interstitial cells of the testis, also known as Leydig cells. These cells are located in the connective tissue between the seminiferous tubules in the testes. The production of testosterone is stimulated by Luteinising Hormone (LH) which is produced in the anterior lobe of the pituitary gland in the brain. However, the pituitary gland does not produce testosterone itself. Hence, the correct answer to your question is C. Interstitial cells of the testis.
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Which of the following tools of DNA technology is incorrectly paired with its use?
a. electrophoresis—separation of DNA fragments
b. DNA ligase—cutting DNA, creating sticky ends of restriction fragments
c. DNA polymerase—polymerase chain reaction to amplify sections of DNA
d. reverse transcriptase—production of cDNA from mRNA
The one that is incorrectly paired is DNA ligase—cutting DNA, creating sticky ends of restriction fragments. The correct option is B.
What is DNA ligase?DNA ligase is a ligase enzyme that facilitates the joining of DNA strands by catalyzing the formation of a phosphodiester bond.
It aids living life forms repair single-strand cuts in duplex DNA, but some forms may specifically repair double-strand breaks.
By connecting cuts in the phosphodiester backbone of DNA that occur during replication and recombination, as well as as a result of DNA damage and repair, DNA ligases play an important role in maintaining genomic integrity.
Electrophoresis is used to separate DNA fragments, DNA polymerase is used to amplify sections of DNA, and reverse transcriptase is used to produce cDNA from mRNA.
Thus, the correct option is B.
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Mosquito Dunks contain protein, produced by Bacillus thuringiensis bacteria, that is toxic to some aquatic insect larvae. Mosquito Dunks may help control which of the following?
a. malaria
b. bubonic plague
c. Legionnaires' disease
d. toxoplamosis
e. all of the above
Answer:
a. malaria
Explanation:
Malaria is caused by a parasite that is transmitted by the bite of infected anopheles mosquitoes, entering the bloodstream, affecting the red blood cells. Therefore, Mosquito Dunks may be a good control of malaria in tropical countries.
On the other hand, mosquito dunk is an inappropriate control method for the other diseases listed, because bubonic plague was transmitted by the fleas of rats, legionellosis is an infectious disease caused by a bacterium (Legionella) that is found mainly in water bodies and toxoplasmosis is caused by a parasite present in undercooked contaminated meat and the feces of cats.
List all the ways in which CO2 is transported in the blood. Of these, how is the majority of the CO2, transported in the blood?
Answer:
Carbon dioxide is the prime waste component of aerobic respiration. Too little or too much carbon dioxide in the blood can result in extreme issues. In order to make this safe and sound, carbon dioxide is transported in the blood in three ways. That is, in the form of carbamino compounds, hydrogen carbonate, and in the form of a dissolved state.
Approximately 30 percent of all the carbon dioxide is mediated in the form of carbamino compounds. At high quantity, carbon dioxide binds directly with amino acids and the amine groups of hemoglobin to produce carbaminohaemoglobin.
Almost 60 percent, that is, the majority of the carbon dioxide is transported in the blood in the form of bicarbonate ions or hydrogen carbonate. The diffusion of carbon dioxide takes place in the RBCs and gets transformed into bicarbonate ions and hydrogen ions with the help of an enzyme known as carbonic anhydrase.
Approximately 10 percent of all the carbon dioxide is transported in the form of a dissolved state in plasma. The concentration of gas dissolved in a liquid relies upon its partial pressure and its solubility. In spite of its solubility, only some of the carbon dioxide in the blood is in reality transported in the form of dissolved state in plasma.
Answer:
CO2 is transported in three forms from tissues to the alveoli: a) In dissolved form through plasma; b) As bicarbonate ions; c) by RBC as HbCO2.
Explanation:
When oxygen reaches the body cells, oxidation of glucose takes place during which CO2 is released along with H2O and energy. CO2 diffuses out of the tissues into the blood capillaries where it is transported to the lungs in three forms:
1. In dissolves form in plasma: About 7% of CO2 gets transported in dissolved form. It gets dissolved in the blood plasma and is carried to the lungs.
2. As bicarbonate ions: about 70% of CO2 is converted into bicarbonate ion and transported in plasma. Within the RBC there is an enzyme called carbonic anhydrase. When CO2 enters the RBC, it reacts with water to form carbonic acid which further dissociates into bicarbonate and hydrogen ion. Both these reactions occur in the presence of carbonic anhydrase enzyme. The HCO3- ion formed in the RBC quickly diffuses into the plasma, where they are carried to the lung.
3. By RBCs as carbaminohemoglobin: About 20-25% of CO2 is transported as HBCO2. CO2 binds with the amino group of globin protein which is a part of Hb.
Absence of bicoid mRNA from a Drosophila egg leads to the absence of anterior larval body parts and mirror-image duplication of posterior parts. This is evidence that the product of the bicoid gene
a. normally leads to formation of head structures.
b. normally leads to formation of tail structures.
c. is transcribed in the early embryo.
d. is a protein present in all head structures.
Answer:
a. normally leads to formation of head structures.
Explanation:
The product of the bicoid gene is a protein that is responsible for establishing the anterior-posterior axis during the embryogenesis phase of Drosophila development.The bicoid protein is localized at the anterior end of the egg of the drosophila and the anterior end it represses the expression of the mRNAs that are translated at the posterior end and helps in the expression of only anterior genes due to which it plays an essential role in the formation of anterior structures and thus the head of the fruit fly.Therefore, from the given question it is very clear that since the absence of anterior larval body parts occurs in the absence of bicoid mRNA, bicoid must be specifying the head structures.The product of the bicoid gene in Drosophila predominantly influences the formation of anterior body parts, suggesting that it primarily leads to the formation of head structures.
Explanation:Based on the provided information, bicoid mRNA's absence from a Drosophila egg causes the absence of anterior (head/front) body parts and duplicates posterior (tail/back) parts. This indicates that the bicoid gene's product primarily influences the formation of anterior parts. It can be inferred that the bicoid gene's product is crucial in the formation of the head structures. So, the correct answer is, (a) the bicoid gene's product normally leads to the formation of head structures.
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How does complement connect the innate and adaptive immune responses?
Answer:
Innate immunity: first line of defense against infectious agents; Most pathogens can be controlled before a declared infection occurs.
Adaptive immunity: takes action when innate immunity fails. Make a specific response for each infectious agent and keep memory of it (it can prevent reinfection).
Explanation:
The immune system has evolved to be able to identify the strangers and develop a protective response to the latter (cognitive and destructive processes).
Innate or natural immunity: it is present at birth, being the first line against invasive microorganisms. Its characteristics are: it is present for life, it is not specific, it lacks memory and does not change intensity with exposure. It is useful against pyogenic microorganisms, fungi and multicellular parasites and includes three components: 1) physicochemical: skin, mucous membranes, secretions and cilia, which perform a washing and continuous cleaning, 2) humoral: complement, lectin binding to mannan and opsonins additional as C-reactive protein and proteolytic enzymes and 3) cellular: neutrophils, eosinophils, mast cells and natural killer lymphocytes.
Complement: it includes a high number of serum proteins that are produced mainly in the liver, form cascades so that each activated component catalyzes the activation of several molecules of the next component, amplifying the response. The consequences are cell lysis, the production of proinflammatory mediators and the solubilization of antigen-antibody complexes. Activation of the complement system occurs through three different pathways (alternating, classical and mannan-binding lectin) that converge in the final common pathway that provides most of the biological activity.
Cellular mechanisms: Neutrophils participate in the destruction of bacteria and fungi. Upon activation, adhesion molecules facilitate their entry into tissues, moving to chemical attractants and phagocytizing microorganisms. Destruction is mediated by oxygen dependent and independent pathways.
acquired immunity: acquired as part of development, increases with age and with repeated exposures, has specificity and memory for what is called adaptive. Its components are antibodies and cells (lymphocytes) and protects against bacteria (including those that produce intracellular infections), viruses and protozoa. In general, innate and acquired immune responses are not activated independently but complement each other.
Antigen-Antibodies: Antigens are structures that generate an anti-response from the immune system that has 3 elements of union and recognition of these;
1) antibodies that are soluble glycoproteins belonging to the group of immunoglobulins produced by B lymphocytes and plasma cells,
2) T-cell receptors that are large glycoproteins that interact with the peptide epitope preserved and presented by the third element
3) which are the major histocompatibility complex (CMH) molecules.
Antibodies perform many functions and have numerous uses as biological and clinical instruments
Which of the following statements about excision repair is correct?
a. base excision repair is initiated by DNA glycosylases that recognize abnormal deoxyriboses in DNA.
b. nucleotide excision repair removes large regions of DNA via an excinuclease which cuts on either side of the damaged bases.
c. e. coli exonuclease activity is carried out by uvrA, uvrB and uvrC, and the resulting gap is filled in by DNA Pol III
d. DNA glycosylases cleave the altered nucleoside (base and sugar) from the DNA backbone creating an apurinic or apyrimidinic site
e. in humans, a mechanism similar to E.coli is carried out where the protein XPA acts as the exonuclease.
Answer:
a. True, b. False, c.True, d. True
Explanation:
a. Base excision repair is started by a DNA glycosylase that recognizes the changes and removes the altered base by cleavage of the glycosidic bond binding the base and the deoxyribose sugar together.
b. Nucleotide excision repair works by a cut-and patch mechanism that removes their heavy lesions, including pyrimidine dimers and nucleotides . Endonucleases are responsible for the lesion of the damaged strand.
c. Nucleotide excision repair is initiated by the proteins namely UvrA, UvrC, and UvrB in Escherichia coli.
-UvrD (helicase II) later removes the damaged strand
-DNA polymerase I (PolI) fills in the resulting gap.
d. DNA glycolases removes the damaged nitrogenous base.
-It leaves the sugar-phosphate backbone intact and thus creating an apurinic/apyrimidinic site, which is commonly referred to as an AP site.
e. Xeroderma pigmentosum complementation group A(XPA)
-This is an essential protein in the nucleotide excision repair pathway.
- It helps to make a pre-incision complex along with other proteins.
What gives DNA its acid property?
a. the parts that make up the steps
b. the sugar
c. the part that alternates with sugars in a strand
d. the hydroxyl group attached to the 4' carbon
e. the covalent bonds that connect the steps to the strands
Answer:
The correct answer will be option-C.
Explanation:
Each DNA molecule exists in helical shape made by two strands of nucleotides. Each nucleotide is composed of a five-carbon sugar, a phosphate group and nitrogenous bases.
The sugar-phosphate unit bond via the phospho-diester bond and form the backbone of the DNA. The phosphate functional group is acidic in nature as it donates the proton while the formation of the phosphodiester bond which provides DNA with the acidic nature.
Thus, Option-C is the correct answer.
How many different types of phenotypes are possible when working various pea seed-color problems?
a. 3
b. 4
c. 5
d. 6
e. 2
Answer:
e. 2
Explanation:
The seed color in a pea plant is a discontinuous trait and is regulated by two alleles of a gene. The dominant allele imparts "yellow" color to the seeds while the recessive allele imparts them "green" color.
If the dominant allele is represented by letter "Y" and the recessive allele is written as the lower case letter "y"; the homozygous dominant and heterozygous dominant genotypes exhibit "yellow" phenotype with respect to seed color while the homozygous recessive genotype express "green" seed color.
In which direction does RNA synthesis proceed?
a. left to right
b. 3' to 5'
c. top to bottom
d. 5' to 3'
Answer:
5' to 3'.
Explanation:
Transcription may be defined as the process of formation of RNA molecule from the template DNA with the help of enzymes and the various transcriptional factors.
The process of RNA synthesis occur in the 5' to 3' direction. The RNA polymerase reads in the 3' to 5' direction and leads to the synthesis in 5' to 3' direction. The transcription factors binds to the DNA promoter during RNA synthesis.
Thus, the correct answer is option (d).
Why does it make sense that the sensors and controller of body temperature reside in the brain?
A. Because the brain enables the body's response to temperature changes
B. Because deep inside the brain is considered the body's core
C. Because temperature changes happen most quickly in the brain
D. Because the brain works best at the correct temperature
E. Because temperature changes affect the brain first