Answer: the coding strand of Henry CYP2C9 aids in the determination of the precise/correct nucleotide sequence of the RNA strand produced after transcription
Explanation: The template strand functions in the production of the RNA strand during transcription ie the RNA strand is complementary to the template strand. RNA produced runs from the 5' to 3' direction. The coding strand differs from the template strand in that it is the strand that contains the exact nucleotide sequence as the RNA strand but differs in that the RNA is replaced with uracil instead of thymine and it also runs in the 5' to 3' direction.
Final answer:
The coding strand of DNA is the sequence that corresponds to the mRNA sequence produced during transcription, while the template strand is the actual template used by RNA polymerase to synthesize mRNA.
Explanation:
The function of the coding strand of DNA is to serve as the basis for the mRNA sequence during transcription, albeit indirectly. The coding strand has the same sequence as the mRNA, with the only difference being that thymine (T) bases in DNA are replaced by uracil (U) in RNA.
On the other hand, the template strand, also known as the non-coding strand, is the direct template for mRNA synthesis.
RNA polymerase uses this template strand to align and add the appropriate RNA nucleotides to form a new mRNA strand that is complementary to the template strand and matches the coding strand (except for T being replaced by U).
Which of the following structures does not exist as a pair? Select one: a. The kidney. b. The sphincter muscle. c. The ureter. d. The urethra.
Structures does not exist as a pair d. The urethra Therefore , d. The urethra is correct .
Explanation:
a. The Kidney:
The kidneys are paired organs in the human body, each individual having two kidneys.
They are essential in the urinary system, primarily responsible for filtering waste products and excess substances from the blood to form urine.
b. The Sphincter Muscle:
Sphincter muscles are circular muscles that surround various openings in the body, including the digestive and urinary tracts.
While there are multiple sphincter muscles, they exist as pairs, serving to regulate the flow of substances through the associated openings.
c. The Ureter:
The ureters are a pair of tubes that carry urine from the kidneys to the urinary bladder.
Humans typically have two ureters, one connected to each kidney, allowing the flow of urine from both kidneys to the bladder.
d. The Urethra:
Unlike the kidneys and ureters, the urethra is a single tubular structure. It is the tube that carries urine from the urinary bladder to the exterior of the body.
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d). The urethra is the correct answer because it is a singular structure, unlike kidneys and ureters, which exist in pairs. Sphincter muscles are specialized muscles and not paired organs.
The question asks us to identify which of the following structures does not exist as a pair: The kidney, the sphincter muscle, the ureter, or the urethra.
Kidneys: Humans have two kidneys, which filter blood to produce urine.Sphincter muscles: Multiple sphincter muscles exist in the human body, but they are not paired in the same way as organs like kidneys. Instead, they are specialized muscles to control the passage of substances through various openings and are found singularly based on their function.Ureters: There are two ureters, one for each kidney, which carry urine from the kidneys to the bladder.Urethra: There is only one urethra, which is a tube that carries urine from the bladder out of the body.Thus, the correct answer is d. The urethra, as it is a singular structure.
Human cells normally have 46 chromosomes. For each of the following stages, state the number of chromosomes present in the particular cell. a) metaphase of mitosis b) metaphase I of meiosis c) telophase of mitosis d) telophase I of meiosis e) telophase II of meiosis
Final answer:
During metaphase of mitosis and metaphase I of meiosis, cells have 46 chromosomes. Following telophase of mitosis, cells also contain 46 chromosomes. However, after telophase I and telophase II of meiosis, cells contain 23 chromosomes.So ,option b) metaphase I of meiosis is correct.
Explanation:
Human cells normally have 46 chromosomes, comprising 23 pairs, with one set from each parent. This diploid state (2n) changes during the process of cell division.
a) metaphase of mitosis: Each cell has 46 chromosomes. The chromosomes align in the center of the cell, but each one still consists of two sister chromatids at this stage.
b) metaphase I of meiosis: Each cell also has 46 chromosomes. However, these are in the form of 23 pairs of homologous chromosomes that will be separated in the next phase.
c) telophase of mitosis: The cell still contains 46 chromosomes, but they are now being divided into two new daughter nuclei.
d) telophase I of meiosis: The two cells created after this phase each have 23 chromosomes, as the homologous pairs have been separated.
e) telophase II of meiosis: After this final phase of meiosis, there are four cells, and each cell has 23 chromosomes—these are the haploid gametes.
Without a plateau extending the refractory period in cardiac muscles sarcomeres might be stimulated so quickly that they would contract and not relax and instead would experience a sustained contraction also called:
1) depolarization
2) Tetany
3) relaxation
4) repolarization
Answer:
Tetany
Explanation
The refractory period in the heart refers to the shortest interval between consecutive contraction conducted impulses out of the cardiac muscle.
It is important because it allows adjustments to stimuli and limits amount of action potential sent per minute.
Without a plateau extending the refractory period in the cardiac muscles.
Sarcomeres might be stimulated to contract instead of relaxing leading to sustained contraction also called Tetany.
Tetany is an involuntary muscle cramp or contraction caused by overly stimulated nerves.
Please describe the environmental issue in Borneo concerning conservation of the rainforest (a global hot spot) and the orangutans. What grassroots tools were used to improve the conservation and allow the regrowth of the rainforest?
Answer:
Borneo Island was once home to one of the most majestic forests in the world. During the 80s and 90s, Borneo underwent a profound transformation. Their forests were demolished at an unprecedented speed. Its rainforests ended in countries like Japan and the United States in the form of garden furniture, paper pulp and chopsticks. Today, the remaining forests are threatened by the new biofuel market, especially palm oil. As a result, large tracts of land are being transformed to oil palm plantations.
Explanation:
The jungles of Borneo were considered one of the wildest and pristine jungles of the planet, home to nomadic tribes and important populations of orangutans, pygmy elephants and rhinos. Currently, the traditions of these tribes have disappeared, rhinos are almost extinct, and orangutans and elephants are in danger. Otherwise, the rainforests of Borneo have gone from being a net carbon sink, which absorbed greenhouse gases, to being a carbon source, thus contributing to climate change along with deforestation and fires.
Conservation is a priority in Borneo, especially in biologically diverse regions that have escaped intensive logging and fires. The initiative called "Heart of Borneo" is an example of what can be achieved. It is essential that forests be restored. The use of native tree species should be encouraged through financial incentives and education programs, especially with the help of external governments, NGOs and private foundations. In addition, there is a possibility that under future climate agreements, reforestation could pay direct dividends stimulating the local economy and entrepreneurship in the villages.
You have some circular DNA in a tube and wish to introduce negative supercoils into it. However, you ran out of topoisomerase. Can you think of a way to do this usingDNA ligase and ethidium bromide
Answer:
First the DNA ligase will join the DNA together.
Secondly, ethidium bromide is added to generate circles after then it is extracted.
After these are done, the DNA becomes negatively supercoiled since
Tw + Wr =constant.
Tw = twist
Wr = writhe
A diploid organism has a somatic chromosome number of 16. The centromeres of the 8 homologous pairs are designated as Aa, Bb, Cc, Dd, Ee, Ff, Gg and Hh, with uppercase letters referring to the centromeres of maternal chromosomes and lowercase letters to the centromeres of paternal chromosomes. Each centromere represents one chromosome. Where n = the haploid chromosome number. How many different combinations of centromeres could be produced during meiosis?
a.) 8
b.) 64
c.) 256
d.) 512
e.) 6561
Answer:
Option C, [tex]256[/tex]
Explanation:
The diploid number of chromosomes are represented as [tex]2N[/tex].
Here the somatic chromosome number is [tex]16[/tex] which can be represented in the "[tex]2N[/tex]" format with N being equal to [tex]8[/tex].
Now, as we know that the total number of chromosomal combination with chromosome number being
[tex]16 \\OR\\2 N(8)[/tex]
is equal to
[tex]2^N\\[/tex]
Substituting the given values in above equation, we get -
[tex]2^8\\= 256\\[/tex]
Therefore, total of [tex]256[/tex] different combinations of centro-meres could be produced during meiosis.
Hence, option C is correct
In a mono hybrid cross of heterozygous pea plant with round (R) yellow (Y) seeds there will be genotype ratio of 9 _____ 3,____ 3,____ and 1____ PLEASE PLEASE HELP IM KIND OF CONFUSED
The genotype ratio for a monohybrid cross of heterozygous pea plants with round (R) and yellow (Y) seeds is 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green. This ratio comes from the dihybrid cross of RrYy plants, reflecting the principle of independent assortment.
Explanation:In a monohybrid cross of heterozygous pea plants with round (R) and yellow (Y) seeds, the genotype ratio you are asking about is derived from the principle of independent assortment and dominance, leading to a phenotypic ratio of 9:3:3:1 in a dihybrid cross. When focusing on one trait, a monohybrid cross, the ratio simplifies to a 3:1 ratio because each trait segregates independently.
For the dihybrid cross of RrYy (round, yellow) × RrYy, the ratios of expected phenotypes are 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green, which corresponds to the same proportions in genotype.
Therefore, the correctly completed sentence would be a genotype ratio of 9 round yellow (RRYY, RRYy, RrYY, RrYy) : 3 round green (RRyy, Rryy) : 3 wrinkled yellow (rrYY, rrYy) : 1 wrinkled green (rryy).
Final answer:
In a mono hybrid cross of heterozygous pea plant with round (R) yellow (Y) seeds, the genotype ratio is 9:3:3:1 for round, yellow seeds : round, green seeds : wrinkled, yellow seeds : wrinkled, green seed.
Explanation:
The mono hybrid cross of a heterozygous pea plant with round (R) yellow (Y) seeds would result in a genotype ratio of 9 round, yellow seeds : 3 round, green seeds : 3 wrinkled, yellow seeds : 1 wrinkled, green seed. This ratio is obtained by applying the product rule to the independent assortment and dominance of the alleles for seed shape and color.
20. Which of the following statements is/are correct?
A. Molluscs can live in marine, and freshwater habitats but do not live in terrestrial habitats.
B. Given the following on the organism:
I. An animal
II. Multicellular
III. has tissues
IV. has a digestive tract
V. exhibits bilaterally symmetry as an adult
This organism could be a member of the Phylum Mollusca
C. Molluscs can produce a shell and the presence of a shell at one point in development is a requirement for inclusion into this phylum.
D. Lobsters are members of the phylum Mollusca
E. Molluscs have three main body parts, a muscular foot, a visceral mass, and a head
Answer: Option C and E are correct.
Explanation:
Phylum molluscs are invertebrates species. They are the second largest phylum in the animal kingdom .Most mollusc live in marine and others live in freshwater and terrestrial habitat. Molluscs are ceoomates and they have three universal structures which are the mantle (breathing and excretion),radula and nervous system. The molluscs include snail, oyster,clams, periwinkles,octopus, squid e.t.c.
The larvae of molluscs are bilateral symmetrical while the adults are asymmetrical. Molluscs can produce shell, the presence of shell is an inclusion in the phylum.
Molluscs have three body part, the head, visceral mass and muscular foot
Answer: Options B and E are correct
Explanation: molluscs are found in freshwater, marine and terrestrial habitat. Some small species such as the land snail lives in dampy and dark places in terrestrial habitats. Molluscs are animal, are multicellular, have tissue, a digestive tract which includes mouth, stomach, anus and they also possess an oral structure called the radula. Some cephalopods do not have shells as they have lost their shells at one point in evolution and examples include octopuses. Lobsters are crustaceans belonging to the phylum arthropoda. Molluscs do have three main body parts head, visceral mass and the muscular foot.
Why do you think that the first thing you learn about in a Biology course is Chemistry? How does learning chemistry apply to Biology?
Final answer:
The fundamental principles of chemistry are critical for understanding biological processes. The composition and reactions of chemicals form the basis of all life, and this knowledge is requisite for grasping the complexity of biology, which is why it is taught first in Biology courses.
Explanation:
One might wonder why the first thing you learn in a Biology course is Chemistry. The reason is that chemistry forms the fundamental basis for understanding the biological processes that occur in all living organisms. Our cells consist of thousands of chemicals, involving elements such as carbon, hydrogen, oxygen, nitrogen, phosphorus, and sulfur. These chemicals combine through chemical reactions to form the building blocks of life. Additionally, the products we use daily, our food preparation methods, and even how our bodies function, involve chemistry at a fundamental level.
In a Biology course, understanding the chemical foundation of life allows students to grasp complex biological processes. For instance, studying the atomic structure of molecules helps in understanding how proteins and nucleic acids work. Similarly, knowledge of how different forms of matter combine is pivotal in areas such as the synthesis and design of drugs—a key aspect of medicine.
The field of genetics has demonstrated that both molecular and transmission genetics characterize the same hereditary processes at different levels. For example, the wrinkled seed phenotype used by Mendel results from a defect in the gene that encodes which of the following products?
(A) gibberellin 3-[beta] hydroxylase, an enzyme required for synthesis of the growth factor gibberellin
(B) an enzyme that catalyzes a step in the breakdown of chlorophyll
(C) a transcription factor that activates the expression of genes required for anthocyanin production
(D) starch-branching enzyme
Answer:
The wrinkled seed phenotype used by Mendel results from starch branching enzyme.
Explanation:
During mono hybrid cross, Mendel took 2 types of pea plant of which one was tall (TT) and another pea plant was (tt). After crossing, he observed that in F1 generation the ratio of tall and dwarf plant was 3:1
He observed that there is a particular enzyme which causes formation of dwarf plant as the enzyme is absent in wrinkle offspring but present in the round offspring,The enzyme is called as starch branching enzyme.
Thus failure in the formation of this enzyme produce complex metabolic consequences on starch, protein and lipid biosynthesis.
Tropism can be defined as the Choose one: A. ability to infect a particular type of cell within the host. B. mechanism of entry into a host cell for bacteriophages. C. emergence of a new type of virus. D. ability to infect a broad range of hosts.
Answer:
A. ability to infect a particular type of cell within the host.
Explanation:
In plants, tropism refers to the movement of a plant part in a particular direction in response to any stimulus. In microbiology, tropism refers to the ability of pathogen to infect a particular type of cell in the host.
The tropism in which a particular cell is infected is called cell tropism. The tropism in which a particular tissue is infected then it is cell tissue tropism. Host tropism is also seen in pathogen when they infect specific host. Stimulus is required for any tropism. So the right answer is A.
Answer:
A) ability to infect a particular type of cell within the host.
Explanation:
Viral tropism refers to the ability of a virus to infect a particular cell, tissue and a host species.
The viruses have to interact with a variety of negative and positive factors in the cell so the viral tropism is determined at different levels of the viral-host interaction that is during replication and viral progeny production.
The viral tropism occurs at two-step and based on this is of two types- the receptor-independent tropism which takes place intracellularly and the receptor-dependent tropism which takes place at the cell surface.
Thus, Option-A is the correct answer.
Below is the DNA strand template for making an mRNA needed to produce a small peptide. a.) Transcribe the sequence into mRNA. (0.5 pt.) b.) Translate the mRNA sequence into protein. (0.5 pt.) c.) Transcribe the mRNA and translate into protein with a mutant sequence that has a transition at position 11. What type of mutation does this cause
Answer:
a) mRNA: AUG ACC GGC AAU CAA CUA UAU UGA
b) Protein: Methionine, Threonine, Glycine, Asparagine, Glutamine, Leucine, Tyrosine, Stop.
Explanation:
The Question was missing the DNA Strand. I have looked up the question and found this to be the DNA strand related to this question. A picture for converting codons to amino acids is also attached. A protein is simply a strand of Amino Acids.
DNA (3 to 5 sequence):
TAC TGG CCG TTA GTT GAT ATA ACT
a). Transcribing to mRNA (5 to 3 sequence):
AUG ACC GGC AAU CAA CUA UAU UGA
b). Related Amino Acids:
1st Codon: Methionine
2nd Codon: Threonine
3rd Codon: Glycine
4th Codon: Asparagine
5th Codon: Glutamine
6th Codon: Leucine
7th Codon: Tyrosine
8th Codon: Stop Codon
c.) DNA with Transition Mutation at Position 11 (5 to 3 sequence):
In genetics, Transition is a point mutation which converts a purine nucleotide to another i.e. A ↔ G, or a pyrimidine nucleotide to another pyrimidine i.e. C ↔ T. Hence, after transition at 11 position, T will be changed to C.
TAC TGG CCG TCA GTT GAT ATA ACT
Transcribing to mRNA (5 to 3 sequence):
AUG ACC GGC AGU CAA CUA UAU UGA
Related Amino Acids:
1st Codon: Methionine
2nd Codon: Threonine
3rd Codon: Glycine
4th Codon: Serine
5th Codon: Glutamine
6th Codon: Leucine
7th Codon: Tyrosine
8th Codon: Stop Codon
If a compound has a number of individual dipoles, then: I. It is polar overall. II. There is an electronegativity difference between the bonded atoms. III. it is ionic. IV. It doesn't have resonance.
Answer:
II. There is an electronegativity difference between the bonded atoms.
Explanation:
In a compound that has a number of individual dipoles(weaker force of attractions between compounds). There exist a weaker intermolecular force of attraction ( Vander waal's forces), these force of attraction arises from fluctuating dipoles in atom molecules brought by the movement of electrons around the atomic nucleus. They possess weak electronegativity difference between the bonded atoms because they have less energy to break these weaker bonds.
Also, individual dipoles can interact with each other in an way which will also increase the boiling point and as boiling point increases there is decrease in electronegativity difference between the bonded atoms.
On the other-hand, A polar compound are ionic compounds and the bonding between them are electrovalent bonding because of their high melting and boiling points. They possess high electronegativity.
Proteins A, B and C bind to each other to form a complex, ABC. Under equilibrium the concentrations of A, B, C and ABC are 10-2 M. The equilibrium constant and the standard free energy of this association reaction at T=300 K are, respectively,
a) 10-6 M2 and -8.3 kcal/mol.
b) 10-4 M2 and -5.5 kcal/mol.
c) 10-3 M2 and -4.1 kcal/mol.
d) 10-2 M2 and -2.8 kcal/mol.
e) 10-1 M2 and -1.4 kcal/mol.
Answer: b) [tex]10^{4}M2[/tex] and [tex]-5.5 kcal/mol[/tex]
Explanation:
Equilibrium concentration of [tex]A[/tex] = [tex]10^{-2}M[/tex]
Equilibrium concentration of [tex]B[/tex] = [tex]10^{-2}M[/tex]
Equilibrium concentration of [tex]C[/tex] = [tex]10^{-2}M[/tex]
Equilibrium concentration of [tex]ABC[/tex] = [tex]10^{-2}M[/tex]
The given balanced equilibrium reaction is,
[tex]A+B+C\rightleftharpoons ABC[/tex]
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[ABC]}{[A][B][C]}[/tex]
Now put all the given values in this expression, we get :
[tex]K_c=\frac{10^{-2}}{(10^{-2})^3}[/tex]
[tex]K_c=10^4M^2[/tex]
[tex]\Delta G^o=-2.303\times RT\times \log K_c[/tex]
where,
R = universal gas constant = 2 cal/K/mole
T = temperature = 300 K
[tex]K_c[/tex] = equilibrium constant = [tex]10^4[/tex]
[tex]\Delta G^o=-2.303\times 2\times 300\times \log (10^4)[/tex]
[tex]\Delta G^o=-5527.2cal/mol=-5.5kcal/mol[/tex]
Thus the equilibrium constant and the standard free energy of this association reaction at T=300 K are [tex]10^4M^2[/tex] and [tex]-5.5kcal/mol[/tex]
Suppose that an arginine residue in the active site of an enzyme was mutated to alanine. As expected, the alanine mutant was inactive, suggesting that the arginine residue was critical to the catalytic mechanism. Which mutation is most likely to restore wild‑type level of activity to the alanine mutant?
Answer Options:O A to EO A to KO A to SO A to MO A to Y
Answer:
O. A to K
Explanation:
Arginine (R) is a positively charged basic amino acid and alanine (A) is a hydrophobic amino acid. Arginine is mostly found on surface of protein to produce hydrogen bonds or other ionic combinations so as to provide stability to the protein or control the activity of the protein. Alanine on the other hand, is hydrophobic. Thus, when arginine is replaced with alanine the catalytic ability of the enzyme are affected.
Lysine (K) is known to be a positively charged basic amino acid like arginine. Therefore, mutating alanine to lysine will confer the chemical abilities of the protein as a result of similarity in chemical attributes of arginine and lysine, which will confer the wild type characters of the enzyme.
What is the difference between biomass and standing crop
Answer:
Biomass: can be defined as the total number of viable organisms, such as plants, animals in a given area.
While,
Standing crop: can be defined as the amount of biomass at a particular time or it is the amount of above-ground plant biomass.
A colony of bacteria originally contains 200 bacteria. It doubles in size every 30 minutes. How many hours will it take for the colony to contain 2,000 bacteria? (Round your answer to one decimal place.
Answer: It will take 1.7 hours. If it isn't correct then please feel free to delete this. If it is however please mark me brainliest. Thanks!
Which of the following is an example of an abiotic factor?A.the number of individuals of a particular species living in a communityB.the interactions between different species in a communityC.the diversity of prey and predator species in a communityD.the climate of the community in which the species mentioned above inhabit
Answer: Option D) the climate of the community in which the species mentioned above inhabit
Explanation:
Abiotic refers to non-living factors. Thus, the abiotic factors in an habitat where living organisms dwells include the atmospheric conditions such as climate, weather; inorganic nutrients such as phosphorus, carbon compounds, water etc present in the soil of such habitat.
Thus, Option D represent the abiotic factor
Individuals III-3 and III-4 are expecting their first child when they become aware that they both have a family history of this recessive condition. As their genetic counselor, you can calculate the probability that they are carriers and that their child will be affected with the condition.1.) The probability that III- 3 is a carrier (Rr) =2.) The probability that III - 4 is a carrier (Rr) =3.) The probability that IV - 1 will be affected (rr) =Options are 1/4, 1/2, 1/16, 1/3, 3/4, 2/3, 1/12, 1/6
Pedigree attached
Answer:
1) 2/3
2) 1/2
3) 1/12
Explanation:
1) Individual III.1 is affected. He inherited one r allele from his carrier mother, but must also have inherited another allele from his father, who's genotype we are not told. His father is unaffected, but because III.1 is affected, must be a carrier with the genotype Rr. Therefore, the parents of III.3 have are Rr x Rr. We can carry out a punnett square to assess probabilities
R r
R RR Rr
r Rr rr
The probability of him inheriting an r from either parent is 1/2 (as there is 50% chance of being Rr in the punnet square). However, we already know that he is not rr (he is unaffected and has a R allele), so that means 2/3 of the options involve him being a carrier
2) The probability III.4 is a carrier can be calculated because we have the genotypes of her parents, RR x Rr
R R
R RR RR
r Rr Rr
The probability III.4 is a carrier is 1/2 but with no possibility that she is affected.
3) If both parents were carriers, there would be a 1/4 chance that individual IV.1 would be affected, see punnet square below:
R r
R RR Rr
r Rr rr
Only rr genotype is affected, so the probability is 1:4. However, there is only 1/2 probability that their mother is affected and 1/2 that their father is. To work out their probability based on this we have to multiply the probabilities: 2/3 x 1/2 x 1/4 = 1/12
The probabilities that the parents III-3 and III-4 are carriers of a recessive genetic trait are both 1/2, while the probability of their child inheriting the condition is 1/4.
Explanation:This question is asking about the probability of individuals III-3 and III-4 being carriers of a recessive genetic condition and the probability of their child being affected. Since we're dealing with a recessive genetic trait, if both parents are heterozygous carriers (Rr), the chance that any one of their children will be a carrier is 50% or 1/2. This applies to the probabilities that III-3 and III-4 are carriers. The chance that their child IV-1 will be affected (rr) is 1/4, as in classical Mendelian inheritance, when both parents are heterozygous carriers, there is a 25% chance for the child to inherit two recessive alleles.
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1. Explain why the communication skills and techniques used within a business unit (department) are not always effective in communicating across business units or up and down the corporate ladder.
Answer:
Communication can be complicated in any organization. This is due to the fact that discussing peer to peer in a corporate office, of the institution, is going to be a lot unlike discussing peer to peer in a low-level office setting.
Also, cultural diversity can take a form in how efficient communication is. It is also very important that varying forms of communication are employed. Businesses that consists of many varying departments won't be dependent wholly on face to face communication. Therefore, it is necessary to have other forms of communication being utilized often. Examples of such Forms: email, phone, radio, etc.
In order to establish an effective route of communication that each sector of the organization feeds from, skilled communication professionals is necessary to be placed at every sector or place of business. This will permit corporate to possess “eyes on the target” per say. In other words, they can possess actual people all through every department or unit that can convey messages back and forth. International seems to do a good job at this. Their communications directors are involved with the relay of messages, permitting corporate’s suggestions which is then tailored to fit their particular department
The solutions listed below are in two arms of a tube separated by a semipermeable membrane. The solution in the left arm is given first. In which case the osmotic flow proceeds from the right to the left arm? Group of answer choices
0.010 M NH3(aq) |
0.010 M NaCl(aq)
0.010 M NaCl(aq) |
0.010 M FeCl3(aq)
0.010 M NaCl(aq) |
0.010 M CH3COOH(aq)
0.010 M NaCl(aq) |
0.010 M CaCl2(aq)
0.010 M CaCl2(aq) |
0.010 M FeCl3(aq)
Answer:
The correct answer is the couple: 0.010 M NaCl | 0.010 M CH₃COOH
Explanation:
A semipermeable membrane allows to pass through solvent molecules but not solute molecules. In this case, all solutions have the same molarity (M) but they do not have the same quantity of solute particles because some of them dissociate in two or more ions (van't Hoff factor i is higher than 1) . Osmotic flow proceeds always from the side with lower concentration of solute (with more solvent molecules) to the side with higher concentration of solute. For each pair of solution, we have to determine the number of particles of solute or the van't Hoff factor (i). If the right side has the lower concentration of solute (higher i), the osmotic flow will proceed from the right to the left.
0.010 M NH3(aq) | 0.010 M NaCl(aq): NH₃ is a nonelectrolyte (i=1) and NaCl has i= 2. Osmotic flow proceeds from left to the right.
0.010 M NaCl(aq) | 0.010 M FeCl3(aq): NaCl has i=2 and FeCl₃ has i=3 (it dissociates in Fe⁺ and 3 Cl⁻). Osmotic flow proceeds from left to the right.
0.010 M NaCl(aq) | 0.010 M CH3COOH(aq): NaCl has i=2 and CH₃COOH is a nonelectrolyte (i=1). The lower concentration is in the right side, so the osmotic flow proceeds from right to left.
0.010 M NaCl(aq) | 0.010 M CaCl2(aq): NaCl has i=2 and CaCl₂ has i=3 (it dissociates in Ca⁺ and 2 Cl⁻). The lower concentration is in the left side so the osmotic flow proceeds from left to right.
0.010 M CaCl2(aq) | 0.010 M FeCl3(aq): CaCl₂ has i=3 and FaCl₃ has i=4 (it dissociates in Fe⁺ and 3 Cl⁻), so the lower concentration is in the left side and osmotic flow proceeds from left to right.
In the given case, the osmotic flow proceeds from the right to the left arm - 0.010 M NaCl(aq) | 0.010 M CH3COOH(aq)
Osmotic flowIt is the spontaneous flow of molecules of the solvent through a semipermeable membrane. The flow would be for the zone of high concentration in order to achieve equal concentration on both sides.
Since we have to apply the pressure on the left arm of the tube to stop osmosis. It indicates that the solution in the left arm is hypertonic to the one in the right arm of the tube.
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Duroc Jersey pigs are typically red, but a sandy variation is also seen. When two different varieties of true-breeding sandy pigs were crossed to each other, they produced F1 offspring that were red. When these F1 offspring were crossed to each other, they produced red, sandy, and white pigs in a 9:6:1 ratio. Explain this pattern of inheritance in terms of number of genes, and any allele or gene interactions.
Answer:
The sandy variation could be as a result of a homozygous recessive allele at one of two varying genes in these two different types of sandy pigs. Let's refer to them as genes A and B.
A kind of sandy pig may be aaBB and the variety AAbb. The F1 generation in this cross produces heterozygotes for both genes to give all red. This indicates that the A and B alleles are dominant. In the F2 generation, 6 out of 16 give us homozygous for either the aa or bb alleles and produces sandy.
Out of 16 will we have one that would be doubly homozygous and be white. The other 9 will possess at least one dominant allele for both genes.
Which of the following is NOT true regarding the sounds associated with the heart? Select one: a. The "lub-dub" sound originates from the physical act of the heart contracting. b. The "lub-dub" sound originates from blood turbulence through the heart.
The 'lub-dub' sound we hear is the sound of the heart valves closing, not the contraction of the heart muscle. The 'lub' is the sound of the mitral and tricuspid valves closing, while the 'dub' is the sound of the aortic and pulmonic valves closing.
Explanation:The statement 'a. The "lub-dub" sound originates from the physical act of the heart contracting' is NOT true regarding the sounds associated with the heart. The "lub-dub" sound we hear is actually made by the closing of heart valves, not the contractions of the heart muscle itself. Specifically, the "lub" is the sound of the mitral and tricuspid valves closing at the beginning of systole (an active phase when your heart pumps blood), while the "dub" is the sound of the aortic and pulmonic valves closing at the beginning of diastole (a resting phase when your heart fills with blood).
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Final answer:
The incorrect statement is that b. the "lub-dub" sound originates from blood turbulence through the heart. These sounds are actually the result of the closing of the AV valves (for "lub") and the semilunar valves (for "dub"), which can be heard with a stethoscope.
Explanation:
The question is asking which statement is NOT true about the origins of the heart sounds. The answer to this is b, "The "lub-dub" sound originates from blood turbulence through the heart." It is not correct because the "lub-dub" sounds, known medically as S1 and S2, are primarily caused by the closing of the heart valves, not by blood turbulence.
The "lub" or the first heart sound, S1, occurs when the atrioventricular (AV) valves close at the beginning of ventricular systole, while the "dub" or the second heart sound, S2, happens when the semilunar (SL) valves close at the end of ventricular systole. When a medical practitioner uses a stethoscope to listen to the heart, these sounds provide critical information regarding the heart's condition.
23. Select all accurate statements
A. All chordates will have a muscular post an*l tail as adults
B. All chordates will have a notochord as adults
C. All chordates have pharyngeal slits or clefts that function in swimming.
D. All chordates are bilaterally symmetrical animals.
E. All chordates have a ventral, hollow nerve cord.
Answer:
E.All chordates have ventral l, hollow nerve cord
D.All chordates are bilaterally symmetrical animals
Beta ( ???? ) sheets are a type of secondary structure in proteins. A segment of a single chain in an antiparallel ???? sheet has a length of 94.5 Å . How many residues are in this segment?
Answer:
The correct answer is 27 residues.
Explanation:
In an antiparallel beta sheet, there is a repeat for every 7 Angstrom units and the number of residues is two for every repeat. In an antiparallel sheet that exhibits a length of 94.5 Angstrom, the number of residues can be determined by using the following formula:
2 residues per every 7 Angstrom,
Length of the given segment is 94.5 Angstrom
Number of residues in this segment are:
= (number of segments) * (residues per segment)
= (94.5/7) * 2
= 13.5 * 2
= 27 residues.
Thus, the number of residues in this segment is 27.
Beta sheets are a type of secondary structure in proteins where segments of the polypeptide chain are aligned side by side. The number of residues in a segment of an antiparallel beta sheet can be calculated using the formula number of residues = length / distance between adjacent residues.
Beta sheets are one of the two major types of secondary structure in proteins, along with alpha helices. In a beta sheet, segments of the polypeptide chain are aligned side by side, forming a sheet-like structure. The length of a segment in an antiparallel beta sheet can be calculated using the formula length = number of residues * distance between adjacent residues. In this case, the length of the segment is given as 94.5 Å, so we can rearrange the formula to find the number of residues: number of residues = length / distance between adjacent residues.
It's important to note that the distance between adjacent residues in a beta sheet can vary, but a common value used for calculations is 3.5 Å. So, using this value, we can calculate the number of residues: number of residues = 94.5 Å / 3.5 Å = 27 residues.
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The incidence of Down syndrome, also known as trisomy 21, increases with increasing maternal age. Which of the following errors most likely produces this condition?A. Nondisjunction during meiosis II in either the male and female gameteB. Nondisjunction during either meiosis I or II in the female gameteC. Nondisjunction during either meiosis I or II in the male gameteD. Nondisjunction during meiosis I in either the male and female gamete
Answer:. Nondisjunction during meiosis I in either the male and female gamete
The failure of homologous chromosome to separate at mitosis, or failure of sister chromatids to separate at Anaphase of meiosis I or ii is called non-disjunction.
In this cell division anomaly the unsegregated chromosomes are separated into cells as single chromosomes, therefore extra copies are therefore inherited.This result is one of the cell having an extra copies of chromosomes than others( chromosomes 21.
If this extra chromosome copies is segregated into a sperm or egg, the resulting zygote cells will have extra copy as chromosome 21, and hence Down syndrome.
→It is more common in Meiosis 1 than ii,because of longer duration of the first stage of division than meiosis ii,given rooms to errors.
→Most common in women(with age increase) than men because of exchange of telomeres in them,which worsen with age. Their Meiotic mechanism which weaken with age, and more prone to errors compare to men,
Explanation:
when DNA mutations occur they can have a variety of effects on an organism if the DNA is TAC CGA ACC TTA create a mRNA strand with a point mutation that would shorten the protein
Final answer:
A nonsense mutation in the mRNA sequence can introduce a stop codon, prematurely ending the translation process and therefore shortening the protein. The mutated mRNA sequence for the given DNA would have a point mutation turning the second codon into a stop codon, thus producing a truncated and likely non-functional protein.
Explanation:
To induce a point mutation that shortens the protein, you would create a nonsense mutation in the mRNA sequence. For the original DNA sequence TAC CGA ACC TTA, the corresponding mRNA sequence is AUG GCU UGG AAU. If we introduce a point mutation in the second codon to change it to a stop codon, we can use the stop codon UAA. The mutated mRNA sequence would be AUG UAA UGG AAU. Here, UAA is a stop codon and will cause the translation process to terminate prematurely, therefore shortening the resulting protein.
Nonsense mutations convert a codon that can specify an amino acid into a stop codon (UAA, UAG, UGA), leading to premature termination of translation. This type of mutation can have a profound effect on the protein, as it will be truncated, likely destroying its normal function. This is because the ribosome will cease to translate the mRNA into protein upon reaching the stop codon, resulting in an incomplete and non-functional protein product.
Mutations can profoundly impact an organism by altering its proteins. Depending on where the mutation occurs, the consequences may be mild or severe. Inducing such mutations intentionally can be a valuable tool for research but can be detrimental when they occur naturally and disrupt the organism's biological functions.
________ produced by microbial fermentation of glucose from cellulose or cornstarch is becoming a more important component of biofuels in the United States, and specialized ________ are needed to make this a commercially available product. A) Biodiesel / biotechnologists B) Biodiesel / industrial microbiologists C) Ethanol / biotechnologists D) Ethanol / industrial microbiologists
Answer:
The correct answer is option D) "Ethanol / industrial microbiologists".
Explanation:
Ethanol is one of the most common components of biofuels in the United States. Ethanol is used in biofuels for its property of producing oxygen when burned, which makes car's engine to burn fuel more efficiently. Ethanol is produced by microbial fermentation of glucose from cellulose or cornstarch, and industrial microbiologists are the professionals responsible of making this product commercially available.
Answer:
D) Ethanol / industrial microbiologists
Explanation:
In this case of bio-fuels we have already prepared bio ethanol that is being produced after a process called as fermentation run by utilization of specially designed strains of micro-organisms. This simple fermentation is on a small scale and is not feasible for let say the whole country. Now, the industrial microbiologists are the only expert people that shall advance in some kind of already present hyper fermentation techniques that'll eventually give us enough quantity of ethanol to be used commercially.
Suppose that two parents are both heterozygous for sickle cell anemia, which is an autosomal recessive disease. They have seven children. Use the binomial theorem to determine the probability that three of the children have sickle cell anemia and four of the children are healthy. Round your answer to the nearest hundredth.
Huntington\'s disease is an inherited autosomal dominant disorder that can affect both men and women. Imagine a couple has had seven children, and later in life, the husband develops Huntington\'s disease. He is tested and it is discovered he is heterozygous for the disease allele, Hh. The wife is also genetically tested for the Huntington\'s disease allele, and her test results show she is unaffected, hh. What is the percent probability that the first child of this couple will have Huntington\'s disease?also..
What is the probability that any two of the seven children will have Huntington\'s disease?
Answer:
Part A
[tex]0.108[/tex]
Part B
B.1
Fifty Percent
B.2
[tex]16.4[/tex] %
Explanation:
Part A
Given -
Both the parents are heterozygous for the sickle cell anemia
Let the genotype of the parents be
Ss
where S is the allele for presence of sickle cell anemia
and s is the allele absence of sickle cell anemia
If the above two parents are crossed, following offspring are produced
Ss * Ss
SS, Ss, Ss, ss
The probability of children with sickle cell anemia is equal to
[tex]\frac{3}{4}[/tex]
The probability of children with sickle cell anemia is equal to
[tex]\frac{1}{4}[/tex]
Probability that three of the children have sickle cell anemia and four of the children are healthy
[tex]\frac{7!}{3! * 4!} * \frac{3}{4}^4 * \frac{1}{4}^3\\ = 35 * \frac{81}{256 *64} \\= 0.108[/tex]
Part B
B.1
Hh * hh
Hh, Hh, hh, hh
The first child of the parent has a probability of fifty percent to have Huntington\'s disease
B.2
[tex]\frac{7!}{2!*5!} * \frac{1}{2}^2 * \frac{1}{2}^5[/tex]
[tex]= 16.4[/tex]
According to existing research, chimpanzees Select one: a. do not use tools. b. have complex grooming and courtship behaviors. c. cannot solve technical problems. d. All of the choices are correct.
Answer:
option b. have complex grooming and courtship behaviors is correct answer.
Explanation:
According to a research paper published in American journal of primatology provided evidences that chimpanzee males, as promiscuous breeders with minimal costs to mating, should show little or no preference when choosing mating partners (e.g. should mate indiscriminately).
Reference: Proctor, D. P., Lambeth, S. P., Schapiro, S. J., & Brosnan, S. F. (2011). Male chimpanzees' grooming rates vary by female age, parity, and fertility status. American journal of primatology, 73(10), 989–996. doi:10.1002/ajp.20964