Answer:
Embedded in a lipid bilayer. (Ans. C)
Explanation:
Fluid mosaic model explains different observations such as the structure of functional cell membranes. According to this model, there is a thin polar membrane composed of a two-layer of lipid molecules called lipid bi-layer or phospholipid bi-layer in which proteins molecules are embedded. This layer provides fluidity and elasticity to the membrane structure.
These membranes form a continuous barrier around all cells and they are a flat sheet in structure. The lipid bilayer is working as a barrier that keeps proteins, ions and other molecules where they are needed to be and preventing diffusing them to other areas. They are impermeable to most hydrophilic molecules. They are particularly impermeable to ions, which allow cells to regulate pH and salt concentration in membrane through the transport ions across their membranes using proteins knows as ion pumps.
According to the fluid mosaic model of membrane structure, proteins of the membrane are mostly embedded in a lipid bilayer and have a specific inside-outside polarity.
Explanation:According to the fluid mosaic model of membrane structure, proteins of the membrane are mostly embedded in a lipid bilayer. They can be found throughout the membrane, both on the inner and outer surfaces, as well as in the hydrophobic interior. However, they are not randomly oriented in the membrane, but have a specific inside-outside polarity.
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All of the following are products or intermediaries in glycolysis except
A) ATP.
B) NADH
C) FADH2.
D) pyruvate.
E) phosphoenolpyruvate.
Answer:
The correct answer for this is C, because the Fadh2 is only involved at the redox reactions.
Explanation:
ATP is just a product from glycolysis. Remember that when you break the glucose, energy is been free as an ATP molecule. NADH2 is a substratum, you need it, to get NAD at the fermentation process in the cytoplasmic matrix. Pyruvate and phosphoenolpyruvate are also products at the glycolysis.
The old-growth forests of the Pacific Northwest are considered to be rainforests, which means it rains a lot. How do the forests help people deal with the rain?
A. They stick up into the clouds and absorb some rain directly.
B. They direct the water into the salmon streams.
C. They provide shelter to people out hiking.
D. They make it seem less gloomy.
E. They retain water to prevent flooding and erosion.
Answer:
The correct answer is E.They retain water to prevent flooding and erosion.
Explanation:
Rain forests are used to be very dense forests where lots of raining takes place. The forest floor can soak up lots of rain and during flood dense forest with woodland trees do not allow water to run fast through them thereby reducing flood's effect significantly.
The roots of the trees in the forest binds to the soil on the floor and prevent soil erosion during flooding. Study shows that water retention is more in summer than in winters. Forest can store lots of water and transfer it to the streams which is helpful in providing clean water to the people in dry season.
Therefore, the correct answer is E. They retain water to prevent flooding and erosion.
A stack of thylakoids are known as
a. Thylakoid discs
b. Grama
c. thylakoid lumen
d. Stroma
A stack of thylakoids in a chloroplast is called a granum, which is the correct answer to the question.
Explanation:A stack of thylakoids within a chloroplast is known as a granum (plural = grana). Thylakoids are disc-shaped, membrane-bound structures where the light-dependent reactions of photosynthesis take place. Each thylakoid disc contains chlorophyll, which is responsible for the initial interaction between light and plant material. The inner membrane space that surrounds the grana is termed the stroma. Therefore, the correct answer to the question is 'b. Grama'. Thylakoids are disc-shaped, membrane-bound structures where the light-dependent reactions of photosynthesis occur.
There are many different types of touch receptors (hot, cold, etc.) in the skin, and these different types of receptors are not distributed throughout the various parts of the body equally.
a. True
b. False
Answer:
It is true.
Explanation:
Throughout the skin, there are sensitivity receptors, no matter what sector we look for.
The difference is that there are areas of higher density.
Places like the hand and fingers are the ones that help us explore and know things. The number of receptors there is much greater than in areas such as the back or neck.
In maize (corn) plants, a dominant allele I inhibits kernel color, while the recessive allele i permits color when homozygous. At a different locus, the dominant allele P causes purple kernel color, while the homozygous recessive genotype pp causes red kernels. If plants heterozygous at both loci are crossed, what will be the genotypic and phenotypic ratios of the offspring?
Answer:
Phenotype ratio= 12 colorless : 3 purple : 1 red
Genotype ratio= 1:2:1:2:4:2:1:2:1
Explanation:
According to the given information, the dominant allele "I" is epistatic to P and p alleles and presence of "I" inhibits the expression of both "P" and "p" alleles.
A cross between two heterozygous plants for both locus would give the F2 progeny in 12 colorless: 3 purple: 1 red ratio.
Here, the F2 genotype with "I-G" and "I-gg" would produce colorless kernels while the ones with "ii-G-" would exhibit purple colored kernels. The F2 genotype "iigg" would beak red kernels.
Hence, the phenotype ratio= 12 colorless : 3 purple : 1 red
Genotype ratio would be same as for the mendelian dihybrid cross= 1 IIPP: 2 IIPp :1 IIpp: 2 IiPP :4 IiPp: 2 Iipp: 1 iiPP :2 iiPp: 1 iipp
When heterozygous maize plants are crossed, the genotypic ratio of the offspring will be 1IIpp:2Iipp:1iipp. The phenotypic ratio will be 1 purple:2 yellow:1 red.
Explanation:In the given scenario, the maize plants have two different gene loci: one for kernel color and the other for kernel color. The genotype for kernel color is represented by the alleles I (dominant, inhibits color) and i (recessive, allows color). The genotype for kernel color is represented by the alleles P (dominant, purple) and pp (recessive, red). When plants heterozygous at both loci are crossed, the genotypic ratio of the offspring will be 1IIpp:2Iipp:1iipp. The phenotypic ratio will be 1 purple:2 yellow:1 red.
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A nucleosome forms hydrogen bonds with what part of the DNA? A nucleosome forms hydrogen bonds with what part of the DNA? only with bases via the major groove with the phosphodiester backbone and with bases via the minor groove only with bases via the minor groove with the phosphodiester backbone and with bases via the major groove
Answer:
A nucleosome forms hydrogen bonds with the phosphodiester backbone and with the bases through the minor groove. The histone-fold hydrogen bonds with both the A: T enriched bases and the phosphodiester backbone. The histone-fold domains' association with the minor groove is responsible for the majority of the associations in the nucleosome.
Final answer:
A nucleosome forms hydrogen bonds primarily with the phosphodiester backbone of DNA, facilitated by interactions at the L1-L2 loops and αl helices, including minor groove entry by arginine side chains from histone folds. These arrangements enable efficient DNA packaging and sequence-independent binding necessary for nucleosome function.
Explanation:
A nucleosome forms hydrogen bonds with the phosphodiester backbone of DNA. These interactions are fundamental for DNA packaging within the nucleus, ensuring DNA is wrapped around histone proteins to form nucleosomes. The L1-L2 loops at the ends of each histone dimer and the αl helices at the center of the DNA binding site are primary contact points, where hydrogen bonds are formed between amino acids of the histones and the phosphate backbone of DNA. Additionally, arginine side chains from histone folds enter the minor groove of DNA, influencing the DNA's structural dynamics within the nucleosome.
These interactions allow nucleosomes to bind to DNA in a non-sequence-specific manner, which is crucial for their role in DNA packaging and regulation. The presence of water-mediated interactions and the flexibility in binding accommodate the DNA's varying sequences and structures, enabling nucleosomes to organize DNA efficiently within the cell nucleus.
Cystic fibrosis is a genetic disorder caused by a recessive allele. If two parents each carry an allele for the disorder but have normal phenotypes, indicate what percentage of their children will be phenotypically normal and what percentage will have cystic fibrosis.
Answer: 75% normal phenotype, 25% cystic fibrosis.
Explanation:
Since cystic fibrosis is caused by a recessive allele, then it needs both affected alleles to develop. If the parents have normal phenotyps but they carry one allele for the disorder, then their genotypes will be Cc for both (they are heterozygous).
C is the dominant allele that codes for the trait, and c is the affected allele.
They only have one copy of the affected allele, thereby they will not develop cystic fibrosis.
The next step is to find out the genotypes of the gametes.
Gametes are sex cells, sperm or eggs produced by the parents. Those cells only have one allele of the gene, that means the gametes produced by them can be either C or c. Those gametes are used in the punnett square as shown in the picture.
There we can see 50% of the offspring is Cc, 25% is cc and 25% is CC.
As it was stated before, individuals who are cc will develop cystic fribrosis. Then, 25% of them will have it. And 75% of them (25% of CC + 50% of Cc) will not.
Would you expect a shrub or dandelion (dispersed by wind-blown seed) to be a more likely pioneer plant species? why? Thank you so much for helping! means a lot !
Answer:
Dandelions may appear quicker after harsh conditions and reproduce at a faster rate. However, both dandelions and shrubs are considered fast-growing plant species that can be categorized as pioneer species.
Explanation:
Secondary succession refers to the changes that take place in a disturbed habitat. Pioneer plant species are those that colonize new habitats after harsh climate conditions and that tend to reproduce at a fast rate.
According to researcher J.W. Darlling (2008), pioneer herbs and shrubs are species that tend to grow faster in comparison to other species, making them excellent pioneer species.
This occurs thanks to plants that are wind-pollinated, such as dandelions, have a higher chance to appear because, as it is a disturbed environment, there are no insects or other fauna present. In addition, shrubs are persistent species that are able to reproduce fast with limited soil availability but a bit slower in comparison to dandelions.
A dandelion (dispersed by wind-blown seed) would be a more likely pioneer plant species because wind-dispersed seeds enable rapid colonization of disturbed or barren environments, while shrubs may take longer to establish.
I would expect a dandelion (dispersed by wind-blown seed) to be a more likely pioneer plant species. Pioneer plant species typically need to colonize disturbed or barren environments quickly, and wind-dispersed seeds like those of dandelions have the advantage of being able to spread over long distances and establish themselves rapidly in new areas.
Additionally, dandelions are known for their ability to grow in a variety of soil conditions and climates, making them well-suited to colonize and stabilize the soil in harsh or unpredictable environments. In contrast, shrubs might take longer to establish themselves and may not have the same capacity for rapid colonization as plants with wind-dispersed seeds.
Which of the following combinations is most likely to result in digestion?
a. pepsin, protein, water, body temperature
b. pepsi, protein, water, body temperature, HCl (hydrochloric acid)
Explain.
Answer:
Option (b).
Explanation:
Digestion may be defined as the process of break down of large food particles into simpler substances that be absorbed by the body. Specific enzymes are required for the breakdown of specific chemicals.
The combination that contains protein, water, HCl (hydrochloric acid), temperature and pepsin result in digestion. HCl (hydrochloric acid) activates the enzyme pepsinogen into pepsin. The pepsin than digest the proteinsat the body temperature in presence of water. Proteins are broken down into simple amino acids that can be seen as a result of digestion.
Thus, the correct answer is option (b).
Luteinizing hormone stimulates testosterone secretion by the leydig cells of the testes.
a. True
b. False
Answer:
True
Explanation:
Testosterone is the primary male sex hormone. It is produced by the Leydig cells of the testis. The testosterone produced by both males and females. The amount is more male than females. In females, testosterone is in the form of androgen hormone.
The anterior pituitary secretes 2 hormones i.e. LH and FSH. The luteinizing hormone from the pituitary gland enters into the interstitial space of the testis. In the interstitial space, Leydig cells are present, which are the target organ for LH. The LH stimulates the Leydig cells to produce testosterone. Hence Leydig cells are also called interstitial cells.
Testosterone secretes from the Leydig cells of the testis and mixes with the bloodstream. It produces in the presence of LH. Then it reaches to different cells of the body by the bloodstream. Testosterone maintains bone and muscle growth, induces the secondary sexual characters.
When testosterone level is high in the blood, it sends a signal to the brain to decrease the secretion of testosterone. This is the negative feedback mechanism of testosterone.
Describe the "blender experiments" of Hershey and Chase and what they revealed about DNA versus protein.
Answer:
Explanation:
Alfred Hershey and Martha Chase showed that DNA is the genetic material by their blending experiment. Earlier it was believed that protein is the genetic material. As it is found in a larger amount in the cell.
Hershey and Chase infected the bacteria E coli with the T2 phage virus. In the bacteria, the viral DNA is replicated as the bacterial DNA. They prepare 2 culture media one contains radiolabeled phosphorus (32P) and another has radiolabeled (35 S)sulfur. Then he cultures the bacteria in separate culture medium and infected with many T2 phages.
After the infection, they isolated the infected cell and centrifuge the cells. When they tested separate supernatant of 2 culture mediums, they found the virus of radiolabeled phosphorus doesn't show any radioactivity. But the virus in the radiolabeled sulfur contains the radiolabeled.
The viruses have 2 biomolecules one is protein and the other is DNA/RNA. When they have centrifuged the cells which are the palette, the phosphorus is found there and it also degrades. But the sulfur which is present in the supernatant is present in the newly replicated viruses.
He also treated the virus with both the culture medium but newly infected cells show radiolabeled sulfur, not the phosphorus. Thus they concluded DNA is the genetic material.
What are some symptoms you might expect an individual suffering from anemia to exhibit?
Answer:
Anaemia is the major blood related disease whose symptoms can include the following.
1- weakness
2- tiredness
3- pale coloured skin
4- cold extremities
5- irritation
Explanation:
Anaemia is defined as the condition which arises due to lack of red blood cells. Red blood cells contain protein, haemoglobin, whose major constituent is iron. Human body uses iron to produce red blood cells in the bone marrow.
This haemoglobin helps red blood cells to transport oxygen to all the parts of human body. Oxygen gets attached to the haemoglobin and is carried all over the body. This oxygen is necessary for the cells of human body to function properly and survive.
Anaemia can be said to occur due to the lack of iron in human body. Lack of iron results in lack of red blood cells which results in decrease in oxygen supply to all the parts of the body.
Lack of oxygen hampers the normal functioning of human body as described below.
1- Capacity to work reduces and causes easy tiredness.
2- Weakness increases due to lack of iron.
3- The skin changes its colour, loses its shine and becomes yellowish showing lack of red blood cells.
4- The body temperature is also affected due to lack of oxygen. Hands and legs become cold.
5- Irritation increases. The patient becomes irritated easily due to weakness and fatigue.
Reasons for risk of anaemia
1- Disorder of intestine affects intake of nutrients by the small intestine. This increases chances of anaemia.
2- Menstrual cycle results in loss of blood. This also increases the chances of getting anaemia.
3- Anaemia can be hereditary. It can be genetically acquired.
4- Improper diet including heavy intake of coffee and tea also leads to Anaemia.
The severity of anaemia can vary from mild to severe. Treatment can be in the form of supplements for mild anaemia to medical procedures in case of severe anaemia.
is the transport of potassiun ions into the cell endergonic
orexergonic and why?
Answer: Endergonic
Explanation:
The sodium-potassium pump (Na+/K+ pump) moves sodium and potassium ions against its concentration gradient, which means in opposite directions through the membrane. Two potassium ions are imported into the cell while three sodium ions are exported.
This pump uses the energy derived from exergonic ATP hydrolysis. This means, ATP releases energy that will be used by the pump. Thereby the use of energy by the pump is an endergonic reaction, because it requires the absorption of energy to function.
So, when ATP is hydrolyzed, it donates free energy to the pump which goes under a conformational change, allowing it to move the ions.
Where would you find a Riboswitch?
Answer:
The correct answer will be- in the 5' untranslated region of prokaryotic mRNA.
Explanation:
Riboswitches are the cis-regulatory RNA elements present in the non-coding segment of the mRNA.The riboswitches are common in prokaryotes but research showed that they are also present in the few eukaryotes. They are located in the non-coding 5' untranslated region of prokaryotic mRNA.
The riboswitches are usually made of 35 to 200 nucleotides which can bind to co-enzymes, small molecules and metabolites and change their conformation to regulated gene expression.
Thus, in the 5' untranslated region of prokaryotic mRNA is the correct answer.
If a mutation occur in a somatic cell, the resulting mutant phenotype will occur:
a. only in the individual cell
b. only in the progeny from that individual cell
c. only in the offspring of that organism
d. in both the progeny of that individual cell and the individual cell itself
e. neither the progeny from that individual cell or the offspring of the organism
Answer:
d. in both the progeny of that individual cell and the individual cell itself
Explanation:
Somatic mutations occur in the somatic cells of the individuals. Since the genetic material of the somatic cells is not passed to the next generation of an organism, the somatic mutations do not appear in the progeny of an individual.
Somatic cells divide by mitosis which in turn maintains the identity of DNA between the parent and the daughter cells. Therefore, the new cells derived from a mutated somatic cell would also carry the same mutation.
A mutation in a somatic cell results in a mutant phenotype in the individual cell itself and the progeny or descendants of that cell, but it will not be passed to the organism's offspring.
Explanation:If a mutation occurs in a somatic cell (i.e., a non-reproductive cell), the resulting mutant phenotype will occur in both the mutant cell itself and the progeny that arises from the cell division of that individual cell. So, the correct answer is. in both the progeny of that individual cell and the individual cell itself.' This happens because the mutation becomes part of the cellular DNA and will therefore be replicated each time that cell divides, spreading to all descendent cells. However, because somatic cells do not participate in sexual reproduction, these mutations will not be passed to the offspring of the organism.
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Which of the following BEST describes transcriptional regulation of the lac operon in E. coli?
a. On in the presence of lactose
b. On in the presence of lactose and presence of glucose
c. Off in the presence of glucose
d. On in the absence of lactose and presence of glucose
e. On in the presence of lactose and absence of glucose
Answer:
E. On in the presence of lactose and absence of glucose
Explanation:
Expression of lac operon synthesizes the enzymes required for catabolism of lactose sugar. When both glucose and lactose are available, glucose is preferred as a nutrient and the lac operon is not expressed.
Lac operon is expressed only when glucose is absent in the medium and lactose is present. If any of the two conditions deviate, the operon is not expressed.
In the absence of glucose and the presence of lactose, the repressor is rendered inactive to bind to the operator. RNA polymerase enzyme is free to bind to the promoter and continue the process of transcription.
The reduced levels of glucose increase the cAMP levels which in turn bind to the Catabolite activator protein (CAP). CAP is a positive regulator that binds to the promoter to facilitate the transcription of the operon by RNA polymerase.
The lac operon in E. coli is 'on' in the presence of lactose and the absence of glucose, allowing the bacteria to respond efficiently to changes in the nutritional environment.
Explanation:The transcriptional regulation of the lac operon in E. coli is a complex process. 'e. On in the presence of lactose and absence of glucose' best describes this regulation. The lac operon is activated, or 'turned on', in the presence of lactose when glucose is not present. If glucose is available, the bacteria preferentially use glucose, and the lac operon is turned off. The lac operon system thus helps the bacteria efficiently respond to changes in the nutritional environment.
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Which of the following statements about DNA structure is true? View Available Hint(s) Which of the following statements about DNA structure is true? The nucleic acid strands in a DNA molecule are oriented antiparallel to each other, meaning they run in opposite directions. Hydrogen bonds formed between the sugar‑phosphate backbones of the two DNA chains help to stabilize DNA structure. Nucleic acids are formed through phosphodiester bonds that link nucleosides together. The pentose sugar in DNA is ribose.
Answer:
The correct answer will be option- the nucleic acid strands in a DNA molecule are oriented antiparallel to each other, meaning they run in opposite directions.
Explanation:
Deoxyribose nucleic acid or DNA is the molecule which acts as the genetic material of the organisms.
The structure of DNA suggests that DNA molecule is made of two strands of nucleotides which are oriented in the opposite direction with respect to each other.
Each nucleotide is composed of a five-carbon sugar called deoxyribose bonded with a phosphate group via ester bond and four types of nitrogenous bases bonded to complementary bases via hydrogen bond.
The sugar-phosphate backbone runs in the opposite direction with a free 5' carbon atom of phosphate group end a 3' OH group of sugar. The orientation of strands is that one strand run from 5' to 3' direction while another runs from 3' to 5' direction.
Thus, the selected option is the correct answer.
DNA consists of two antiparallel strands forming a double-helix structure. Each strand is made of nucleotides linked with phosphodiester bonds, and the pentose sugar in DNA is deoxyribose.
Explanation:The correct statement about the structure of DNA is that the nucleic acid strands in a DNA molecule are oriented antiparallel to each other, meaning they run in opposite directions. DNA is composed of two strands that are twisted to form a double helix; each strand composed of nucleotides that include a nitrogenous base, a five-carbon sugar (deoxyribose), and a phosphate group. The two DNA strands are antiparallel, such that the 3' end of one strand faces the 5' end of the other, resulting in the nitrogenous bases of each strand facing inward and forming hydrogen bonds, which stabilizes the double helix structure. Nucleic acids are indeed formed through phosphodiester bonds that link nucleotides together, but the pentose sugar in DNA is not ribose, but deoxyribose.
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Twogenes,
A and B, are located 10 maps units fromeach
other. A third
gene,
C, is located 15 map units from B and 5 mapunits
from A. A parental
generation
consists of AAbbCC and aaBBccindividuals. The F1
are then test-
crossed to
aabbcc individuals. What percentage ofthe offspring would
you
expect to be
AaBbCc?
Answer:
5%
Explanation:
We have the following loci map:
C/c -------------A/a--------------------------B/b 5 m.u. 10 m.u.The parental cross was between the individuals:
CCbbAA, which can be written as CAb/CAb.ccBBaa, which can be written as caB/caB.Each parental individual can produce only 1 type of gamete, so the F1 will be homogeneous with the genotype: CAb/caB.
The F1 are test crossed to cba/cba individuals.
CAb/caB X cab/cabThe homozygous recessive can only produce cba gametes.
The F1 can produce 8 types of gametes:
The parentals: CAb and caBThe crossovers between the genes C/c and A/a: CaB and cAbThe crossovers between the genes A/a and B/b : CAB and cabThe double crossovers: Cab and cABThe question is asking about the percentage of offspring that will have the genotype CAB/cab
The cab chromosome comes from the homozygous recessive individual with a probability of 1.
The CAB chromosome comes from the F1 individual, and was a result of crossing over between the genes A/a and B/b.
The formula to relate genetic distance with recombination frequency is:
Genetic Distance (m.u.)= Recombination Frequency X 100.
In our problem, Genetic distance between A/a and B/b loci is 10 map units.
Therefore:
10 m.u. = (Recombination Frequency between A/a and B/b) x 100.
0.1 = Recombination Frequency between A/a and B/b
When crossing over happens between the A/a and B/b genes, both CAB and cab are generated, so each of them will appear in a frequency of half the total recombination frequency between those genes, to add a total of 0.1.
The gamete CAB will appear with a frequency of 0.05.
The gamete cab will appear with a frequency of 1.
The percentage of the offspring that will be CAB/cab is:
0.05 x 1 x 100% = 5%In patients infected with nonresistant strains of the tuberculosis bacterium, antibiotics can relieve symptoms in a few weeks. However, it takes much longer to halt the infection, and patients may discontinue treatment while bacteria are still present. How might this result in the evolution of drug-resistant pathogens?
When patients infected with nonresistant strains of tuberculosis discontinue treatment before completing the full course, it can lead to the evolution of drug-resistant bacteria. This is because some bacteria may survive and develop resistance to the antibiotics.
Explanation:Patients infected with nonresistant strains of the tuberculosis bacterium may discontinue treatment once their symptoms are relieved, but before the planned course of treatment is complete. This can result in the evolution of drug-resistant pathogens. The reason for this is that when patients stop taking antibiotics prematurely, there is a higher chance that some bacteria may survive and develop resistance to the drugs. These drug-resistant bacteria can then continue to spread and cause infections that are more difficult to treat.
The discontinuation of antibiotic treatment in patients infected with non-resistant strains of the tuberculosis bacterium can indeed result in the evolution of drug-resistant pathogens through a process known as natural selection.
1. Initial Effectiveness of Antibiotics: When antibiotics are administered to a patient with a non-resistant strain of tuberculosis, the drugs begin to kill the bacteria. This is because the antibiotics are specifically designed to target and disrupt essential processes in the bacteria, leading to their death.
2. Persistence of Bacteria: Despite the effectiveness of the antibiotics, a small number of bacteria may survive. This could be due to various reasons, such as the bacteria being in a dormant state, residing in parts of the body that are less accessible to the antibiotics, or having genetic mutations that confer some level of resistance.
3. Incomplete Treatment Course: If a patient stops taking the antibiotics prematurely, the surviving bacteria are given an opportunity to multiply and grow in number. Since the antibiotics are no longer present at therapeutic levels, there is no pressure to suppress the bacterial population.
4. Selection for Resistance: Among the surviving bacteria, some may have acquired mutations that make them less susceptible to the antibiotics. These mutations might have arisen randomly or as a response to the antibiotic pressure. When the patient discontinues treatment, these less susceptible bacteria have a survival advantage and are more likely to reproduce and pass on their resistance genes.
5. Spread of Resistance: As the resistant bacteria replicate, they can accumulate additional mutations that further enhance their resistance. These resistant strains can then be transmitted to other individuals, spreading the drug-resistant form of tuberculosis.
6. Evolution of Drug-Resistant Pathogens: Over time, and with repeated cycles of incomplete treatment and transmission, the resistant strains become more prevalent in the population. This leads to the evolution of tuberculosis strains that are resistant to one or more of the antibiotics that were previously effective.
To prevent the evolution and spread of drug-resistant tuberculosis, it is crucial for patients to complete the full course of antibiotics as prescribed by healthcare professionals. This ensures that all bacteria are eliminated, thus preventing the survival and proliferation of resistant strains. Public health measures, including infection control practices, regular monitoring, and the development of new antibiotics and treatment regimens, are also essential in combating the rise of drug-resistant pathogens.
The bases of one of the strands of DNA in a region where DNA replication begins are shown here. What is the sequence of the primer that is synthesized complementary to the bases in bold? (Indicate the 5' and 3' ends of the sequence.) 5' AGGCCTCGAATTCGTATAGCTTTCAGAAA 3'
Answer:
Complementary primer- 3' TCCGGAGCTTAAGCATATCGAAAGTCTTT 5'
Explanation:
The synthesized primer will have base pair complementary to the given strand and also the leading and lagging ends will be opposite to the given strand.
As per the base pair rule for DNA
Guanine binds to cytosine & vice versa
Adenine always binds to thymine & vice versa
Given Sequence - 5' AGGCCTCGAATTCGTATAGCTTTCAGAAA 3'
Complementary primer- 3' TCCGGAGCTTAAGCATATCGAAAGTCTTT 5'
Of the 64 possible nucleotide codon triplets, how many specify polypeptide chain termination?
a. 61
b. 1
c. 2
d. 3
e. 64
Of the 64 possible mRNA codons, three are designated for polypeptide chain termination, also known as stop codons. The correct answer to the multiple choice question is d. 3.
In the universal genetic code, of the 64 possible mRNA codons, which are triplet combinations of the nucleotide bases Adenine (A), Uracil (U), Guanine (G), and Cytosine (C), a total of three specify polypeptide chain termination. These are often referred to as stop codons or termination codons. They include the codons UAA, UAG, and UGA, with UGA sometimes serving a dual function to encode selenocysteine—a rare 21st amino acid—given the presence of a SECIS element. The remaining 61 codons correspond to the addition of amino acids to the growing polypeptide chain during translation, with the codon AUG also doubling as the start codon for initiation of translation.
To answer the multiple choice question: Of the 64 possible nucleotide codon triplets, three specify polypeptide chain termination. So, the correct answer is d. 3.
Explain how DNA stores complex information.
Answer: Via four types of smaller molecules (adenine, cytosine, guanine, and thymine) called nucleotides.
Explanation:
DNA is the main molecule of life on Earth, it is present in the cells of all living beings, being responsible for storing the information necessary for its formation and reproduction. DNA is a double strand of nucleotides that twist to form a double helix with a rotational sense on the right.
Basically, the binding between two single strands of DNA, forming the double helix, occurs following a single rule, adenine always binds to thymine and cytosine always binds to guanine and vice versa. The DNA molecule is made up of smaller molecules called nucleotides. There are four types of nucleotides that make up DNA, they are adenine, cytosine, guanine, and thymine, represented by their first letter {A, C, G, T}, forming the DNA alphabet.
Which of the following statements about eating disorders is false?
a. Demineralization of teeth tfrom exposure to stomach acid is a health issue associated with purging
b. Prolonged anorexia often leads to a reduced metabolic rate due to underproduction of thyroid hormones
c. Prevention of eating disorders involves weighing often.
d. The primary goal of nutrition therapy in anorexia nervosa is to have the patient slowy increase oral food intake
e. Bulimia most commonly occurs in adolescent and college age individuals
The correct answer is C. Prevention of eating disorders involves weighing often.
Explanation:
Eating disorders include multiple disorders that involve unhealthy eating habits as well as thoughts and emotions related to them. Individuals who suffer from these disorders constantly worry about their weight and appearance or have an unhealthy relationship with food, for example, individuals with anorexia or bulimia tend to believe they are "fat" even if they had low weight.
Due to this, weighing often is not a way of preventing eating disorders as this behavior just supports an unhealthy obsession with weight and appearance that can lead to eating disorders, instead, a balanced diet should be promoted and risk factors such as depression, anxiety, low self-esteem, etc should be addressed. Thus, the false statement is "Prevention of eating disorders involves weighing often".
Which of the following is
true aboutimprinting?
It may betriggered by visual or chemical
stimuli.
It happens tomany adult animals, but not to their
young.
It is a type oflearning that does not involve
innate behavior.
It results inbehaviors that cease after the
critical period.
Answer: It may be triggered by visual or chemical stimuli
Explanation:
Imprinting in pyschobiology is a phenomena of learning in animals. In this the young animals are exposed to one object or stimulus that can be visual, tactile, auditory and later on experience the other object or stimulus. This is tested on some birds such as chickens, geese, ducks also in some fishes, mammals and insects.
During _____________ ______________, oxygen enters the blood and carbon dioxide leaves the blood, and enters the alveoli.
During respiration oxygen enters the blood and carbon dioxide leaves the blood
You have discovered a new species of microbe. This microbe is unicellular, has a cell wall, and possess ribosomes. This microbe is most likely which of the following?
a. a bacterium
b. a protist
c. a fungus
d. it could be any of these three types of microbes
Answer:
d. it could be any of these three types of microbes
Explanation:
Cell wall made of peptidoglycan is a characteristic feature of bacteria that are otherwise unicellular prokaryotes.
Fungi have a chitinous cell wall and may be unicellular or multicellular. The example of unicellular fungi is yeast.
Protists are unicellular eukaryotes and have a cell wall made of cellulose.
Ribosomes are the site of protein synthesis and are present in all the organisms. Bacteria have 70S type of ribosomes while the eukaryotic fungi and protists have 80S ribosomes.
Which of the following statements about the basic structural features of DNA are true? Select all true statements. Select all true statements. The major and minor grooves prevent DNA binding proteins from making contact with nucleotides. In a DNA macromolecule, the two strands are complementary and antiparallel. The twisting of the DNA double helix is attributed to the tight packing of DNA bases and base-stacking. The major and minor grooves form in the DNA helix because the DNA strands are antiparallel.
Answer:
Option (2) and (3).
Explanation:
DNA is the genetic material of all organisms except some viruses. DNA is composed of the nitrogenous base, pentose sugar and the phosphate group. DNA strands are complimentary to each other.
The DNA strands run in the opposite direction, one strand in 5' to 3' direction and other strand in 3' to 5' direction thus they are anti parallel with each other. The DNA strand can twist with each other which result in the tight packing of DNA base and base stacking. DNA consists of major and minor grooves.
Thus, the correct answer is option (2) and (3).
The true statements about the basic structural features of DNA include the complementary and antiparallel nature of the two strands, the twisting of the double helix due to base packing, and the formation of major and minor grooves.
Explanation:The true statements about the basic structural features of DNA are:
1. In a DNA macromolecule, the two strands are complementary and antiparallel. This means that one strand of DNA runs in the 3' to 5' direction, while the other strand runs in the 5' to 3' direction.
2. The twisting of the DNA double helix is attributed to the tight packing of DNA bases and base-stacking. The hydrogen bonds between the nitrogenous bases hold the two strands together and contribute to the helical structure of DNA.
3. The major and minor grooves form in the DNA helix because the DNA strands are antiparallel. These grooves are binding sites for DNA binding proteins during processes such as transcription and replication.
Distinguish between sister chromatids and non-sister chromatids.
Answer:
Sister chromatids:
The chromatids of replicated chromosome that are joined through a centromere is known as sister chromatids. These chromatids are identical to each other. They contains the same allele at similar loci. They are formed at the synthesis phase of cell cycle.
Non-sister chromatids:
The chromatids of the different homologous chromosomes are known as non-sister chromatids. These chromatids are non- identical to each other. They contains the different allele at similar loci. They are formed at the prophase I of phase of meiosis.
All animals must get oxygen for respiration. Which of the following is NOT involved in this critical function
A. Diffusion
B. Vascularized gills
C. Trachea found in crustaceans and insects
D. Lungs of lungfish
E. Corpus luteum
Answer:
E
Explanation:
Corpus luteum is not involved in respiration, it is a temporary structure (mass of cells) formed in the female body, precisely in the ovary, after the ovulation. If an oocyte is fertilized, It is responsible for the production progesterone during the first stages of pregnancy, if the oocyte is not fertilized the corpus luteum will break down.
Explain how human cells compensate for the X-Iinked gene dosage difference in XX and XY nuclei.
Answer:
Dosage compensation in human includes inactivation of one X chromosome in all the cells of human females.
Explanation:
Human females have two copies of X chromosomes while the human males have only one X chromosome. To balance the X linked genes among human males and females, one of the two X chromosomes in the human females is inactivated randomly during early embryonic development.
The inactivated X chromosome is present in the form of a dark spot of chromatin near the nuclear envelope in each cell of human females. This inactivated X chromosome is called the Barr body.
Inactivation of one X chromosome in human females balances the total number of X linked genes between human males and females.