Answer:
C) they lack lignified vascular tissue. ( Since they do not have vascular tissues, they are not able to develop vertically like the trees we know.)
Explanation:
A) Plants do not have "sperm". They have sporophytes or gametophytes.
B) No specific parasitic bacteria of bryophytes are known.
D) In fact, they have. Although they require water for their reproduction, some species of bryophytes are found in deserts.
E) Rhizoids are not weak, otherwise they would not have the ability to keep bryophytes attached (mosses).
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.
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.
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.
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'
1. Plants carry out chemical reactions that make sugars, which contain carbon, hydrogen, and oxygen, from carbon dioxide (CO2) and water (H2O). Researchers want to know whether the oxygen atoms in sugar come from the carbon dioxide or the water. How could they use radioactive tracers to find out? Design an experiment that investigates this question. Be sure to include the following elements in your experimental design: hypothesis, procedure, control group, experimental group.
Answer:
Radioactive labeling is a procedure used to monitor the path followed by a chemical element within a biological system to demonstrate the source.
Hypothesis:
The free oxygen produced during photosynthesis comes from water.
Procedure.
Single-celled algae were placed in four petri dishes containing carbon dioxide, glucose dissolved in Water.
In each box the oxygen was radioactively marked.
1 petri dish with CO2 marked. Experimental groups
2 petri dish with H2O marked. Experimental groups
3 petri dish with Glucose marked. Experimental groups (control)
4 petri dish with CO2 + H2O + Glucose, all marked. (control).
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 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.
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.
How might a protein kinase select the correct residues
tophosphorylate given the availability of serine and
threonineresidues on the surface of globular proteins?
Answer:
The flanking residues facilitate recognition and specificity.
Explanation:
The kinase is able to to specifically phosphorylate serine and threonine due to the recognition of the flanking residues. Altogether, these form a consensus sequence across the substrate.
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
How many different types of genotypes are possible when working various pea seed color problems?
a. 3
b. 4
c. 5
d. 6
e. 2
Answer:
Option A, 3
Explanation:
Primarily all pea seeds have two colors one is yellow and other is green.
Let green color allele be represented by "G" and yellow color allele be represented by "g".
When we start working on pea seed color problem, there can be following possible genotypes -
a) Heterozygous green color - Gg
b) Homozygous yellow color - gg
c) Homozygous green color - GG
Hence, option A is correct.
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).
Explain how genetic information is stored in DNA and how this information is used to make functional proteins.
Answer:
The correct answer will be- codons and each codon specific for amino acids.
Explanation:
Deoxyribose nucleic acid is the genetic material of the organism which provides instructions for the organism. DNA is made up of nucleotide monomers which are composed of five-carbon sugar deoxyribose, a phosphate group and four types of nitrogenous bases (adenine, guanine, cytosine and thymine).
It is the arrangement of these nitrogenous bases which provide codes to the organism as it forms mRNA molecule through transcription. The sequence of the nitrogenous bases in mRNA is read by the ribosome during translation.
The ribosome reads the bases in triplets called "codon" which code for a specific amino acid. If the sequence of the base changes, therefore, the amino acid also changes. These amino acids bond to each other by peptide bond and form a protein molecule.
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.
Learn more about lac operon here:https://brainly.com/question/33360857
#SPJ3
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.
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.
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.
Learn more about DNA Structure here:https://brainly.com/question/36469737
#SPJ11
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.
A column on the periodic table is called a group. Which statement would not support the conclusion that two elements are in the same group?? A) The two elements both have similar atomic radii B) The two elements both have very high melting points C) The two elements both are highly reactive in oxygen D) The two elements both bonds with chlorine to form a salt in a 1 to one ratio
Answer: Option (A) is the correct answer.
Explanation:
A group in a periodic table is represented by a column. And, when we move down a group then there occurs an increase in atomic size or atomic radii of the elements.
This is because number of electrons add into new shells which leads to increase in atomic radii of the element.
Therefore, if we say that two elements in a group have similar atomic radii then it is false.
Whereas the two elements of a group might have a high melting point.
Also the statements, the two elements both are highly reactive in oxygen and the two elements both bonds with chlorine to form a salt in a 1 to one ratio can be true for elements of the same group.
Thus, we can conclude that the statement two elements both have similar atomic radii would not support the conclusion that two elements are in the same group.
The statement that would not support the conclusion that two elements are in the same group is B) The two elements both have very high melting points.
Explanation:The statement that would not support the conclusion that two elements are in the same group is B) The two elements both have very high melting points. Grouping elements in the same group is based on their similar chemical properties, not on their physical properties such as melting points. For example, elements in the same group may have different melting points due to varying atomic structures and bonding.
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.
Learn more about Mendel's experiments here:https://brainly.com/question/35864275
#SPJ3
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.
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.
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.
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.
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.
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.
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.
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.
What are the advantages of being a respiratory parasite (the tongue worms) over one in the digestive system (if any)?
Answer:
Explanation:
The parasites living at the respiratory system or tongue worms will hide under the soft tissues of the oral cavity, beneath the tongue and even beneath the throat or esophagus. These worms remain undetected during diagnosis. These worms are not subjected to the treatment of acids which worms in the digestive system are exposed to. Thus these worm parasites in the respiratory system or tongue survive comparatively for long as compared to the worms in the digestive system.
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.