A positive tuberculin skin test is an example of delayed cell-mediated immunity.
The tuberculin skin test (TST) is a screening test that aids in the diagnosis of latent tuberculosis (TB) infection. The test is also known as the Mantoux test. The test includes injecting a little quantity of a substance called tuberculin into the skin of the forearm.
Tuberculosis is transmitted through the air and affects the lungs in most instances. The bacteria may cause a latent tuberculosis infection, in which the immune system is capable of halting its development. This means that the bacteria are alive but not active in the body, and a person with latent tuberculosis infection does not have TB disease symptoms.
Delayed cell-mediated immunity, which is a specific immune reaction, is responsible for keeping the bacteria under control. This immunity is indicated by a positive tuberculin skin test. A reaction occurs at the site of the injection after 48-72 hours if a person has this immunity.
A medical professional reads the test reaction, and the results determine the degree of immunity.
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which of these will most likely happen if a plant population grows larger than the carrying capacity of its ecosystem?
If a plant population grows larger than the carrying capacity of its ecosystem, there will be a shortage of resources like water, nutrients, and space that are necessary for survival and reproduction.
This will lead to increased competition among individuals for these limited resources, which will result in a decrease in the overall health and vitality of the population. The plants may also become more susceptible to disease and predation, and their growth and reproduction rates may decline.
Over time, the population may experience a decline in numbers as individuals die off or fail to reproduce successfully, until it reaches a size that is sustainable by the available resources in the ecosystem.
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What is the role of a fungus’s fruiting body?absorb nutrientsmovementproduce sporesfood production
organisms, such as plants or phytoplankton that carry out photosynthesis would cause a decrease of which gas in the atmosphere?
The overall effect of photosynthesis on atmospheric gases is complex and depends on a variety of factors, including the amount and distribution of photosynthetic organisms, as well as other natural and human-made processes that affect the atmospheric composition.
Organisms such as plants or phytoplankton that carry out photosynthesis are able to absorb carbon dioxide (CO2) from the atmosphere and convert it into organic compounds through the process of photosynthesis. As a result, the presence of these organisms can cause a decrease in the concentration of CO2 in the atmosphere.
Photosynthesis is the process by which carbon dioxide and water are converted into oxygen and organic compounds such as glucose, which is used by plants and other organisms as a source of energy. This process is responsible for removing a significant amount of carbon dioxide from the atmosphere and converting it into organic matter.
The net effect of photosynthesis is to reduce the concentration of carbon dioxide in the atmosphere while increasing the concentration of oxygen. This is why plants and other photosynthetic organisms are often referred to as "carbon sinks" since they absorb more carbon dioxide than they release.
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To which taxa do all four organisms belong
All four organisms belong to phylum Chordata, animalia kingdom.
The taxonomic group of creatures known as the Phylum Chordata is distinguished by the development of pharyngeal gill slits, a dorsal nerve cord, and a notochord. Animals classified as chordates differ widely in kind, from fish and amphibians to birds and mammals like humans.
A flexible structure resembling a rod that runs the length of the body, the notochord offers support and facilitates mobility. The animal's back is covered in a ribbon of nerves called the dorsal nerve cord, which carries messages from the brain to the rest of the body.
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the percentage of floating leaf disks is a reasonable measure of photosynthetic rate because the leaves float due to ________ production.carbon dioxide oxygen water bicarbonate cresol red
Answer:
Production
Explanation:
the percentage of floating leaf disks is a reasonable measure of photosynthetic rate because the leaves float due to Production. Carbon dioxide , oxygen, water , bicarbonate cresol red.
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which types of fats are discussed as being more susceptible to rancidity?
Polyunsaturated fats, such as omega-3 and omega-6 fatty acids, are more susceptible to rancidity than saturated and monounsaturated fats.
This is because they have more double bonds in their structure, which are susceptible to oxidation by free radicals, heat, and light. The oxidation process can cause the fats to become rancid, producing off-flavors and odors, and reducing their nutritional value.
Rancid fats are also potentially harmful to health as they can produce harmful compounds that may contribute to inflammation and chronic diseases. Therefore, it is important to store and handle polyunsaturated fats carefully to prevent rancidity.
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assume that 2.5 atps are generated per nadh and 1.5 atps per fadh2. what is the total number of atps generated from 7 acetyl-scoa molecules?
The total number of ATPs generated from 7 acetyl-CoA molecules is 131.5 ATPs.
The breakdown of one acetyl-CoA molecule in the citric acid cycle generates 3 NADH molecules and 1 FADH2 molecule, which can be used to produce ATP through oxidative phosphorylation. Based on the given information, 2.5 ATPs are generated per NADH and 1.5 ATPs are generated per FADH2.
Therefore, the total ATPs generated from 7 acetyl-CoA molecules can be calculated as follows:
7 acetyl-CoA molecules produce 7 x 3 = 21 NADH molecules7 acetyl-CoA molecules produce 7 x 1 = 7 FADH2 moleculesTotal ATPs generated from NADH = 21 x 2.5 = 52.5 ATPsTotal ATPs generated from FADH2 = 7 x 1.5 = 10.5 ATPsTotal ATPs generated from 7 acetyl-CoA molecules = ATPs generated from NADH + ATPs generated from FADH2 + ATPs generated from substrate-level phosphorylation (7 x 10 ATPs) = 52.5 + 10.5 + 70 = 131.5 ATPs.
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Which best represents the overall equation for photosynthesis?
6 CO2 + 12 H2O + Light energy → C6H12O6 + 6 O2
6 CO2 + 6 H2O + Light energy → C6H12O6 + 6 O2
6 CO2 + Light energy → C6H12O6 + 6 O2 + 6 H2O
C6H12O6 + 6 O2 + 6 H2O + Light energy → 6 CO2 + 12 H2O
The equation that best represents the overall equation for photosynthesis is: 6 CO2 + 6 H2O + Light energy → C6H12O6 + 6 O2
Photosynthesis is the process in which green plants, algae, and cyanobacteria convert light energy into chemical energy, which is stored in the form of glucose or sugar, which is utilized for nourishment or energy at a later time. The process of photosynthesis occurs in the chloroplasts of the plant cell, which contains chlorophyll pigment that traps light energy and converts it into chemical energy, which is then utilized for various metabolic activities.
The process of photosynthesis involves two types of reactions: light-dependent reactions and light-independent reactions. In the light-dependent reactions, the chlorophyll pigment traps the light energy and converts it into chemical energy, which is utilized for various metabolic activities such as ATP production and NADPH formation.In the light-independent reactions or Calvin cycle, the carbon dioxide and water molecules are utilized to form glucose or sugar, which is the end product of photosynthesis.
The overall equation for photosynthesis is as follows: 6 CO2 + 6 H2O + Light energy → C6H12O6 + 6 O2. The reactants of photosynthesis are carbon dioxide, water, and light energy, while the products are glucose or sugar and oxygen.
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the conducting zone of the respiratory system ends when terminal bronchioles divide into respiratory bronchioles. which structure of subsequent divisions is the main site of gas exchange?
The conducting zone of the respiratory system ends when terminal bronchioles divide into respiratory bronchioles. The main site of gas exchange is in the alveoli.
These tiny sacs, which are at the end of the respiratory bronchioles, are surrounded by capillaries. Oxygen from the air inhaled into the lungs passes through the walls of the alveoli and into the capillaries, where it binds to hemoglobin and is transported to the body's cells.
Carbon dioxide from the cells diffuses into the capillaries and is carried to the alveoli, where it is exhaled out of the body. This process is known as gas exchange and is essential for proper respiration.
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What does the biological approach to psychological disorders focus on?
The biological approach to psychological disorders focuses on the biological factors that contribute to the development of mental illnesses. These biological factors may include genetics, brain structure, and function, and neurochemical imbalances.
The biological approach to psychological disorders is a theory that suggests that psychological disorders are caused by biological factors, including genetics, brain structure and function, and neurochemical imbalances. According to this approach, psychological disorders are not caused by purely environmental factors, but rather by a complex interaction between environmental and biological factors. The biological approach to psychological disorders is supported by a large body of research, which has demonstrated that certain genetic and neurochemical factors are associated with an increased risk of developing mental illnesses. For example, studies have found that people with a family history of mental illness are more likely to develop these disorders themselves, suggesting a strong genetic component to mental illness.
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the nucleosome core includes two each of four histones named , , , and .
The four histones that make up the nucleosome core are:
H2AH2BH3H4NucleosomeThe nucleosome core particle, which is the fundamental component of chromatin, is created when these small, positively charged histones attach to negatively charged DNA strands. Within the nucleosome, the histone proteins are organized in a precise manner, with two copies of each protein producing an octamer around which DNA is wrapped. The DNA molecule is compressed by this structure and shielded from harm while still being available for numerous biological functions including transcription and replication.Finally, there are two copies of each of the four H2A, H2B, H3, and H4 histones in the nucleosome core, for a total of eight histones. The fundamental structural component of chromatin is an octamer formed by these histones and the DNA it surrounds.learn more about nucleosomes here
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After you isolated DNA, you measured its absorbance using a Nanodrop spectrophotometer. From this measurement, you can estimate how many copies of the hypervariable region you have added to the PCR. To do this calculation, you will first need to calculate the weight of 1 copy of the haploid human genome. (Remember that there is one copy of the hypervariable region in each copy of the haploid genome). Information you will need to use: - The haplold human genome contains 3. 2 x 10^9 base pairs - The average molecular mass of a base pair is 660 - Avogadro's number 6. 02 x 10^23. 1. Work step-wise through these calculations Step 1: What is the molecular mass of the haploid human genome? 6. 02*10^23*3. 5*10^-12-21. 1-10^11 ( 3. 5*10^-12 is one copy of human haplold genome weight noe in step 2) Step 2: How much does 1 copy of the human haploid genome weigh (g)? the normal haploid genome is 3,2-10 Sbp at 660g/mol bp and Avogadro's number 6. 02 x 1023 so one copy genome contains 3. 2M049bp -(6. 6*10*14*3. 210^9)/ 6. 02M0-23. 3. 5-10^-12g/haploid genome Step 3: If your isolated DNA solution had a concentration of 90 ng/HL how many copies of the VNTR, DIS80 would you be adding to your PCR? Remember you are adding 5uL of the isolated DNA to your PCR when Isolating DNA there 900*5=450ng DNA added, as 3. 5*10^-12g/haploid genome, 450ng=450*10^-9mg, so there are 450*10-9/3. 5-10-12 1. 2*10^4 copies added in PCR Now, repeat for your isolated DNA: how many copies of D1S80 did you theoretically add to your PCR tube? 3. Each D1580 allele is double stranded DNA. Will both strands be amplied? Explain your answer. Yes 2 strands of DNA have specific bp to primers to amplified, when forward and reverse primers attached to both strands, they will amplified with region
4. How many copies of the primers do you add to the PCR? If you had 100 copies of the template DNA would there be enough primers to support 30 cycles of PCR? Remember the amount of template doubles (in theory) with each cycle. After 1 cycle there will be 2xthe starting amount, after 2 cycles 4 x the starting amount. 3ul of a 5 um primers, there are coples: 3*5*660*10^-12/3. 5*10^-12-2828
can you help me with step1-3 and question3 and 4?
The molecular mass of the haploid human genome is:1.99 x 10¹² g, the weight of 1 copy of the human haploid genome is, 1.99 x 10¹² g, the number of copies of the VNTR, D1S80 added to the PCR is 1.29 x 10¹¹ copies, and Both strands of the D1S80 allele can be amplified during PCR. Because the PCR process involves the use of two primers.
Step 1: Molecular mass of haploid human genome
The haploid human genome contains 3.2 x 10⁹ base pairs
The average molecular mass of a base pair is 660 g/mol
Avogadro's number is 6.02 x 10²³
Therefore, the molecular mass of the haploid human genome is:
3.2 x 10⁹ base pairs × 660 g/mol × 6.02 x 10²³ = 1.99 x 10¹² g
Step 2: Weight of 1 copy of the human haploid genome
The weight of one copy of the haploid human genome is equal to the weight of the entire haploid genome divided by the number of copies (which is 1)
Therefore, the weight of 1 copy of the human haploid genome is:
1.99 x 10¹² g / 1 = 1.99 x 10¹² g
Step 3: Number of copies of the VNTR, D1S80 added to PCR
The isolated DNA solution has a concentration of 90 ng/µL
When 5 µL of the isolated DNA is added to the PCR, the amount of DNA added is:
90 ng/µL x 5 µL = 450 ng
We know that 1 copy of the haploid human genome weighs 3.5 x 10⁻¹² g, so we can calculate the number of copies of the VNTR, D1S80 added to the PCR:
450 ng / (3.5 x 10⁻¹² g/copy) = 1.29 x 10¹¹ copies
Both strands of the D1S80 allele can be amplified during PCR. This is because the PCR process involves the use of two primers, one for the forward strand and one for the reverse strand. The primers are designed to bind specifically to the ends of the target DNA region, and both strands will be replicated during the amplification process. Therefore, both strands of the D1S80 allele will be amplified.
The number of copies of the primers added to the PCR depends on the amount of primer solution added. If 3 µL of a 5 µM primer solution is added to the PCR, the number of copies of the primers added is:
3 µL x 5 µM x 6.02 x 10²³ molecules/mol x 660 g/mol / 3.5 x 10⁻¹² g = 2.83 x 10⁹ copies of each primer
If you had 100 copies of the template DNA, there would be enough primers to support 30 cycles of PCR. In theory, the amount of template DNA doubles with each cycle of PCR. After 30 cycles, the amount of DNA would have doubled 30 times, which is 2³⁰ times the starting amount. This is approximately 10⁹ times the starting amount, which is more than enough to support PCR amplification.
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what would you predict would happen in an area of the body where blood was relatively acidic (low ph)?
In an area of the body where blood was relatively acidic (low ph), you can predict that the enzymes will be affected or deactivated because of the change in the pH value.
pH is a scale that measures how acidic or basic (alkaline) a substance is. The pH range varies from 0 to 14, with 7 being considered neutral. The lower the pH, the more acidic the substance is, and the higher the pH, the more basic or alkaline it is. When the pH of the blood is disturbed, it can cause a variety of issues. In an area of the body where blood was relatively acidic (low ph), you can predict that the enzymes will be affected or deactivated because of the change in the pH value. The pH of the blood can affect enzyme function. The activity of an enzyme can be significantly affected by small changes in pH. Enzymes are most effective at a specific pH, and if the pH is too low or too high, they may be inactivated. In addition, blood pH is essential for maintaining the body's metabolic processes, and minor pH changes can have a significant impact. Blood pH must be maintained between 7.35 and 7.45, and anything outside of this range might lead to issues with the body's essential functions. Therefore, an area of the body with a relatively acidic environment may cause a variety of issues.Learn more about enzymes https://brainly.com/question/14577353
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------ ATP produced during glycolysis ------- ATP used by glycolysis = ------ net gain of ATP during glycolysis
HURRTYYYY NEEEDDD ITTT
During glycolysis, one glucose molecule produces a total of four ATP molecules. Since the first stage of glycolysis uses up 2 ATP molecules, there is a net gain of 2 ATP molecules.
The glycolysis process uses two ATP molecules. When glucose is transformed to glucose-6-phosphate, one ATP molecule is used, and when fructose-6-phosphate is changed to fructose-1,6-bisphosphate, the other ATP molecule is used. So, during glycolysis, there is a net gain of 2 ATP molecules.
In the metabolic route known as glycolysis, glucose molecules are broken down to release energy. This energy is then used by the cell to produce ATP, which is the currency of life, from ADP and phosphate.
This information demonstrates that the glycolysis metabolic pathway is a well-known chemical reaction that produces ATP by dissolving glucose molecules rather than simple carbohydrates.
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a peach breeder has an orchard of peach trees with an average fruit mass of 150 g. the breeder selects the top 10% of plants to reproduce for the next generation. the selected plants have an average fruit mass of 170 g. the next generation has an average fruit mass of 155 g. what is the narrow-sense heritability for fruit weight in this orchard? enter your answer as a fraction, not as a percentage.
The narrow-sense heritability for fruit weight in the orchard in the next generation selected by the breeder is 0.25.
Heritability refers to the amount of phenotypic variation that is inherited by offspring from their parents. The narrow-sense heritability for fruit weight can be calculated using the formula below:-
h² = R/S
where; R = response to selection, S = selection differential
Average fruit mass of the orchard = 150 g
Selected plants' (by the breeder) average fruit mass = 170 g
Next generation average fruit mass = 155 g
The selection differential (S) can be calculated by subtracting the average fruit mass of the orchard from the selected plants' average fruit mass.
S = 170 g - 150 g = 20 g
The response to selection (R) can be determined by subtracting the average fruit mass of the next generation from the average fruit mass of the selected plants.
R = 155 g - 170 g = -15 g
Substituting these values into the formula, we get,
h² = R/S= (-15)/20= -0.75
However, we need to convert the heritability from a negative value to a positive value by squaring it.
h² = (-0.75)²= 0.5625
The narrow-sense heritability for fruit weight in the orchard is 0.5625. However, the answer should be expressed as a fraction and not as a percentage.
Therefore,
h² = 0.5625 can be written as h² = 9/16 or 0.5625/1
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how is genetic information conserved during dna replication?
By using the existing DNA strands as templates, semi-conservative replication ensures that the genetic information is conserved and passed on to the daughter cells.
Genetic information is conserved during DNA replication through a process of semi-conservative replication. In semi-conservative replication, the two strands of the original DNA double helix separate, and each strand serves as a template for the synthesis of a new complementary strand.
The process of DNA replication involves several steps:
Initiation: DNA replication begins at specific places along the DNA molecule, called origins of replication.
Unwinding: To produce a replication fork, an enzyme known as helicase unwinds and divides the two strands of the double helix.
Priming: On each of the template strands, a short RNA primer is created by the enzyme primase. The primer offers a starting point for the synthesis of the new DNA strand.
Elongation: An enzyme called DNA polymerase adds nucleotides to the growing DNA strand in a 5' to 3' direction, using the template strand as a guide. The complementary base pairing rules ensure that the new strand is complementary to the original template strand.
Proofreading: DNA polymerase has the ability to check for and correct any base pairing mistakes as they happen.
Termination: DNA replication is complete when the polymerase reaches the end of the DNA molecule or encounters another replication fork.
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the microorganism that was first used to prove the germ theory was a spore former. what is the name of this microorganism?
which of the following is considered the major function of calcium: promotes tooth decay. fat cell metabolism. homeostatic balance. it is a component of mineralized bone.
The major function of calcium is homeostatic balance. Calcium is an essential mineral for maintaining homeostasis in the body.
Homeostasis is the body's ability to maintain a stable internal environment even when there are external changes. The regulation of calcium levels in the body is a vital function of homeostasis.
Calcium ions play an essential role in muscle contractions, nerve function, blood clotting, and bone and teeth development.
Calcium's functions in the human body include:
Assists in the formation and growth of teeth and bones.
Helps with blood clotting.
Maintains proper nerve function.
Helps muscles contract and relax.
Assists in maintaining healthy blood pressure levels.
Calcium is a mineral that our bodies need to function correctly. Calcium is not made by the body, so we must acquire it from our diet.
Dairy products, leafy green vegetables, and other calcium-fortified products are excellent sources of calcium.
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how is genetic information conserved during dna replication?
Explanation: Each newly formed DNA strand joins with an original DNA strand to form a new double helix. DNA is a large polymer molecule made up of repeating nucleotides.
explain the pressure changes that occur during the cardiac cycle and their relationship to flow through the heart and blood vessels.
During the cardiac cycle, pressure changes occur within the heart chambers and blood vessels. During systole, ventricular contraction leads to an increase in ventricular pressure. This results in the ejection of blood from the ventricles into the pulmonary artery and aorta.
What happens during cardiac cycle?During diastole, the ventricles relax, and ventricular pressure decreases. The aortic and pulmonary valves close, while the atrioventricular valves open. This allows blood to flow from the atria to the ventricles, refilling them for the next cycle.
Pressure changes in the heart and blood vessels are important for regulating blood flow. The pressure gradient created by ventricular contraction and relaxation drives blood through the circulatory system.
The amount of blood that flows through the heart and blood vessels is directly proportional to the pressure difference between the two ends of the system.
Higher pressure gradients result in faster blood flow, while lower pressure gradients result in slower blood flow. Therefore, the pressure changes that occur during the cardiac cycle are essential for maintaining adequate blood flow through the heart and blood vessels.
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In eukaryotic cells the pre-mRNA undergoes processing. Check the following options that are a type of mRNA processing. A. 5'G Cap B. 3' G Cap C. 3' Poly A tail D. 5' Poly A tail E. removal of exons F. removal of introns
The following which are the type of the mRNA processing are 5' G Cap, 3' Poly A tail and Removal of introns option A, C and F.
Messenger RNA (mRNA) is a cellular molecule that transports codes from DNA in the nucleus to the cytoplasmic locations where proteins are synthesised (the ribosomes). Scientists Elliot Volkin and Lazarus Astrachan initially characterised the molecule that would later be known as mRNA in 1956. Ribosomal RNA (rRNA) and transfer RNA are the other two main forms of RNA in addition to mRNA (tRNA).
Post-transcriptional modifications are the processes by which pre-mRNA undergoes some chemical modifications to produce a mature and functional mRNA that can synthesize protein.
Three major modifications are:
RNA splicing - Non-coding introns are removed, only the expressed exons are joined together to form functional mRNA.
Poly 'A' tailing - Multiple adenosine monophosphates are added to the 3' end of mRNA.
5' Capping - Addition of methylated Guanine nucleotides to the 5' end of mRNA.
Both Poly A tailing and 5' capping,
Protects from enzymatic digestion of nucleases
Helps in exporting mRNA from nucleus to cytoplasm
Helps in translation process.
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Organisms..................are single-celled organisms that.......................have circular DNA, while they have linear DNA and can be unicellular or multicellular.
'Organisms prokaryotes are single-celled organisms that only have circular DNA, while eukaryotes have linear DNA and can be unicellular or multicellular.
What is the difference between eukaryotes and prokaryotes in terms of chromosomes?The difference between eucaryotes and prokaryotes in terms of chromosomes is based on the fact that eukaryotes are more complex and therefore they have more genetic material that may be divided into one or more linear chromosomes.
Therefore, with this data, we can see that differences in eukaryotes and prokaryotes in terms of chromosomes depend on the shape and number of pdf chromosomes.
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how do forces change the growth of plants or fungi according to gravitropism?
High quantities of auxin flow towards the cells on the bottom side of the root as it gravitropistically grows in the direction of gravity forces. In contrast to the top of the plants root, where cell elongation is permitted, this inhibits growth on this side.
Shoots and roots are directed higher and downward by the key plant development response known as gravitropism, allowing each organ to reach surroundings that are suitable for carrying out its basic duties.
Roots spread away from the light and downward, or towards the center of the Earth. Tropisms are the names for these reactions to outside stimuli. The growth response of plants to light and gravity is referred to as phototropism and gravitropism, respectively. Plant growth hormones regulate both tropisms.
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How do prokaryote phylogenies differ from mammal phylogenies?
O The phylogenetic reconstruction process is much simpler for prokaryotes than for mammals because prokaryotes are much smaller.
O There is a high degree of lateral gene transfer in mammals, but not in prokaryote phylogenies.
O There is a high degree of lateral gene transfer in prokaryotes, but not in mammals.
O Mammal phylogenies are constructed from evidence based in DNA, whereas prokaryote phylogenies are based in RNA.
O Mammal phylogenies are constructed from evidence based in RNA, whereas prokaryote phylogenies are based in DNA.
The option that correctly identifies how prokaryote phylogenies differ from mammal phylogenies is: There is a high degree of lateral gene transfer in prokaryotes, but not in mammals.
Phylogeny refers to the evolutionary history of an organism or group of organisms. Phylogenies indicate the connections between distinct organisms or groups of organisms. It describes a cluster of organisms linked by a single ancestor from the past.A prokaryote phylogeny, on the other hand, differs significantly from mammal phylogenies. In a mammalian phylogeny, there is a low degree of lateral gene transfer. Lateral gene transfer is defined as the movement of genetic material between two unicellular organisms that aren't direct descendants.
As a result, when constructing a mammalian phylogeny, one can rely solely on the organisms' genomic DNA sequence as evidence.However, in a prokaryote phylogeny, there is a high degree of lateral gene transfer. In this case, constructing phylogenies using DNA sequences can be challenging because some genes may come from other bacterial cells. Instead of DNA, RNA sequences are used to create prokaryote phylogenies. Thus, the correct option is: There is a high degree of lateral gene transfer in prokaryotes, but not in mammals.
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In fruit flies, allele for long wing is dominant to the allele for vestigial wing. When heterozygous long winged flies were crossed with vestigial winged flies, 192 offsprings were produced. Of these 101 had long wing & 91 had vestigial wings.
If an exact Mendelian ratio had been obtained, then the number of each phenotype would have beenO Long winger - 64Vestigial winged - 128O Long winger - 96
Vestigial winged - 96O Long winger - 128
Vestigial winged - 64O Long winger - 192
Vestigial winged - 0
In fruit flies, the long wing allele is dominant over the vestigial wing allele. Of them, 128 had long wings and 128 had vestigial wings, with long wings making up 101 of the total. Option 2 is Correct.
Long wings result from the dominant V allele, whereas vestigial wings result from the recessive V allele. We refer to a trait's inheritance pattern as recessive when it requires two copies of the same allele to display a certain phenotype. For instance, the vestigial phenotype is transmitted via recessive genes. A vestigial fly must have the genotype vg/vg.
In this 192 progeny were created when heterozygous long winged flies and vestigial winged flies were bred. Due to a deficiency in their "vestigial gene," located on the second chromosome, flies with vestigial wings are unable to fly. Option 2 is Correct.
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Correct Question:
In fruit flies, allele for long wing is dominant to the allele for vestigial wing. When heterozygous long winged flies were crossed with vestigial winged flies, 192 offsprings were produced. Of these 101 had long wing & 91 had vestigial wings.
1. If an exact Mendelian ratio had been obtained, then the number of each phenotype would have been Long winger - 64Vestigial winged - 128 with Long winger - 96
2. Vestigial winged - 96 and Long winger - 128
3. Vestigial winged - 64 and Long winger - 192
4. Vestigial winged.
1 ptWhich best describes the amount of Earth's total water supply that is usable freshwater?less than 1%less than 25%
Less than 1 percent of Earths total water supply is usable freshwater.
Although there is a lot of water on the planet, just a very small portion (approximately 0.3 percent) is even useful to people. The remaining 99.7% is dispersed throughout the earth's oceans, soils, icecaps, and atmosphere. Yet, a large portion of the usable 0.3 percent is inaccessible. Rivers provide the majority of the water that people utilise. Surface water refers to the visible water bodies. Indeed, the majority of fresh water is found underground as soil moisture and in aquifers. A river can continue to flow even when there has been no precipitation because groundwater can feed the streams. Both surface and groundwater are useful to people.
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which of the following is not a cephalopod? group of answer choices octopus nautilus clam cuttlefish
The correct answer is clam. Clam is not a cephalopod. The correct answer is option c.
What are cephalopods?Cephalopods are a class of marine animals that are classified as Mollusca's third-largest class, after Gastropoda and Bivalvia. All cephalopods are carnivorous predators that prey on a variety of prey items ranging from plankton to fish and mammals. Cephalopods have a complex nervous system and an advanced brain compared to other mollusks.
What are the examples of cephalopods?The following are some examples of cephalopods:
NautilusesSquidsCuttlefishesOctopusesIn contrast to other mollusks, cephalopods have numerous tentacles extending from their heads. They have been successful because of their sophisticated sensory systems, intelligence, and the capacity to alter their body shape, texture, and coloration in response to environmental stimuli.
The correct answer is option c.
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Complete question
which of the following is not a cephalopod? group of answer choices
a. octopus
b. nautilus
c. clam
d. cuttlefish
post transcriptional control includes regulation of mrna degradation. explain how this affects translation
Post-transcriptional control includes the regulation of mRNA degradation. This directly affects translation because mRNA degradation leads to reduced protein synthesis, meaning that fewer proteins are produced as a result of the mRNA not being present.
Here are some ways that mRNA degradation can impact translation:1. Reduced stability of mRNA can lead to premature degradation, which reduces the amount of mRNA available for translation.2. RNA-binding proteins can bind to specific sequences in mRNA, promoting degradation by exonucleases. This is known as deadenylation-dependent mRNA decay.3. Endonucleases can cleave mRNA internally, degrading it before translation can occur.
4. Certain mRNA molecules contain elements that promote their decay, including AU-rich elements (AREs). Proteins such as tristetraprolin can bind to these elements and promote mRNA decay.5. Other factors that can affect mRNA degradation include the presence of other RNA molecules, such as miRNAs, as well as changes in the cellular environment such as pH and temperature.
Overall, mRNA degradation plays a crucial role in regulating protein synthesis by controlling the amount of mRNA available for translation. This process is tightly regulated by a variety of factors and can be influenced by both external and internal stimuli.
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the innovation of pollen allowed this plant group to colonize low-water environments. called___
The plant group that is characterized by the innovation of pollen and allowed them to colonize low-water environments is called the gymnosperms.
Gymnosperms are a group of plants that includes conifers, cycads, Ginkgo, and gnetophytes. They are called gymnosperms because they do not have enclosed seeds like angiosperms, instead, their seeds are exposed on the surface of cones or other structures.
The innovation of pollen eliminated the need for water to transport sperm to the egg, making it possible for these plants to colonize drier environments. This adaptation was critical in the evolution and diversification of gymnosperms, making them one of the most successful and widespread plant groups on Earth.
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Name several ways to reduce greenhouse gas emissions
Greenhouse gases are gases in the Earth's atmosphere that trap heat and contribute to the greenhouse effect, which is the process by which the Earth's atmosphere keeps the planet's temperature within a certain range that is habitable for life. The most common greenhouse gases are carbon dioxide (CO2), methane (CH4), nitrous oxide (N2O), and fluorinated gases.
Several ways to reduce greenhouse gas emissions are as follows:
Reducing fossil fuel usage: One of the most important ways to reduce greenhouse gas emissions is by reducing the usage of fossil fuels. This includes decreasing the consumption of petroleum, natural gas, and coal.
Renewable Energy: Another way to reduce greenhouse gas emissions is by using renewable energy such as solar, wind, hydroelectric, and geothermal energy sources. The energy generated from these sources is clean and does not emit greenhouse gases.
Reducing deforestation: Forests play a significant role in reducing the amount of carbon dioxide in the atmosphere. However, due to deforestation, carbon dioxide is released into the atmosphere. Therefore, reducing deforestation is crucial in reducing greenhouse gas emissions.
Reducing waste: The management of waste is also crucial in reducing greenhouse gas emissions. Waste produces methane which is a potent greenhouse gas. Therefore, reducing waste through recycling, composting, and other methods will help reduce greenhouse gas emissions.
Reducing water usage: Reducing water usage is also important in reducing greenhouse gas emissions. The energy used to pump, treat, and heat water results in the emission of greenhouse gases. Therefore, reducing water usage can reduce greenhouse gas emissions.
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