Around 10% of global warming is caused by the destruction and loss of forests. If we don't put an end to deforestation, there is just no way that we can combat the climate issue. Thus, option A is correct.
What mature trees in a forest extend the life of the forest?By cutting down a few trees, you can gain access to timber and other resources, while also fostering diversity in your neighbourhood. Lack of tree removal might cause some animal species to lose their habitats. Many ecosystems can be found in forests, and certain creatures can only survive in young forests.
Carbon dioxide is absorbed and stored by trees. Releases of carbon dioxide and other greenhouse gases occur when trees are destroyed or even slightly disturbed.
Therefore, cutting down only mature trees in a forest extend the life of the forest, and It maintains forest habitats.
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Process that moves any material from high concentration to low concentration is called ??
Answer:
diffusion
Explanation:
Diffusion is the process that transports materials from a high-concentration area to a low-concentration area.
This fundamental mechanism controls the spontaneous movement of ions, molecules, and even gasses as they attempt to reach an equilibrium state. Substances travel along their concentration gradient during this process, propelled by particles' innate propensity to disperse and reach a balanced distribution. In biological systems, diffusion is pervasive and essential to many activities, including gas exchange in the lungs and food absorption in cells. Furthermore, it transcends the boundaries of biology, influencing physics and chemical phenomena, demonstrating its universal significance in comprehending the normal movement and distribution of matter.
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A certain cell type has existed in the blood and tissue of its vertebrate host's immune system for over twenty years. One day, it recognizes a newly arrived antigen and binds to it, subsequently triggering a secondary immune response in the body. Which of the following cell types most accurately describes this cell?
Answer:
The options are
A) thyroid cells
B) macrophage
C) memory cell
D) plasma cell
C) memory cell is the ANSWER
Explanation:
B lymphocytes is one of the important cells of the immune system. It's the building block for the memory cells which acts as a means of recognizing pathogens that have previously infected the body system in order to boost the production of antibodies against such pathogen for faster immune response in future infections
Organize the steps of the dihydroxyacetone phosphate/glycerol-3-phosphate shuttle.
a. an electron pair is transferred from FADH2 to coenzyme Q and down the electron chain.
b. DHAP is reduced by NADH in the glycerol to glycerol-3-phosphate.
c. DHAP returns back to the cytosol.
d. flavin-dependent glycerol-3-phosphate dehydrogenase converts glycerol-3-phosphate back to DHAP and reduces FAD to FADH2
Answer:
The correct answer is - b, d, a, and c is correct order.
Explanation:
The correct steps of the dihydroxyacetone phosphate/glycerol-3-phosphate shuttle is as follows -
DHAP or dihydroxyacetone phosphate is reduced by NADH in the glycerol-3-phosphate than flavin-dependent glycerol-3-phosphate dehydrogenase transform glycerol-3-phosphate back to DHAP and reduced to FAD to FADH2.
Which is follows by the an electron pair is transport to the coenzyme Q from electron after that the DHAP returns to the back to the cytosol.
Vision can be affected by drugs and alcohol making, especially difficult to distinguish
A) LightB) ColorsC) DistanceD) Shapes
Answer: B) Colors
Explanation:
The alcohol and drugs can blur one's vision. This is going to affect one's driving ability or skills. These substances can slow down the ability to focus, causes double vision, and the person is not able to distinguish color. The person will not be able to observe and understand the traffic signals, signs and roadways markings. Hence, alcohol and drug consumer in heavy dose are susceptible to accident.
A group of scientists determines that a mitochondrial DNA sequence shared by two species has a constant mutation rate. The sequences of mitochondrial DNA for the two species are shown below.Species A: GACTGGGCATTASpecies B: GACCGGACATGAUsing the principles of molecular clocks and a known mutation rate for this sequence, the scientists determine that these species diverged from a common ancestor 30 million years ago. Based on this information, how much longer do you predict it would take these species to differ by a total of seven base pairs?A. 60 million yearsB. 20 million yearsC. 140 million yearsD. 40 million years
Answer:
D. 40 million years
Explanation:
from the sequence, there is a difference in three base pairs and this took 30 million years. it can thus be concluded that it took 10 million years for a difference in a base.
thus, to make it up to 7 base pairs will require another mutation of 4 base pairs. this will take 40 million years.
Answer:
D) 40 Million years
Explanation:
I took the test
Earth formed about __________ years ago, and the oldest known living organisms date to __________ years ago.
A. 4.6 billion years ago ... 3.5 billion years agoB. 6,000 years ago ... 6,000 years agoC. 13.7 billion years ago ... 4.6 billion years agoD. 3.9 million years ago ... 2.0 billion years ago
Answer:
The answer is A.
Explanation:
According to the researchs and studies, the earth formed about 4.54 billion years ago and the oldest known living organism date to around 3-3.5 billion years ago which are the first bacteria that formed in the water.
I hope this answer helps.
Earth was formed about 4.6 billion years ago and the oldest known living organisms are believed to date back to about 3.5 billion years ago.
Explanation:The Earth formed about 4.6 billion years ago. This is established based on scientific research and evidence gathered from various fields like Geology and Astrophysics. The oldest known living organisms on Earth are believed to date back to about 3.5 billion years ago. The evidence of life from this period comes from the discovery of fossilized microorganisms.
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An ancient plant called horsetail contains 216 chromosomes. How many homologous pairs of chromosomes does it contain? How many chromosomes are present in its cells during metaphase?
Answer:
The answers respectively are 108 and 216.
Explanation:
According to the information given in the question, the plant called horsetail has cells that contain 216 chromosomes. They must contain 108 homologous pairs of chromosomes and provided that they are not reproductive cells, during metaphase the replication of the DNA is yet to start so they contain 216 chromosomes in the cells.
I hope this answer helps.
The center of the retina is a pit called the ____________ . The cells that are photoreceptors (detect color) are ____________ . In low light conditions, only rods are activated, and visual acuity is best in the ____________ of the eye. In bright light conditions, visual acuity is best when light is focused on the ____________ of the eye. A single ganglion cell outside of the fovea receives input from ____________ rod(s). Each cone synapses with ____________ ganglion cell(s).
Answer:
The center of the retina is a pit called the fovea centralis The cells that are photoreceptors (detect color) are cones
In low light conditions, only rods are activated, and visual acuity is best in the Corners(as peripheral vision)of the eye.
In bright light conditions, visual acuity is best when light is focused on the center of the macula of the eye.
A single ganglion cell outside of the fovea receives input from _Multiple rod(s). Each cone synapses with one ganglion cell(s).
The fovea is the center of the retina. Photoreceptors, which detect color, are known as cones. In low light, visual acuity is best in the eye's periphery, while in bright light it is best when focused on the fovea.
Explanation:The center of the retina is a pit called the fovea. The cells that are photoreceptors (detect color) are cones. In low light conditions, only rods are activated, and visual acuity is best in the periphery of the eye. In bright light conditions, visual acuity is best when light is focused on the fovea of the eye. A single ganglion cell outside of the fovea receives input from multiple rods. Each cone synapses with one ganglion cell.
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The ________ muscle originates on the inner surface of the ilium and the lower borders of the transverse processes (L1-L5), sides of the bodies of the last thoracic vertebra (T12), lumbar vertebrae (L1-L5), intervertebral fibrocartilages, and base of the sacrum.
Answer:
The iliopsoas muscle originates on the inner surface of the ilium and the lower borders of the transverse processes (L1-L5), sides of the bodies of the last thoracic vertebra (T12), lumbar vertebrae (L1-L5), intervertebral fibrocartilages, and base of the sacrum.
Explanation:
The iliopsoas is a muscle that acts in very important functions and is often the cause of low back pain, being a muscle that is not very accessible. It originates on the inner surface of the ilium, base of the sacrum and sides of the bodies of the last dorsal vertebra (known thoracic vertebra as well) (T12) and of all the lumbar vertebrae.
Anatomically the iliopsoas is formed by two muscles: the iliac and the psoas, which in turn we could divide into psoas major and psoas minor. The psoas major is inserted proximally in the body of the 12th dorsal vertebra, in all the lumbar vertebrae and through thinner fascicles, in the lumbar transverse processes. On the other hand, we have the psoas minor that is absent in 41-50% of people. Its path includes from the anterolateral face of the 12th dorsal vertebra and the first two lumbar, until reaching the pectineal line.
Finally, the iliac muscle, which is the second muscular component, originates in the upper two thirds of the internal surface of the iliac fossa, until it reaches the lesser trochanter. It's here that the majority of the fibers join the psoas major forming the tendon of the iliopsoas.
A power lifter has been advised by his coaches that he needs to lose some weight while still maintaining his rigorous training schedule. When evaluating this athlete's diet, which macronutrient will likely need to be monitored closely and possibly limited?A. FatB. CarbohydrateC. ProteinD. Water
Answer:
A. Fat
Explanation:
An athlete Fat intake should be between 20-35% of total everyday calories. A 10% of fat intake each from monounsaturated sources, polyunsaturated sources and saturated fat.
Excess weight can have varying negative effects on a athlete body, which ranges from increase in heart rate to high oxygen consumption, body temperature during exercise will increase, followed by high energy usage and a rise in the bones and joints pressure.
Note:
A power lifter with a lower body weight has improved rate of endurance, coupled with a rise in speed and strength when compared with a power lifter with a higher body weight
Organisms can adapt to cold temperatures by increasing the amount of unsaturated fatty acids in their membranes, which help maintain fluidity as temperatures decrease.
A. True
B. False
Answer:
Answer is A. True.
Explanation:
Organisms have developed an adaptive property which will enable them to adapt to cold temperature, through increase in their membranes fatty acid. This process allows them to keep the liquid-crystalline sate of their cell membranes at a cold temperature. This process is known as the homeoviscous adaptation.
This process is done by organisms especially the mammals to maintain functional membrane phenotypes.
Final answer:
True, organisms indeed adapt to cold by increasing unsaturated fatty acids in membranes to maintain fluidity, essential for proper cell function and survival in such environments.
Explanation:
The statement that organisms can adapt to cold temperatures by increasing the amount of unsaturated fatty acids in their membranes, which help maintain fluidity as temperatures decrease, is true. Unsaturated fatty acids contain kinks due to double bonds in their hydrocarbon tails, which prevent the molecules from packing too closely together. This spacing allows for greater fluidity in the cell membrane, especially crucial in cold environments where saturated fatty acids press against each other, forming a rigid structure. Adaptation to cold environments is essential for maintaining membrane fluidity, preventing rupture, and ensuring the proper function of the cell. Examples include poikilothermic organisms like fish, which increase their cell membrane's unsaturated fatty acid content in response to lower temperatures, maintaining the necessary fluidity for survival.
An increase in the amount of carbon dioxide in the blood stimulates the respiratory center of the brain. As a result, impulses are sent from the(1) medulla to the diaphragm, increasing the rate of breathing(2) cerebrum to the chest muscles, decreasing the rate of breathing(3) medulla to the trachea, causing it to constrict(4) cerebrum to the alveoli, causing them to activelytransport oxygen
Answer:1) medulla to the diaphragm, increasing the rate of breathing
Explanation:
Basically involuntary respiration is controlled by the Respiration centers in the brain Stem. They are group of respiratory neurons ; the dorsal and ventral respiratory groups located in the medulla oblongata , and the pontine respiratory group located in the Pons varolii.
The primary function of this center is for creation and maintenance of respiration rhythms and adjustment of the sequences to fluctuation in body homeostatic balance .
In order to regulate respiration rate and its depth, it collects its input signals from chemoreceptors ( cells that generate biological signals from chemical substances), mechanoreceptors( receptors which relay external physical stimuli of touch, and pressure to internal stimulus by regulation of gated ion channels), the hypothalamus and the cerebral cortex. The fluctuation in the levels of plasma C02, oxygen, and blood pH, stimulated the input progression.
Therefore an elevated C02 blood levels, marked by low pH send input signals to the respiratory centers in the medulla.The latter sends signals to the respiratory muscles (and the diaphragm,) to initiate mechanism of rapid breathing to exhaled more C02 so that the acidity of the blood is reduced..
The correct choice is the medulla sending signals to the diaphragm to increase the rate of breathing in response to high levels of carbon dioxide. This process involves deepening respiration to bring CO2 levels back to normal.
Explanation:The major factor that stimulates the respiration centers in the medulla oblongata and pons is the concentration of carbon dioxide (CO2) in the blood. An increase in CO2 prompts the medulla to send impulses to the respiratory muscles. When CO2 levels rise, it diffuses across the blood-brain barrier and results in increased hydrogen ion concentration, which in turn lowers the pH of the blood.
This acid-base effect is sensed by central chemoreceptors, which initiate a response to increase the contraction of the diaphragm and intercostal muscles. Consequently, the rate and depth of respiration increase, and more CO2 is expelled, bringing levels in the blood back to normal.
To answer the student's question directly, the correct choice is: (1) medulla to the diaphragm, increasing the rate of breathing. This process helps regulate the levels of CO2 and maintain a stable pH, ensuring proper function of the body's systems.
The most high-quality images of the soft tissues of the brain are provided by _________ scans.
Answer:
The answer is MRI.
Explanation:
The method used to get the best quality images of the brain's soft tissue is MRI (Magnetic Resonance Imaging) scans. One of the most common benefits of using MRI is because it does not use x-rays therefore does not emit radiation that is harmful to the subject, which is one of the reasons that it provides high-quality images.
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Bile salts are synthesized from cholesterol and are used for the emulsification of dietary lipids. What is the primary fate of bile salts during digestion?
-They are reabsorbed in the ileum
-They are excreted in the feces
-They are digested by pancreatic lipase
-They are packaged into chylomicrons
Answer:
The correct answer is ''They are reabsorbed in the ileum.''
Explanation:
Bile salts are the fundamental constituent of bile, solubilizing dietary lipids so that they can be absorbed. Bile salts are the main products of cholesterol degradation and are poorly absorbed by passive diffusion in the proximal portion of the small intestine and most of the intestinal bile salts are reabsorbed in the terminal ileum, which actively absorbs 90 % of bile salts to the portal venous circulation, which returns them, again, to the hepatocytes that conjugate and excrete, again, the bile into the gallbladder. Bile salts circulate through this pathway from the liver to the intestine and again to the liver (enterohepatic circulation) between 10 and 12 times a day.
The Shine-Dalgarno sequence in bacteria ________. a) is a purine-rich consensus sequence found in the 5ʹ UTR of the mRNA b) is a pyrimidine-rich consensus sequence found in the 3ʹ UTR of the mRNA c) is a purine-rich consensus sequence found
Answer:
The correct answer is a) is a purine-rich consensus sequence found in the 5ʹ UTR of the mRNA.
Explanation:
The mRNA of bacteria contains Shine- Dalgarno sequence and initiation codon which together make up the ribosomal binding site. The Shine-Dalgarno sequence(5'- AGGAGGU-3') is a short polypurine sequence that is found at 5' end of UTR of the mRNA.
Shine-Dalgarno sequence is found 3-10 nucleotides upstream to start codon(AUG). This sequence is complementary to a region of the 16S rRNA component of small ribosomal units found in bacteria.
So Shine-Dalgarno sequence play an important role in the attachment of small ribosomal subunit of bacteria and recognition of start codon. So the right answer is a.
Final answer:
The Shine-Dalgarno sequence is a purine-rich sequence found in the 5' UTR of bacterial mRNA that is essential for initiating protein synthesis by aligning the ribosome with the start codon. Hence, the correct answer is a. is a purine-rich consensus sequence found in the 5ʹ UTR of the mRNA.
Explanation:
The Shine-Dalgarno sequence in bacteria is a purine-rich consensus sequence found in the 5' UTR (untranslated region) of the mRNA. The sequence is crucial for the recognition and binding by the small ribosomal subunit, which is essential for the initiation of protein synthesis. The Shine-Dalgarno sequence aligns the ribosome with the start codon AUG by base pairing with a complementary pyrimidine-rich sequence at the 3' end of the 16S rRNA, which is part of the 30S ribosomal subunit. This interaction is vital for precise positioning of the ribosome so that translation can begin at the correct start codon.
A population of animals feeds entirely on plants. Some of these animals are good at digesting Plant A, while others are good at digesting Plant B, and others are good at digesting Plant C. What might be an advantage of a single population of animals having a large amount of variability surrounding certain traits?
There’s less scarcity of food for the population is an advantage of a single population of animals having a large amount of variability surrounding certain traits
Explanation:
The population of animal is eating three different types of plants. One of them is good in digesting plant A, one of them is good in digesting plant B and other group is good in digesting plant C.
Thus, the population although has a huge number, but each of the individual is divided into one of the 3 groups which is different in feeding habits. Thus there’s no scarcity of food in the population. Thus the population can live well and reproduce without any nutritional strain.
Answer:
It increases the chance that some members of the population will survive under changing environmental conditions.
Explanation:
Variation of traits between organisms within and among species increases the chance that some members of the population will survive under changing environmental conditions.
If, for example, a disease reduces the amount of Plant A and Plant B in the environment, some members of the animal population will still survive. Since the population can survive on a variety of plants, those that can eat Plant C will be less affected by the disease and will still be successful.
A species of malaria-carrying mosquito lives in a forest in which two species of monkey, A and B, coexist. Species A is resistant to malaria, but species B is not. The malaria-carrying mosquito is the chief food for a type of bird in the forest. If all these birds were suddenly eliminated by hunters, which of the following would be an immediate observable consequence?
Answer:
(B) increased mortality in monkey species B
Explanation:
The following deduction can be made from the question:
Species B is susceptible to malariaThe more the population of malaria-carrying mosquito, the more the incidence of malaria in the population of species BA species of bird has been limiting the population of malaria-carrying mosquito by feeding on them. Consequently, this bird species has been indirectly reducing the incidence of malaria in species BHence, if all the birds were suddenly eliminated by hunters,
Malaria-carrying mosquito population will thrive in the forest,Incidence of malaria will increase in species BThis will consequently lead to increased mortality in monkey species BThe correct option is B.
A technology for collecting data to improve farming practices and increase efficiency is called __________.
Answer:
Precision agriculture
Explanation:
Precision agriculture is the act of using new technologies to increase crop yield and profit, while reducing the amount of the usual input like water, fertilizer, herbicides, insecticide etc. That is to say, farmers cultivates crops and yield more crops for less.
Enhanced genetic seeds,Gps on tractors automatically timed irrigation and using laser to clear fields are all examples of precision agriculture. This gives farmers the boom they need to produce more crops, make profit and promote food availability.
Drosophila melanogaster, the fruit fly, has a 2n chromosome number of 8. Assuming that a somatic G2 nucleus contains about 8.0 picograms of DNA, how many picograms of nuclear DNA would you expect in:
a) Mitotic metaphase
b) a primary oocyte (metaphase)
c) a secondary oocyte (metaphae)
d) a first polar body (metaphase)
Answer:
a) Mitotic metaphase
8 picograms
b) a primary oocyte (metaphase)
8 pecograms
c) a secondary oocyte (metaphae)
4 pecograms
d) a first polar body (metaphase)
4 pecograms
Explanation:
a) In mitotic phase the chromosome numbers remains same as 8 because in this phase duplicated chromosomes align in center and separate to other poles. Therefore the amount of DNA will be 8 pecograms.
b) after the mitotic meta phase the cells divide into daughter cell with equal proportion of DNA which means each cell have 8 chromosomes. therefore they both contain 8 pecograms of DNA. The new daughter cells called as primary oocytes.
c) These primary oocytes (8 chromosomes) divides into two secondary oocyte each contain 4 chromosomes which means each cell have 4 pecograms of DNA.
d) Duplication of chromosomes occur in secondary oocyte and cell divides into polar body (4 chromosomes) and ovum (4 chromosomes). Therefore first polar body have 4 pecograms of DNA in it.
See the attachment for clarification.
During mitotic metaphase and primary oocyte metaphase in Drosophila, there are 8.0 picograms of DNA. In a secondary oocyte during metaphase and the first polar body, there are 4.0 picograms of DNA, as these cells are haploid postcompletion of meiosis I.
Explanation:In the context of Drosophila melanogaster, which has a diploid chromosome number of 8 (2n=8), and a somatic G2 nucleus containing about 8.0 picograms of DNA, we can determine the amount of DNA present at different stages of cell division. During the G2 phase, a cell has duplicated its DNA in preparation for mitosis, therefore, a G2 nucleus contains twice the amount of DNA compared to a G1 nucleus.
(a) Mitotic metaphase: In mitotic metaphase, the cell contains 2 copies of each chromosome, just as it does during G2 phase, so you would expect to see 8.0 picograms of DNA, because the chromosomes are aligned and condensed but not yet separated.(b) Primary oocyte (metaphase): A primary oocyte in metaphase I of meiosis also contains 8.0 picograms as it is still diploid until the completion of meiosis I.(c) Secondary oocyte (metaphase): After the completion of meiosis I, a secondary oocyte is in meiosis II, and is already haploid, but since the chromosomes have not yet divided, it would contain 4.0 picograms of DNA, half of the amount in the primary oocyte before meiotic division.(d) First polar body (metaphase): The first polar body, being a byproduct of meiosis I, would similarly contain 4.0 picograms of DNA.
An individual recovers from the common cold, which is caused by rhinovirus A. The person then becomes infected with the avian influenza virus, which causes the bird flu. Which statement best describes what will most likely happen to this person?
Answer: what will most likely happen to this person is that He will have the symptoms of the bird flu because he is not immune to the avian influenza virus.
Explanation: rhinovirus A is the virus that causes common cold and is different from the virus that cause bird flu which is called avian influenza virus. Although common cold is a symptom of the bird flu disease, he will still show signs of it since he is not immune to the avian influenza virus.
Most of the exchange surfaces of multicellular animals are lined with _____.
Answer:
D) epithelial tissue
Explanation:
The exchange surfaces of the organisms are the site for an exchange of substances or gases. These surfaces mostly lie within the body and protect the exchange tissues. The exchange surfaces of multicellular organisms are lined with epithelial tissues. These tissues have tightly packed cells with little or no intercellular spaces. The epithelial tissues serve to protect the exchange surfaces. Sometimes, these epithelial linings are modified to perform an additional function. The exchange surfaces are finely branched or folded to have a very large area. The digestive, respiratory, and excretory systems all have such exchange surfaces.
A tree community consists of 4 maples, 3 oaks, and 1 alder. The species richness of this community is:________.A. 0.42.B. 0.97.C. 1.5.D. 3.
Answer:
D (3)
Explanation:
Species richness is the number of different species within an ecological community, area, ecosystem or sample etc. It is the numerical characteristics of an ecosystem. Species richness is a measure of the variety of species and it is done by counting the number of different species in that given region. Species richness does not take into consideration the abundance of the different species or how widely distributed they are.
In this case, a tree community is involved. The community consists of three different tree species: Maple, Oak, and Alder. Hence, the species richness is simply the number of different species in the tree community, which is three(3).
HELP!! Please answer immediately! Answer is for 30 points!
It is possible that a recessive trait in a survey such as this one may have a greater total number than its dominant counterpart. Explain how that might happen in an investigation like this.
Answer:
In such an investigation like this, it might happen that the recessive trait is more than dominant trait making the recessive trait to be the dominant one due to its high rate of occurrences and the dominant one will be the recessive one die to low probability of occurring.
This situation can also happen where there is higher occurrences of recessive alleles than dominant ones as in autosomal dominance and autosomal recessiveness and / or x-linked recessiveness and x-linked dominance.
Describe the relationship and movement between Temperature and Density in a convection cell. Make sure you identify the direction of travel.
Answer:
There is an inverse correlation.
Explanation:
A convection cell can be described as an area of which it's boundries can be stated where two substances either fluid or gas can move due to the temperature changes.
Temperature and density in a convection cell have a relationship which is inversely correlated meaning that if the temperature increases, the density decreases and the matter which has a higher temperature and also lower density moves towards the part with the greater density. The direction of travel of the warmer substance is upwards and vice versa.
The mantle of the earth or the atmospheric movements between the land and the sea can be given as examples to convection cells.
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The pancreas is responsible for the elimination of 95% of ingested alcohol from the body through metabolism.
Answer:
FalseExplanation:
The human body mechanisms try to excrete out the ethanol from the body before it gets circulated in the body through blood. The body does so by converting the ethanol to certain metabolic products which get filtered and excreted out through urine.
The metabolism of alcohol takes place in the liver in two steps- Ethanol is converted to acetaldehyde by the alcohol dehydrogenase enzyme. The acetaldehyde is toxic to cells and is the compound responsible for the hangover after drinking.
The acetaldehyde is then converted to acetic acid by an enzyme called acetaldehyde dehydrogenase. The acetic acid through respiration gets converted to water and carbon dioxide and rest is excreted out through urine.
Thus, the liver is the organ responsible for the metabolism of ethanol and not pancreas.
Which of the following statements about olfactory receptors is false?
Select one:
a. They are bulb-like structures at the tip of the frontal lobe where the olfactory nerves begin.
b. They are cells.
c. They are located in a mucous membrane at the top of the nose.
d. They have small, hair-like extensions.
Answer:
They are bulb-like structures at the tip of the frontal lobe where the olfactory nerves begin. (Ans A)
Explanation:
Olfactory receptors which are also known as smell receptor, and present in the cell membrane of olfactory receptor neurons. They are the proteins which are capable of binding with odour molecules which play an important role in the sense of smell.
Each receptor of olfactory cell contains a single external process which extends to the surface which is found in the epithelium, and gives rise long, slender extensions know as cilia. Mucus of the nasal cavity covered the cilia, and facilitating the detection of odour molecules.
Answer:
a
Explanation:
The base height of the dingdong plant is 10 cm. Four genes contribute to the height of the plant, and each dominant allele contributes 3 cm to height. If you cross a 10 cm plant (quadruply homozygous recessive) with a 34 cm plant, how many phenotype classes will there be in the F2?
There will be 5 phenotype classes in the F2 generation.
Explanation:To determine the number of phenotype classes in the F2 generation, we need to consider the possible combinations of alleles from the parent plants. The quadruply homozygous recessive plant has 4 recessive alleles, while the 34 cm plant is a result of 4 dominant alleles. Each dominant allele contributes 3 cm to the plant's height. When the plants are crossed, the F1 generation will be heterozygous with 2 dominant and 2 recessive alleles.
In the F2 generation, the possible genotype combinations are as follows: 1 plant with 4 dominant alleles (34 cm), 4 plants with 3 dominant alleles (31 cm), 6 plants with 2 dominant alleles (28 cm), 4 plants with 1 dominant allele (25 cm), and 1 plant with 0 dominant alleles (22 cm). Therefore, there will be 5 phenotype classes in the F2 generation.
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Which pair do you predict would have the largest difference in mass-specific basal metabolic rate?
Answer:
The pair's options needed to complete the question are
A. dog and cat
B. lion and dog
C. lion and cat
D. cat and mouse
E. they are all likely to be the same
D. CAT AND MOUSE IS THE ANSWER
Explanation:
Mass-specific basal metabolic rate (mass-specific BMR), is the resting energy expenditure per unit body mass per day.
mass-specific BMR mammals decreases as weight increases i.e
Metabolic rates of animals are allometric functions of body size
D. Cat and mouse is the answer
Acetylcysteine may be used as a mucolytic agent. What is an additional indication for the drug?
Answer:
Apart from the role as mucolytic agent( breakdown of mucus in cystic fibrous and chronic obatructive pulmonary diseases (COPD)
IT CAN BE USED TO TREAT OVER DOSE OF PARACETAMOL.
Explanation:
It performs this role by increasing the availability of antioxidant GLUTATHIONE.The latter neutralizes the toxic effect that paracetamol over dose would produce in the blood plasma.
Which adjustment knob is used for bringing the specimen into focus
Answer:
The coarse adjustment knob is for focusing of the slide.
Explanation:
Coarse Adjustment Knob- The coarse adjustment knob built on the arm of the microscope is for moving the stage up and down to bring the specimen into focus.
Note:
Fine Adjustment Knob - is inside the coarse adjustment knob and is used for sharp focusing of the specimen under low power and is used for all focusing when high power lenses is used.