Answer:Left Lobar pneumonia
Explanation:
Since the smear test with acid acid fast bacteria is negative,then the test is tuberculosis The respiration, chills,and fever pointed to pneumonia and the consolidation in the left suggested;left lobar.
If Chargaff's equivalence rule is valid, then hypothetically we could extrapolate this to the combined genomes of all species on Earth (as if there were one huge Earth genome).
In other words, the total amount of A in every genome on Earth should equal the total amount of T in every genome on Earth.
Likewise, the total amount of G in every genome on Earth should equal the total amount of C in every genome on Earth.Calculate the average percentage for each base in your completed table.
Do Chargaff's equivalence rules still hold true when you consider those six species together?
Answer:
yes . A approximately equals to T and G approximately equals C in average.
Explanation:
according to Chargaff's equivalence rule
the number and concentration of adenine is similar to the number of thymine and number of guanine is similar to that of cytosine in the DNA.
so according to conditions given in the question i.e average amount of A should be equal to average amount of T and total average amount of G in every genome on Earth should equal the total average amount of C . so Chargaff's equivalence rules still hold true when you consider those six species together.
Ringing in the ears is called:
a. otitis externa.
b. Meniere's syndrome.
c. tinnitus.
d. otosclerosis.
Answer:
c. tinnitus
Explanation:
Tinnitus (a hearing impairment) is a form of hearing of sound when no external sound is present Which is often described as a Ringing,Tinnitus may also sound like a clicking or roaring. i.e usually unclear voices or music are heard. The sound may be soft or loud, low or high pitched, and appear to be coming from one or both ears. Ranging from one person to another, the sound may causes depression or anxiety and can interfere with concentration.
Otitis externa is a condition that causes inflammation , the inflammation also covers (redness and swelling) of the external ear canal, i.e the tube between the outer ear and eardrum.
Meniere's disease popularly called (MD), is a disorder of the inner ear that is characterized by episodes of feeling like the world is spinning (vertigo), hearing loss, and a fullness in the ear. The cause of MD involves both genetic and environmental factors. Some of the factors include constrictions in blood vessels, viral infections, and autoimmune reactions.
Otosclerosis is a condition where one or more foci of irregularly laid spongy bone replace part of normally dense enchondral layer of bony otic capsule in the bony labyrinth. This condition affects one of the ossicles (the stapes) resulting in hearing loss, vertigo or a combination of symptoms.
Therefore from the foregoing we can conclude that Tinnitus is the correct anwser.
Answer: option C
Explanation:
Ear ringing and strange ear noises is medically called tinnitus. People who have tinnitus hear buzzing and roaring sounds which are actually absent.
Tinnitus can be caused by problems in any of the four areas which are responsible for hearing: the outer ear, the middle ear, the inner ear, or in the brain.
Clover is a plant that is found in many fields in Europe. There is a genetic polymorphism present in clover in Europe associated with the release of cyanide from the leaf tissue, when the leaf tissue is damaged. Individuals of type AA release lots of cyanide when the leaf is damaged, Aa individuals an intermediate amount, and aa individuals NO cyanide. The release of cyanide when the leaf is damaged prevents or reduces other leaves of the plant from being eaten by herbivorous animals, such as slugs, cattle, etc.
Answer:
Absolutely, Genetic Polymorphism is observed
Explanation:
In genetics, Polymorphism refers to the regular existence of two or more different genotypes within a given species or population.
Definitely, the Clover leaf as described above exhibit genetic Polymorphism, bearing that the existence of the 3 genotypes: AA, Aa, and aa in Same species of Clover has been observed.
Note that this condition occurs because that natural selection is unable to pick a single trait above another; thus diverse phenotype of Clover leaves are observed.
HIV is classified as a retrovirus because _____.
(A) it reverts to an inactive form when it infects B lymphocytes
(B) this virus is composed of two cells surrounded by a lipoprotein coat
(C) it makes a DNA copy of its RNA once inside the host cell
(D) it infects only cells with a CD4 receptor
(E) it causes the production of HIV antibodies
Answer: C
Explanation:
HIV is classified as a retrovirus because its genetic material is composed of single-stranded RNA nucleotide. And also Retroviruses have the enzyme reverse transcriptase, which is capable of making a DNA copy of its RNA once inside the infected host cell
Answer:
The answer is C; it makes a DNA copy of its RNA once inside the host cell
Explanation:
HIV is called a retrovirus because common to all retroviruses, they store their genetic information using RNA instead of DNA, and they need to ‘make’ or convert their RNA to DNA when they enter a human cell in order to make new copies of themselves.
Why do you believe you need, or should receive, financial assistance, such as a Houston Livestock Show and Rodeo Scholarship to attend the college/university of your choice? This should include details about family situations such as:
a. care of family members,
b. medical situations,
c. family financial burdens and/or how important it will be for you to work while in college.
Houston Livestock Show and Rodeo Scholarship to attend the college/university of your choice because it is and will be.
What is Houston Livestock?One of the most lucrative professional rodeo events for the regular season is included. Since 2003, it has been held at NRG Stadium in Houston, Texas, with the exception of 2021 because of the COVID-19 pandemic's effects.
The Astrodome served as the previous venue. Similar to Mardi Gras in New Orleans, the Texas State Fair in Dallas, Comic-Con in San Diego, and New Year's Eve in Times Square in New York City, it is regarded as the city's "signature event."
A record-breaking 2,611,176 attendees and 33,000 volunteers showed up in 2017. The rodeo was dubbed "the year of the volunteer" in 2007.
Therefore, Houston Livestock Show and Rodeo Scholarship to attend the college/university of your choice because it is and will be.
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Which accessory eye structures function to produce the tears that cleanse and protect the eye?
a. lacrimal glandsb. conjunctivac. Meibomian glandsd. medial canthi
Answer:
a. lacrimal glands
Explanation:
Lacrimal glands are paired, almond-shaped exocrine glands, situated at each eye. Their main function is to secrete the aqueous layer of the tear film.This lacrimal gland produce the tears that flow into canals that connect to the lacrimal sac. and also cleanse and protect the eye. They can be located at the upper lateral region of each orbit, in the lacrimal fossa of the orbit formed by the frontal bone.
The major function of the conjuctiva is to Produce mucus to prevent the eyes from drying out.
Meibomian glands are holocrine type exocrine glands, along the rims of the eyelid inside the tarsal plate. They produce meibum, (an oily substance).This oily substance prevents evaporation of the eye's tear film.
Medial canthi is the corner of the eye where the upper and lower eyelids meet.
This basic explanation clearly depicts that the Lacrimal glands is the correct answer.
Using the patch clamp technique, a researcher measures a single channel in a membrane. The receptor channel passes 3 pA (picoamperes) of ionic current over a period of 8 msec at -60 mV; 1 pA = 10 -12 ampere.Based on your knowledge of nerve transmission, this movement of ions willA. stimulate a change in a postsynaptic membrane if 50 channel events are combined.B. stimulate a change in a postsynaptic membrane if three channel events are combined.C. not stimulate a change in a postsynaptic membrane.D. generate a current resulting in membrane depolarization.E. stimulate a change in a postsynaptic membrane
Answer:
(C)
Explanation:
Not stimulate a change in a postsynaptic membrane because it is preset in inner membrane.
You cross a true breeding tomato plant with large fruits to a true breeding plant that has small fruits. All of the F1 offspring have medium sized fruits, with a variance in fruit size of 0.06 cm2 . When you create F2 plants, the variance in fruit size is 0.60 cm2 . What is the broad sense heritability (H2 ) of fruit size?
Answer:
H² = 0.9
Explanation:
Phenotypic variance (VP) = genetic variance (VG) + environmental variance (VE)
The observed VP in the F1 is composed only of VE, because the parents were true breeding (thus VG=0).
The F2 variance is due to both genetic and environmental variation. If the tomato plants are in the same environment, the VE wil be the same, and then we can calculate VG as:
VG = VP - VE
VG = 0.60 cm² - 0.06 cm²
VG= 0.54 cm²
Broad sense heritability is calculated as: H² = VG/VP. It is used to describe how much of the observed phenotypic variation is due to genetic variation (comprised by additive, dominance and epistasis variance).
In this case,
H² = VG/VP
H² = 0.54 cm² / 0.60 cm²
H² = 0.9
The broad sense heritability (H^2) of fruit size in the cross breeding experiment involving tomato plants with large fruits and small fruits is 0.90.
Explanation:The question is about determining the broad sense heritability (H2) of fruit size in a cross breeding experiment involving tomato plants with large fruits and small fruits. Broad sense heritability (H2) is a statistic used in genetics to estimate the proportion of the phenotypic variance in a population that is due to the genetic variance. It is given by the formula:
H2 = VG/VP
Where, VG is the genetic variance, and VP is the phenotypical variance. The phenotypic variance is made up of genetic variance (VG), environmental variance (VE), and possibly the interaction between genetic and environmental variance (VGxE).
In this case, we need to find out the VG and VP to find H2. Since the F2 variance (VF2) is provided (0.60 cm2) and the F1 variance (VF1) is provided (0.06 cm2), we can assume that the F1 variance represents the environmental variance (VE) and the F2 variance minus the F1 variance gives the genetic variance (VG) since F1 was produced by crossing two true breeding plants. Hence:
VG = VF2 - VF1
VG = 0.60 cm2 - 0.06 cm2 = 0.54 cm2
Substituting the values in the formula:
H2 = VG/VP = 0.54 cm2/0.60 cm2 = 0.90
So, the broad sense heritability of fruit size in this experiment is 0.90.
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After an exceptionally rainy spring, there is no longer a barrier between the populations, and some individuals migrate inland from the coast. Consequently, 20 % of individuals in the new inland population are originally from the coastal population. Calculate the spotted allele frequency for the inland population after migration, q combined . Round the answer to the nearest hundredth.
Answer
After inland population after migration allele frequency is 0.62 or 62%
Explanation:
Given,
Coastal striped phenotype freq. = 0.22
ss = 0.22
[tex]q_{coastal} \times q_{coastal} = 0.22[/tex]
[tex]q_{2 coastal}[/tex]= 0.22
Similarly, inland striped phenotype freq. = 0.43
[tex]q_{2inland}[/tex] = 0.43
[tex]q_{coastal} = \sqrt{q_{2coastal}}[/tex]
= [tex]\sqrt{0.22}[/tex]
= 0.4690
[tex]q_{coastal}[/tex]= 0.47 i.e. 47%
[tex]q_{inland} = \sqrt{0.43}[/tex]
= 0.655
[tex]q_{inland}[/tex] = 0.66 i.e. 66%
the migration range (m) is given as 20%
m= 0.2
allele freq. after migration = pre migration + ∆q
here,
∆q = change in the allele frequency
or
migration of allele freq. from coastal to inland
=[tex]m(q_{coastal} - q_{inland})[/tex]
= 0.2 (0.47 – 0.66)
=[tex]0.2 \times (- 0.191)[/tex]
= - 0.0382
∆q = -0.04 i.e. 4%
Final answer:
In population 1, genetic drift likely caused the change in allele frequencies, while in population 2, natural selection likely caused the change.
Explanation:
In population 1, the change in allele frequencies is likely caused by genetic drift. Genetic drift is the random change in allele frequencies due to chance events, and it is more likely to occur in smaller populations, where chance can have a larger impact. The small population size and the fact that only half of the individuals reproduce suggest that genetic drift played a role in the change in allele frequencies in population 1.
In population 2, the change in allele frequencies is likely caused by natural selection. Natural selection is a process where certain traits or alleles are favored due to their beneficial effects on survival and reproduction. The larger population size and the fact that there is differential reproduction depending on the traits suggest that natural selection played a role in the change in allele frequencies in population 2.
The first organisms on Earth appeared approximately 3.8 billion years ago and were all unicellular. Multicellular organisms have only existed for about one billion years.
What evolutionary milestone made it possible for more complex, multicellular organisms to exist?
The development of specialized cells and cellular cooperation made it possible for more complex, multicellular organisms to exist.
Explanation:The evolutionary milestone that made it possible for more complex, multicellular organisms to exist was the development of specialized cells and cellular cooperation. As cells started to live together in colonies, some cells began to specialize in performing different functions, making the colony more efficient. This specialization allowed the cells to remain small while still carrying out specialized tasks, leading to the evolution of multicellular organisms. These organisms were bigger, more efficient, and capable of performing more complex functions than single-celled organisms.
Final answer:
The development of eukaryotic cells, cell specialization within colonies, and sexual reproduction were pivotal milestones in the evolution of multicellular organisms which led to the diverse array of life we see today.
Explanation:
The evolutionary milestone that made it possible for more complex, multicellular organisms to exist was the development of eukaryotic cells and later the advent of colonial organisms and cell specialization. Eukaryotic cells, which emerged about 2 billion years ago, have more complex structures like a nucleus, allowing for greater control over cellular functions. Another key development was the emergence of sexual reproduction over 1 billion years ago, which led to increased genetic variation and the potential for greater evolutionary adaptation.
As single-cell eukaryotes started forming colonies, some cells in these colonies began specializing in different functions. These specialized cells could perform tasks more efficiently, benefitting the organism as a whole and setting the stage for the evolution of true multicellularity. True multicellular plants appeared about 1 billion years ago, and true animals followed roughly 100 million years later.
Ultimately, these developments laid the foundation for the Cambrian Explosion, a period of rapid evolution around 540 million years ago during which most major animal phyla appeared. Multicellular life evolved over time through an increase in biodiversity and complexity. Organisms like Volvox illustrate the progression from solitary eukaryotic cells to more complex multicellular structures.
Prokaryotes and eukaryotes share some mechanisms for controlling gene expression, but all of the following are uniquely used in eukaryotes EXCEPT:a) Enhancers, silencers, cell-type specific factorsb)Operons for coordinated gene expressionc)chromatin remodelingd)RNA processing
Answer:
b) Operons for coordinated gene expression
Explanation:
Operon refers to a unit of genetic function and is mainly found in bacteria and phages. An operon consists of a promoter, an operator, and a coordinately regulated cluster of structural genes. The genes of an operon code for proteins or enzymes that mostly function in a common pathway. It allows a single promoter and operator or any other regulatory sequence to regulate the expression of functionally related genes.
For instance, lac operon consists of three structural genes that code for the enzymes involved in the catabolism of lactose sugar. The operon is expressed only when lactose is available in the medium. This pattern of gene regulation is not found in eukaryotes.
Your are a microbiologist. The owner of a restaurant hires you to examine a sample of preserved vegetables that appears to have spoiled in its can. The metal can only contains vegetables and water. It appears to have been sealed tightly, but the sides of the can have bulged. Could bacteria in the can have survived the canning process and caused the sides to bulge?
Answer:
Yes bacteria can survive if heating process is inadequate.
Explanation:
Most of the canned food is spoiled by thermophilic and mesophilic bacteria. These bacteria can survive the canning process if the can is inadequately heated. Most of them are able to produce gas which can cause bulging of can.
Clostridium species are mainly involved in the spoilage of canned food. C. thermosaccharolyticum is a thermophilic bacteria which spoil can food and evolve CO2 gas that cause bulging of can. Example of mesophilic bacteria that cause spoilage and swelling of can are C. botulinum, C. puterifaciens, etc.
Therefore thermophilic and mesophilic bacteria can survive the canning process if heat treatment and the temperature are not adequate.
Final answer:
Bacteria such as Clostridium botulinum could survive improper canning procedures, germinate in an anaerobic environment, and produce gas, causing the can to bulge, indicating possible botulinum toxin presence.
Explanation:
Yes, bacteria in the can could have survived the canning process and caused the sides to bulge. This is often due to the presence of anaerobic bacteria such as Clostridium botulinum, which can survive in low-oxygen environments like a sealed can. If the canning process was not properly conducted, the spores of such bacteria could remain viable, and when conditions inside the can become anaerobic, these spores can germinate and produce gas as a byproduct of their metabolism, resulting in the bulging of the can. This scenario is also a potential sign of the presence of botulinum toxin, which is highly toxic. It's important that the restaurant owner disposes of these cans safely and reviews their canning procedures to prevent potential foodborne illnesses.
Day-to-day choices can help reduce the risk of heart disease. One of the major risk factors for development of heart disease is elevated LDL, which can be affected by the types and amounts of dietary fat consumed as well as other dietary factors. Read the statements below and select all of the correct statements regarding how various dietary fats affect LDL cholesterol levels. Select all that apply. a. Typically, the higher your consumption of unsaturated fats, the higher the LDL cholesterol levels in your blood. b. Trans fats are worse for heart health than saturated fats because they raise LDL cholesterol and lower HDL cholesterol. c. Increasing intake of plant foods may be one of the easiest ways to decrease LDL cholesterol. d. Typically, the higher your consumption of saturated fats, the higher the LDL cholesterol levels in your blood. e. Dietary cholesterol does not affect blood cholesterol levels.
Answer:
b. Trans fats are worse for heart health than saturated fats because they raise LDL cholesterol and lower HDL cholesterol.
c. Increasing intake of plant foods may be one of the easiest ways to decrease LDL cholesterol.
d. Typically, the higher your consumption of saturated fats, the higher the LDL cholesterol levels in your blood.
Explanation:
Generally, the consumption of saturated fat increases the amount of LDL cholesterol in the blood system while the consumption of unsaturated fat can reduce the amount of LDL cholesterol in the blood system. In addition, the consumption of plant food can reduce the amount of LDL in the blood system. Trans fats are more harmful to one's heart than saturated fat because they can increase the amount of cholesterol in the blood system.
Answer:b,c,d
Explanation:a. Unsaturated fats are mostly beneficial to health, e.g- omega 3 fatty acids, and are not known causes of LDL increase.
b. Trans fats are known to increase LDL cholesterol while also decreasing HDL cholesterol levels. We do not find this in saturated fats which are known to only increase LDL cholesterol.
c. Plants possess "soluble fiber" that drastically help to reduce cholesterol levels. They are known to be very low in saturated fats, and also free of cholesterol.
d. Saturated fats are common causes of elevated LDL in blood stream.
e. Dietary cholesterol as opposed to serum cholesterol are fats found in our diets(food) that may contain high levels of cholesterol that increase LDL levels in blood.
Small vesicles containing pigment inside of pigmented fish epidermal cells aggregate or disperse in response to treatment with certain chemicals. When nocodazole is added to cells in which the pigment granules have been induced to aggregate, the granules cannot disperse again. Small vesicles containing pigment inside of pigmented fish epidermal cells aggregate or disperse in response to treatment with certain chemicals. When nocodazole is added to cells in which the pigment granules have been induced to aggregate, the granules cannot disperse again. Pigment granules can either aggregate or disperse. Once aggregated the granules cannot disperse again, and vice versa. Intermediate filaments have a large effect on the pigment granule dispersal process and can stabilize resulting aggregates. Pigment granule dispersal is a microtubule-dependent process.
Answer:
Pigment granule dispersal is a microtubule-dependent process.
The question discusses how nocodazole, by disrupting microtubules, prevents the dispersion of pigment granules in fish epidermal cells, demonstrating the role of microtubules and intermediate filaments in pigment granule movement.
Explanation:The question pertains to the behavior of pigment granules in pigmented fish epidermal cells in response to chemical treatment. Nocodazole, a chemical inhibitor that disrupts microtubules, prevents the dispersion of pigment granules once they have aggregated. This indicates that pigment granule dispersal is a microtubule-dependent process. Furthermore, the question notes that intermediate filaments significantly impact the dispersal and stabilization of these pigment granules.
This process is akin to those observed in eukaryotic cells, where microtubules, actin filaments, and intermediate filaments form the cytoskeleton and play a critical role in various cellular functions, including intracellular transport and cell division. Observations like this are often visualized using fluorescent dyes, highlighting specific cellular structures under a fluorescence microscope.
Choose the key terms listed below with the phrase that is the best match for it.a. Reinforcement b. Cryptic species c. Biological species concept d. Morphospecies concept e. Species f. Phylogenetic species concepti. The smallest evolutionarily independent unit. ii. Distinguishing species based on their phenotypes. iii. Distinguishing species based on whether or not they interbreed regularly with each other and are reproductively isolated from other such groups iv. Distinguishing species by finding the smallest monophyletic group on a phylogeny v. Species that are evolutionarily independent from each other, but ARE NOT distinguishable based on morphology vi. Selection for reproductive isolation by means of reduced fitness of hybrids
Answer:
a) Reinforcement - Selection for reproductive isolation by means of reduced fitness of hybrids.
b) Cryptic species - Species that are evolutionarily independent from each other, but ARE NOT distinguishable based on morphology.
c) Biological species concept - Distinguishing species based on whether or not they interbreed regularly with each other and are reproductively isolated from other such groups.
d) Morphospecies concept - Distinguishing species based on their phenotypes.
e) Species - The smallest evolutionarily independent unit.
f) Phylogenetic species concept - Distinguishing species by finding the smallest monophyletic group on a phylogeny.
Explanation:
Species is a group of organisms which are closely related and very similar to each other. It is the smallest evolutionarily independent unit that are able to interbreed and produce fertile offspring.
The inability of related species to breed due to behavioral, geographical, morphological or genetic differences is known as reproductive isolation. When two populations have come back into contact after being separated, the complete reproductive isolation between them results in speciation and the incomplete reproductive isolation causes the production of hybrids, which are of low fitness or sterile.
The process of speciation in which natural selection increases the reproductive isolation between two populations of species by acting against the production of hybrids of low fitness is known as reinforcement.
Cryptic species are the species which are evolutionarily independent from each other, but are morphologically indistinguishable and cannot interbreed.
The different species concept are; biological species concept, ecological species concept, morphospecies concept, phylogenetic species concept, evolutionary species concept etc.
Biological species concept is a concept of species in which the individuals of a group can interbreed and create fertile offspring, but cannot breed with other groups (reproductively isolated).
Morphospecies concept or morphological species concept distinguish species based on the differences in their phenotypes (morphological characters).
Phylogenetic species concept distinguish species by finding the smallest monophyletic group (group of all organisms that are the descendants of a common ancestor) on a phylogeny (evolutionary history of genetically related group of organisms or species). This group have a shared and unique evolutionary history because they are descended from a common ancestor.
Which of the following is true about Gran Dolina adult hominids? a. They were more modern than Homo erectus, and like later Homo sapiens, had a wide nasal aperture. b. They had a larger cranial capacity than later Homo sapiens. c. They were able to produce spectacular art, similar to later Homo sapiens. d. They were similar to modern humans but with a narrow, Homo erectus–like nasal aperture.
Answer:d. They were similar to modern humans but with a narrow, Homo erectus–like nasal aperture.
Explanation:
Gran Dolina adult hominids were more advanced than Homo erectus, with a wide nasal aperture similar to that of Homo sapiens supporting this fact. Their cranial capacity was not larger than that of Homo sapiens, and there is no evidence suggesting they produced art like later Homo sapiens. Moreover, their nasal aperture was wider rather than narrow like that of Homo erectus.
Explanation:The correct answer would be option 'a'. Gran Dolina adult hominids were indeed more advanced than Homo erectus, resembling later Homo sapiens in having a wide nasal aperture. This is consistent with data suggesting that these individuals were an intermediary between Homo erectus and later Homo sapiens. Such anatomical features place Gran Dolina adult hominids closer in affinity to modern humans than to Homo erectus.
However, they were not as modern as Homo sapiens in other aspects. For example, their cranial capacity was not larger than that of later Homo sapiens (option b), and there is no evidence suggesting that they were capable of producing art as spectacular as that of later Homo sapiens (option c).
Moreover, while they were similar to modern humans in many respects, they did not have a narrow, Homo erectus-like nasal aperture (option d). Instead, as mentioned earlier, they had a wide nasal aperture.
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Which of the following is an acceptable definition of evolution?
a. a change in the genotypic makeup of an individualb. a change in the genetic makeup of a populationc. a change in the phenotypic makeup of a populationd. a change in the environmental conditionse. a change in the species composition of a community
Answer:
a change in the species composition of a community
Explanation:
Answer:
B: A change in the genetic makeup of a population
Explanation:
Evolution can be defined as any net directional change or any cumulative change in the characteristics of organisms or populations over many generations. It explicitly includes the origin as well as the spread of alleles, variants, trait values, or character states.
Why do mountain climbers carry oxygen?
Answer:
The atmospheric pressure goes on decreasing as we go up above the sea-level. Therefore, the amount of oxygen also decreases at higher altitude. Due to this reason, mountaineers carry oxygen cylinders with them, while climbing high mountains
Explanation:
Imagine you are studying a population of plants and you have determined that they are all genetic clones of one another. You are interested in a particular gene found in the population that you have named LLT (low-light tolerance), that allows it to live in shady areas.
What is the frequency of this allele in the population you study?
Answer:
The frequency of the allele is high in the population
Explanation:
The selective pressure in this population is low light. Thus only plants with high tolerance for low light will be able to survive in this population. LLT gene plant has this trait.
Therefore the LLT gene plants will have higher rate of survival compare to other clones at very low light intensity .
The frequency of expression of these gene will be high due to high survival rates in the population.
U937D cells express high levels of creatine kinase (CK‑B) mRNA but do not translate the mRNA into protein. Ribosomes bind the 5' end of the CK‑B mRNA; however, translation into protein is repressed in these cells. U937D cells synthesize the CK‑B enzyme when researchers introduce numerous short segments of RNA containing 3' UTR consensus sequences into the cells. The total amount of CK‑B mRNA does not change after adding RNA containing 3' UTR sequences. Introducing short RNA segments without the 3' UTR consensus sequences does not stimulate CK‑B synthesis. Which of the statements explains how the introduction of short RNA containing 3' UTR sequences allows CK‑B translation in U937D cells?.Explain how the introduction of short segments of RNA containing the 3'UTR sequences might remove the inhibition.
Answer:
Basically the translation of CK-B protein is inhibited in the U937D cells, despite the fact that the CK-B protein is bounded to the ribosomes. This is because(mechanisms) the translation is inhibited by the binding of translational repressors to the 3’UTR of the CK-B mRNA rather than the actual CK-B mRNA 3'UTR.
Furthermore, the soluble protein inhibitions is due to the reaction of the U937 cells to the short RNA sequences with the 3’UTR.
The introduction of these sequences(shot segement of RNA) into the U937D cells leads to CK-B synthesis. This makes 3’UTR sequences to bind to the translational repressor proteins, thus preventing them from binding to the CK-B mRNA .
COMPLETED QUE.
A common feature of many eukaryotic mRNAS is the presence of a rather long 3' UTR, which often contains consensus sequences. Creatine kinase B (CK-B) is an enzyme important in cellular metabolism. Certain cells—termed U937D cells—have lots of CK-B mRNA, but no CK- B enzyme is present. In these cells the 5’ end of the CK-B mRNA is bound to ribosomes, the mRNA is apparently not translated. Something inhibits the translation of the CK-B mRNA in these cells. Researchers introduced numerous short segments of RNA containing only 3’UTR sequences into U937D cells. As result, the U937D cells began to synthesize the CK-B enzyme, but the total amount of CK-B mRNA did not increase. The introduction of short segments of other RNA sequences did not stimulate the synthesis of CK-B; only the 3’UTR sequences turned on the translation of the enzyme. Based on these results, purpose a mechanism for how CK-B translation is inhibited in U937D cells. Explain how the introduction of short segments of RNA containing the 3'UTR sequences might remove the inhibition.
Explanation:
What organisms are responsible for decomposing organic matter? Why are they essential? A. heterotrophs; they return biotic material to the abiotic component of the Earth B. social insects; they carry out fertilization of many crops C. decomposers; they return biotic material to the abiotic component of the Earth
Answer: Option C - decomposers; they return biotic material to the abiotic component of the Earth
Explanation:
Decomposers are also known as SAPROPHYTES. They feed on and break down dead and decaying remains of plants and animals, thus enabling the release of certain compounds like ammonia, methane gas etc into the environment that then enrich/improve the soil fertility.
Final answer:
Correct option is C. Decomposers, mainly bacteria and fungi, are responsible for decomposing organic matter and are essential for recycling nutrients back into the ecosystem, supporting the stability of food webs.
Explanation:
Organisms responsible for decomposing organic matter are known as decomposers, which include bacteria and fungi. These decomposers are essential because they are involved in the process of breaking down dead materials and waste products, thereby recycling nutrients such as nitrogen and phosphorus back into the environment. Without decomposers, nutrients would remain locked within dead organic matter, preventing them from being accessible to living organisms that require them for growth.
Decomposers carry out the crucial task of returning biotic material to the abiotic component of the Earth, which ensures the stability of ecosystems. They break down organic substances, which are then termed as biodegradable, into inorganic nutrients that can be utilized by primary producers like plants. This cycle is fundamental to the maintenance of the food web and the overall health of the ecosystem.
Which is the class of cnidarian that is characterized by having a dominant medusa stage and four gastric pouches?
Answer:scyphozoa
Explanation:cniderians are aquatic animals.they have radial symmetry, that is they can be divided into two halves through any plane.
Cniderians may have nematocyst,which are stinging cells.
They have two forms,which are polyp and Medusa .
There are four classes of cniderians;hydrozoa, scyphozoa,cubozoa and anthozoa.
Scyphozoa are predominantly medusa,which is umbrella shaped.they are jelly fishes.
They have four or eight oral arms.
They possess four gonads and four gastric pouches .an example is the Aurelia.
If the normal nucleotide sequence was TACGGCATG, what type of gene mutation is present if the resulting sequence becomes TAGGCATG?A. additionB. deletionC. substitutionD. inversion
Answer:
B. Deletion
Explanation:
the 3rd letter (C) is missing (deleted)
In the normal nucleotide sequence C is missing. So, the type of mutation is deletion. Thus, option B is correct.
What is mutation?The term mutation is defined as a change in the normal sequence of DNA at a particular gene locus and mutations are seems to be harmful as it results into serious life taking disorders such as cancer.
Mutation also affect the function and division of cell that causes cancer and other serious disorders. Mutation in a single gene causes the body to produce thick and sticky white substance known as mucus that blocks lungs and digestive organs.
There are mainly three types of DNA mutations and these are deletions, insertions, and base substitutions.
By nature mutation is recessive but in some condition it can be dominant and these are harmful for the individual. It is random and recurrent event.
Therefore, the normal nucleotide sequence C is missing. So, the type of mutation is deletion. Thus, option B is correct.
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How does the balance of populations in the ecosystem change when the rattlesnake population increases? Choose the two correct answers. A. The bison population increases. B. The grasshopper population increases. C. The grasshopper population decreases. D. The blue jay population increases. E. The prairie dog population decreases.
Answer: The correct options are C and E.
Explanation: When the rattlesnake population increases, the population of grasshopper and prairie dog are also decreases.
Both grasshopper and prairie dog are herbivores and eaten by rattlesnake. When rattlesnake population increases they feed on grasshopper and prairie dog and decrease occurs in their population.
In guinea pigs, the gene for black fur, B, is dominant over the gene for white fur, b. Complete the Punnett square below to show the results of one possible cross between two guinea pigs with black fur.
Answer:
In guinea pigs, the gene for black fur, B, is dominant over the gene for white fur, b. Complete the Punnett square below to show the results of one possible cross between two guinea pigs with black fur.
BB X Bb = BB, Bb, BB, Bb
Explanation:
From the analogy shown, it is obvious that both parents are gene with black fur. During such crossing, the offspring are two homozygous dominant black fur and two heterozygous dominant black fur
Which of the following statements about viruses is FALSE?a. Vaccines are available to prevent several diseases caused by viruses. b. If a virus mutates, the immune system may not recognize the virus. c.The widespread use of antibiotics has led to resistant strains of viruses. d. People can get colds and flus again because the viruses are constantly mutating.
Answer: option C - The widespread use of antibiotics has led to resistant strains of viruses.
Explanation:
Antibiotics are substances (usually drugs) that can destroy or inhibit the growth of BACTERIA and similar microorganisms.
Do note that viral infection DO NOT respond to antibiotic treatment.
So, it is FALSE to say that the widespread use of antibiotics has led to resistant strains of viruses
The statement C- 'The widespread use of antibiotics has led to resistant strains of viruses.' is false. Antibiotics are used for bacterial infections, not viral ones.
Explanation:The statement that is FALSE among those given about viruses is: c. The widespread use of antibiotics has led to resistant strains of viruses. Antibiotics are used to treat bacterial infections, not viral infections. While bacteria can indeed develop resistance to antibiotics, viruses do not. This is due to the different nature of bacteria and viruses. Vaccines do help prevent diseases caused by viruses (a), and it's also true that the immune system may not recognize a mutated virus (b), and that constant mutation is the reason why people can get colds and flus more than once (d).
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Arsenate is a toxic ion that can interfere with both glycolysis and oxidative phosphorylation. Arsenate resembles inorganic phosphate (P) and can replace it in many enzymatic reactions. One such reaction is catalysed by glyceraldehyde 3-phosphate dehydrogenase in glycolysis. Arsenate competes with phosphate for entry into the active site of glyceraldehyde 3-phosphate dehydrogenase and upon completion of the reaction, instead of the normal product, 1,3-bisphosphoglycerate, the mixed anhydride 1-arsenato- 3-phosphoglycerate is formed: this undergoes rapid spontaneous hydrolysis into arsenate plus 3-phosphoglycerate, the latter being a normal intermediate in glycolysis, so glycolysis is able to proceed from the step at 3-phosphoglycerate. What would be the effect of arsenate in glycolysis? (a) Arsenate activates glycolysis so that more ATP and NADH is generated for every glucose molecule. In the presence of arsenate there is no net formation of ATP from glycolysis, but NADH generation is not directly affected Arsenate brings glycolysis to an abrupt stop with immediate lethal consequences In the presence of arsenate glycolysis produces fewer ATP and NADH molecules per glucose molecule. Arsenate does not affect the number of ATP or NADH molecules generated per glucose molecule. (b) (c) (d) (e)
Answer:
In the presence of arsenate glycolysis produces fewer ATP and NADH molecules per glucose molecule.
Explanation:
it is supposed to be 2 molecules of 3-phosphoglecerate with each producing 2 ATP making a total of 4 ATP gain but since it is just one and arsenate, only 2 ATP will be generated instead of four, thus ATP yield is reduced.
The vagus nerve does not innervate the ________.
a. kidneys
b. gallbladder
c. pancreas
d. parotid gland
Answer: D.the parotid gland
Explanation:
The innervation of the parotid gland is by auriculotemporal nerve and great auricular nerve as sensory innervation , and
the parasympathetic innervation originated from glossopharyngeal nerve;which synapse with otic ganglion;from this end point(the ganglion) the auriculotemporal nerve conduct parasympathetic fibres to the parotid gland.
The superior cervical ganglion , innervated the parotid gland as the sympathetic innervation. It travels along the external carotid artery to reach the parotid gland.
therefore PAROTID GLAND IS NOT INNERVATED BY VAGUS NERVE.
All other options in the questions are innervated by the vagus nerve.
Vagus nerve has the longest branch of all the cranial nerves, which extends from the brain to the abdomen. As it exist the jugular foremen , it innervates the head, the neck and abdomen to supply the above options a-c
The vagus nerve, which is part of the parasympathetic nervous system, does not innervate the parotid gland; instead, the parotid gland is innervated by the glossopharyngeal nerve.
Explanation:The vagus nerve does not innervate the parotid gland. The vagus nerve is a crucial part of the parasympathetic nervous system, which controls the function of many internal organs. The vagus nerve innervates structures such as the kidneys, gallbladder, and the pancreas. In particular, to the pancreas, the autonomic nervous system controls the secretion of insulin and glucagon, indicating that both sympathetic and parasympathetic nerve endings are found in pancreatic islets. Parasympathetic neurons, which are part of the peripheral nervous system and include the vagus nerve, leave the central nervous system via cranial nerves or sacral regions of the spinal cord. These neurons synapse in ganglia that are very close to the organs they innervate. The parotid gland, however, is innervated by the glossopharyngeal nerve, not the vagus nerve.
Jenner's successful use of cowpox virus as a vaccine against the smallpox virus was due to the fact that ______. (A) the cowpox virus made antibodies in response to the presence of smallpox(B) the immune system responds nonspecifically to antigens(C) cowpox and smallpox are caused by the same virus(D) there are some epitopes (antigenic determinants) common to both pox viruses
Answer: option C) cowpox and smallpox are caused by the same virus
Explanation:
Cowpox is a skin disease that affects cattle. It is caused by an Orthopoxvirus, with lesions occurring principally on the udder and teats of the animals.
Human infection may occur from touching infected cows, and thus giving immunity to smallpox (an acute infection caused by the same poxyvirus, in HUMANS)
So, Jenner's successful use was because cowpox and smallpox are caused by the same virus
_______________ occurs when coal is burned and sulfur oxides are released into the atmosphere, causing acid-forming particles to accumulate.
Answer:
Acid rain
Explanation:
So2 +H2O = HSO3+