The given question is not complete. The complete question is:
Watch the film “Liz Hadly Tracks the Impact of Climate Change in Yellowstone (Links to an external site.)Links to an external site.” from the Scientists at Work series. Prior to watching the film, read the questions below and think about how you might answer them. You do not need to turn in your answers from before you watch the film.
Questions
Explain what is happening to the whitebark pine in Yellowstone National Park as a result of climate change
Researchers often study ecosystems for a long period of time. Dr. Hadly has studied Yellowstone’s ecology for 30 years and the amphibians for 20 years. What is the value of long- term studies to advancing scientific understanding?
Answer:
1. As a consequence of climate change, the whitebark pine is been targeted by pine beetles, which can thrive in winter conditions. An attack led by various beetles for one to two days has generated holes in the pine leading to its destruction.
2. The objective of long term studies is to find the connections between animals and plants and to witness how they are associating with each other in their micro-environment. Like in the documentary it was the association between the squirrel, pine trees and bear.
The beer needs to consume high content nutritious seeds of the pine tree, which safeguarded them in the winters due to the high amount of fat present in them that offered warmth. However, the beers were not able to reach the pine trees as they were far from their reach and thus, they took help from squirrels, which serve the objective for the bears.
Thus, it can be comprehended that all three creatures, that is, squirrel, tree and bear are associated with each other and developing a food web. All these studies need an ample amount of time and observation. Hence, Dr. Hardley took thirty years examining them and also took twenty years examining the influence of climate on amphibians living in the pond. Their species diversity declined substantially due to global warming that resulted in the drying up of the pond.
An mRNA molecule has a sequence 5'- CAGAUCUAAUGCUUAUCGGAU-3'. When translated in a laboratory setting in which translation can be initiated anywhere along the molecule, how many reading frames are possible?
Answer:
Three reading frames
Explanation:
Translation of mRNA always happen in one direction from the 5' end to the 3' end of the RNA strand.
Reading frame refers to the grouping of three consecutive bases to form a codon that can constitute an amino acid.
There are six possible reading frames in any nucleotide sequence.
Three from the 3' to the 5' end and three possible reading frames from the 5' to the 3' end.
As mentioned earlier, translation in mRNA happens in one direction therefore the three possible reading frames are;
5'- C AGA UCU AAU GCU UAU CGG AU-3'. 5'- CA GAU CUA AUG CUU AUC GGA U-3'. 5'- CAG AUC UAA UGC UUA UCG GAU-3'.An mRNA sequence can be translated into a protein in three possible reading frames, therefore, for the given sequence that can be initiated anywhere, there are three possible reading frames.
Explanation:In the realm of molecular biology, an mRNA sequence can be translated into a protein in three possible reading frames. Each reading frame will read the sequence from a different starting point, thereby creating a unique series of codons, and potentially, a different protein sequence. Hence, for an mRNA sequence such as 5'- CAGAUCUAAUGCUUAUCGGAU-3' that can be translated anywhere in a laboratory setting, it means there are 3 possible reading frames. Each frame begins from a different nucleotide within the first three (frame 1 from C, frame 2 from A, frame 3 from G) and continues in triplets from there on.
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Snapdragons occur in nature as either green or yellow plants. A green snapdragon is homozygous and has the genotype CC. A yellow snapdragon is heterozygous and has the genotype Cc. Suppose that a gardener crosses two yellow snapdragons, and one-third of the offspring are green and two-thirds of the offspring are yellow.
What type of allele could be responsible for the 2:1 offspring ratio seen when two yellow snapdragons are crossed?
A. dominant allele
B. polymorphic allele
C. recessive lethal allele
D. codominant allele
Answer:
Snapdragons occur in nature as either green or yellow plants. A green snapdragon is homozygous and has the genotype CC. A yellow snapdragon is heterozygous and has the genotype Cc. Suppose that a gardener crosses two yellow snapdragons, and one-third of the offspring are green and two-thirds of the offspring are yellow.
What type of allele could be responsible for the 2:1 offspring ratio seen when two yellow snapdragons are crossed?
dominant allele
Explanation:
As a result of two yellow snapdragons crossed, the dominat yellow allele gives the reason why the reult reflects in such a way that there is yellow dominates over green and leads to why yellow has two-third while green has one-third
Victoria needs to work through the night to finish a project for class. Victoria is concerned she will be too sleepy so she drinks energy drinks throughout the night to keep her alert and awake.What are the affects of energy drink?
Answer:
Victoria will intially felt active, reduced sleepiness but consuming too much caffeine may be risky.
Explanation:
Victoria will felt more active and alert after 30-45 minutes of drinking the enrergy drink because energy drink contains caffeine and its concentration in at peak after 30-45 minutes. Caffeine will reduce sleepiness.
Caffeine will block the adenosine pathway for short period of time. Adenosine is a chemical becuase of which we felt tired and Caffeine will allow feel good molecules to be released in brain such as dopamine. That why Victoria will felt more active, concentrated and good.
But consuming too much energy drinks mean victoria will be consuming too much caffeine that may be risky.Excess amount of caffeine can cause vomiting, nausea, convulsions and high blood pressure even can cause death.Gas exchange that occurs at the level of the tissues is called. Select one: a. Pulmonary ventilation b. External respiration c. Internal respiration d. Interpulmonary respiration
Answer: Option C.
It is called internal respiration.
Explanation:
Respiration is the process of inhaling oxygen from the outside environment and exhaling carbon dioxide ,which lead to the release of energy.
Respiration can be divided into internal , external, ventilation and cellular respiration.
Internal respiration or peripheral gas exchange occurs in the tissue where oxygen moves out of the blood and carbon dioxide is transported out of the cells.
External respiration occurs in the lungs where oxygen move into the blood and carbon dioxide diffuses into the alveolar air.
Internal respiration is the gas exchange that occurs at the level of the tissues. It involves the release of oxygen and pickup of carbon dioxide within the body tissues, occurring via simple diffusion without the need for energy. Option (c) is correct.
Explanation:The gas exchange that occurs at the level of the tissues is called internal respiration. This is the process where oxygen is released and carbon dioxide is picked up within the body tissues. This process alongside external respiration, which occurs in the alveoli of the lungs, forms an integral part of the respiratory system. While external respiration refers to the exchange of gases with the external environment, internal respiration refers to the exchange of gases with the internal environment.
These exchanges occur due to simple diffusion, following pressure gradients, without need for energy to move the gases across membranes. The anatomy of the lungs with its thin respiratory and blood capillary membranes, as well as large surface area, maximizes this diffusion process, thus promoting efficient gas exchange.
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If the diastolic blood pressure at the heart level is 80.0 mm Hg, what is the diastolic pressure at the height of the head, which is 0.300 m above the heart? Ignore any pressure drop due to the viscosity of the blood. g
Answer:
56.6 mm of Hg
Explanation:
Given that;
the diastolic blood pressure at the heart level is 80.0 mm Hg = [tex](P_0)[/tex]
height = 0.300 m
the diastolic pressure at the height of the head, which is 0.300 m above the heart can be determined using the formula
P= [tex](P_0) - (pgh)[/tex]
where; [tex](pgh)[/tex] is the hydrostatic pressure applied by the column of the liquid (Blood) of height (h) and average density [tex]p[/tex], also the g = gravitational acceleration.
the average density [tex]p[/tex] of a human blood = 1060 kg/m³
gravitational acceleration. ( g ) = 9.81 m/s²
h = 0.300 m
∴ the [tex](pgh)[/tex] = 1060 kg/m³ × 9.81 m/s² × 0.300 m
= 3119.58 Pascal (Pa)
From the standard conversion rate, 1 mm of Hg(mercury) = 133.322 Pa
∴ the amount of mm of Hg(mercury) that can be gotten from 3119.58 Pascal (Pa) will be; [tex]\frac{3119.58}{133.322}[/tex]
= 23.40 mm of Hg(mercury)
P= [tex](P_0) - (pgh)[/tex]
P= (80.0 - 23.4) mm of Hg
P= 56.6 mm of Hg
Diastolic pressure at the height of the head, which is 0.300 m above the heart is 56.6 mmHg.
Given here,
Diastolic blood pressure at the heart level = 80.0 mm Hg
0.300 m above, diastolic pressure = ?
The Diastolic pressure can be calculated using the formula
[tex]\bold {P_d = h \times g \times \rho }[/tex]
Where,
[tex]\rho[/tex] = average density of a human blood = 1060 kg/m³
g - gravitational acceleration. = 9.81 m/s²
h = Height = 0.300 m
Diastolic pressure,
= 1060 kg/m³ × 9.81 m/s² × 0.300 m
= 3119.58 Pascal (Pa)
Since, 1 mmHg = 133.322 Pa
So, 3119.58 Pascal = 23.40 mmHg
Thus ,
Pd = (80.0 - 23.4) mmHg
Pd = 56.6 mmHg
Therefore, diastolic pressure at the height of the head, which is 0.300 m above the heart is 56.6 mmHg.
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A researcher identified a bacterial enzyme that is essential in the breakdown of glucose. The researcher wants to test a potential antibiotic that acts on the newly identified enzyme. She finds that glucose is indeed broken down at a slower rate when the potential antibiotic is present. Addition of a higher concentration of glucose does not have any impact on Vmax. Which of the following properly characterizes this finding?
a. The potential antibiotic may be a competitive or noncompetitive inhibitor of the enzyme. These possibilities could be tested by adding more enzyme.
b. The potential antibiotic is a competitive inhibitor of the glucose-converting enzyme and could be an effective treatment for bacterial infections.
c. The potential antibiotic is a noncompetitive inhibitor of the enzyme and likely changes the shape of the active site.
d. The potential antibiotic reduces the number of available enzymes and is, therefore, a competitive inhibitor.
Answer:
The potential antibiotic is a competitive inhibitor of the glucose converting enzyme and could be an effective solution for bacterial infections.
Explanation:
Enzyme kinetics mainly deals with the study of the rate of reaction of the enzyme. Different factors that can affects the enzyme rate are also studied in enzyme kinetics.
The competitive inhibitors of the enzyme do not show any effect on the Vmax value of the enzyme and binds to the enzyme only. These inhibitors do not have ability to binds with enzyme substrate complex. The increase in the enzyme and substrate concentration can remove the inhibitor affect.
Thus, the correct answer is option (b).
The potential antibiotic is acting as a noncompetitive inhibitor since higher glucose concentrations do not affect the Vmax, suggesting the antibiotic changes the enzyme's conformation rather than competing with glucose for the active site. So the correct option is c.
Explanation:The finding that the potential antibiotic slows down the breakdown of glucose by a bacterial enzyme but that a higher concentration of glucose does not affect the Vmax (maximum rate of enzyme activity) indicates the antibiotic is likely acting as a noncompetitive inhibitor. A competitive inhibitor would have a different effect: the Vmax could potentially still be reached if sufficient additional substrate (in this case, glucose) were added, as competitive inhibitors can be outcompeted by high concentrations of the substrate. Since adding more glucose did not change the Vmax, this suggests that the potential antibiotic is not a competitive inhibitor but rather implies that the antibiotic binds at a site other than the active site, thereby altering the enzyme's conformation and affecting its activity without directly blocking the substrate from binding. Therefore, the correct characterization of this finding would be option c: The potential antibiotic is a noncompetitive inhibitor of the enzyme and likely changes the shape of the active site.
Gibson Assembly – Several enzymes are present in the Gibson assembly reaction. For each of the following enzymes, define the role of the enzyme at room temperature (when the reaction is set up) and at 50 °C (the temperature for the 30 minute incubation after set up).
a. Phusion DNA polymerase ,
b. T5 exonuclease ,
c. Taq DNA ligase
Answer with explanation:
Gibson Assembly is a method of molecular cloning which join multiple DNA fragments in a single reaction. It was created by Daniel G. Gibson, Chief Technology Officer and co-founder of SGI-DNA. (See attached picture)
a. Role of Phusion DNA Polymerase
It bring the enzymes closer to DNA fragment and help enzymes to make PCR product with speed and more accurately. Moreover it has the ability of amplifying long templates. (See attached picture)
a. Role of T5 exonuclease
It is an exonuclease enzyme which means it remove the nucleotide from DNA stand in 5´ to 3´ direction. It create nicks in the double stranded DNA for the incorporation of other fragments. Furthermore, it also work fine in single stranded DNA. (See attached picture)
a. Role of Taq DNA ligase
It is a thermostable enzyme which catalyzes the phosphodiester bond formation between 5´-phosphate and the 3´-hydroxyl of two adjacent DNA strands. (See attached picture)
Enzymes are proteins that change energy-requiring reactions to energy-releasing reactions. true or false
Answer: True
Explanation:
Enzymes are biological molecules or proteins that speed up chemical reactions or processes in living cells but do not get used up and can be use over again.
Enzymes are very important in the body in that they help in digestion( I.e breakdown of food particles into small pieces) and metabolism. They change energy requiring actions to energy releasing reactions.
The statement that "enzymes are proteins that change energy-requiring reactions to energy-releasing reactions" is false.
Explanation:Enzymes are proteins that act as biological catalysts to speed up both energy-requiring and energy-releasing reactions by lowering the activation energy, but do not change the type of reaction.
Enzymes are indeed proteins, but they do not change the type of reaction. Instead, they function as biological catalysts to speed up chemical reactions, including both energy-requiring (endergonic) and energy-releasing (exergonic) reactions. They achieve this by lowering the activation energy needed for reactions to proceed.
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Which of the following statements concerning transcription is true?
A. All types of RNA in the cell are synthesized by transcription, which uses a portion of DNA as a template for copying.
B. Promoter sequences signal the end of a gene and mark the place where transcription stops.
C. During transcription, entire chromosomes are copied because the starting position of genes is unknown.
D. Transcription is the process whereby identical copies of DNA are made in preparation for cell division.
Answer: Option A) All types of RNA in the cell are synthesized by transcription, which uses a portion of DNA as a template for copying.
Explanation:
All types of RNA:
- viral RNA,
- ribosomal RNA,
- transfer RNA,
- messenger RNA and
- double-stranded RNA, have a sequence that is the same as 'antisense' strand of the DNA.
Thus, it is true that all types of RNA in the cell are synthesized by transcription, which uses a portion of DNA as a template for copying.
Option A is the correct statement regarding transcription: All types of RNA in the cell are synthesized by transcription using a portion of DNA as a template.
Explanation:The correct statement concerning transcription is option A: All types of RNA in the cell are synthesized by transcription, which uses a portion of DNA as a template for copying. Transcription is the process by which an RNA molecule is synthesized using a DNA template. Specific sequences called promoters signal the start of a gene, and terminator sequences signal the end of gene transcription, not the other way around as mentioned in option B. The entire chromosome is not copied during transcription; only the specific gene being transcribed is copied.
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A cross is done between two mutant T4 phage (a- b+ c+) and (a+ b-c)(no order implied by these genotypes). The genotypes of the progeny from this cross and their relative frequencies are given below: a-b+ c+ (30%) a+D c. (30%) a+ b+ c-(2%) a+ b+ c+ (8%) a- b-e+(2%) a-b-c-(8%) e+b-c+ (10%) a-b+ c-(10%) What is the distance between "a" and "b? Please give your answer in % recombination, but L number please! EAVE OUT the units (NO % sign), such as 20, or 3,
Since, the relative frequency is incorrectly mentioned in the question. The correct relative frequency table is attached below.
Answer:
20.
Explanation:
The genotype with the least recombinant frequency will represent the double cross overs.
The double cross overs progeny are a+ b+ c-(2%) and a- b-c+(2%).
The percentage of the recombinant frequency determined the distance between the genes. In the double crossovers, the b gene gas been exchanged and present in the middle.
The single recombinant crossovers as compared with the parents are a+ b+ c+ (8%) and a-b-c-(8%).
Distance between the gene a and b = Single cross overs + double cross overs
Distance between the gene a and b = 8 + 8 +2 +2 = 20.
Therefore, the distance distance between the gene a and b is 20 centi morgan.
Meselson and Stahl used density labeling of DNA to show that DNA replication occurs via a semiconservative mechanism. In their experiment, they started with an organism grown in a heavy density label (15N). After two generations of growth in light medium (the more common 14N isotope), if the DNA is isolated and separated by density, how many bands would be observed and how would their density compare with the starting DNA
Answer:
The organism previously used 15N for replication so all the DNA molecules were of 15N15N type. Then the organism is shifted to a medium where only 14N is available for replication. According to semi conservative mode of replication, a newly synthesised DNA molecule consists of one new strand and one parental strand. So after the first round of replication, All the 15N strands will synthesise new DNA strands using 14N resulting into intermediate 15N14N DNA molecules. Hence, only one band would be observed (15N14N) above the original 15N15N band since 15N14N has lighter isotope too so it will be lighter than 15N15N molecules and will lie above it.After second round of replication, 15N strand from 15N14N would synthesise another 14N strand. 14N strand from 15N14N molecules will also synthesise another 14N strand. So now, 50% of the DNA molecules will be of 15N14N intermediate type and 50% of them will be of 14N14N type.Two bands will be observed above the original 15N15N band. One band of 15N14N molecules will be right above it and other band of 14N14N molecules will be even higher because it is the lightest band since it has only the lighter isotope of nitrogen.Final answer:
Two bands would be observed after isolation and ultracentrifugation of DNA from the Meselson and Stahl experiment following two generations in 14N: one intermediate density band (indicative of one 15N and one 14N strand) and one light density band (indicative of double 14N strands), proving semiconservative DNA replication.
Explanation:
In the Meselson and Stahl experiment which aimed to understand the mechanism of DNA replication, they observed the effects of consecutive generations of bacterial growth in media with different nitrogen isotopes. Initially, E. coli was grown in a heavy nitrogen isotope, 15N, followed by growth in a lighter isotope, 14N. After two generations in 14N, when the DNA was isolated and subjected to density gradient ultracentrifugation, two bands were observed. One band was of intermediate density, indicating it contained one 15N-labeled strand and one 14N-labeled strand. The second band was of lighter density, corresponding to DNA composed solely of 14N-labeled strands. This provided strong evidence for the semiconservative model of replication where each daughter DNA molecule consists of one parental and one new strand.
An index fossil is:________.a. a fossil found in a particular site. b. the ideal specimen of that species to which all later descriptions must refer. c. the type specimen of a species. d. a fossil used to categorize a stratigraphic layer.
Answer:
Option d, a fossil used to categorize a stratigraphic layer
Explanation:
Fossils that are typical of a specific time range in the course of history of evolution of earth are generally termed as index fossil. These fossils can be used to determine the age of the rock layers and fossils with in which they are found. An index fossil must have a unique identity so that it can be easily recognized.Along with this they must have a lived for a short span of time in horizontal rock layer which must be geographically widespread for matching up to huge distances.
Hence, option D is correct
Answer:
Option (D)
Explanation:
An index fossil is usually defined as those fossil that appeared for a short geological time and were widely distributed over the surface of the earth. These fossils are extremely rare, and it plays an important role in determining the age of a rock, and it also helps in the correlation of rocks. A noticeable number of index fossil species were observed in different places on earth which were deposited on the rock sequences in the geological time.
It enables a geologist to categorize the different stratigraphic layers.
Thus, the correct answer is option (D).
When giving a talk about the theories of Marshall McLuhan, Sandy first made sure to thoroughly explain the term "media ecology." This is an example of_________.
Answer:
Giving a definition.
Explanation:
Giving a definition deals with properly explaining the term(s) concerned to avoid further confusion and to make sure your audience ain't lost. It is not merely to copy what was printed in other dictionaries that gives a definition. Choose terms that your understanding or that personal observations has affected. By defining its roles, structure or existence, you can define a phrase.
What name is given to the rigid structure, found outside the plasma membrane, that surrounds and supports the bacterial cell?
The rigid structure that supports and envelopes the bacterial cell outside the plasma membrane is the cell wall, primarily composed of peptidoglycan. This provides the cell with strength, maintains its shape, and shields it from environmental threats.
Explanation:The rigid structure that surrounds and supports the bacterial cell, which is found outside of the plasma membrane, is called the cell wall. Bacterial cell walls are primarily composed of peptidoglycan, a large polymer of amino sugars. These sugars are linked together in a mesh-like structure providing the cell with strength and rigidity. The cell wall helps maintain the shape of the cell and also protects it from environmental threats such as changes in osmotic pressure.
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The rigid structure found outside the plasma membrane that surrounds and supports the bacterial cell is called the "cell wall."
The cell wall is a crucial component of bacterial cells, providing structural support and protection. It is a tough, rigid layer composed of peptidoglycan, a unique macromolecule that consists of sugar chains cross-linked by short peptides. The cell wall maintains the cell's shape and prevents it from bursting due to changes in osmotic pressure. It also plays a significant role in bacterial pathogenicity, as differences in cell wall structure are used to classify bacteria into two major groups: Gram-positive and Gram-negative.
Gram-positive bacteria have a thick layer of peptidoglycan, while Gram-negative bacteria have a thinner layer of peptidoglycan surrounded by an outer membrane. The cell wall is an essential target for antibiotics like penicillin, which disrupt its integrity and ultimately lead to bacterial cell death.
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In a study where participants rated the pleasantness of T-shirt odors, what gene alleles influenced their odor preference? a. Serotonin transporter b. Phosphatase enzyme c. Blood type d. Major histocompatibility complex
Answer:
Major histocompatibility complex
Explanation: In a variety of animals, including humans, there is a correlation between odor preference, and genetic similarity at the Major Histocompatibility Complex (MHC).
MHC is a highly polymorphic group of genes that play an important role in the immunological self/non self recognition. Its products have been recognised to take part on the numerous compounds and reactions that build up an individual's body odor.
The cell responsible for secreting the matrix of bone is the__________a. osteoclast. b. chondroblast. c. chondrocyte. d. osteoblast.
Answer
D.Osteoblasts
Explanation:
Appositional growth is the increase in the diameter of bones by the addition of bony tissue at the surface of bones. Osteoblasts at the bone surface secrete bone matrix, and osteoclasts on the inner surface break down bone. The osteoblasts differentiate into osteocytes.
The cell responsible for secreting the matrix of bone is the osteoblast, which supports the growth, maintenance, and repair of bones.
Explanation:The cell responsible for secreting the matrix of bone is the osteoblast. Osteoblasts are a type of bone cell that form new bone, or 'osteo', tissue. They do this by secreting a matrix that later mineralizes to become hardened bone tissue. This process is critical for the growth, maintenance, and repair of bones in the body.
In contrast, osteoclasts are involved primarily in the breakdown and resorption of bone tissue, while chondroblasts and chondrocytes are associated with cartilage formation and maintenance, not with bone tissue formation.
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Which of the following statements is/are accurate?
A. Horses and Mules can be bred but their offspring is typically sterile. This could be an example of hybrid breakdown.
B. One sponge species releases its gametes during the night and another species releases its gametes during the day. This is an example of ecological isolation
C. One species of sea turtle mates during the early spring and closely related species mates during late spring. This is an example of temporal isolation
D. One species of sea urchins can not fertilize another closely related species because the eggs do not have a receptor for the sperm. This is a good example of gametic isolation.
E. Damselflies have sensory receptors that are sensitive to touch (tactile cues). Two related species of damselflies are unable to mate because their touch cues are not compatibale. This is a good example of mechanical isolation.
Answer:
These statements refer to various mechanisms that act as a reproductive or hybridization barriers.
In nature there are different mechanisms that prevent the crossbreeding between different species, what in biology is called reproductive barriers. Some mechanisms that act by preventing hybridization between different species are:
Hybrid breakdownGametic isolation.Mecanical isolation.Temporal isolation.Ethological aislaminet.Ecological insulation.These mechanisms are responsible for preserving the genetic integrity of each species by preventing hybridization between different species.
Explanation:
A. Horses and Mules can be bred but their offspring is typically sterile. This could be an example of a hybrid breakdown.This is accurate. Horses and donkeys belong to two different species, with a different chromosomal load:
Horses have 32 pairs of chromosomes.Donkeys have 31 pairs of chromosomes.Both species can be bred, but their descendant, mules (Equus africanus x ferus), have an odd number of chromosomes (63) and are infertile. This represents an exact example of hybrid breakdown.
B. One sponge species releases its gametes during the night and another species releases its gametes during the day. This is an example of ecological isolation.This is not accurate. In sponges, like some coral species, periods of release of gametes and fertilization vary throughout the day, with some synchrony between individuals of the same species.
The fact that some sponges release their gametes by day and other species do it at night is an example of the reproductive barrier called temporal isolation.
C. One species of sea turtle mates during the early spring and a closely related species mates during late spring. This is an example of temporal isolation.This is accurate. When two related but different species - such as turtles - have their mating period at different times of the year, there is talk of temporal or seasonal isolation.
Temporal isolation is a reproductive barrier that prevents crossing between different species, due to their mating habits at different times.
D. One species of sea urchins can not fertilize another closely related species because the eggs do not have a receptor for the sperm. This is a good example of a gametic isolation.This statement is accurate. In the case of sea urchins, the encounter of gametes requires two chemical mechanisms:
The first mechanism is called chemotaxis, which consists of the presence of a chemical signal on the surface of the egg, for which only sperm has a receptor. Another mechanism is that - once the sperm and egg are found - the membrane of the egg releases substances that interact with receptors in the sperm, allowing the sperm to enter it.These two chemical mechanisms ensure that gametes of two different species cannot be joined and fertilized, which is an example of gametic isolation.
E. Damselflies have sensory receptors that are sensitive to touch (tactile cues). Two related species of damselflies are unable to mate because their touch cues are not compatible. This is a good example of mechanical isolation.This is accurate. The sensitivity to the touch of damselflies is specific to individuals of the same species, preventing mating between male and female of different species.
This, like courtship, is a mechanism that prevents crossbreeding between different species, establishing an example of mechanical isolation.
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If Meselson and Stahl had used CsCl gradient analysis and identified DNA molecules with two distinct densities after the first generation, which model of DNA replication would have been supported by this data?
O conservative replication model
O semiconservative replication model
O dispersive DNA replication model
Answer:
conservative replication model
Explanation:
In their experiment, Meselson and Stahl found that the DNA duplexes present after the first round of replication had intermediate density. This occurred since the semiconservative process of DNA replication of one DNA molecule (having both strands with 15N) forms two DNA duplexes each with one parental strand (15N) and one newly formed strand (14N).
If they had found two duplexes of two different density, this would have supported the conservative replication model. In this case, the DNA replication of the parent DNA molecule would have formed two DNA duplexes. One DNA molecule would have both the parental DNA strands with 15N while the other DNA duplex would have both newly formed strands with 14N.
If Meselson and Stahl had observed two distinct densities after the first generation using CsCl gradient analysis, it would have supported the conservative replication model. However, their actual findings supported the semiconservative replication model.
If Meselson and Stahl had used CsCl gradient analysis and identified DNA molecules with two distinct densities after the first generation, the model of DNA replication that would have been supported by this data is the conservative replication model. In this model, the parental DNA strands (blue) would remain associated in one DNA molecule while the new daughter strands (red) would remain associated in newly formed DNA molecules. However, Meselson and Stahl actually found a single band after one generation, which disproved the conservative model and supported the semiconservative replication model.
DNA from a newly discovered virus was purified, and UV light absorption was followed as the molecule was slowly heated. The absorbance increase at the melting temperature was only 10%. What does this result tell you about the structure of the viral DNA? Please be sure to make your initial post and respond to at least two classmates with additional questions and comments to further the discussion.
Answer: The viral DNA has a sharp thermal transition.
Explanation:
Due to more guanine to cytosine (G-C) bonds in viral DNA, and the extra stability its presence confers, the double stranded Viral DNA is more stable to heat and have a higher melting point.
Hence, even when denatured by heat, the unstacked base pairs in its single strands only absorb ultraviolet light at a minimal rate such as 10% increase or less, before it gets renatured into double stranded DNA again.
Thus, the ability of viral DNA to adjust to thermal (heat) denaturation is the reason.
The low absorbance increase at the melting temperature suggests that the structure of the viral DNA may be more resistant to the heating process, which could mean it has a more complex or compact form than typical DNA.
Explanation:The UV light absorption data suggests that the newly discovered viral DNA has a unique structure differing from typical DNA. When DNA is heated to its melting temperature, the two strands 'unzip' or separate, leading to an increase in absorbance. A typical increase in absorbance at the melting temperature for ordinary DNA is around 37%. In this case, the increase was only 10%, indicative that the structure may be more resistant to the heating process, perhaps due to additional base-pairing or tighter coils. This suggests that the viral DNA might have a more complex or compact structure compared to typical DNA.
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Which of the following is the best explanation for why cells are considered the smallest units of living things. Cells have the ability to reproduce identical copies of themselves in a process called mitosis. Cells cannot be seen with the naked eye and are considered microscopic. Cells are the simplest structure to fit all of the characteristics necessary to be considered alive. Cells are highly ordered and complex.
Answer: Cells are the simplest structure to fit all of the characteristics necessary to be considered alive.
Explanation:
The cell is regarded as the smallest unit of all living things because it can carry out all life activities such as:
- feeding
- reproduction
- excretion
- growth
- adaptation
- respiration
- definite life span
- sensitivity, and
- movement
All these characteristics possessed by cells are the characteristics of living things.
Answer: Cells are the simplest structure to fit all the characteristics to necessary to be considered alive.
Explanation:
Cell is the smallest and basic units of life I.e it is the building blocks of living organisms. Cells can exist on their own. Cells are the smallest unit of life because they are the smallest components of living organisms and have simplest structure to fit characteristics to be considered alive.
These characteristics are exhibited by living cells and they are;
Reproduction, feeding,respiration, movement, sensitivity, excretion ,growth and death.
The type of synapse that occurs between the terminal end of the presynaptic cell and the terminal end of the postsynaptic cell is referred to as which kind of synapse?
Answer:
Synaptic cleft
Explanation:
Synaptic cleft is the gap between the terminal end of the presynaptic cell and the terminal end of the postsynaptic cell. It helps in transportation the neurotransmitters from a synapse to another.
The resurrection fern (Pleopeltis polypodioides) grows on the bark of southern live oak trees (Quercus virginiana). Resurrection ferns that grow on the underside of branches receive less water than those growing on the topside and, therefore, have a fitness of only 0.28. How many of these ferns will be non-reproducing?
a. 52%
b. 72%
c. 80%
d. 28%
Answer:
b. 72%
Explanation:
Survival rate of underside of branches & topside of branches resurrection ferns are equal. Let say p + q = 1
Resurrection ferns that grow on the underside of branches receive less water than those growing on the topside and, therefore, have a fitness of only 0.28
That shows and indicate that they vary in reproducing factor.
let the reproducing factor be (p) = 0.28
let the no-reproducing factor be (q) = ???
0.28 + q = 1
q = 1 - 0.28
q = 0.72
q = 72 %
Interneurons don't conduct signals from one structure to another; a. they integrate activity within a single brain structure. b. have two short axons but no dendrites. c. have one long axon and one short dendrite. d. have several short axons and no dendrites. e. have bipolar axons and no dendrites.
Answer:
a. they integrate activity within a single brain structure.
Explanation:
This can be local interneurons and relay interneuron based on function and structure.
Based on function the interneuron synapse with the sensory and motor neurons,This structural characteristic is related to their functions. Thus as sensory neurons synapse with the Interneurones, the recieved message from sensory neuron of the spinal cord first branched off to the brain for processing and analysis, and output (response) from the brain returns to the interneuron , which synapse this with the motor neurons, for transmission to the effectors.
This type of information integration and processing is especially important when processing information from high brain centers, and complex tasks,and it involved the relay interneuron. This explains the integration function within the brain structure.
However if the information is simple the local interneuron mediated this/ a sensory information is not sent to the brain. Rather the information is transmitted from the sensory to the motor , coordinated by the spinal cord and response is transmitted to the effectors.
25. Select all accurate statements
A. All chordates will have a notochord as adults
B. All chordates have a ventral, hollow nerve cord
C. All chordates have pharyngeal slits or clefts
D. All chordates are bilaterally symmetrical animals.
E. All chordates will have a muscular post an*l tail
Answer:
A, B, and D
Explanation:
Food couldn't reach the stomach without the blank and the blank
Answer:
Food could not reach the stomach without the "esophagus and the throat".
Explanation:
Esophagus is like an elastic pipe, about 25 cm in length. It passes food from the throat back towards the stomach. There's the trachea at the back of the mouth which enables air to come in and out of the body. Once food in the form of a small ball of mushed-up food or liquids is ingested, the epiglottis slips down the opening of the windpipe to ensure that the food reaches the esophagus and not the windpipe.
Food is transported to the stomach via the esophagus with the help of the pharynx through peristalsis. Once in the stomach, food is stored, chemically digested, and mechanically broken down into chyme before moving into the small intestine.
Explanation:Food couldn't reach the stomach without the involvement of the esophagus and the pharynx. The esophagus is a muscular tube that connects the pharynx to the stomach, and its main function is to transport food from the throat down into the stomach. The pharynx, also part of the digestive tract, is the area at the back of the throat that receives food from the mouth. Together, the coordinated contractions of these structures move food along in a process called peristalsis. After passing through the esophagus, the food reaches the stomach, where it is stored and chemical digestion begins in earnest, starting with the conversion of food into a semi-liquid mixture called chyme. The stomach's muscular movements help in further breaking down the food before it enters the small intestine for continued digestion and absorption.
Some traits do not obey Mendel's law. For example, people with red hair tend to also have pale skin. Why might this be the case?
Answer: Polygenic inheritance
Explanation:
Polygenic inheritance is a phenomenon that explains how a character like skin color show a range of more or less continuous variation due to many genes controlling it. And this is unlike Mendel traits that are controlled by single genes.
So, skin color in an individual is expressed as red hair and pale skin.
A scientist discovers a DNA-based test for one allele of a particular gene. This and only this allele, if homozygous, produces an effect that results in death at or about the time of birth. Which of the following statements describes the best use of this discovery?
a. Screen all newborns of an at-risk population.
b. Design a test for identifying heterozygous carriers of the allele.
c. Follow the segregation of the allele during meiosis.
d. Introduce a normal allele into deficient newborns.
Answer:
b. Design a test for identifying heterozygous carriers of the allele.
Answer:
A scientist discovers a DNA-based test for one allele of a particular gene. This and only this allele, if homozygous, produces an effect that results in death at or about the time of birth. Which of the following statements describes the best use of this discovery?
Screen all newborns of an at-risk population.
Explanation:
When all newborns are screened, it enable the researcher to identify areas with lapses in order to correct such either by introducing an heterozygous to correct the abnormality
Choose the best answer.
Her repeated pulmonary infections have weakened the right side of her heart, so it is enlarged.
a) Since blood is moving between her atria, the blood in her ventricles and arteries is staying where it is, distending those structures.
b) Since she has a pulmonary infection, more blood is being diverted to her lungs; this distends her pulmonary trunk and leaves less blood in the systemic circuit.
c) Since blood is moving from her systemic circuit into her pulmonary circuit, the pulmonary circuit is distended and the systemic circuit is low on blood.
Answer:
Option B
Explanation:
Failing of right side is usually caused by failure of left side of heart. Malfunctioned right side of the heart loses its power of efficient pumping and as a result of this the blood is pumped back into the lungs. This backward flow backs up in the veins thereby causing the fluid to swell and hence the swelling in various body parts such as legs, liver, GI tracts, abdomen etc.
Hence, option B is correct
Answer: Option B.
Explanation:
Pulmonary infection is also lung infection. It can be caused by virus, bacteria or fungi.
A person can be infected when he breathing the pathogens from the air. A person with pulmonary infection can have right sided heart failure.
Right sided heart failure occur when because of left sided heart failure. The left sided failure occur when the left ventricle loses power to pump blood to the rest of the body. As a result, blood is pump to the lungs which weaken the right side of the heart and this lead to right side heart failure.
Which of the following is/are true?
A. Redheaded females in a population will only breed with red headed males. The yellow-headed females will only breed with yellow-headed males. This population lives in the same geographic area. The red and yellow individuals could breed and produce fertile offspring, but they normally do not.
This would be an example of sympatric speciation.
B. Sympatric speciation is best described as a random event that disrupts the allele frequencies in a population
C. Redheaded females in a population will only breed with red-headed males. The yellow-headed females will only breed with yellow-headed males. This population lives on separate continents and rarely met in nature. The red and yellow individuals could breed and produce fertile offspring, but they normally do not.
This would be an example of sympatric speciation
D. Sympatric speciation is never due to sexual (mate) selection
E. Sympatric speciation requires geographic isolation
Answer: option B) Sympatric speciation is best described as a random event that disrupts the allele frequencies in a population
Explanation:
Sympatric speciation is an event/situation whereby organisms of the same species:
- live in the same territory or nearby territories ( i.e do not live in geographical isolation)
- DO NOT interbreed, but select a sexual mate from a much diverse territory to yield new species or offsprings.
This sexual selection then results in generations of offsprings that are genetically different from the rest of the same species due to uneven gene flow or disruption of alleles among the population of same species.
Thus, only option B is true.
A demonstration is performed during a lecture on muscle
physiology in which a student is asked to fully extend his
right arm with the palm up. Two large textbooks are placed
on his palm, one at a time. Which of the following facilitates
the maximum amount of tension that allows the student to
keep his arm extended in place under the increasing weight of
the books?
(A) Amount of Ca2+ released from the sarcoplasmic
reticulum
(B) Amount of muscle phosphocreatine
(C) Amplitude of the action potential
(D) Number of motor units recruited
(E) Rate of cross-bridge recycling
Answer:
The answer is D
Number of motor units recruited
Explanation:
The force a muscle generates is dependent on the length of the muscle and its velocity which is shortened.
A motor unit is the grouping of muscle fibers supplying nerves by the neuron and consists of the motor neuron.
The number of muscle fibers within a motor unit varies, and is a function of the muscle’s ability for accurate and refined motion.
Precision can be determined to be inversely proportional to the size of the motor unit. As a result, small motor units can exercise greater precision of movement compared to larger motor units. For example, thigh muscles which are responsible for large powerful movements, can have a thousand fibers in each unit, while eye muscles which requires small precise movements, might only have about ten.
Groups of motor units are supplied with nerves to coordinate contraction of a whole muscle and generate appropriate movement; all of the motor units within a muscle are referred to as a motor pool.
As a result, the number of motor units with are recruited will facilitate the maximum amount of tension that allows the student keep his arm extended in place under the increasing weight of the books.