Answer:
The unearthed Humerous bones don't belong to the species A.
Explanation:
Hello!
You are studying the Humerus bones species A, who is known to have a mean ratio of 8,5. This value corresponds to the population mean of the length-to-with ratio of bones of species A, symbolized as μ.
The hypothesis you want to study is "The Humorous bones unearthed belong to the species A" if you assume this to be true, then the mean of the length-to-with ratio should be equal to the known population mean of the length-to-with ratio.
Symbolized:
H₀: μ = 8,5
H₁: μ ≠ 8,5
Significance level: α: 0,05
You are asked to use a Z-test, since you don't know the value of the population variance, but have the sample values, the sample size is big enough (more than n=30). Assuming that the sample values are independent, the statistic test of choice is the approximation:
[tex]Z= \frac{x_bar-μ}{S\sqrt[]{n} }[/tex]≅N(0;1)
The critical region, in this case, it's a two-tailed test (remember the type is determined by the null hypothesis) so you'll have two critical values.
Left value [/tex][tex]Z_{\alpha/2} = Z_{0,025} = -1,96[/tex]
Right value [tex]Z_{1-\alpha/2} = Z_{0,975} = 1,96
So you'll reject the null hyphotesis if the calculated [tex]Z_{obs}[/tex] value is ≤-1,96 or ≥1,96 and you'll support it if -1,96<[tex]Z_{obs}[/tex]<1,96
Now we calculate the statistic by replacing the formula with the data:
x_bar = 9,26
S = 1,20
n = 41
[tex]Z_{obs}= \frac{x_bar-μ}{S\sqrt[]{n} }[/tex]
[tex]Z_{obs}= \frac{9,26-8,5}{1,20\sqrt[]{41} }[/tex]
[tex]Z_{obs}[/tex]= 4,0553
Since the calculated value falls in the rejection region, this means, you have statistically significant results. In other words you can reject the null hipothesis (H₀: μ = 8,5) and asume that the unearthed Humerous bones don't belong to the species A.
I hope you have a SUPER day!
Reject null hypothesis; mean ratio of fossil bones significantly greater than Species A mean, based on z-test at 0.05 level.
To check if the fossil humerus bones belong to Species A, we can perform a z-test for a population mean.
Given:
- Population mean [tex](\( \mu \))[/tex] of Species A = 8.5
- Sample mean [tex](\( \bar{x} \))[/tex] of the fossil humerus bones = 9.26
- Sample standard deviation [tex](\( s \))[/tex] of the fossil humerus bones = 1.20
- Sample size [tex](\( n \))[/tex] = 41
- Significance level [tex](\( \alpha \))[/tex] = 0.05
The null hypothesis ([tex]\( H_0 \)[/tex]) is that the mean length-to-width ratio of the fossil humerus bones is equal to the mean ratio of Species A [tex](\( \mu = 8.5 \))[/tex].
The alternative hypothesis ([tex]\( H_1 \)[/tex]) is that the mean length-to-width ratio of the fossil humerus bones is greater than the mean ratio of Species A [tex](\( \mu > 8.5 \))[/tex].
We'll use the z-test formula:
[tex]\[ z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \][/tex]
Substituting the given values:
[tex]\[ z = \frac{9.26 - 8.5}{\frac{1.20}{\sqrt{41}}} \][/tex]
[tex]\[ z \approx \frac{0.76}{\frac{1.20}{\sqrt{41}}} \][/tex]
[tex]\[ z \approx \frac{0.76}{\frac{1.20}{\sqrt{41}}} \][/tex]
[tex]\[ z \approx \frac{0.76}{0.187} \][/tex]
[tex]\[ z \approx 4.07 \][/tex]
Now, we compare the calculated z-value with the critical z-value at [tex]\( \alpha = 0.05 \)[/tex] for a one-tailed test.
From the z-table, [tex]\( z_{\alpha} = 1.645 \)[/tex] (approximately).
Since the calculated z-value (4.07) is greater than the critical z-value (1.645), we reject the null hypothesis.
Therefore, we can conclude that the mean length-to-width ratio of the fossil humerus bones is significantly greater than the mean ratio of Species A, at the 0.05 level of significance. Thus, we cannot conclude that these bones belong to Species A.
According to the phylogeny presented in this chapter, which protists are in the same eukaryotic supergroup as land plants?
a. green algae
b. dinoflagellates
c. red algae
d. both a and c
Answer: d. both a and c
Explanation:
Green algae are present in the most diverse environments. The vast majority of species, approximately 90%, are freshwater, with a cosmopolitan distribution, that is, they have a wide distribution on the planet. It is the predominant group of freshwater plankton. BASIC CHARACTERISTICS: Eukaryotic, Chlorophyll a and b, Xanthophylls (mainly lutein) and Carotenes (mainly β-carotene), Reserve: starch, Cell wall: mainly cellulose, Presence of flagella at some stage of the life cycle.
Red algae: Cell wall - consists basically of two parts, one internal and rigid, formed by cellulose microfibrils (most red algae), and the other outer, mucilaginous, formed by galactan polymers, such as agar and carrageenans. Certain groups of red algae have calcium carbonate deposition on the wall, giving the stalk great rigidity. This deposition may be in the form of aragonite or calcite. BASIC CHARACTERISTICS: Eukaryotic, Chlorophyll a and Phycobiliproteins (b, re-phycoerythrin, allophycocyanin and echo-cytocyanin), Xanthophylls (zeaxanthin, lutein, etc.) and carotenes (mainly β-carotene), Reserve: Cellular starch: cellulose, agar and carrageenan, Absence of flagella at all stages of life, including gametes and spores.
At the end of the Pleistocene, there are a series of extinctions in which the majority of large mammals go extinct. Why of these is a hypothesized cause of the extinction?
a. meteor impact
b. increased volcanism
c. glacial-interglacial cycles
d. human induced climate change
e. all of the above are hypothesized causes
Answer:
The correct option is: c. glacial-interglacial cycles
Explanation:
Pleistocene, also known as the Ice Age, is a geological epoch that is dated about 2.588 million to 11,700 years before present.
At the end of Pleistocene, the last glacial period ended and it also corresponds to the extinction of large mammals.
The cause of the extinction during Pleistocene is considered to be a combination of many factors such as human predation, climate change, interspecific competition, and unstable population dynamics.
Mendel's second law is the Law of Independent Assortment. According to that law and according to other facts that Mendel discovered, what statement or statements are true in the following list?
a. genes behave like fluids rather than like separate particles
b. genes behave like separate particles rather than like fluids
c. the alleles of the different genes separate independently of one another during meiosis
d. both a. and c. are correct
e. both b. and c. are correct
Answer:
Option (c).
Explanation:
The Mendel's give the two most important laws to explain the transmission of the characters in the organism. Mendel is known as father of genetics and explained his work on the pea plant Pisum sativum.
The Mendel's law of independent assortment explains that in the combination of different genes the alleles of the genes separate or assort independently to one another during the meiosis process. The genes must be present on different chromosome to assort independently.
Thus, the correct answer is option (c).
Mendel's second law, the Law of Independent Assortment, supports that genes behave as separate particles and alleles of different genes separate independently during meiosis, making the answer both b. and c. are correct.
The question pertains to Mendel's second law, known as the Law of Independent Assortment. This law highlights how alleles of different genes separate independently of one another during the process of meiosis, leading to the gametes containing a mix of alleles. It refutes the concept of blending inheritance, where genes were thought to behave like fluids, and instead, supports the idea of particulate inheritance, where genes behave as discrete units or particles.
Therefore, the statements that are true according to Mendel's second law and his other discoveries are that genes behave like separate particles rather than like fluids, and the alleles of different genes separate independently of one another during meiosis. Hence, the correct answer is: both b. and c. are correct.
Summarize the evidence showing that the sequence of nucleotides in a gene is colinear with the sequence of amino acids in a protein.
Answer:
A DNA is the genetic material of the organism which gets transferred to progeny controlling their traits. The DNA contains a sequence of nucleotides which codes for a protein called genes which provide instruction of the trait.
The DNA forms an mRNA molecule which by the process of Transcription and this mRNA molecule gets translated into the protein by Translation. Translation proceeds in the ribosomes where mRNA bases are studied by the ribosomes in the form of triplets called codons. Each codon codes for a specific amino acid which gets bonded to form a polypeptide.
Since the sequence of bases form codons which form the proteins in the same sequence as they are arranged on the mRNA, therefore nucleotides of the gene are collinear with the amino acids if the proteins.
Answer:
n
Explanation:
s
The decline of MPF activity at the end of mitosis is due to
a. the destruction of the protein kinase Cdk.
b. decreased synthesis of Cdk.
c. the degradation of cyclin.
d. the accumulation of cyclin.
Answer:
c. the degradation of cyclin is the correct answer.
Explanation:
The decline of MPF activity at the end of mitosis is due to the degradation of cyclin.
The MPF stands for maturation promoting factor is an enzyme promotes the passage into M phase from the growth phase (G2 )
During the mitosis process the enzyme which breaks the cyclin gets activated and due to this level of cyclin gets decrease.
The decrease in the levels of the cyclin leads to a decline in the levels of MPF at the end of mitosis.
MPF is composed of cyclin and kinase. It lets the cell go from G2 stage to M stage. When the cell is going to anaphase, MPF inactivates. The decline of MPF activity at the end of mitosis is due to the degradation of cyclin. Option (C).
----------------------------
Many molecules regulate the cell cycle during the different stages of the mitosis process to make it possible.
There are too many factors interacting, activating, and deactivating to pass from one stage to another.
Cyclins are cell cycle regulator proteins that control the cell during the process of division. Different types of cyclins act on a different stages.
When the cell is going through one of the stages, the only cyclin in high levels is the one that is regulating that part of the cycle.
Once that stage is over, that cyclin concentration decreases, and the following one increases.
Kinases, Cdk, are enzymes that depend on cyclins and that remain inactive until cyclins are present.
The enzyme becomes functional when the cyclin binds it. Then, the kinase can accomplish its functions.
When the cyclin and the kinase are together, they compose the Maturation Promotor Factor, MPF.
MPF is composed of cyclin and kinase. MPF is in charge of letting the cell go from the G2 stage to the M stage.
The inactivation of MPF is due to the degradation of cyclin. As cyclin is destroyed, the MPF gets inactive.
Steps:
1) During the whole cell cycle, cyclin M remains at low levels.
2) When the cell is ready to pass from G2 to M phases, cyclin M increases its concentration and binds Cdk.
3) Together, these proteins compose the MPF that makes this transition possible.
4) Once the cell is in the M stage, MPF needs to decrease.
5) The factor autoregulates by activating another complex. The anaphase-promoting complex/cyclosome, APC/C.
6) When the cell is ready to go from metaphase to anaphase, the APC/C complex begins the cyclins M elimination process. The complex also activates separase to separate chromatid sisters while destroying cyclins M.
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Explain how the ears of a human help control balance.
Answer:
Human ear is not only responsible for hearing by also control balance.
Human ear has three sections the outer; middle; and inner ear.
The outer ear consist of pinna and ear lobe, middle ear consist of three tiny bones the malleus, incus and stapes, and inner ear contains cochlea and the vestibular system.
The inner ear carry vestibular system which consist of three semicircular canals, saccule and utricle, responsible for maintaining balance.
The outer ear collects sound waves from outside and guide it to middle ear, and the middle ear changes the sound waves into vibrations and pass it to inner ear. the inner ear utilizes vibrations and send it to nerve impulses. The semicircular canals in the inner ear contains hair cells and fluid and hair cells. When anyone moves their head the fluid in semicircular canals moves along. the motion is then detected by hair cells, which are then send nerve impulses to the brain and functions in maintaining balance.
Answer:
The semicircular canals of the inner ear help you with balance. When you move your head, fluid inside the semicircular canals moves as well. This movement of the fluid moves the hairs of the canals, creating nerve impulses that travel up to your brain and let it know that your head is off balance.
Explanation:
Why do women's rights matter when it comes to controlling population growth?
A. Men always want more children than women do, and if women have no power, they don't get a choice.
B. Women with an education and career won't have children.
C. Empowering women gives them control over their own reproduction and they typically will chose to postpone having children.
D. Women generally make the major family decisions.
E. There is no way for men to control their own reproduction.
Answer:
The correct answer is option C.
Explanation:
Empowering women provides them with the power to decide over their own reproduction and they usually should have the choice to postpone having children. All women should possess the right to autonomy and choice of reproduction.
Women should possess the right and social accountability to decide about when and how to have children and should possess the right to decide about how many to have, no women can be forced to bear a child or should be inhibited from doing so against her desire.
A patient is admitted to hospital with a suspected bacterial illness. Describe two immunological techniques that may be used in diagnosing the patient.
Answer:
For the diagnosis of patients having bacterial infection, many immunological tests can be performed by the doctor's like agglutination test and enzyme immunoassays.
Agglutination test: This test confirms the bacterial infection based on antigen-antibody interaction. In this test, patient serum is taken which contains antibody, and latex beads with attached specific antigens are added into it. Agglutination of blood serum confirms the infection of that bacteria whose antigen is used in the test.
Enzyme immunoassays: In this test, the antibody is used which has attached enzyme. This antibody binds to the antigen and the enzyme converts some substrate into a product which have fluorescence. Higher the fluorescence higher is the amount of antigen.
Infants between the ages of six and 12 months are especially at high risk for contracting measles because they are too young to receive the vaccine. Infants are often protected until they are about 6- months-old by maternal antibody. Their mothers either had measles, or were vaccinated against it, before they became pregnant, but as these babies get older, that protection fades. Which of the following is the maternal antibody that protects infants for the first six months of life?
a. IgA
b. IgD
c. IgE
d. IgG
e. IgM
Answer:
The correct answer is d. IgG
Explanation:
During pregnancy the fetus is attached to the mother body by placenta and IgG is the only antibody which can cross the placenta and reach the fetus. This antibody is very important for an infant to get protection from infections.
Measles is a viral disease that usually infects infants who are not vaccinated against it. Until the vaccination is done the infant is protected by the passive immunity given by his mother through passing the IgG antibody.
This maternal antibody is capable of protecting infants for the first six months of life after which vaccination is required to protect the infant from measles.
Escherichia coli cells are about 2.85 μm long and 0.650 μm in diameter. i) How many E. coli cells laid end to end would fit across a pinhead of diameter 0.5 mm? Number of cells = cells ii) Calculate the volume of an E. coli cell, assuming the cell to be cylindrical in shape. The volume of a cylinder is: Cell volume = m3 iii) Calculate the surface area and surface area to volume ratio of an E. coli cell. The surface area of a cylinder is: Surface area = m2 Surface area:Volume = m-1
Answer:
i) 769 E. coli cells should be laid end to end would fit across a pinhead of diameter 0.5 mm.
ii) [tex]4.1466\times 10^{-18} m^3[/tex] is the volume of an E. coli cell.
iii) The surface area of E coli cell is [tex]1.8578\times 10^{-11} m^2[/tex] and surface area to volume ratio of an E. coli cell is [tex]4,480,297.110 m^{-1}[/tex].
Explanation:
Diameter of the Escherichia coli cell ,d= 2.85 μm = 0.00285 mm
1 μm = 0.001 mm
Length of the Escherichia coli cell ,h= 0.650 μm = 0.000650 mm
Let the number of Escherichia coli cell laid end to end to total length of 0.5 mm be x
[tex]x\times h =0.5 mm[/tex]
[tex]x=\frac{0.5 mm}{0.000650 mm}=769.23\approx 769 [/tex]
769 E. coli cells should be laid end to end would fit across a pinhead of diameter 0.5 mm.
Radius of the Escherichia coli cell r = d/2 = [tex]\frac{0.00285 mm}{2}=0.001425 mm[/tex]
Length of the Escherichia coli cell ,h = 0.000650 mm
Volume of the cylinder = [tex]\pi r^2h[/tex]
Volume of the E coli cell: V
[tex]V=3.14\times (0.001425 mm)^2\times 0.000650 mm[/tex]
[tex]V=4.1466\times 10^{-9} mm^3=4.1466\times 10^{-18} m^3[/tex]
[tex]1 mm^3=10^{-9} m^3[/tex]
Total surface area of cylinder :[tex]2\pi r (r+h)[/tex]
Area of the E coli cell: A
[tex]A=2\times 3.14\times 0.001425 mm\times (0.001425 mm+0.000650 mm)[/tex]
[tex]A=1.8578\times 10^{-5} mm^2=1.8578\times 10^{-11} m^2[/tex]
[tex]1 mm^2=10^{-6} m^2[/tex]
Surface area to volume ratio of an E. coli cell:
[tex]\frac{A}{V}=\frac{1.8578\times 10^{-11} m^2}{4.1466\times 10^{-18} m^3}[/tex]
[tex]=4,480,297.110 m^{-1}[/tex]
Approximately 175 E. coli cells would fit across a pinhead. An individual cell has a volume of approximately 0.971 μm³, a surface area of approximately 5.77 μm², and a surface area to volume ratio of approximately 5.95 m-1.
Explanation:Let's tackle these calculations one at a time:
i) Number of E. coli cells that can fit on a pinhead
To answer this, we need to convert the pinhead's diameter from millimeters (0.5 mm) to micrometers (500 µm) because the length of an E. coli cell is given in micrometers (2.85 µm). If we divide the diameter of the pinhead by the length of the E. coli cell, it will give us the number of cells that can fit across a pinhead. Therefore: 500 µm / 2.85 µm = approximately 175 cells.
ii) Volume of an E. coli cell
Given the E. coli cell can be represented as a cylinder, the formula for the volume of a cylinder is V=πr²h. The radius (r) is the diameter divided by 2, and the height (h) is the length of the cell. Calculating the volume, we get: V = π * (0.325 µm)² * 2.85 µm = approximately 0.971 µm³.
iii) Surface Area and Surface Area to Volume Ratio of an E. coli cell
For a cylinder, the Surface Area(A) = 2πrh + 2πr². Substituting the values: A = 2π*0.325 µm * 2.85 µm + 2π*(0.325 µm)² = approximately 5.77 µm². Finally, the ratio of surface area to volume is 5.77 µm² / 0.971 µm³ = approximately 5.95 m-1.
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In cellular respiration, the steps following glycolysis depend on whether oxygen is present. Select the BEST explanation:
A
If oxygen is present, production of acetyl-CoA, the citric acid cycle, and electron transport chain follow in order. If no oxygen is present, photosynthesis occurs starting with Photosystem II.
B
If oxygen is present, production of acetyl-CoA, the citric acid cycle, and electron transport chain follow in order. If no oxygen is present, either lactic acid fermentation or alcoholic fermentation follows.
C
In the presence of oxygen, carbon fixation occurs during the Calvin cycle, when a carbon atom from atmospheric carbon dioxide is added to a 5-carbon sugar.
D
Regardless if oxygen is present or not, the production of acetyl-CoA, the citric acid cycle, and electron transport chain follow in order.
SECOND QUESTION is the picture
here are the answer choices if you can only see the picture.
A. destroyed.
B. absorbed.
C. reflected.
D. converted to heat.
Answer:
For the first question the answer is B. For the second question the answer is C.
Explanation:
The first question: If oxygen is present, production of acetyl-CoA, the citric acid cycle, and electron transport chain follow in order. If no oxygen is present, either lactic acid fermentation or alcoholic fermentation follows.
The second question: Leaves appear green because the green portion of visible light that strikes them is reflected.
Hope this helps
-Amelia
Cellular respiration is a type of respiration ( metabolic pathway )that involves the breaking down of glucose into ATP. and it follows four ( 4 ) stages which are ;
glycolysis, pyruvate oxidation, Krebs cycle, oxidative phosphorylation.Glycolysis is the stage ( metabolic pathway) in cellular respiration when glucose is converted into two molecules of pyruvate in the presence of oxygen or the conversion of glucose into two molecules of lactate if no oxygen is present during the process
Hence we can conclude that The best explanation on the steps following glycolysis is If oxygen is present, production of acetyl-CoA, the citric acid cycle and electron transport chain follow in order. if no oxygen is present either lactic acid fermentation or alcoholic fermentation follows.
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What functions do membrane proteins serve? synthesize phospholipids for the maintenance and growth of the cell membrane transport nutrients into the cell that cannot otherwise cross the phospholipid bilayer transport wastes out of the cell that cannot otherwise cross the cell membrane prevent small, uncharged substances from crossing the phospholipid bilayer synthesize DNA within the cell using raw materials found outside the cell
Answer:
transport nutrients into the cell that cannot otherwise cross the phospholipid bilayer transport wastes out of the cell that cannot otherwise cross the cell membraneExplanation:
Cell membrane integral proteins especially trans-membrane proteins facilitate and regulate the movement of particular molecules across the cell membrane. Examples of these molecules are glucose and sodium ions. These molecules are either charged and cannot pass through the hydrophobic lipid layer of the cell membrane and/or are too large to pass through the cell membrane pores (like the aquaporins).Other types of cell membrane proteins are peripheral proteins. Collectively these proteins can have several other functions include cell signaling, enzymatic activity, cell-to-cell recognition , and etcetera.
Membrane proteins serve various functions including the transport of nutrients into and out of the cell, excretion of wastes, signal transduction, and serving as anchors for structural components. Integral proteins form channels or pumps for transport, while peripheral proteins function as enzymes or recognition sites. Together, they are essential for maintaining cellular homeostasis.
Explanation:Functions of Membrane ProteinsMembrane proteins are crucial for a variety of cellular processes. Firstly, they are involved in the transport of nutrients across the cell membrane. Such nutrients cannot cross the phospholipid bilayer on their own due to their polar nature or size. Similarly, membrane proteins assist in the excretion of waste products like CO₂, which need to be regulated within the cell. Another key role of these proteins is in transmitting signals; acting as signal transducers, they allow the cell to respond to external cues. Additionally, they also serve as anchors for the cell's structural framework, interacting with the cellular cytoskeleton and extracellular matrix.
Specifically, these proteins can be classified into two major classes: integral proteins, which are typically embedded within the lipid bilayer, and peripheral proteins, which are associated with the bilayer's surface. Integral proteins can form channels or pumps that facilitate active and passive transport, while peripheral proteins can function as enzymes or cell recognition sites that aid in immune responses or cell adhesion.
Overall, membrane proteins are indispensable for maintaining the cell's internal environment, allowing for a controlled exchange of substances to sustain cellular life.
In prokaryotes, a search for genes in a DNA sequence involves scanning the DNA sequence for long open reading frames (that is, reading frames uninterrupted by stop codons). What problem can you see with this approach in eukaryotes?
Final answer:
Scanning for long open reading frames to find genes works in prokaryotes due to their less complex genome organization and lower noncoding DNA content. This method is not as effective in eukaryotes, which have more noncoding DNA, individual promoters for each gene, and complex chromatin structures that hinder straightforward gene identification.
Explanation:
In prokaryotes, scanning a DNA sequence for long open reading frames (ORFs) is a logical method for gene identification because prokaryotic genomes have less noncoding DNA and often organize genes encoding proteins of related functions into operons. This means that genes are grouped together, have a single promoter, and can be transcribed into polycistronic mRNA, making it easier to identify coding sequences.
However, using this approach in eukaryotes imposes challenges. Eukaryotic DNA contains a much higher percentage of noncoding DNA, which means that scanning for long ORFs would yield many false positives, as noncoding introns and other noncoding elements could be mistakenly identified as genes. Additionally, eukaryotic genes are typically monocistronic, each gene having its own promoter, and exist in a complex chromatin structure, which makes simple scanning insufficient.
Another complexity in eukaryotic genomes is that promoters can be located within genes or far away from them, both upstream or downstream. This means that the regulatory sequences are not always neatly placed before genes, as seen in prokaryotes. Also, eukaryotic DNA is wrapped around histone proteins, forming nucleosomes, which complicate direct access to the DNA sequence for transcription and therefore, gene identification.
In which step of the light independent reaction is G3P producod?
a. CO2 production
b. CO2 reduction
c. CO2 fixation
d. Regeneration of RuBP
Answer:
The correct answer will be option-B.
Explanation:
Photosynthesis process takes place in two phases: light-dependent reaction and light-independent reaction.
The light-independent reaction takes place in the fluid-filled space called stroma of a chloroplast and is involved in the formation of glucose molecules. The light-independent reactions take place in cycles and are known as the Calvin cycle.
Calvin cycle proceeds in three phases: carboxylation or carbon fixation, reduction and regeneration. During reduction, the ATP and NADPH provides energy and hydrogen atom to the 3-PGA which gets converted to glyceraldehyde-3-phosphate (G3P).
Thus, Option-C is the correct answer.
What is the nature of "second signal molecules" in signaling pathways?
Answer:
The target receptor is expressed in the membrane, the activity of the receptor can instigate the production of small molecules known as the secondary messengers that coordinate and initiate the intracellular signaling pathways. For example, cyclic AMP is a secondary signal molecule.
When the signal molecules stimulate the receptor, the signal is carried into the cell generally by means of secondary messengers.
Second signal molecules, also known as second messengers, are crucial components of cell signaling pathways. They relay the signal from the cell surface receptor to the interior of the cell, helping to propagate or even amplify the signal for a larger cellular response. Examples include cyclic AMP, inositol trisphosphate, and calcium ions.
Explanation:The nature of "second signal molecules" in signaling pathways refers to a significant part of cell communication. These molecules are often termed second messengers because they relay the signal from the receptor on the cell surface, which is first to identify the signaling molecule, to the interior of the cell. The function of these intermediates is to propagate the signal and sometimes amplify the signal to produce a larger cellular response. Biologically essential second messengers include cyclic AMP (cAMP), inositol trisphosphate (IP3), and calcium ions.
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What
is the only amino acid with a substituted alpha amino group?the
presence of this amino acid influences protein folding byforcing a
bend in the chain- draw its structure.
Answer:
Proline
Explanation:
Proline's unique characteristic it's that it is the only amino acid where the side chain is connected to the backbone twice forming a nitrogen-containing ring, this means it has a substituted alpha-amino group. This characteristic is important to the structure of proteins, because, contrary to glycin which adopts many main chain conformations, proline if most commonly found in tight turns because of its rigidity.
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All of the following statements regarding the pathogens that cause tetanus, anthrax, and botulism are correct except:
(If all of the statements are correct, choose the final answer.
a. all three pathogens are soil-bacteria.
b. all three pathogens are obligate anaerobes.
c. all three pathogens produce endospores.
d. all three pathogens produce exotoxins
e. all of the above are true for tetanus, anthrax, and botulism
Answer:
b. all three pathogens are obligate anaerobes.
Explanation:
Tetanus is caused by an exotoxin produced by the bacterium Clostridium tetani, which is normally found in the soil, although it may be present in the intestines of mammals. It is an anoxic and spore-forming bacteria. It enters the body through wounds or deep punctures, finding in the wound the anoxic conditions to germinate its spores, producing its potent toxin.
Anthrax is caused by the bacterium Bacillus anthracis, which is a stationary and endospore-forming bacteria that is resistant to high temperatures and chemicals. It is a strictly aerobic bacterium and can be found naturally in the soil. Both animals and humans can breathe or ingest spores of this bacterium by consuming meat or contaminated water. When spores get into the body, their turn into active cells and produce exotoxins
Botulism is caused by the pathogen Clostridium botulinum, a strict anaerobic bacillus, producing endospores and 8 different types of toxins. Being a soil-borne bacterium, it usually appears in the food of animal and vegetable origin.
What advantage does crossing over provide for organisms that undergo meiosis over those at only undergo mitosis throughout their life-cycle?
Answer:
Explanation:
Crossing over is a complex mechanism which allows the exchange of the DNA sequence during meiosis.
Crossing over serves many purposes:
1. form genetically different daughter cells from parents.
2. leads to genetic diversity
3. leads to variation in the organism which proves useful.
The crossing over does not take place in the mitosis type of cell division which produces genetically similar daughter cells.
What would happen to the hormone levels in a man's bloodstream if his hypothalamus continuously produces releasing hormones?
a. LH and FSH levels would be low
b. Testosterone would be low
c. LH levels would be high and FSH levels would be low
d. FSH levels would be high and LH levels would be low
e. LH, FSH and testosterone levels would be high
Answer:
e. LH, FSH and testosterone levels would be high
Explanation:
Releasing hormones produced by the hypothalamus stimulate the anterior pituitary gland to release the tropic hormones which in turn stimulate the release of hormones from other glands.
The gonadotropin-releasing hormone produced by the hypothalamus makes anterior pituitary to release gonadotropins (LH and FSH).
The LH stimulates the Leydig cells of testes to release the androgens (testosterone). Therefore, continuous release of releasing hormone from the hypothalamus in a male would increase the LH, FSH and testosterone levels.
If a man's hypothalamus continuously produces releasing hormones, it would lead to high levels of LH, FSH, and testosterone in the bloodstream owing to a successive chain of hormonal stimulation leading up to increased testosterone production in the testes.
Explanation:The hypothalamus is a small region in the brain that plays a crucial role in many important functions, including sending signals to the pituitary gland to release hormones into the bloodstream. When the hypothalamus continuously produces releasing hormones, it stimulates the pituitary gland to secrete luteinizing hormone (LH) and follicle-stimulating hormone (FSH). These hormones, in turn, stimulate the testes to produce more testosterone. Thus, the answer is e. LH, FSH, and testosterone levels would be high.
Here's a breakdown of this process:
The hypothalamus produces gonadotropin-releasing hormone (GnRH).GnRH signals the pituitary gland to produce and release LH and FSH.LH and FSH travel via the bloodstream to the testes.LH prompts the testes to produce and secrete testosterone, while FSH is involved in sperm development.The increased levels of these hormones in the bloodstream then lead to the various effects on the body and behavior.Learn more about Hypothalamic-pituitary-gonadal axis here:https://brainly.com/question/29869336
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What are the components of the CNS? What is the function of each?
Answer:
Two main component of Central nervous system are brain and spinal cord.
Brain:
Brain acts as the main coordinating center of the body. The brain is protected in the skull. The brains control the emotions, thought process and learning memory of the organism. The brain interprets the information received by the five different sense organs.
Spinal cord:
The spinal cord is involved in the reflex action of the body. Spinal cord is involved in the somatosensory organization of the body. The motor pathways are originated from the spinal cord. Body's proprioceptive information travels through the spinal cord.
Describe how the population of the Earth increase if the number of children being born is less?
Answer:
As this ratio of land and man was given by demographer Thomas Malthus who said population on grew in an arithmetic and earth production in a geometric fashion.
Explanation:
The population or the production capacity of the earth will double and increase as long as it's kept stable i.e no interference from the outside world or human world. Even when there are lower birth and population is still decreasing the earth still will produce less as comparatively due to the slow growth of resources. This population may increase due to the number of medical aid facilities provided or may increase due to the population reaches its maturity and gets fixed or stabilized. Population and growth in resource distribution are always linked to the demographic changes in the population distribution around the world. \As the country of Africa has large scale childbirth and is next after Asia in terms of the population even after several years when its population stabilizes there still going to be pressure on the natural resources as these are finite in numbers.How does ADH work and what does it do?
A. It binds to the collecting tubules in the kidney, opening pores to allow water to flow back out of the kidney into the blood.
B. It binds to blood cells, making them more porous to water so it more easily diffuses back into them and is not lost in urine filtration.
C. It binds to water molecules, making them too large to diffuse out of the blood stream, retaining water when the body needs it.
D. It binds to nerve cells in hypothalamus, signaling that the body is dehydrated and should retain water.
E. It binds to the kidney tubules, making them less porous to water so less water filters out.
Answer:
The correct answer will be option-A.
Explanation:
ADH or anti-diuretic hormone is a chemical messenger produced by the hypothalamus in response to the water level in the blood, maintain blood pressure.
ADH is released in response to the osmolarity of the blood that is a low amount of water which is perceived by the osmorecpetors in neurons. The low level of water leads to the production of ADH from neurons which binds to the ADH receptors in the collecting tubules of the kidneys.
They allow insertion of aquaporins-2 in the membrane which allow re-absorption of the water to maintain the water level of the body.
Thus, option-A is the correct answer.
State what you would expect to see in a urine sample that was positive for laxative abuse.
Answer:
Laxatives, which enhance bowel movements, are abused by the individuals who desire to lose weight and are suffering from eating disorders. Laxative abuse leads to extreme symptoms like hypokalemia, diarrhea, abdominal pain, acid-base disturbance, and weight loss.
The majority of the laxatives are captivated by the colon and are discharged in feces or urine as metabolites. In the analysis of urine, the volume of urine gathered within a 24 hours duration is more than two liters. The level of sodium and chlorine reduces in urine and the levels of potassium will enhance in the urine.
A change in electrolyte balance and changes in acid-base results, hypokalemia is the clinical characteristic of laxative abuse. The pH of urine will decline on the basis of the extremity of the abuse. The levels of creatinine will increase, and one can witness elevation in the secretion of renin and aldosterone because of the loss of fluid.
What do we mean when we refer to a cell as differentiated?
Answer:
We mean that a cell changes to another type of cell.
Explanation:
Stem cells are a type of cell that has the potential to become a specialized cell, this change or differentiation, is possible thanks to the activation or deactivation of certain genes that promote (or inhibit) the expression of certain proteins that origins different types of cells (fmuscle cells, osteocytes, neurons). This differentiation happens when the cells receive cues internally ( through signals or contact between a group of cells and another or through transcription factors) or externally.
Then, a differentiated cell is a cell that had gone under the previously described process.
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A sex linked trait can be defined by which genotypes. Choose all which apply.
a. Coded for on the X and Y chromosomes
b. Coded for on the X but not Y chromosome
c. Coded for on the Y but not X chromosom
d. a dominant allele on any chromosome
Answer:
a. Coded for on the X and Y chromosomes
b. Coded for on the X but not Y chromosome
c. Coded for on the Y but not X chromosome
Explanation:
The X and Y chromosomes are the sex chromosomes as they are involved in sex determination. All the genes that are present on the X and Y chromosomes are inherited along with these chromosomes and therefore, exhibit sex-linked inheritance.
X and Y chromosomes share a homologous region. Therefore, the sex-linked genes are defined as the ones present on both X and Y chromosomes as well as the ones that are present on either X or Y chromosomes.
For example, the gene for hemophilia is present on the X chromosome; but not on the Y chromosome. The SRY gene is present only on the Y chromosome.
In the history of Earth and the evolution of cells, aerobic cells appeared first and anaerobic cells appeared later.
a. True
b. False
Answer:
False
Explanation:
Scientist believe that the primitive atmosphere lacked free oxygen. Oxygen is a highly reactive molecule that would have made difficult for complex macromolecules to appear and then create life. Rocks from the precambrian period support this theory.
Anaerobic cells appeared first, then photosynthesis appeared filling the atmosphere with oxygen and the aerobic organism develop cellular respiration to obtain energy, a process more efficient than anaerobic respiration.
The respiratory system branches like the circulatory. Oxygen and carbon dioxide diffuse between the end of these branches and the bloodstream. What are the ends of those branches called?
What is the purpose of the sphincters where the arterioles branch into the capillary beds?
A. To slow down blood flow so the capillaries don't burst under pressure
B. To block blood from coming back out of the capillary bed too quickly to allow for more complete diffusion of oxygen
C. To control blood flow to different areas of the body depending on activities
D. To speed up blood flow into the capillary bed for faster oxygen exchange
E. They remove carbon dioxide from blood passing through.
Answer:
1. Ends of the respiratory branches are called alveoli.
2. C. To control blood flow to different areas of the body depending on activities
Explanation:
1. The trachea divides into left and right primary bronchi which in turn divide multiple times upon entering the lungs and make the bronchial tree.
The final branches of the bronchial tree are the terminal bronchioles that lead to alveoli. The alveoli are the balloon-shaped structures and serve as the site of gas exchange between the blood and inhaled air.
2. The opening and closing of sphincters of capillary beds regulate the direction of blood flow. The opening of sphincters allows the blood to flow into associated branches of capillary beds while closed sphincters direct the blood from arterioles to venules via thoroughfare channel.
This local change in blood flow is responsible for the autoregulation of blood flow to different tissues to match their respective metabolic demands. For example, during physical activity, more blood is directed to skeletal and cardiac muscles.
Contrast meiosis with mitosis, mentioning homologous chromosomes, chromatids, tetrads and crossing over.
Answer:
Explanation:
Mitosis takes place in somatic cells of the body while the meiosis takes place in the germ cells. The mitosis is the equational division while the meiosis is reductional division.
The meiosis has 2 cell divisions i.e. meiosis I and meiosis II. The mitosis is a single division.
The pairing between the homologous chromosome/synapsis occurs in meiosis. It is not found in mitosis.
During the pachytene stage of meiosis, crossing over takes place. The homologous chromosome exchanges its genetic materials in crossing over.
In mitotic cell division, the chromatids do not exchange their genetic material.
The chromosome numbers are always the same as the parent cell in the mitotic division. In meiosis, the number of the chromosome turns to half of their parent cells.
There are 4 haploid chromosomes/tetrads forms at the end of meiosis. In meiotic division 2 chromosomes form from the parent cells.
Final answer:
Meiosis includes two nuclear divisions leading to four haploid genetically distinct cells, with homologous chromosomes pairing, crossing-over, and lining up as tetrads. Mitosis results in two identical diploid cells without these features. A diploid human cell has 46 chromosomes, whereas a haploid cell has 23.
Explanation:
In comparing meiosis and mitosis, several key differences are noteworthy. Both processes are preceded by one round of DNA replication, but meiosis involves two nuclear divisions that result in four genetically distinct haploid daughter cells, whereas mitosis results in two identical diploid daughter cells.
During meiosis I, a crucial aspect is the pairing of homologous chromosomes, each consisting of two sister chromatids. They become bound together with the synaptonemal complex, leading to the formation of chiasmata during crossing-over. Crossing-over is the exchange of genetic material between non-sister chromatids of homologous chromosomes, bringing about new combinations of genes. After this genetic recombination, homologous chromosomes line up along the metaphase plate in groups of four, known as tetrads, with kinetochore fibers from opposite poles attached to each homolog's kinetochore.
A diploid human cell, such as a skin or liver cell, has 46 chromosomes. In contrast, a haploid human cell, like a sperm or egg cell, has 23 chromosomes. Genetic variation arises from crossing-over, the independent assortment of chromosomes during meiosis, and random fertilization of gametes.
Which of the following sequences correctly represents the flow of electrons during photosynthesis?
a. NADPH -> O2 -> CO2
b. H2O ->S NADPH -> Calvin cycle
c. H2O -> photosystem I -> photosystem II
d. NADPH -> electron transport chain -> O2
Answer: b. H2O ->S NADPH -> Calvin cycle
Explanation:
Photosynthesis can be compartmentalized into two phases: one that depends directly on the light - photochemical phase and one that does not depend on the light, called chemical phase.
The first uses H2O to produce ATP and a reduced electron carrier (NADPH + H +), the second uses ATP, NADPH + H + and CO2 to produce sugar. In the photochemical phase, light energy is used to produce ATP from ADP + Pi, through a set of reactions mediated by groups of molecules - the photosystems - in a cycle called photophosphorylation.
There are two types of photophosphorylation: one non-cyclic that produces NADPH and ATP and one cyclic that produces only ATP. In the chemical phase, which is not directly dependent on light, non-cyclic photophosphorylation products - NADPH and ATP - and CO2 are used to produce glucose, in the so-called Calvin-Benson cycle. Although it is also called the dark phase, it is not independent of light, since for the enzyme responsible for fixing CO2, RuBisCo, requires light to be reduced and to be in its active state.
In DNA, which base is complementary to guanine?
a. andenine
b. thymine
c. cytosine
d. guanine
Answer:
Guanine is complementary to Cytosine.
Explanation:
The DNA molecule is made up of different components one of them is pairing of nitrogenous bases. The nitrogenous bases are the A, G, C, T. The A , G are the purine and C, T is the pyrimidine.
The A pairs with T by double hydrogen bond, and C pairs with G by a triple hydrogen bond. Thus the helix is arranged in antiparallel order.
The pairing of nitrogenous bases obeys Chargaff's rule. According to this rule, the combination of adenine and thymine is approximately equivalent to the combination of guanine and cytosine.
If the DNA bases do not obey this rule then the helical structure can't be possible. The pairing of purine, purine, and pyrimidine, pyrimidine results too short and too broad helix. Thus proper order of the helix can't be possible.
This is the very reason A double bonds with T and C triple bonds with G.
Answer:
cytosine
Explanation: