Answer:
X: Carbon Dioxide
Y: Oxygen
Explanation:
Answer:
X: carbon dioxide
Y: oxygen
What areas are most important to wildlife in wildlife management areas
Answer:
These are lands which are set aside for the purpose of increasing wildlife numbers by protecting wildlife and key habitat. One of the major goals of a wildlife management areas is to protect at least a minimal number of animals so the population can increase.
Explanation:
Hope this helps
Answer:
It entirely depends on location and what wildlife you are trying to protect. The obvious response would be the land that best suits the wildlife you are trying to protect. For example, the most important areas to southeastern beach mice are the dune areas at beaches they are found.
Explanation:
Which statements best describe the first stage of cellular respiration? Check all that apply
Answer:
the answer is the stage happens in the cytoplasm and glucose is broken down. IG....
Explanation:
What are the two most common reasons for modifying the land in South and Southeast Asia?
I will give brainiest to first answer
THE TWO MOST COMMON REASON FOR MODIFYING THE LAND IN SOUTH AND SOUTH ASIA IS
(i) first is significantly, the area of lands categorized as under forest/woodland and wetlands declined .At the same time, cultivated area increased .
(ii) another is requirement of additional land due to increase in population.
Explanation:
Between 1880 and 1980, the South and Southeast Asian landscape underwent dramatic modification.The 81% of the lost forest and wetland vegetation appears to have been converted into expanded agricultural land.
Human population in this region more than tripled between 1880 and 1980, producing an enormous demand for additional land for cultivation.
Answer:
C-
agriculture and forestry
Explanation:
The bacterium responsible for gastric ulcers, Helicobacter pylori, survives the acidic pH of the stomach surprisingly well. It does so in part by synthesizing and excreting large amounts of the enzyme urease, which hydrolyzes urea to bicarbonate and ammonia. Explain how these reaction products lead to a less acidic environment for H. pylori.
Answer:
H. pylori uses the enzyme urease to breakdown urea into ammonia (NH3) & carbon dioxide (CO2), where NH3 can act as a buffer to the acidic solution in the stomach.
Explanation:
H. pylori is a bacteria that has the enzyme urease to breakdown urea into ammonia (NH3) & carbon dioxide (CO2). The compound of interest here would be ammonia, or NH3. NH3 is a base, although relatively weak to other stronger bases, which means it has a pH above 7. In the stomach, the pH is acidic, or below 7. By synthesizing ammonia, H. pylori is able to buffer the stomach solution in a manner so that it isn't entirely acidic, but more toward the basic side, thereby allowing for its survival.
Nutrients from digested food enter the bloodstream through the process of
(1) absorption (3) respiration
(2) elimination (4) secretion
Answer:
1. Absorption is the answer
Answer:
absorption
Explanation:
the food gets into the blood stream by absorbing the tiny cells of food
1. What kingdom is Phylum Echinodermata classified in?
Answer:
Explanation:
Radiata
In the early part of the 19th century, Echinodermata was recognized as a distinct group of animals and was occasionally associated with the cnidarians and selected other phyla in a division of the animal kingdom known as the Radiata; the concept of a superphylum called Radiata is no longer valid.
1.
What does your liver do? Ist two major roles and briefly explain each
role.
Select the reasons why organisms are classified. Check all that apply.
to name them
to determine their lifespan
to form groups
to allow identification
to show relatedness
to identify their habitats
Answer:
A, C, D, E
Explanation:
i took the test on edge
Answer:
its A C D E, and if its not them you messed up something
Explanation:
Pseudomonas sp. has a mass doubling time of 2.4 h when grown on acetate. The saturation constant using this substrate is 1.3 g/l (which is unusually high), and cell yield on acetate is 0.46 g cell/g acetate. If we operate a chemostat on a feed stream containing 38 g/l acetate, find the following:
a. Cell concentration when the dilution rate is one-half of the maximum
b. Substrate concentration when the dilution rate is 0.8 Dmax
c. Maximum dilution rate
d. Cell productivity at 0.8 Dmax
Answer:
a) Cell concentration when the dilution rate is one-half of the maximum is 0.598g cell/L
b) the substrate concentration when the dilution rate is 0.8 [tex]D_{max}[/tex] is 5.2g/l
c) the maximum dilution rate is : 0.41 h⁻¹
d) the cell productivity at 0.8 [tex]D_{max}[/tex] is 2.40g cell/L
Explanation:
Given data :
Mass doubling time of Pseudomonas sp. = 2.4 hr
Saturation constant = 1.3 g/L
Cell yield on acetate = 0.46g cell/g acetate
We are to find;
a. Cell concentration when the dilution rate is one-half of the maximum.
Here, cell yield =amount of cell produced / amount of substrate consumed.
[S] at 0.5D max is determined using the Monod's equation.
Using the formula :
[tex]D = \frac {D_{max}[S] }{ks+[S]}[/tex]
, where D is the dilution rate,
[S] is substrate concentration; &
Ks is the saturation constant.
By replacing the values, we get :
[tex]0.5 = \frac{S}{1.3+[S]}[/tex]
[tex]\\\0.65=0.5[S][/tex]
[S]=1.3g/L
The cell concentration at 0.5Dmax= cell yield x substrate consumed at 0.5Dmax.
=0.46×1.3
= 0.598g cell/L
b)
Substrate concentration when the dilution rate is 0.8 [tex]D_{max}[/tex] is calculated as:
[tex]D = \frac {D_{max}[S] }{ks+[S]}[/tex]
0.8=[S]/1.3+[S]
1.04+0.8[S]=[S]
[S]= 5.2g/L
Therefore , the substrate concentration when the dilution rate is 0.8 [tex]D_{max}[/tex] is 5.2g/l
c)
Maximum dilution rate is calculated using the expression [tex]D_{max} = \frac{1}{time}[/tex]
=1/2.4
=0.41 h⁻¹
So, the maximum dilution rate is : 0.41 h⁻¹
d)
The cell productivity at 0.8 [tex]D_{max}[/tex] can be calculated by multiplying the amount of the cell yield with the amount of the substrate consumed at 0.8[tex]D_{max}[/tex]
Cell yield = [tex]\frac {cell \ productivity \ at \ 0.8Dmax}{amount \ of \ substrate\ consumed \at\ 0.8 \D_{max}}[/tex]
Cell productivity at 0.8 [tex]D_{max}[/tex] = 0.46 × 5.2
=2.40g cell/L
Therefore, the cell productivity at 0.8 [tex]D_{max}[/tex] is 2.40g cell/L
To find the cell concentration at half the maximum dilution rate, calculate the biomass concentration using the Monod equation. Use the same equation to find the substrate concentration when the dilution rate is 0.8 Dmax. The maximum dilution rate can be determined using the saturation constant and cell yield on acetate. Cell productivity at 0.8 Dmax is found by multiplying biomass concentration by the dilution rate.
Explanation:To find the cell concentration when the dilution rate is one-half of the maximum, you need to calculate the biomass concentration in the chemostat. This can be done using the Monod equation, which relates the specific growth rate of the bacteria to the substrate concentration. When the dilution rate is one-half of the maximum, the specific growth rate is also half of the maximum growth rate. Using the Monod equation, you can then find the corresponding biomass concentration.
To find the substrate concentration when the dilution rate is 0.8 Dmax, you can use the Monod equation again. This time, you need to rearrange the equation to solve for the substrate concentration, given a specific growth rate. The specific growth rate at 0.8 Dmax can be determined by multiplying the maximum growth rate by 0.8.
The maximum dilution rate is the dilution rate at which the specific growth rate is at its maximum. To find it, you need to determine the maximum specific growth rate. This can be done using the saturation constant and cell yield on acetate. Once you have the maximum growth rate, you can determine the dilution rate at which it occurs.
The cell productivity at 0.8 Dmax can be calculated by multiplying the biomass concentration at 0.8 Dmax by the dilution rate at that point.
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16-16: In a series of experiments, genes that code for mutant forms of an RTK are introduced into cells. The cells also express their own normal form of the receptor from their normal gene, although the mutant genes are constructed so that the mutant RTK is expressed at considerably higher concentration than the normal RTK. What would be the consequences of introducing a mutant gene that codes for an RTK (A) lacking its extracellular domain, or (B) lacking its intracellular domain
Answer:
Kinase-connected receptors or receptor tyro-sine kinases react for the most part to protein and chemical go between. A solitary trans membrane helix interfaces the extracellular restricting area to the intra-cellular space. e.g. insulin, development factors. The official of the ligand triggers the commencement of a few succession of occasions related with phosphorylation of proteins, this is called protein kinase course.
For instance, the official of development hormone to the receptor in the plasma layer causes dimerization (by the actuation of Janus kinase 2, JAK2) of the receptor (conformation change) that bring about auto-phosphorylation of tyro-sine buildups. The official of SH2-space (src homology) protein (Grb-2) to the phosphorylated tyro-sine buildups invigorates cell development through a course of protein phosphorylation.
a). RTKs are the trans-membrane receptors, which have a ligand restricting site on the extracellular area and tyro-sine authoritative on the intra-cellular space. In the event that it comes up short on the extracellular area, the ligand can't tie to the receptor site, so no cell reaction happens.
b). On the off chance that it does not have the intra-cellular space, the ligand can tie to the receptor site, the receptor can't impart signs tot eh intra-cellular area because of the absence of intra-cellular space.
Answer:
(A) lacking its extracellular domain:
RTK receptors can bind to the ligand, this phenomenon leads to a decrease in the rate of molecular interaction between receptors and ligands.
(B) lacking its intracellular domain:
RTK receptors can not bind to the ligand and therefore they are inactive, this phenomenon doesn't have effects on the receptor-ligand interaction mechanism
Explanation:
These types of experiment are very important in molecular biology since they enable to determine the function of transmembrane proteins in molecular signaling pathways
Which of the following is NOT a function of lymphatic system?
a) production of red blood cells
b) return of excess tissue fluid to the blood
c) transport of fats and lipids from the small intestine
d) defend the body against pathogens
Final answer:
The lymphatic system does not produce red blood cells; this process occurs in the bone marrow. The lymphatic system is instead responsible for functions such as fluid drainage, immune response, and lipid transport.
Explanation:
Function of the Lymphatic System
The lymphatic system has several key functions, such as draining body fluids and returning them to the bloodstream, circulating lymphocytes (a type of white blood cell essential for the immune system), and transporting fats and lipids from the small intestine to the bloodstream. However, production of red blood cells is not a function of the lymphatic system; this occurs in the bone marrow, part of the hematopoietic system.
To answer the student's question directly: The option that is NOT a function of the lymphatic system is a) production of red blood cells.
Mr. Walker plants a new plant next to a tall tree in his yard. It is partly in the sun, but is mostly in the shade He notices that it is
growing with all of its leaves on one side. Why is the plant growing this way?
A.It is dying
B.It is trying to get sunlight.
C.It is trying to stay in the shade.
B.It wants to be closer to the tree.
Answer:
B. it's trying to get sunlight.
(D. is plausible too)
Explanation:
Plants need sunlight for photosynthesis, so they tend to act towards the sunlight (phototropism)
Mr. Walker's plant is growing with all its leaves on one side to reach more sunlight, which is a response called phototropism, influenced by the hormone auxin aiding in maximizing photosynthesis.
Mr. Walker's plant is exhibiting a common response to its light environment known as phototropism, which is the growth of a plant in the direction of its light source. Since the plant is mostly in the shade due to a tall tree, it grows toward the area where it can receive more light. In more scientific terms, this is caused by the plant hormone auxin, which redistributes on the shaded side of the plant, causing those cells to elongate more and thus bending the plant toward the light. This adaptation helps the plant maximize its photosynthesis capabilities by increasing the exposure of its leaves to sunlight. Plants in different light conditions, such as full light or shade, may also exhibit different growth characteristics, with sun-loving plants having reduced leaf area and thicker leaves, and shade plants showing the opposite traits.
What hypotheses have been proposed to explain how prions evolve? Select the TWO answers that are correct.
a. When the environment changes, one of the less common variants in a prion population may be able to better spread and become the dominant type of prion.
b. Prions respond to environmental changes by bonding with other prion variants in the population and potentially forming new variants able to handle the change.
c. Changes in the environment change the shape of some prions, possibly producing a variant that is better able to tolerate the change than the original type of prion.
d.Environmental changes cause some prions to break into small segments that may be better able than larger prions to survive and reproduce in the new environment.
Answer:
b. Prions respond to environmental changes by bonding with other prion variants in the population and potentially forming new variants able to handle the change.
d.Environmental changes cause some prions to break into small segments that may be better able than larger prions to survive and reproduce in the new environment.
Explanation:
The first hypothesis that tried to explain how prions replicate in a protein-only manner was the heterodimer model. This model assumed that a single PrPSc molecule binds to a single PrPC molecule and catalyzes its conversion into PrPSc. The two PrPSc molecules then come apart and can go on to convert more PrPC.
prion hypothesis states that the misfolded prion protein, rather than virus or bacteria, is the infectious agent that results in the transmission of prion disease.
Which disease can destroy red blood cells?
EOS
Which disease infects T cells?
Which disease produces a scaly, stinging, and itchy rash on the feet?
Which disease can destroy red blood cells?
I) Hemolytic anemia is a decease which destroys red blood cells .
Which disease infects T cells?
ii) chromosomal breakage syndromes (CBSs) is one one of those decease that infect T cells.
Which disease produces a scaly, stinging, and itchy rash on the feet?
iii) Eczema , it's a skin disorder .
thanks
Make chart with Microsoft Office that describes at least 3 major events that occur in development of the fetus during each month of pregnancy. The must include the month, the size of the embryo or fetus, and a description of 3 major events that occur that month. You can cut and paste or simply attach this to your submission.
Answer:
Check the explanation
Explanation:
Month 1
Fertilization
Embryo formation
Blastocyst
Fetus development
Month 2
Fetus growth
External organs begin to differentiate
Month 3
Eyelids form
Nose begins to develop
Face starts to form
Month 4
Organ systems complete
Fetal circulation complete
Placenta complete
Distinct fingers and toe
Month 5
Fetus size 10 – 17 cm
Heartbeat is present
Sex differentiated
Rudimentary kidneys formed
Month 6
Fetus size 25 cm
Weight 260gms
Fetal movements felt by mother
Langua covers
Month 7
Fetus size 28 – 36cm
Weight 680gms
Eyebrows and finger nails develop
Wrinkled skin
Month 8
Fetus size 35 – 38 cm
Weight 1200 – 1800gm
Eyes open and close
Skin is red
Month 9
Fetus size 42 – 49 cm
Weight 1900 – 2700 gms
Eyelids open
Amniotic fluid decreases
Month 10
Fetus 48 – 52 cm
Weight 3000g
Smooth skin
Bones osstifies
Chymotrypsin has three residues in the active His-57, Asp-102, and Ser 195. To evaluate the importance of Ser-195 for the enzymatic activity, this amino acid residue is replaced by another amino acid. Predict what changes you expect in the activity of the enzyme in the following three different mutant enzymes. In every case, explain your answer (5 p each).
The question given is incomplete, below is a complete questiongotten from google:
Chymotrypsin has three residues in the active site: His-57, Asp-102, and Ser 195. To evaluate the importance of Ser-195 for the enzymatic activity, this amino acid residue is replaced by another amino acid. Predict what changes you expect in the activity of the enzyme in the following three different mutant enzymes. In every case, explain your answer (5 p each).
A) Ser-195 to Thr
B) Ser-195 to Cys
C) Ser-195 to Ala
Answer:
Explanation:
A serine protease that breaks the peptide bond in the protein at the carboxy terminal of aromatic amino acids tyrosine, tryptophan and phenylalanine is known as Chymotrypsin.
Three amino acids form a catalytic triad: aspartate, histidine and serine.
Histidine acts as a general acid base catalyst, and it is stabilized by Aspartate, while serine provides the nucleophile in form of oxygen that attacks on the carbonyl carbon of the peptide bond to form a tetrahedral intermediate.
Since serine has a hydroxyl (-OH) group and the bond between oxygen and hydrogen is removed by histidine which acts a general base catalyst and accepts the proton. This makes the oxygen in the hydroxyl group nucleophilic and it binds with carbonyl carbon of the peptide bond. This function can be replaced by any other amino acid which can produce a nucleophile.
A) Threonine has a hydroxyl group. Just like serine, the bond between oxygen and hydrogen is broken due to the acceptance of protons by histidine. Here, also oxygen in the hydroxyl group become negatively charged species, hence a nucleophile that can attack on the carbonyl carbon of the peptide bond.
B) Cysteine has thiol (-SH) group. Here, similar to serine and threonine, when histidine, acting as a general base, will accept the protons, the Sulphur (S) in cysteine will become negatively charged and act as a nucleophile. It can then attack the carbonyl carbon of the peptide bond to be cleaved.
Similar to serine, threonine and cysteine also have polar side chain so they can produce nucleophile and can replace the serine residue in the active site.
C) Alanine has a non polar side chain containing a methyl group. This methyl group being non polar is not able to produce a nucleophile specie when histidine acting as a general base catalyst tries to extract proton from methyl. No nucleophile is produced and it fails to attack the carbonyl carbon of the peptide bond, therefore no peptide bond is broken.
While boogie boarding, a woman did a "face plant" in the sand, scraping the skin from her face, and causing bleeding in both eyes. While her face healed, she was left with jaw pain, visual problems, mental confusion, memory loss, and personality changes. Explain what brain regions and/or cranial nerves might have been affected for each symptom caused by the accident.
Answer:
The mentioned symptoms exhibited by the woman in the given question show that damage has occurred in the occipital and the frontal lobe, which is, primarily in the cerebral cortex. The cerebral cortex is the place where memory processing takes place. Thus, any kind of damage to the mentioned parts of the brain will eventually result in personality changes, loss in memory, and mental confusion.
In the given case, the cranial nerves getting affected will be the one taking part in the visual activity, that is, optic nerves and oculomotor nerves. Damage to these cranial nerves would have resulted in the visual conditions as mentioned in the given case.
Place the steps of circulation described below in the correct order, beginning with C.
13. _C_
14. ___
15. ___
16. ___
17. ___
A. De-oxygenated blood exits through the pulmonary artery and travels to the lungs to pick up oxygen and get rid of carbon dioxide
B. Oxygenated blood in the left ventricle leaves through the body’s main artery, the aorta, and travels throughout the body
C. De-oxygenated blood from the body enters the heart through the right atria, then right ventricle
D. Blood delivers oxygen needed for cellular respiration to all cells in the body and removes the CO2 that is created as a result.
E. Newly oxygenated blood is brought back to the heart through the pulmonary veins into the left atria and then down to the left ventricle
Answer:
B,A,E,D
Explanation:
hope this helped ;-}
The correct order of the steps of circulation, beginning with C, is: C, A, E, B, D
Explanation:C. De-oxygenated blood from the body enters the heart through the right atria, then right ventricleA. De-oxygenated blood exits through the pulmonary artery and travels to the lungs to pick up oxygen and get rid of carbon dioxideE. Newly oxygenated blood is brought back to the heart through the pulmonary veins into the left atria and then down to the left ventricleB. Oxygenated blood in the left ventricle leaves through the body’s main artery, the aorta, and travels throughout the bodyD. Blood delivers oxygen needed for cellular respiration to all cells in the body and removes the CO2 that is created as a result1. Flowering plants and bees have influenced each other's evolution, providing an example of
2. The idea that life can develop from non-living things, has been postulated in the
Oparin-Haldane hypothesis and demonstrated to be possible in the Miller-Urey experiment.
3. The idea that eukaryotes developed when prokaryotes were engulfed by other cells, called the
is supported by the fact that mitochondria and chloroplasts have their own DNA.
Answer:
coevolution
abiogenesis
Endosymbiont Theory
It's estimated that over 99 percent of the species that have existed on Earth at some point in time are extinct today.
Coevolution implies that the evolution of one species is dependent on and works in relation to the evolution of another species. This may cause positive or negative impacts and could be beneficial to both organisms or only to one.
This theory states that the first building block of life that allowed for reproduction of organisms was the development of self-replicating RNA. This hasn't been able to be fully demonstrated in any science experiment but is based on the idea that there are RNAs that can catalyze biochemical reactions on their own without proteins.
This process points to the development of structures through the envelopment of smaller cells that perform specific functions. This is how eukaryotic cells evolved from prokaryotic cells.
This experiment was important because it showed that, in the right primordial soup, organic compounds could develop from inorganic compounds.
Explanation:
penn foster
1. coevolution
2. abiogenesis
3. Endosymbiont Theory
What do you mean by coevolution, abiogenesis and endosymbiont theory?Coevolution occurs when two or more species reciprocally affect each other's evolution through the process of natural selection. The term sometimes is used for two traits in the same species affecting each other's evolution, as well as gene-culture coevolution.
Abiogenesis or the origin of life is the natural process by which life has arisen from non-living matter, such as simple organic compounds.
The endosymbiotic theory states that some of the organelles in eukaryotic cells were once prokaryotic microbes. Mitochondria and chloroplasts are the same size as prokaryotic cells and divide by binary fission. Mitochondria and chloroplasts have their own DNA which is circular, not linear.
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As energy is transferred through
an ecosystem, not all trophic
levels receive the same amount.
The diagram shows a marine
ecosystem. Which leopard seal
prey is the most efficient energy
transfer for this ecosystem? *
In a marine ecosystem, the leopard seal's most efficient prey from an energy transfer viewpoint would be the organism closest to it in the trophic level - likely fish or penguins. This is because less energy is lost to the environment each time it moves up a trophic level.
Explanation:In a marine ecosystem, it's important to remember that energy is transferred from one trophic level to the next with a general efficiency of about 10%. This means that for a leopard seal, the prey that would provide the most efficient energy transfer would likely be the one closest to it in the food chain.
Let's say the leopard seal's potential prey includes krill, fish, and penguins. The krill, being primary consumers that eat producers (like phytoplankton), would have the most energy available. Next would be the fish that eat the krill, and then the penguins - secondary and tertiary consumers respectively.
Therefore, the most efficient prey for a leopard seal, from an energy transfer perspective, would be the one closest to it in trophic level - likely the fish or the penguins. This is because less energy is lost to the environment as it moves up each trophic level.
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Homologous recombination occurs in a heterozygote in which alleles D and d differ by a single base pair. The D allele has a G (a GC base pair) at one position, whereas the d allele has a C (a CG base pair) at the same position. If branch migration causes heteroduplex formation across this position, what is the expected outcome?
a. Both D and d alleles will remain unchanged because G can base pair with C.
b. Mismatch repair will identify abnormal G-G and C-C base pairs and may convert the D allele to a d allele or the d allele to a D allele.
c. Mismatches will cause the Holliday junction to be unstable and to resolve by the noncrossover pathway.
d. Mismatch repair will identify abnormal G-G and C-C base pairs and gene conversion will always occur in situations like this one when mismatched bases exist within the heteroduplex region.
e. Mismatch repair will identify an abnormal C-G base pair and will ensure that the cell has two copies of each allele.
Answer:
b. Mismatch repair will identify abnormal G-G and C-C base pairs and may convert the D allele to a d allele or the d allele to a D allele.
Explanation:
The expected outcome if branch migration occurs is that mismatch repair will identify abnormal G-G and C-C base pairs and may convert the D allele to a d allele or the d allele to a D allele.
Final answer:
The correct option is b. Mismatch repair will identify abnormal G-G and C-C base pairs and may convert the D allele to a d allele or the d allele to a D allele. In homologous recombination involving a heterozygote with alleles differing by a single base pair, mismatch repair following branch migration may convert the D allele to a d allele or vice versa, leading to gene conversion.
Explanation:
When homologous recombination occurs in a heterozygote where the alleles D and d differ by a single base pair, the Holliday junction and branch migration can lead to heteroduplex formation across the differing positions. If the D allele has a G (a GC base pair) and the d allele has a C (a CG base pair) at a particular position, branch migration across this position may result in mismatches of G-G and C-C in the heteroduplex region. The process of mismatch repair will recognize these mismatches and can potentially convert a D allele to a d allele or vice versa. This could result in gene conversion, where one allele is changed to match the sequence of the other without affecting the flanking sequences.
Therefore, the correct option is b. Mismatch repair will identify abnormal G-G and C-C base pairs and may convert the D allele to a d allele or the d allele to a D allele.
Statement: Allopatric models of speciation follow Occam’s Razor in the sense that they require minimal assumptions to allow populations to evolve separate traits. For example, once populations are separate, only a single simple process like drift is necessary to drive differentiation (phenetic concept), evolution of novel characters (phylogenetic concept), or even permanent breeding barriers (biological species concept). It doesn’t mean that other factors don’t participate (e.g., selection, mutation), but it doesn’t require them.Sympatric models (which don’t physically separate populations) require more assumptions, usually needing to invoke at least two additional assumptions. These are usually:A) That sister species are each other's closest relative.B) Selection against hybrids and assortative mating.C) Ploidy level shifts
Answer:
B) Selection against cross breeds and assortative mating.
C) Ploidy level movements
Explanation:
Sympatric speciation is typically known as troublesome determination, a kind of common choice wherein two particular varieties in an attribute are supported rather than a transitional type of the characteristic. This speciation mostly happens through polyploidy, in which a posterity or gathering of posterity will be delivered with double the ordinary number of chromosomes.
Science is seldom advanced in isolation. Watson and Crick could not have arrived at their model of the helix without discovery of the structure of and how they are linked together, X-ray crystallography data, or biochemical analysis of the molar proportions of in double-stranded .
Answer:
Correct question= What does the history behind the elucidation of secondary structure of DNA tell you about the role of the scientific community in advancing science? Were Watson and Crick solely responsible?
Explanation
The earlier discovery of nucleotide sequence and base pairing by Levene formed the basis of later discovery and formulation of double Helix DNA structure by Waston and Crick. Thus Levene work contributed to discovery of DNA double helix structure of Waston and Crick
In addition earlier work of Franklin on X-ray crystallography data, and comprehensive research on biochemical analysis of ,molecular proportion of nucleotide by Chargaff, were another supporting bedrock on which Waston and Crick built their discovery of double helix structure of DNA.
Evidently most scientific discoveries were usually arrived at from the earlier work or discoveries, and rarely in isolation.
Which of the following statements is true about living things?
a. The organism’s development is completed before the life cycle begins.
b. Living things are based on a universal genetic code.
c. Cells may change in number but never differentiate.
d. A cell is the smallest unit of an organism that is considered alive.
Answer:
The answer is D) A cell is the smallest unit of an organism that is considered alive.
Explanation:
The statement that a cell is the smallest unit of an organism that is considered alive is true about living things.
What is living things?" Any organism or a life form that possesses or shows the characteristics of life or being alive."The characteristics include: it can grow, move, reproduce, eat food and perform metabolic activities.What is a cell?"The cell is a structural and functional unit of life."All organisms are made up of one or more number of cells.The cell has various organelles that performs various functions like: protein synthesis, replication, formation of energy, etc.
Hence, the correct option is d. A cell is the smallest unit of an organism that is considered alive.
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A change over time in the inherited traits of living things is called????
Please help
Answer:
Evolution
Explanation:
Evolution consists of changes in the heritable traits of a population of organisms as successive generations replace one another
Which of these energy resources is both renewable and sustainable?
A. Coal
B. Biofuel
C. Solar
D. Oil
Answer:
C. Solar
Explanation:
Solar energy doesn't run out as long as the sun is around and it doesn't affect the environment.
Answer:
C. Solar
Explanation:
Suppose that you have just started a summer internship working for a cooperative service, where you will collect soil samples, conduct laboratory and field tests, and make recommendations on soil conservation and agricultural practices.
(i) Identify one biome that is characterized by soil that is rich in humus. Describe how humus originated in the soil of this biome and TWO ways that humus improves conditions for plant growth.
(ii) Describe TWO soil conservation practices that are designed to decrease soil erosion. *
Answer: seen below
Explanation:
I) Deciduous forest: is an example of a biome (a biome is a large naturally occurring community of flora and fauna occupying a major habitat) it's seasonal/tropical and temperate.
Humus which is an organic component of soil originated here because of plant and animal materials decomposition on the forest. Two ways humus improves soil conditions includes;
a) by providing fertility or nutrients for plants and soil organisms.
b) humus also helps the topsoil hold water.
II) Two soil conservation practices that decreases erosion include;
a) Conservation tillage, minimizes the number of tillage passes, where the soil aggregate disruption is reduced, and a minimum of 30% of the soil surface covered with residues, and this aims to reduce by disturbing the soil as little as possible when planting crops.
b) Terracing or division terrace, this helps by shaping the land to create level shelves of earth to decrease soil and water runoff, thereby decreasing erosion.
What is the true about smooth muscle cells?
Answer:
Smooth cells are involuntary non-striated muscle cells.
Explanation:
They line the hollow organs like stomach,intestine,urinary bladder, uterus, arteries. They work automatically. They are of two types:
1. Visceral smooth muscle
2. Multi unit smooth muscle
What does DNA stand for?
Answer:
it stands for deoxyribonucleic acid,
Explanation:
HOPE I HELPED
PLS MARK BRAINLIEST
DESPERATELY TRYING TO LEVEL UP
✌ -ZYLYNN JADE ARDENNE
JUST A RANDOM GIRL WANTING TO HELP PEOPLE!
PEACE!
The pyruvate dehydrogenase (PDH) complex catalyzes the oxidative decarboxylation of pyruvate to acetyl−CoA and CO 2 . CO2. Multiple copies of pyruvate dehydrogenase (E1), dihydrolipoyl transacetylase (E2), and dihydrolipoyl dehydrogenase (E3) along with five cofactors form the PDH complex. Biochemists have studied the PDH complex for decades, in part due to its interesting use of substrate channeling during catalysis. What is the benefit of substrate channeling?
The PDH active site forms in the hydrophobic core of the complex instead of a surface-exposed region.
Reaction progress is not limited by the diffusion constant.
Intermediates of a multistep reaction sequence do not dissociate from the enzyme complex.
The PDH complex sequesters excess substrate to use at later time.
Every intermediate or product made by the PDH complex enters the citric acid cycle as a substrate.
Answer:Intermediates of a multistep reaction sequence do not dissociate from the enzyme complex.
Explanation:
Substrate channeling is the moving of intermediate metabolic product of one enzyme to another enzyme without it be lost to another enzyme.
Channeling makes metabolic pathway more fast and efficient that it will be when the enzyme are at random.
It avoid the use up of intermediate formed by other reaction catalyze by another enzyme.
For pyruvate dehydrogenase channeling prevent the intermediate from dissolving and been used up by other reaction.
Substrate channeling in the PDH complex improves reaction speed, minimizes side reactions, and sequesters reactive intermediates. It raises efficiency by avoiding the limitations of diffusion constants and reducing the dissociation of intermediates.
Explanation:Substrate channeling in the Pyruvate Dehydrogenase (PDH) complex provides several advantages in biochemical reactions. The benefit of substrate channeling lies mainly in three aspects: improvement of reaction speed, minimization of side reactions, and sequestration of reactive intermediates.
Since the active sites of different enzymes are closely aligned in substrate channeling, the intermediates do not need to diffuse out into the aqueous solution before they are used by the next enzyme. This prevents reaction progress from being limited by the diffusion constant and increases the overall reaction speed. Furthermore, it prevents the intermediates of a multi-step reaction sequence from dissociating from the enzyme complex, thus minimizing the chance of side reactions.
Another advantage is that reactive or unstable intermediates are sequestered within the enzyme complex, preventing unwanted reactions with other cellular components. Therefore, substrate channeling enhances the efficiency and control of metabolic pathways.
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