Answer:
It would be C
Explanation:
The most likely explanation for aquatic animals excreting ammonia, and land animals excreting urea or uric acid, is due to ammonia's high toxicity which requires significant amounts of water to dilute, something that is readily available in aquatic environments but not on land.
Explanation:The most likely explanation for the difference in nitrogen waste products excreted by aquatic animals (ammonia) versus land animals (urea or uric acid) is that ammonia is very toxic and requires large amounts of water to dilute it. Aquatic animals are ammonotelic; they excrete ammonia directly into the water, which is a less energy-intensive process and utilizable in an environment where water is abundant. Terrestrial animals, on the other hand, are either ureotelic or uricotelic. Mammals excrete urea, and birds, reptiles, and some terrestrial invertebrates excrete uric acid, primarily because these compounds are less toxic than ammonia and can be safely transported in the body until they can be eliminated. The conversion of ammonia to urea or uric acid requires more energy, but this trade-off is essential for survival on land where water is not as readily available for dilution purposes.
A young couple went to see a genetic counselor because each had a sibling affected with cystic fibrosis. (recessive disease, neither member of the couple is affected. Their four parents are also not affected))(a) What is the probability that the female of this couple is a carrier?(b) What are the chances that their child will be affected with cystic fibrosis?(c) What is the probability that their child will be a carrier of the cystic fibrosis mutation?
Answer: A: I would say a 50% chance that she's the carrier SF
B: 25%
C. The child will most likely be a carrier
Explanation: I'm not really sure if i'm correct but how I figured it out was. Mapping out the family tree and shading the circles if the had SF and if they did not have SF and was a carrier I only shaded half of it. After I finished shading everything out I made a punnet square to predict the probability of the child.
Scientists theorize that all living things are fundamentally alike at the cellular and molecular level. These fundamental similarities are the basis of evolutionary theory: all life shares a common ancestor. If we compare the two cell types, prokaryotic and eukaryotic, there is evidence of these similarities and that evidence includes all EXCEPT:
A) the presence of DNA containing the same four nitrogen bases.
B) the presence of ribosomes capable of preforming protein synthesis.
C) the compartmentalization of the cell due to the presence of organelles.
D) that all life is composed of one or more cells, although the organization of cells does vary.
Prokaryotic and eukaryotic cells share fundamental similarities at the cellular and molecular level. The evidence of these similarities is seen in the presence of DNA, ribosomes, and the fact that all life is composed of cells. The exception is the compartmentalization of cells due to organelles, which is a feature exclusive to eukaryotic cells.
Explanation:The question asks for evidence of fundamental similarities between prokaryotic and eukaryotic cells, with the exception of one option. The options are:
The presence of DNA containing the same four nitrogen basesThe presence of ribosomes capable of performing protein synthesisThe compartmentalization of the cell due to the presence of organellesThat all life is composed of one or more cells, although the organization of cells does varyThe correct answer is C) the compartmentalization of the cell due to the presence of organelles because this is a characteristic unique to eukaryotic cells and not found in prokaryotic cells.
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A transcription factor that binds to a gene first and facilitates binding of other transcription factors is called a(n) ________. A transcription factor that binds to a gene first and facilitates binding of other transcription factors is called a(n) ________. pioneer factor repressor transcription factor transcription factor II D activator transcription factor
Answer:
A transcription factor that binds to a gene first and facilitates binding of other transcription factors is called an activator transcription factor.
Explanation:
Transcription factors are proteins that regulate the transcription of genes.Transcription is the process where a gene's DNA sequence is transcribed into an RNA molecule. Transcription is a key step in using information from a gene to make a protein.
The enzyme RNA polymerase, which makes a new RNA molecule from a DNA template, must attach to the DNA of the gene. It attaches at a spot called the promoter.In eukaryotes, RNA polymerase can attach to the promoter only with the help of basal (general) transcription factors. They are part of the cell's core transcription toolkit, needed for the transcription of any gene.
A typical transcription factor binds to DNA at a certain target sequence. Once it's bound, the transcription factor makes it either harder or easier for RNA polymerase to bind to the promoter of the gene.
Some transcription factors activate transcription, other transcription factors repress transcription.
The transcription factor that binds to a gene first and facilitates binding of other transcription factors is called a Transcription factor II D (TFIID)
To start transcription, a transcription factor (TFIID) is known to be the first to bind to the TATA box. The Binding of TFIID bring up also other transcription factors.
Transcription factors are referred too as proteins that help turn specific genes "on" or "off" by binding to close DNA. They act as activators that boost a gene's transcription.
Transcription factor II D (TFIID) is one of different general transcription factors that set up the RNA polymerase II preinitiation complex.
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Vascular plants have vascular tissues called xylem and phloem that transport materials throughout the plant. Classify these materials according to whether they are transported by the xylem or the phloem. Xylem Phloem
Answer:
Xylem transports water from roots and stems to leaves and Phloem transports food produced from photosynthesis from leaves to roots and stems.
Explanation:
The xylem and the phloem make up the vascular tissue of a plant and transports water, sugars, and other important substances around a plant.Phloem and xylem are closely associated and are usually found right next to one another. One xylem and one phloem are known as a ‘vascular bundle’ and most plants have multiple vascular bundles running the length of their leaves, stems, and roots.
The phloem carries important sugars, organic compounds, and minerals from the leaves to the non photosynthesized part of plant such as roots and stems . The phloem is made from cells two cells
1.sieve-tube members
2.companion cells.
The xylem is responsible for keeping a plant hydrated, it transports water from stems and roots to leaves. Two different types of cells are known to form the xylem
1. tracheids
2. vessel elements.
Xylem transports water and minerals, while phloem transports organic nutrients (such as sugars).
Xylem and phloem are specialized vascular tissues in vascular plants responsible for the transportation of essential substances throughout the plant. Xylem primarily functions in the upward transport of water and minerals absorbed by the roots from the soil. This process, known as transpiration, is driven by evaporation from the leaves, creating a negative pressure that pulls water upward. Xylem vessels, composed of specialized cells called tracheids and vessel elements, facilitate this transport, providing mechanical support to the plant as well.
On the other hand, phloem is responsible for the transport of organic nutrients, primarily sugars produced through photosynthesis in the leaves, to other parts of the plant. This process, called translocation, occurs bidirectionally, allowing for the distribution of sugars from sources (usually leaves) to sinks (areas of active growth or storage). Phloem consists of sieve tubes and companion cells, forming a continuous network throughout the plant.
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Discuss the effects of population size on both theeventual fate of an allele (fixation, loss) and the time to fixation. Describe evidence from your simulations, citing specific examples.
Explanation:
In populace hereditary qualities, obsession is the adjustment in a genetic supply from a circumstance where there exists in any event two variations of a specific quality (allele) in an offered populace to a circumstance where just one of the alleles remains. Genetic drift is an arbitrary procedure that can prompt large changes in populations over a brief time frame. It is produced the repeating little populace sizes by the Random drift , serious decreases in populace size called "bottlenecks" and founder events where another populace begins from few people. There are two significant sorts of hereditary float populace bottlenecks and the organizer impact. A populace bottleneck is the point at which a populace's size turns out to be little rapidly.How does hemoglobin function as a pH buffer?a.Hemoglobin binds hydrogen ions when carbon dioxide exits the red blood cell. b.Hemoglobin releases hydrogen ions after carbon dioxide enters the red blood cell. c.Hemoglobin binds hydrogen ions after carbon dioxide enters the red blood cell. d.Hemoglobin releases hydrogen ions when oxygen exits the red blood cell.
Answer: Option C.
Haemoglobin binds Hydrogen ion after carbondioxide enters red blood cells.
Explanation:
Haemoglobin is the protein in the red blood cells that help to transport oxygen in the blood. It is an iron compound. Haemoglobin acct as buffer by binding to acid or hydrogen ion in the blood when carbondioxide enters the blood, to remove the acid in the blood before it changes the blood pH.
Final answer:
Hemoglobin acts as a pH buffer by binding free hydrogen ions when carbon dioxide is converted into bicarbonate ions within red blood cells, preventing significant pH changes. So, the correct option is c : Hemoglobin binds hydrogen ions after carbon dioxide enters the red blood cell.
Explanation:
Hemoglobin functions as a pH buffer in the bicarbonate buffer system, which plays a crucial role in the body's regulation of blood pH. Carbon dioxide (CO2) enters red blood cells and is converted into carbonic acid (H2CO3) by the enzyme carbonic anhydrase (CA). This unstable intermediate quickly dissociates into bicarbonate ions (HCO3-) and hydrogen ions (H+). The key to buffering is that hemoglobin within the red blood cells binds to these free H+ ions, thereby preventing a drastic shift in pH levels. This buffering capability keeps the blood's pH within a narrow and healthy range. When carbon dioxide enters the red blood cell and is converted into bicarbonate ions, hemoglobin binds hydrogen ions. Therefore, the correct answer to the original question is that hemoglobin functions as a pH buffer by binding hydrogen ions after carbon dioxide enters the red blood cell (option c).
Bacteria are grouped into two catgories reflecting structural features of their cells. Bacteria are classified as Gram-positive or Gram-negative based on whether or not they retain the crystal violet dye used in the Gram stain procedure. Cell wall structure determines the ability to retain the dye, thus cell wall structure is the basis of categorization into Gram-postive (G+) and Gram-negative (G-).
Which of the following occurs only in Gram-negative bacteria?
(A) Peptidoglycan
(B) Integral proteins
(C) Lipoteichoic acid
(D) Lipopolysaccharide
(E) Phospholipids
Answer:
A) Peptidoglycan
D) Lipopolysaccharide
Explanation:
Gram-negative bacteria have a cytoplasmic membrane, a thin peptidoglycan layer, and an outer membrane containing lipopolysaccharide. There is a space between the cytoplasmic membrane and the outer membrane called the periplasmic space or periplasm.
Final answer:
Lipopolysaccharide (LPS)(D) is the compound that occurs only in Gram-negative bacteria, indicating the difference in their cell wall structure compared to Gram-positive bacteria.
Explanation:
Bacteria are classified as Gram-positive or Gram-negative based on their cell wall structure which determines their reaction to the Gram staining procedure. The distinction lies in the presence or absence of certain cellular components. Specifically, lipopolysaccharide (LPS) is a compound that is found only in the outer membrane of Gram-negative bacteria, distinguishing them from Gram-positive bacteria which do not have an outer membrane containing LPS.
Many bacteria and fungi that are parasitic of plants face the daunting task of finding and infecting a new host by airborne spore dispersal followed by germinating upon and then penetrating the leaf surface of their host. What are some of the specific adaptations possessed by some of these parasites to gain access to leaf tissue by entering through stomata thereby evading the plant leaf cuticle? (Hint: search on rust fungi, Uromyces.)
Answer:
Explanation:
The parasites are those organisms which invade the host cell machinery to obtain nutrition, shelter and to reproduce their own progeny.
The fungi are the saprophytic organisms which obtain their food from the dead and decaying organic matter. But the fungi can also be parasitic in nature. The rust fungi produce spores which get attached to the plant surface. These spores germinate over the plant surface. They produce their germ tubes in the stomata of the leaves.
Some of the fungal strains damage the cuticle, and enter into the cells of the host plant. Some of the fungal strain also produce special structures called as haustoria which helps in deriving the nutrients from the plants.
Final answer:
Parasitic fungi and bacteria possess adaptations such as specialized structures called haustoria and appressoria to penetrate plant tissues and evade plant defenses. They infect through natural openings, wounds, and directly through plant cuticles, utilizing the plant's nutrients for their own survival.
Explanation:
Specific Adaptations of Parasitic Fungi and Bacteria
Bacteria and fungi have evolved various adaptations to parasitize and infect plant hosts. Notably, parasitic fungi like rust fungi produce specialized structures known as haustoria which penetrate the cell walls of their host plants. These haustoria make it possible for the fungi to access the cytosol materials such as sugars and amino acids, essential for their survival. The Uromyces species is a prime example of rust fungi that utilizes such mechanisms. Furthermore, these organisms can enter plant tissue through natural openings, wounds, or even directly through the cuticle. In addition to haustoria, pathogens and herbivores may have specialized cells called appressoria which help the fungi to attach and breach the epidermal cells of the plant. Additionally, certain bacteria enter through natural plant openings like stomata or wounds and multiply rapidly within the plant tissues causing cell death. Parasitic adaptation is an evolutionary process, which allows these organisms to resist the host's defense mechanisms and thrive within their living tissues.
Frank has Klinefelter syndrome (47, XXY). His mother has normal skin, but his father has anhidrotic ectodermal dysplasia, an X-linked condition where the skin does not contain sweat glands. Frank has patches of normal skin and patches of skin without sweat glands. Which of the following options is correct? a. Frank received the mutant chromosome from his mother. Nondisjunction occurred in his mother during the second meiotic division b. Frank received the mutant chromosome from his father. Nondisjunction occurred in his father during the first meiotic division. c. Frank received the mutant chromosome from his father. Nondisjunction occurred in his father during the second meiotic division d. Frank received the mutant chromosome from his mother. Nondisjunction occurred in his mother during the first meiotic division. e. Frank received the mutant chromosome from his mother. Nondisjunction occurred in his father during the second meiotic division f. Frank received the mutant chromosome from his father. Nondisjunction occurred in his mother during the first meiotic division.
Answer:
b. Frank received the mutant chromosome from his father. Nondisjunction occurred in his father during the first meiotic division.
Explanation:
As you can see in the question above, Frank has Klinefelter syndrome which causes him to have normal skin patches and skin patches without sweat glands. Her mother has completely normal hair, which may indicate that the defective gene was not supplied by her. In addition, Frank's father has anhydrotic ectodermal dysplasia, an X-linked condition where the skin does not contain sweat glands.
Although Frank's father's defective gene is linked to the X chromosome, it is likely that Frank inherited the defective gene from his country. This may have occurred because during meiosis I, his father's genes did not show disjunction. As a result, Frank presents a mosaic of his phenotype, because an inactivation of the X chromosome occurred.
In avocets, curved beaks result from a Z-linked recessives gene. A curve-beak male is mated with a flat-beaked (normal) female. The F1 are interbred to produce the F2. What is the probability of producing a curve-beak female in the F2
Answer:
50%
Explanation:
Z O
z Zz zo
z Zz zo
The males will have two copies of the alleles of the genes present on the Z chromosomes. As females have only one Z chromosome, they will have only one allele for the genes present on the Z chromosome.
The results from the punnet square show that there will be chance that the females of the F2 progeny will have curved-beaks.
Final answer:
The probability of producing a curve-beak female in the F2 generation when a curve-beak male is mated with a flat-beaked female, and their F1 offspring are interbred, is 25%. This calculation assumes the trait for a curved beak is Z-linked recessive, and it uses the basic principles of genetics and the specific sex chromosome composition of birds.
Explanation:
The question asks about the inheritance of a Z-linked recessive trait in avocets, specifically the probability of producing a curve-beak female in the F2 generation when a curve-beak male is mated with a flat-beaked female. In birds, males have two Z chromosomes (ZZ) and females have one Z and one W chromosome (ZW). A Z-linked recessive trait, like the curved beak in avocets, will manifest in females (ZW) if the Z chromosome carries the recessive allele since there is no corresponding allele on the W chromosome to mask it.
First, consider the parental generation (P): The male is curve-beaked and therefore has the genotype ZZc where c indicates the recessive allele for the curved beak. The flat-beaked female, assuming she does not carry the recessive allele (as it's not expressed and she's normal), would have the genotype ZW.
The F1 generation, resulting from their mating, would consist of males (ZcZ) and females (ZcW), all of which would have flat beaks because the curved beak is recessive.
When these F1 individuals interbreed, the F2 generation could have the following genotypes:
ZZ (flat-beaked males)
ZcZ (flat-beaked males carrying the recessive gene)
ZW (flat-beaked females)
ZcW (curve-beaked females)
The only way to produce a curve-beaked female is from an F1 male (ZcZ) mating with an F1 female (ZcW), resulting in the ZcW genotype. Therefore, the probability of this outcome is 1/4 or 25% since there are four possible combinations (ZZ, ZcZ, ZW, ZcW) and only one produces a curve-beaked female.
"Assume that a scientist is working on a device to route sounds directly to the brain to provide a type of hearing for people who are completely deaf. Which principle of sensory functioning would be most useful to the scientist in achieving her goal?"
Answer:
Sensory localization
Explanation:
Sensory Localization is a principle that explains the ability of humans or animals to determine the origin of a sensory input.
One of the well developed abilities that humans and other animals possess is the ability to determine where a sensory input originates.
The ability to localize a sound, for example, depends on two general mechanisms. The first is relevant for low frequency (i.e low pitch) sounds and includes the fact that sound coming from a particular source arrives at the ear at slightly different times or period . The second mechanism applies to high frequency (i.e high pitch) sounds; if this sound comes from one side, one ear hears it more loudly than the other and one can detect location based on differences in the loudness of the sound at each ear.
Answer: Sensory localization.
Explanation:
The useful sensory functioning to use is sensory localization.
Sensory localization is the ability of humans to know where sensory input or signals comes from. Humans have the ability to to determine the origin of sound. There are two mechanisms, sound localization operate on. The first mechanism is that low frequency or low pitch sounds would arrive at the ear at different times, when it arrives the location can be determined.
The second mechanism is that high pitch sounds when it come from one side, one ear hears it louder than the other and will determine the location.sound localization will be useful to achieve the goal
The evolutionary ancestor of fruit flies had two pairs of wings. Modern fruit flies have one pair of wings and one pair of halteres, an organ involved in balance. A mutation in the modern fruit fly results in conversion of the halteres to wings. The mutant fly has two pairs of wings, so it resembles its evolutionary ancestor. What is the most likely reason the mutation converts one organ to another?
Answer:
Genetic silencing.
Explanation:
The program that commands the expression and development of an organ can be silenced by interruption of a molecular switch, for instance a blocking by a protein of a promoting region. This way it is possible to convert one organ initially silenced to a functional organ.
When _______________ occurs between nonsister chromatids genetic exchange between chromosomes provides new combination of genes that are different from either parent.
Answer:
crossing over
Explanation:
Crossing over occur during meiosis and this involves an exchange of genetic materials between non-sister chromatids in a tetrad. This exchange of genetic materials ensures that the daughter cells produced at the end of meiosis are genetically different from either of the parental cells.
When crossing over occurs between non-sister chromatids genetic exchange between chromosomes provides a new combination of genes that are different from either parent.
Crossing Over:
It is the process by which non-sister chromatids of homologous chromosomes, exchange their genetic material.
Crossing over occurs during prophase I.Crossing over results in the genetic diversity between gametes.Therefore, when crossing over occurs between non-sister chromatids genetic exchange between chromosomes provides a new combination of genes that are different from either parent.
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Suppose that a scientist performed an mRNA mixed copolymer experiment to decode the nonstandard genetic code of a recently discovered species of archaea. The mixture used to synthesize the mRNA contained four parts uracil (U) and one part cytosine (C) nucleotides. The polypeptide translated from this mRNA contained only phenylalanine, serine, leucine, and proline. From previous experiments, the scientist already knew that the organism's genetic code is a triplet code, like the genetic code of humans. Furthermore, the scientist knew that all codons starting with UU code for phenylalanine, UC code for serine, CC code for proline, and CU code for leucine. Using this information, calculate the percentage of amino acids in the polypeptide expected to be composed of proline.
Answer:
The percentage of amino acids in the polypeptide expected to be composed of proline is 4%
Explanation:
Uracil: 80%= 4/5
Cytosine: 20%= 1/5
Proline: CC: 1/5 x 1/5: 1/25= 4%
"Suppose two independently assorting genes are involved in the pathway that determines fruit color in squash. These genes interact with each other to produce the squash colors seen in the grocery store. At the first locus, the W allele codes for a dominant white phenotype, whereas the w allele codes for a colored squash. At the second locus, the allele Y codes for a dominant yellow phenotype, and the allele y codes for a recessive green phenotype. The phenotypes from the first locus will always mask the phenotype produced by the second locus if the dominant allele (W) is present at the first locus. This masking pattern is known as dominant epistasis. A dihybrid squash, Ww Yy, is selfed and produces 320 offspring. How many offspring are expected to have the white, yellow, and green phenotypes
Answer:
White:120 Yellow:30 Green: 10Explanation:
Given;
W codes for a dominant white phenotype;
w codes for a colored squash;
the Y codes for a dominant yellow phenotype;
the y codes for a recessive green phenotype.
The phenotypes from the first locus mask the phenotype produced by the second locus.
A dihybrid squash, of Ww Yy, is self crossed
WwYy * WwYy
F1 are:
WWYY,WWYy, WWyy, WwYy, Wwyy,wwYy, Wwyy and wwyy etc
Due to dominant epistasis, the progeny are:
White:120
Yellow:30
Green: 10
These many offspring are expected to have the white, yellow, and green phenotypes
White: 160, Yellow: 160, Green: 0
The given scenario involves the interaction of two independently assorting genes in determining fruit color in squash. At the first locus, the W allele confers a dominant white phenotype, while the w allele codes for a colored squash. At the second locus, the Y allele results in a dominant yellow phenotype, and the y allele produces a recessive green phenotype. The key aspect is the dominance relationship between the alleles: W masks the expression of w, and Y masks the expression of y.
In the dihybrid squash Ww Yy, the alleles segregate independently, and the possible combinations are WWYY, WWYy, WwYY, WwYy, and so on. However, due to the dominant epistasis described, the presence of at least one dominant allele (W) at the first locus masks the effect of the alleles at the second locus. Therefore, individuals with the genotype Ww Yy and WW Yy will both exhibit the white phenotype, while those with ww Yy and ww yy will display the yellow and green phenotypes, respectively.
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Constants | Periodic Table Suppose your biking partner claims that hyperventilating at the bottom of steep hill climbs is a good idea because it will "increase the O2 saturation of the blood" and thereby provide "more O2 for leg muscles during the climb." Hyperventilation reduces the CO2 content of blood.
Given this fact, is hyperventilation likely to increase O2 delivery to muscles (as your biking partner claims) or not?
Match the words in the left column to the appropriate blanks in the sentences on the right.
Likely, not likely, reduce, reduced, less, increase, more, increased
It is ? to result in increase oxygen delivery to muscles because a drop in CO2 will ? the stability of the T state, which results in a ? P50 and ? O2 release from hemoglobin.
Answer:
It is NOT LIKELY to result in increase O2 delivery to muscles because a drop in CO2 will REDUCE the stability of the T state, which results in a REDUCED P 50 and LESS O2 release from hemoglobin.
Explanation:
Hyperventilation can be defined as a process whereby a person breathes out more Carbon dioxide that the Oxygen they breathe in.
Hyperventilation is caused by being anxious, fear, panic attacks, stress, underlying heart conditions, drug abuse e.t.c.
A person who hyperventilates has a minimal amount of Carbon dioxide present in their blood.
During hyperventilation as well, lower amount of Oxygen would be released by the red blood cells.
Describe one structure or metabolic process that purple bacteria or cyanobacteria might have dispensed with once they became endosymbionts in a eukaryotic cell. Describe one structure or metabolic process that purple bacteria or cyanobacteria might have dispensed with once they became endosymbionts in a eukaryotic cell. One of the most important features the bacteria might have dispensed with is the cell wall. One of the most important features the bacteria might have dispensed with is the gas vacuole. One of the most important features the bacteria might have dispensed with is the flagellum. One of the most important features the bacteria might have dispensed with is cytoplasmic membrane.
Answer:
The answer is photosynthesis or thylakoid region of plasma membrane.
Explanation:
Cyanobacteria are photosynthetic, prokaryotic organisms which are green-blue in colour. They possess a simple structure at sub-cellular level and lack a nucleus. Whereas, eukaryotic cells are more complex, multicellular and have membrane-bound organelles; this includes a nucleus that holds their DNA.
During endosymbyosis with bacteria, the eukaryotic cell develops an organelle called chloroplast, which is then responsible for photosynthesis.
Hence on metabolic process the bacteria might dispense would be photosynthesis, or more specifically its thylakoid region (infolded regions of the plasma membrane responsible for photosynthesis).
In the bacteria that carry this out, dissimilatory nitrate reduction can lead directly to the production of ___________ in these cells. A. amino acids and nucleotides. B. NO2- C. N2 D. NH3 E. None of the above.
Answer: D. NH3
Explanation:
The dissimilatory nitrate reduction is a process which is conducted by the anaerobic bacteria. In this process the nitrate is to reduce to the ammonia. This process is conducted by the bacteria or chemoorganoheterotrophic microbes. These organisms use nitrate as an electron acceptor for the respiration. These organisms organisms oxidize the organic matter and utilize the nitrate as an electron acceptor. The nitrate is further reduced to nitrite and then to ammonia. The ammonia can be found in the cells of these organisms.
Which three statements may correctly explain why the population size increases after time point C?a. Bacteria that acquired a mutation that conferred drug- resistance had a growth advantage over non-resistant bacteria.b. The population increase just after time point C indicates that antibiotic use was discontinued.c. Between time points C and D, drug-resistant bacteria were reproducing faster than non-drug resistant bacteria were dying.d. The few drug-resistant bacteria in the population reproduced, quickly leading to a large drug- resistant population.e. Because the population grew more rapidly after time point C, the bacteria must have acquired a second drug- resistance mutation.
Answer:
A,C, D
Explanation:
Bacteria that acquired a mutation that conferred drug- resistance had a growth advantage over non-resistant bacteria.
Between time points C and D, drug-resistant bacteria were reproducing faster than non-drug resistant bacteria were dying.
The few drug-resistant bacteria in the population reproduced, quickly leading to a large drug- resistant population.
There is no indication that a second antibiotic was used, so it is not possible to assume a second drug-resistance mutation occurred. There is also no indication that antibiotics were discontinued.
Resistant bacteria are the one which fight against the drug or immune system. It is the capacity of bacteria to withstand the effect of antibiotic which are intended to kill them.
The correct statements are :
a. Bacteria that acquired a mutation that conferred drug- resistance had a growth advantage over non-resistant bacteria.
c. Between time points C and D, drug-resistant bacteria were reproducing faster than non-drug resistant bacteria were dying.
d. The few drug-resistant bacteria in the population reproduced, quickly leading to a large drug- resistant population
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What is the genus of plant that is easily identified by having no true leaves, dichotomous branching as well as lobed sporangia (usually yellow in color). a. Marchantia b. Cardiospermum c. Psilotum Ginkgo d. Equisetum e. None of these
Answer:
Option C.
Psilotum.
Explanation:
Psilotum is a genus of whisk ferns. They are vascular plants. They lack true stem and true leaves, their stems is the organ that contain the conducting tissues. They have stems with many branches and synangium with three lobes. The synangia is is fused sporangia that produce spores. They have rhizome which form rhizoids and help to anchor the psilophyte sporophyte.
Answer: The Genus is Psilotum.
Explanation: Psilotum is the genus of fern-like vascular plants, commonly known as whisk ferns. It is one of two genera in the family Psilotaceae. They are easily identified by having no true leaves, dichotomous branching as well as lobed sporangia.
Image of a Psilotum attached below.
50.Botox, which is used to reduce wrinkles for a limited time interval, inhibits membrane fusion by targeting a protein that allows overcoming the repulsive force of the membranes. Based on this information, what is the protein that is inhibited by Botox?
Answer: Botox functions by inhibiting the function of SNARE protein.
Explanation:
The primary function of the SNARE protein is to mediate the vesicle fusion, which shows the fusion of the target membrane bound compartment with the vesicle.
The botox functions by inhibiting the function of this protein. This is a drug which weakens or paralyzes the muscles.
in small doses it is used to reduce the wrinkles on the face. This drug is made of bolulinum toxin which acts on the SNARE protein.
An example of anabolism is A. glucose oxidation to pyruvate. B. breakdown of starch or glycogen to glucose. C. oxidation of fats to carbon dioxide and water. D. NONE of A-D are examples of anabolism E. DNA replication.
Answer: Option E. DNA replication
Explanation:
DNA replication is described as anabolism because it involves the synthesis of newly synthesized strands of DNA from the double stranded parental DNA units.
Thus, since complex molecules are formed from simpler unit, DNA replication is an example of anabolism.
If selfing starts in a previously randomly mating population composed of 0.50 heterozygotes, what is the frequency of heterozygotes after three generations of selfing
Answer:
So after the third generation of selfing, 0.0625 population will be hetero zygous.
Explanation:
Selfing or in another term self-breeding lead to the loss of heterozygosity which ultimately decreased the genetic variation of any population. Selfing in each successive generation cause the loss of 50% heterozygosity from parental one
if a randomly mating population start selfing with 50% heterozygosity then after three generations following result will be obtained
F1-0.25%
F2-0.125%
F3-0.0625%
Final answer:
After three generations of selfing, starting with a heterozygote frequency of 0.50, the frequency of heterozygotes drops to 12.5%, illustrating the effect of self-fertilization on reducing genetic diversity by halving the proportion of heterozygotes each generation.
Explanation:
The question looks into the concept of self-fertilization and its impact on the proportion of heterozygotes within a population over generations. Given a starting frequency of 0.50 heterozygotes in a randomly mating population, the effect of three generations of selfing (self-fertilization) is quantified. Through selfing, the proportion of heterozygotes is halved each generation, significantly altering the genetic makeup of the population.
Using the formula (½)ⁿ for determining the frequency of heterozygotes after n generations of selfing, where n is the number of generations, we can calculate the expected proportion of heterozygotes after three generations. For three generations (½)³ = 0.50³ = 0.125 or 12.5%. This calculation demonstrates how self-fertilization dramatically decreases heterozygosity, increasing homozygosity.
Functional groups confer specific chemical properties to the molecules of which they are a part. In this activity, you will identify which compounds exhibit certain chemical properties as well as examples of those six different compounds. Drag one molecule and one chemical property to each bin. If one property can apply to more than one functional group, choose the best answer for each functional group.
Answer:
1) ALCOHOL: compound with O-H,
chemical property: is highly polar and may act as weak acid
2)CARBOXYLIC ACID: compound with COOH
chemical property:acts as an acid
3)ALDEHYDE: compound with C=O
chemical property: maybe a structural isomer of a ketone
4)THIOL: compound with S-H
chemical property: forms disulphide bonds.
5)AMINE: compound with N bonded to two H
chemical property: acts as a base
6)ORGANIC PHOSPHATE: compound with P bonded to four oxygen.
chemical property: contributes negative charge.
Functional groups, which impart specific properties to organic molecules, include Alcohol, Aldehyde, Ketone, Carboxylic Acid, Amine, and Ester. These groups influence characteristics such as reactivity, acidity/basicity, hydrogen bonding, and hydrolysis.
Explanation:The functional groups determine the properties of organic molecules. Let's look at six examples:
Alcohol (-OH) - Alcohols have the property of forming hydrogen bonds, leading to higher boiling points. Aldehyde (-CHO) - Aldehydes are reactive due to the presence of a carbon-oxygen double bond. For instance, they readily undergo oxidation to form carboxylic acids. Ketone (-CO-) - Like the aldehydes, ketones are also characterized by a carbon-oxygen double bond. But they're less reactive as it is not at the end of the chain. Carboxylic Acid (-COOH) - These are acidic because they can lose a proton from the -OH group. Amine (-NH2) - Amines can accept a proton and thus be basic in nature due to the lone pair of electrons on the nitrogen atom. Ester (-COO-) - Esters are generally neutral compounds, but they can be hydrolyzed to form alcohols and carboxylic acids. Learn more about Functional Groups here:
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In the lungs, oxygen diffuses into the blood and is loaded onto hemoglobin for transport. In the tissues, oxygen is unloaded from hemoglobin and diffuses from the blood into nearby cells. What drives the diffusion of oxygen?a. concentration of ozone b. blood pH c. body temperature d. partial pressure of oxygen e. partial pressure of carbon dioxide
Answer:
D. partial pressure of oxygen
Explanation:
Partial pressure refers to the force applied by an individual gas per unit area in a mixture of gases which is measured in mm of Hg. The partial pressure determines the movement of the gas molecule in the system.
In the given scenario, the oxygen transport through the blood along with other gases like carbon dioxide and other minute concentration but the gaseous exchange between the lungs and the blood, between the blood and the tissues depends on the partial pressure of the oxygen. The oxygen moves from the higher pressure to lower pressure therefore partial pressure is the determining factor.
Thus, Option-D is correct.
The driving force behind the diffusion of oxygen in the lungs and tissues is the partial pressure of oxygen. Oxygen diffuses from areas of higher concentration to areas of lower concentration, loading onto hemoglobin in the lungs and unloading from it in the cells.
Explanation:In the lungs and tissues, the driving force behind the diffusion of oxygen is the partial pressure of oxygen. Oxygen diffuses from an area of higher partial pressure to an area of lower partial pressure. In the lungs, where oxygen concentration is high, it diffuses from the air into the blood. In the tissues, where oxygen concentration is low, oxygen detaches or is unloaded from hemoglobin and diffuses into the cells. The same principle applies to carbon dioxide, but in reverse - it diffuses from areas of higher concentration in the cells to areas of lower concentration in the blood, and then is exhaled from the lungs.
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Consider a locus with two alleles - A and a. These alleles are codominant, meaning that the fitness of the heterozygote is halfway between either homozygote. Consider further a population of randomly mating green frogs where the genotype counts are AA = 500, Aa = 250, and aa = 250. In this population the relative fitnesses of each genotype are AA = 1.00, Aa = 0.80, and aa = 0.60. What is the mean realtive fitness within this population? Please give your answer to two decimal places.
Consider a locus with two alleles - A and a. These alleles are codominant, meaning that the fitness of the heterozygote is halfway between either homozygote. Consider further a population of randomly mating green frogs where the genotype counts are AA = 500, Aa = 250, and aa = 250. In this population the relative fitnesses of each genotype are AA = 1.00, Aa = 0.80, and aa = 0.60. What is the expected allele frequency change for A after one generation with selection? Please give your answer to two decimal places.
Final answer:
The mean relative fitness of the population is 0.85. The expected allele frequency change for allele A after one generation with selection is a decrease to approximately 0.62 due to the relative fitness differences among the genotypes.
Explanation:
The question at hand requires an understanding of genetic structure, allele frequency, and the Hardy-Weinberg principle to calculate both the mean relative fitness within a population and the expected allele frequency change for allele A after one generation with selection.
Mean Relative Fitness Calculation:
To calculate the mean relative fitness (w), we multiply the relative fitness of each genotype by its proportional representation in the population and sum the results:
w(AA) = 1.00 × (500/1000) = 0.50,w(Aa) = 0.80 × (250/1000) = 0.20,w(aa) = 0.60 × (250/1000) = 0.15.The sum is the mean relative fitness: wμ = 0.50 + 0.20 + 0.15 = 0.85.
Expected Allele Frequency Change:
The next step is to calculate the expected change in allele frequency using the Hardy-Weinberg equation. First, determine the frequency of the alleles (p for A and q for a). The current frequency of A (p) = (2×500 + 250) / (2×1000) = 0.625. The frequency of a (q) = 1 - p = 0.375. To calculate the expected frequency of allele A after one generation of selection, we must account for the relative fitness of each genotype:
p' = (p² × w(AA) + pq × w(Aa)) / wμ,p' = (0.625² × 1.00 + 0.625 × 0.375 × 0.80) / 0.85 ≈ 0.62.Hence, the expected frequency of A after one generation is approximately 0.62, indicating a slight decrease due to selection.
The mean relative fitness within the population is 0.85, rounded to two decimal places.
To find the mean relative fitness within the population, we need to calculate the weighted average of the relative fitnesses of each genotype, considering their frequencies.
Given:
- AA genotype count [tex](n_AA)[/tex] = 500
- Aa genotype count [tex](n_Aa)[/tex] = 250
- aa genotype count [tex](n_aa)[/tex] = 250
- Relative fitness of AA genotype [tex](w_AA)[/tex] = 1.00
- Relative fitness of Aa genotype [tex](w_Aa)[/tex] = 0.80
- Relative fitness of aa genotype [tex](w_aa)[/tex] = 0.60
First, calculate the total number of individuals in the population (N):
[tex]\[ N = n_{AA} + n_{Aa} + n_{aa} \][/tex]
[tex]\[ N = 500 + 250 + 250 \][/tex]
[tex]\[ N = 1000 \][/tex]
Next, calculate the contribution of each genotype to the total fitness:
[tex]\[ Contribution_{AA} = n_{AA} \times w_{AA} = 500 \times 1.00 = 500 \][/tex]
[tex]\[ Contribution_{Aa} = n_{Aa} \times w_{Aa} = 250 \times 0.80 = 200 \][/tex]
[tex]\[ Contribution_{aa} = n_{aa} \times w_{aa} = 250 \times 0.60 = 150 \][/tex]
Now, sum the contributions of all genotypes:
[tex]\[ Total \, Contribution = Contribution_{AA} + Contribution_{Aa} + Contribution_{aa} \][/tex]
[tex]\[ Total \, Contribution = 500 + 200 + 150 = 850 \][/tex]
Finally, calculate the mean relative fitness (w_mean):
[tex]\[ w_{mean} = \frac{Total \, Contribution}{N} \][/tex]
[tex]\[ w_{mean} = \frac{850}{1000} \][/tex]
[tex]\[ w_{mean} = 0.85 \][/tex]
So, the mean relative fitness within this population is 0.85, rounded to two decimal places.
3. The peppered moth provides an example of , as the predominant color of moths
changed over time as pollution dictated which color was best camouflaged for protection from
predators.
Answer:
Natural selection
Explanation:
When there was no pollution, the light color of tree bark provided excellent camouflage for the peppered moth, which was predominantly light in color. Some black peppered moths did exist, although they were rare.
During the industrial revolution when there was lots of soot around, the trees became a darker color. This means that now they were a better camouflage for the black moths. As a result, the black moths had a survival advantage, and were much more likely to survive than the lighter colored moths.
As a result, they were more likely to reach reproductive age and pass their genes on to the next generation, leading to an increase in the frequency of black moths. This is an example of natural selection, and illustrates how a species can adapt to its environment,
What is the function of rooting hormone?
A Encouraging a cutting to develop roots
B Encouraging successful budding
C Encouraging successful grafting
D Encouraging bulbs to grow
Answer:
Option a
Explanation:
For cuttings used in propagation to grow root faster, rooting hormones are usually used.due to the fact that the plant root hormones have not been produced, the rooting hormones added facilities the growth of root till the plant root hormones are produced.this lead to strong root and faster growth of plants. Rooting hormones which has auxis present inside of it helps in cell length growth. Rooting hormones which also help increase auxin concentration in plants leads to faster and stronger plant growth.using when it is placed or applied into the cuttings, it will help produce root cell via(through) the replacing of stem cell in plant with undifferentiated cell ( that form the root cell)
For each of the following characteristics, indicate whether the trait is common to Phylum Arthropoda or specific to certain classes of arthropods: wings, chewing mouthparts, jointed appendances, number of legs, segmented bodies, exoskeleton.
Answer: wings: Insecta class
Chewing mouthparts: some orders in the insects class;
Jointed appendages: common to all athropods
Number of legs: not all the same in some class and even orders
Segmented bodies: common to all
Exoskeleton: common to all
Explanation:
Memeberss in the phylum athropoda all have exoskeleton with itsain component being chitin. They all have segmented bodies; some head, thorax and abdomen like insects, some head, cephalothorax and abdomen like in the crustaceans. All have jointed appendages. Some class possess wings eg insecta and some possess chewing mouthparts egg grasshopers in the class insecta
Wings and chewing mouthparts are traits specific to certain classes within Phylum Arthropoda (e.g., Insecta). Jointed appendages, segmented bodies, exoskeleton, and a fixed number of legs (typically eight) are characteristics common to all members of Phylum Arthropoda.
The characteristics common to Phylum Arthropoda and those specific to certain classes of arthropods are as follows:
1. Wings: Wings are not common to all members of Phylum Arthropoda. They are specific to certain classes within the phylum, such as the Insecta (insects) and Pterygota (winged insects). Therefore, this trait is specific to certain classes of arthropods.
2. Chewing mouthparts: Chewing mouthparts are a trait found in many arthropods, but not all. For example, insects like grasshoppers have chewing mouthparts, while other arthropods like mosquitoes have piercing-sucking mouthparts. This trait is more common in certain orders and families within the class Insecta, making it specific to certain classes of arthropods.
3. Jointed appendages: Jointed appendages are a defining characteristic of the Phylum Arthropoda. All arthropods, regardless of their class, possess jointed appendages. This trait is common to the entire phylum.
4. Number of legs: The number of legs can vary widely among arthropods. While insects typically have six legs, arachnids have eight, and myriapods can have many more. Therefore, the number of legs is specific to certain classes of arthropods.
5. Segmented bodies: A segmented body is a common trait among all arthropods. The degree of segmentation and the function of those segments may vary, but the presence of a segmented body is a characteristic of the Phylum Arthropoda as a whole.
6. Exoskeleton: An exoskeleton is a trait common to all members of Phylum Arthropoda. The exoskeleton provides support and protection and is a key feature that defines the phylum.
Wild-type bacteria can grow on minimal medium. Four mutants that cannot grow on minimal medium but can grow on minimal medium supplemented with the nutrient "H" are isolated. It is suspected that metabolites T, P, and A are in the biochemical pathway for synthesis of H, so each mutant is tested for the ability to grow on minimal medium supplemented with these metabolites:
A. Mutant 1: can grow on minimal medium supplemented with T, but not P or A
B. Mutant 2: is unable to grow on minimal medium supplemented with T, P, or A
C. Mutant 3: is able to grow on minimal medium supplemented with A or T, but not P
D. Mutant 4: can grow on minimal medium supplemented with T, P, or A.
Answer:
Mutant 4: can grow on minimal medium supplemented with T, P, or A.Explanation:
In minimal medium, those bacteria who lacks some enzyme which are necessary for the synthesis of all their required compounds can not grow. They require these metabolites supplemented in minimal medium. In this case , if any of these 3 metabolites T,P,A which are required for the synthesis of H are absent in minimal medium,then there will be no bacteria growth.
The use of selective media to test growth of mutants with specific nutritional requirements helps to understand the biosynthetic pathways and elucidate the genetic defects in metabolic genes. This resembles Beadle and Tatum's experiments and illuminates the concept of auxotrophs and the one gene-one enzyme hypothesis.
Explanation:The question revolves around the principles of biochemical genetics, where wild-type bacteria that can grow on minimal medium are contrasted with four mutant strains that require a nutrient (H) supplementation for growth. The growth behavior of these mutants on media supplemented with metabolites T, P, and A allows us to deduce the order of steps in the metabolic pathway for H synthesis. Mutants that grow on media with a specific supplement likely have a functional gene for the metabolic step before the supplemented compound but have a mutation in the gene responsible for utilizing the compound to make the subsequent metabolite in the pathway.
For instance, based on the given data:
Mutant 1: Can grow on minimal medium with T suggests it has a downstream block after T.Mutant 2: Unable to grow on any supplemented medium implies a mutation early in the pathway before T.Mutant 3: Can grow on A or T, but not P, indicating a block just before P in the pathway.Mutant 4: Able to grow on all supplements, so the block is after A in the pathway.This assessment of mutant growth on selective media is reminiscent of Beadle and Tatum's experiments, which contributed to the one gene-one enzyme hypothesis. Such experiments are essential for understanding metabolic pathways and identifying specific genetic defects in auxotrophs, organisms that have lost the ability to synthesize certain compounds due to mutations.