Answer:
The correct answer is: aerobic respiration -the root cells metabolize the sugar produced during photosynthesis
Explanation:
From the germination of seeds, roots will depend exclusively on energy in the form of photosynthates supplied from the aerial parts of the plant. Photosynthesis takes place in the leaves. Sugar is produced which is transported to all the other plants via phloem tube.Roots absorbs air from the air spaces present between the soil particles and is able to carry out aerobic respiration. The sugar is metabolized and energy is produced.
Plant roots obtain energy for their metabolic needs primarily through aerobic respiration, a process where the glucose produced from photosynthesis is metabolized using oxygen to produce ATP.
Explanation:The roots of plants obtain energy for their metabolic needs primarily through aerobic respiration. While photosynthesis primarily occurs in leaf cells where sunlight is accessible, the energy stored in the form of glucose travels to all parts of the plant, including roots. During aerobic respiration in the root cells, this glucose is metabolized, or broken down, to generate ATP, the energy currency of cells. This process requires oxygen, which is delivered to the roots from air spaces in the soil and from aboveground parts of the plant.
Beyond aerobic respiration, plants do not typically use oxidative phosphorylation to directly convert soil phosphate to ATP. Anaerobic respiration and fermentation processes can occur in the absence of oxygen, but these are not the main source of energy for root cells. Instead, these processes tend to be more common in microorganisms and in plant or animal cells under conditions of oxygen deprivation.
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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%
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.
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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Information transfer is fundamental to all living organisms. For the following examples, explain in detail how the transfer of information is accomplished.
1. The genetic material from one bacterial cell enters another via transformation, transduction, OR conjugation?
2. The genetic material in one eukaryotic cell is copied and distributed to two identical daughter cells.
3. A gene in a eukaryotic cell is transcribed and translated to produce a protein.
Answer:
Explanation:
We are to explain in details how
genetic material from one bacterial cell enters another via transformation, transduction, OR conjugation.
For transformation, here, there is genetic alteration of a cell. It is a process of horizontal gene transfer where bacteria take up foreign genetic material from the environment.
Transduction is genetic recombination in bacteria where genes from a host cell are carried into the genome of a bacterial virus and then carried to another host during infection cycle.
Conjugation. In this, DNA plasmid is transferred from one bacterium to another through pilus.
2. We explain also in details how genetic material in one eukaryotic is copied and distributed.
Genetic material is copied during DNA replication and genetic material is distributed during mitosis.
The followings take place during DNA replication.
• DNA is copied during interphase, S phase of cell cycle.
• Site origin of DNA is known.
• Unwinding of DNA at the origin
• Synthesis of new strands with enzyme called helicase.
• Formation of replication forks.
• Proteins associated with replication forks help in the initiation and continuation.
• DNA polymerase synthesizes new strands by adding nucleotides that complement each strand.
Mitosis.
• Chromosomes are doubled
• Prophase: phase of condensation of chromosomes, formation of spindle.
• Metaphase: Alignment of chromosomes
• Anaphase: Chromatids are separated
• Telophase: reformation of nuclear membrane, division of cells, cell cycle control.
3. We also explain how transcription and translation of gene in eukaryotic.
Transcription
• Here we have DNA sequence to RNA sequence.
• Occurs in the nucleus
• DNA is carried into nucleosomes.
• RNA polymerase: add complementary RNA nucleotides to DNA template.
• Growth of new strands
• Leads to mRNA processing.
Translation
• Here we have mRNA base sequences to amino acids initiation
• Sequence of events such as complexes, small unit of ribosomes.
• Also we have the first tRNA
• Structure of ribosomes formed.
• Complete description of two subunits, 2 action sites, rRNA and proteins.
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.
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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Mutations in which of the following genes lead to transformations in the identity of entire body parts?A) morphogensB) segmentation genesC) egg-polarity genesD) homeotic genesE) inducers
Answer: Option A
Explanation:
A morphogen gradient is an important concept in case of the developmental biology. It produces the signals from the part of embryo which decides the fate of development of the parts.
The differentiation of the cell takes place by the help of the signals produced by the embryo in order to decide where will which part of the embryo grow.
The surrounding cells receive the signals and the development of entire body parts takes place.
So, the mutation in this gene will transform the entire body part.
Answer: Option A.
Explanation:
Morphogens are signaling molecules that originate from a specific location that spread from their origin via diffussion to form a concentration gradients acting at long distances to induce responses in cells base on the gradients the cells interact with. It determines the pattern of tissue development and morphogens gradients cause the differentiation of unspecialized stem cells into different cell types.
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.
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.
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.
"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
What is the expected Hardy-Weinberg frequency of heterozygotes in a population of green anoles, where the locus under investigation has two alleles and the frequency of one allele is 0.005 (give your answer to five decimal places)?
Answer:
0.00995
Explanation:
If the population is in Hardy Weinberg Equilibrium, the sum total of frequencies of the alleles of a locus would be one. The frequency of one allele for a locus is 0.005. So, the frequency of the other allele for the same locus would be=1-0.005= 0.995
According to the given information, there are only two alleles for the locus in the population. So, the frequency of heterozygous genotype in a population= 2 x frequency of one allele x frequency of other allele= 2 x 0.005 x 0.995= 0.00995
Make a hypothesis about how you think the two species of Paramecium sp. will grow alone and how they will grow when they are grown together.
I hypothesize that Paramecium species will grow faster alone due to reduced competition. When together, resource competition might slow growth.
I expect that when each species of Paramecium is cultured alone, their growth rates will be relatively higher due to the absence of interspecific competition for resources. In a mixed culture, where both species coexist, I anticipate a potential decrease in growth rates compared to individual cultures.
This could be attributed to the sharing of limited resources, leading to heightened competition between the two species. Consequently, the presence of interspecific interactions might result in a more constrained growth environment, affecting the growth trajectories of both Paramecium species when compared to their growth in isolation.
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When grown separately, both Paramecium aurelia and Paramecium caudatum will probably proliferate successfully. However, when they share the same environment and compete for resources, Paramecium aurelia is likely to outcompete Paramecium caudatum. This hypothesis relies on previously observed behaviors, but must still be validated by experimental testing.
Explanation:A hypothesis about the growth of two species of Paramecium, specifically Paramecium aurelia and Paramecium caudatum, could be: When each species is grown separately, they will both proliferate successfully. However, if they are grown together and forced to compete for the same resources, Paramecium aurelia is likely to outcompete Paramecium caudatum. This hypothesis is based on the observations shared in figures 45.24, 36.25, and 19.19 that reveal this competitive dynamic.
Beyond competition, these species might engage in symbiosis, which represents different types of close, long-term interactions between species. Depending on the conditions and specific interactions, these could be commensal (one species benefits, the other is unaffected), mutualistic (both species benefit), or parasitic (one species benefits at the expense of the other).
The interaction between these Paramecium species could potentially mirror other observed species interactions, such as between different groups of minnows in a shared environment or different mussel cultures in a lab. It's important to note however that any hypothesis must be tested experimentally to verify its accuracy and be subjected to further refinement based on findings.
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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).
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.
Which term is not correctly matched with its proper definition? Prototroph: possesses the same metabolic/growth capabilities as the parent, wild type strain. The prefixes chemo- and photo-: define the electron source Macronutrient: phosphorous and sulfur are in this category Chemoorganotroph: a heterotroph could be considered this Growth factor: components, either defined or undefined, that promote/support growth
Answer:
The prefixes chemo- and photo-: define the electron source
Explanation:
The prefix 'chemo' indicates something that is a “chemical” or “chemically induced” while the prefix "photo" relates to light. An electron source is that which produces electrons. The prefixx chemo and photo do not relate to sources of electron production.
The term 'Prototroph' is incorrectly matched with its description. A Prototroph is a microorganism that can synthesize all compounds needed for its growth and does not specifically relate to maintaining the same metabolic/growth capabilities as the parent, wild type strain. Other terms are correctly matched with their definitions.
Explanation:In the list of terms and their corresponding definitions, the term that isn't aptly matched is Prototroph. The definition of Prototroph usually refers to a microorganism that, unlike an auxotroph, does not require any additional nutrient other than a source of organic carbon for growth. It does not specifically relate to maintaining the same metabolic/growth capabilities as the parent, wild type strain.
Specifically, a Prototroph is a strain of microorganism that possesses a complete set of genes, enabling it to synthesize all compounds needed for its growth, thereby not requiring any growth factors. Similarly, a wild strain usually refers to the 'normal' or 'natural' condition of an organism, but it may or may not always have the same growth capabilities as its parent strain.
On the other hand, the terms and definitions of Macronutrient, Chemoorganotroph, Growth factor, and terms with the prefixes chemo- and photo- are correctly matched. These terms are standard in the field of Microbiology.
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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.
Saturated fatty acids are degraded by the stepwise reactions of beta-oxidation, producing acetyl-CoA. Under aerobic conditions, how many ATP molecules would be produced as a consequence of removal of each acetyl-CoA?
Answer:
4 ATP molecules
Explanation:
Normally, about 11 ATP molecules are generated as the result of Beta-Oxidation of saturated fatty acid in Kreb's Cycle. But the total removal of acetyl-CoA under certain aerobic condition decreases the overall yield and approximately 4 ATP molecules comes out from each removal of acetyl-CoA.
Final answer:
Each acetyl-CoA molecule produced from the stepwise degradation of saturated fatty acids through beta-oxidation can yield around 10 ATP molecules.
Explanation:
Saturated fatty acids are degraded through beta-oxidation, which produces acetyl-CoA. Each acetyl-CoA molecule can yield a certain number of ATP molecules when it enters the citric acid cycle and oxidative phosphorylation. One mole of acetyl-CoA metabolized by the citric acid cycle yields about 10 ATP molecules. Therefore, the energy produced by the removal of each acetyl-CoA molecule from the beta-oxidation of saturated fatty acids can be estimated to be around 10 ATP molecules.
The genetic condition Xeroderma pigmentosum, which can lead to skin cancer, results from A. inability to correct UV induced lesions B. inability to process phenylalanine. C. inability to produce functional hemoglobin. D. inability to correct transitions E. breaks in the X chromosome
Answer:
Option A
Explanation:
Xeroderma pigmentosum arises as a result of the cell being unable to correct lesions induced by UV. This can be as a result of mutations in the enzymes which include XP A-E needed for correction of the lesions. Failure to correct these lesions leads to their accumulation and then damage to the cell.
An article in the Los Angeles Times (Dec. 3, 1993) reports that 1 in 200 people carry the defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the approximate distribution of the number who carry this gene
Answer: 5
Explanation:
Since 1 out of 200 people carry the defective gene that causes inherited colon cancer.
In a sample of 1000 individuals, we say
1 = 200
y = 1000
cross multiply
y x 200 = 1000 x 1
200y = 1000
y = 1000/200
y = 5
Thus, in a sample of 1000 individuals, approximately 5 will be carriers
One of Mendel’s Laws says that the probability of inheriting a maternal allele for a gene does not affect the probability of inheriting the maternal allele for another gene. If each gene has two alleles, what is the probability that one individual inherited the paternal allele at three different genes?
Answer:
There's 2 alleles for every quality and there are 3 qualities. The likelihood of acquiring the fatherly allele of those 3 qualities would be 1/2 x 1/2 x 1/2. This is on the grounds that you have a 50/50 possibility of getting the allele from your mom or your dad for three unique alleles. Since you are making sense of the likelihood of acquiring 3 qualities from your dad you duplicate at that point to get 1/8.
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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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)
The cornea is the transparent outer layer of the human eye. Because it must be transparent to light, it does not normally contain blood vessels. Therefore, it must receive its nutrients via diffusion. Oxygen from the surrounding air diffuses to the cornea through the surface tears whereas other nutrients diffuse to the cornea from the inner parts of the eye, such as the vitreous humor and lens.
During operation, the cornea produces waste in the form of CO2 gas that must be expelled to keep the eye healthy and functioning. This is accomplished by the simultaneous diffusion of CO2 from the cornea to the surrounding atmosphere, which generally features a low CO2 concentration.
It is therefore critical that modern contact lens materials allow sufficient diffusion rates of oxygen and carbon dioxide. Without oxygen, the cornea will warp, loose transparency, and become susceptible to scarring. The body may also react by growing additional blood vessels into the eye, which can damage the cornea.
If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, which of the following contact lens / ambient condition modifications is not likely to be useful?
Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.
(a) Increase the contact lens thickness
(b) Increase the diffusivity of oxygen gas by increasing the contact lens porosity
(c) Increase the ambient temperature
(d) Increase the ambient partial pressure of oxygen gas
(e) All of the suggestions (a-d) are useful for increasing the flow rate of oxygen
the answer is not c
Increasing the thickness of a contact lens would decrease the flow rate of oxygen to the cornea, not increase it. Thus, this modification would not be useful if the goal is to increase the steady-state flow rate of oxygen to the cornea.
Explanation:The best answer is (a) Increase the contact lens thickness. In the context of diffusion, the flow rate of a molecule is inversely proportional to the thickness of the layer it has to go through. Increasing the contact lens thickness would effectively reduce the rate at which oxygen reaches the cornea, thereby reducing the flow rate, not increasing it as desired. On the other hand, solutions (b), (c), and (d) could potentially increase the flow rate of oxygen to the cornea. Increasing the diffusivity, ambient temperature, or ambient partial pressure of oxygen gas would all potentially increase the rate of oxygen diffusion to the cornea.
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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.
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.
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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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).
Catecholamines mediate their effects throughI. α-adrenoreceptors.II. β-adrenoreceptors. III. receptor tyrosine kinases.IV. monomeric G protein receptors
Explanation:
Catecholamines are hormones made by your adrenal organs, which are situated on your kidneys. Models incorporate dopamine; norepinephrine; and epinephrine (this used to be called adrenalin or adrenaline). Our adrenal organs send catecholamines into our blood when you're genuinely or sincerely focused. They cause you to inhale quicker, raise your pulse, and send more blood to significant organs, similar to your cerebrum, heart, and kidneys.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.