Answer:
C
Explanation:
The energy from sunlight is used to split water molecule into H+ and O- and the emanating electron used to replace the lost electron at the reactive center of the Photosystem I & II. The lost electron is a high energy electron that is transferred from pigment to pigment in the Photosystem. As it does, its energy is harnessed and used to pump H+ protons into the lumen of the thylakoid from the stroma (creating a proton motive force that will be used to make ATPs). Ultimately this electron will then reduce NADP to NADPH.
The process that is most directly driven by light energy is ; ( C ) Transfer of energy from pigment molecule to pigment molecule
The light energy derived from the sunlight or a corresponding light source splits the H⁺ and O⁻ contained in water molecules. The split O⁻ replaces the electron which is lost at the center of photosystem I and II.
The lost electron at the reactive center contains energy while it is been transferred from one pigment molecule to another pigment molecule and this driven directly by the light energy.
Hence we can conclude that the process that is most directly driven by light energy is Transfer of energy from pigment molecule to pigment molecule.
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Are microfilaments and lybosomes in both plant and animal cells?
Answer:
NO
Explanation:
The animal cell and plant cell are both eukaryotic cells. Eukaryotic cells are the type of cells that contains membrane-bound organelles, such as the nucleus, mitochondria, Golgi apparatus and a cytoskeleton.
The cytoskeleton, which is composed of microfilaments, internal filaments, and microtubules, is present in both the animal and plant cells.
However, some of the organelles that are not present in both the animal cell and the plant cell are centrosome and lysosomes.
The lysosome is present only in the animal cell.
Prokaryotes have their chromosomes located in an area called the nucleoid.
a. True
b. False
Answer:
True
Explanation:
Unlike eukaryotic cells, which have a nucleus that contains the genome and is separated from the cytoplasm by a membrane, the prokaryotic nucleoid is not membrane-bound and is not considered an organelle. The nucleoid is simply the area within a prokaryiotic cell where its DNA is located.
A female cat in heat
urinates moreoften and in many places. Male cats congregate near
the urinedeposits and fight with each other. Which of the following
is aproximate cause of this behavior of increasedurination?
It announces tothe males that she is in heat.
Female cats thatdid this in the past attracted
more males.
It is a resultof hormonal changes associated with
her reproductivecycle.
All of the aboveare possible proximate causes of
the behavior.
Answer:
All of the above are possible proximate causes of the behavior
Explanation:
Cats have reproductive organs, these organs go through a normal cycle, called heat cycle, that let reproduction to occur. During this cycle, they have hormonal fluctuations that mean that their body is ready to breed.
During the heat cycle, cats have several behaviors and signs such as spraying urine, rolling on the floor and scratching at doors begging to go outside. When a female cat is in heat, she is likely to urine more frequently and even spray urine on objects, to let any nearby male cats know that she is ready to be mated with. The urine contains pheromones and hormones, signals of her reproductive status to other male cats.
In the wild, females cats pee on trees or objects to find a male cat to mate with. This is why one might infer that female cats did this in the past and the domestic cats inherited this behavior.
Irradiated food may be dangerous because it contains low levels of radioactive compounds.
a. True
b. False
Answer: False
Explanation:
Food irradiation is a process in which the food is exposed to the treatment of radiations. This helps in increasing shelf life and decreasing the colonization of the pathogens on the food. This process does not make the food radioactive.
Imagine you have three test tubes containing identical solutions of purified, double-stranded human DNA. You expose the DNA in tube 1 to an agent that breaks the sugar-phosphate (phosphodiester) bonds. You expose the DNA in tube 2 to an agent that breaks the bonds that attach the bases to the sugars. You expose the DNA in tube 3 to an agent that breaks the hydrogen bonds. After treatment, how would the structures of the molecules in the three tubes differ?
Answer:
DNA is the genetic material of all the living organism except some viruses. The structure of DNA consists of nitrogenous base, pentose sugar and the phosphate group.
Tube 1 - The nucleotide are linked together by phosphodiester bond. As these bonds are broken, the structure of DNA contains the individual complementary nucleotide are linked together by hydrogen bonds.
Tube 2 - The bond between the sugar and the bases are broken. The DNA structure consists of the phosphate group chain only with any pairing with the base.
Tube 3 - The DNA nitrogenous bases are adenine, guanine, thymine and cytosine that are linked together by hydrogen bonds. The agents break the hydrogen bonds. Now, the DNA structure consists of the single strand only.
A functional group on an amino acid that is polar and can become positively charged: ___________
Answer:
Amino group (NH3)
Explanation:
The amino acids have a central carbon atom to which four functional groups are bonded. These are namely a carboxyl group (COOH), an amino group (NH3), a hydrogen atom and an R group (it varies for different amino acids).
Amino group (NH3) is common to all the standard 20 amino acids. NH3 is a polar group since higher electronegativity of nitrogen atom allows it to pull the shared electrons of the covalent bonds towards itself. It makes the nitrogen atom of amino group partially negative and the hydrogen atoms carry a partial positive charge (a dipole is present).
Amino group (NH3) accepts one proton and becomes positively charged (NH4+).
The polar functional group that can become positively charged on an amino acid is typically found in basic amino acids like lysine and arginine.
Explanation:The functional group on an amino acid that is polar and can become positively charged is often found in basic amino acids. Basic amino acids such as lysine and arginine have side chains that contain a positive charge under physiological conditions. These amino acids are essential in the structure and biological activity of proteins. For example, in the classification of amino acids, those like lysine (Lys) and arginine (Arg) are recognized for the basic groups in their side chains which can participate in hydrogen bonding and other important biochemical interactions.
The functional group on an amino acid that is polar and can become positively charged is the amino group (-NH2). This group contains a nitrogen atom, which can accept a hydrogen ion (H+) to form a positively charged ammonium group (-NH3+). Amino acids that have an amino group in their side chain, such as lysine and arginine, are considered basic amino acids because they can donate a proton and become positively charged.
Discuss the evolutionary significance of increasing complexity from unicellular to multicellular organisation?
Answer:
Explanation:
A unicellular organism has only a single cell thus is unable to perform the diversity of functions. The complexity in multi-cellular organisms is more in terms of structure as well as in function thus the evolution of the multi-cellular organism is significant as as the time passes the changes in the morphology, physiology and genetic make up of the organisms will become appreciable.
Lab Report: Using a Microscope Answer the questions below. When you are finished, submit this assignment to your teacher by the due date for full credit. Total score: Click or tap here to enter text. of 10 points (Score for Question 1: Click or tap here to enter text. of 3 points) 1. How do you calculate magnification on a microscope? Answer: Type your answer here. Click or tap here to enter text. (Score for Question 2: Click or tap here to enter text. of 4 points) 2. Explain how to focus a microscope using the high power lens. Include any safety issues you need to be aware of.
To calculate magnification on a microscope, multiply the magnification of the objective lens by the magnification of the eyepiece lens. To focus a microscope using the high-power lens, start by locating the specimen with the low-power lens and then switch to the high-power lens for clearer focus. Safety precautions include proper handling, avoiding touching the lenses, and caution with the fine focus knob.
Explanation:To calculate magnification on a microscope, you need to multiply the magnification of the objective lens by the magnification of the eyepiece lens. For example, if the objective lens has a magnification of 10x and the eyepiece lens has a magnification of 20x, the total magnification would be 10x multiplied by 20x, which equals 200x.
To focus a microscope using the high-power lens, start by placing the slide on the stage and using the low-power lens to locate the specimen. Then, adjust the fine focus knob to bring the specimen into sharp focus. Once the specimen is in focus, switch to the high-power lens and use the fine focus knob again to make any necessary adjustments for a clearer image. Safety issues to be aware of include ensuring proper handling of the microscope to prevent damage, avoiding touching the lenses with fingers, and using caution when manipulating the fine focus knob to avoid breaking the slide.
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To calculate the magnification on a microscope, you multiply the power of the objective lens by that of the eyepiece. When focusing a microscope using a high power lens, adjust the coarse and fine focus knobs and avoid physical contact with the lenses to prevent damage.
Explanation:To calculate magnification on a microscope, you multiply the magnification power of the objective lens by the magnification power of the eyepiece lens. For instance, if the objective lens is 10x and the eyepiece is 40x, the total magnification is 400x.
Focusing a microscope using the high power lens, requires adjusting the coarse and fine knobs while observing through the eyepiece. Start with the lowest power lens and gradually switch to the high power lens. Ensure that you don't touch the lens with your fingers and avoid knocking the microscope to prevent damage. Additionally, use oil immersion with high power objectives for enhancing the resolving power.
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We think that dwarfism in river birch might be a simple Mendelian trait. We have taken a pure breeding normal plant and crossed it with a a pure breeding dwarf plant. The resulting F1 plants were normal height. The F1 plants were selfed and the F2 data is presented below: Normal River Birch 811 Dwarf River Birch 261 Perform a chi square analysis on this data using the following hypothesis: The mode of inheritance for height in river birch is simple Mendelian. What is the calculated chi-square value? Round to three decimal places.
The calculated chi-square value for the analysis is approximately 0.244.
To perform a chi-square analysis, we need to compare the observed frequencies in the F2 generation with the expected frequencies based on the Mendelian inheritance hypothesis. Let's proceed with the calculations:
Observed Frequencies (O):
Normal River Birch (observed) = 811
Dwarf River Birch (observed) = 261
Expected Frequencies (E):
Based on Mendelian inheritance, if we cross a pure breeding normal plant (NN) with a pure breeding dwarf plant (nn), all F1 plants would be heterozygous (Nn), and in the F2 generation, we would expect a phenotypic ratio of 3 normal (N-) : 1 dwarf (nn).
Total F2 plants = 811 + 261 = 1072
Expected normal = (3/4) * 1072 ≈ 804
Expected dwarf = (1/4) * 1072 ≈ 268
Calculating the Chi-Square Value:
The chi-square test formula is: χ² = Σ((O - E)² / E)
Calculating for normal:
χ²[tex]_{normal}[/tex] = ((811 - 804)² / 804) = 0.061
Calculating for dwarf:
χ²[tex]_{dwarf}[/tex]= ((261 - 268)² / 268) = 0.183
Adding both chi-square values:
χ²[tex]_{total}[/tex] = χ²[tex]_{normal}[/tex] + χ²[tex]_{dwarf}[/tex] = 0.061 + 0.183 = 0.244
Rounding the calculated chi-square value to three decimal places:
χ²[tex]_{total}[/tex] ≈ 0.244
The calculated chi-square value for the analysis is approximately 0.244.
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The calculated chi-square value for the inheritance of height in river birch, based on the given F2 data, is 0.245. This value supports the hypothesis that height follows a simple Mendelian inheritance pattern.
Explanation:To determine if the mode of inheritance for height in river birch is a simple Mendelian trait, we perform a chi-square analysis on the F2 generation data. Given that all F1 plants are of normal height, the trait for dwarfism is likely recessive.
If dwarfism follows Mendelian inheritance, we would expect a 3:1 ratio of normal to dwarf plants in the F2 generation. Using the observed numbers of normal (811) and dwarf (261) river birch plants, we calculate the expected values based on a total of 1072 F2 plants (811+261). Expected normal plants would be 3/4 of 1072, and expected dwarf plants would be 1/4 of 1072.
Expected normal = 1072 * 3/4 = 804
Expected dwarf = 1072 * 1/4 = 268
Now we apply the chi-square formula:
χ² = Σ[(Observed - Expected)² / Expected]
For normal = (811 - 804)^2 / 804 = 0.061
For dwarf = (261 - 268)^2 / 268 = 0.184
Calculated chi-square value = 0.061 + 0.184 = 0.245 (rounded to three decimal places).
With respect to angiosperms, which of the following is incorrectly paired with its chromosome count?
a. egg—n
b. megaspore—2n.
c. microspore—n
d. zygote—2n
Answer:
Megaspore—2n.
Explanation:
Angiosperms are the fruit bearing plants and reproduce by the process of sexual reproduction. The chromosome number are specific at each stage of the cell cycle of the angiosperms.
Microspores , egg are haploid. Zygote is diploid in nature. Megaspores get germinate into the female gametophytes and these are haploid in nature. Megaspores are also haploid in angiosperms.
Thus, the correct answer is option (b).
The description 'megaspore—2n' is INCORRECTLY paired with its chromosome count.
In plants (including angiosperms) the egg cell is the haploid (n) female gamete.
After fertilization, the egg cell forms a diploid (2n) zygote that subsequently develops the embryo inside the ovule.
A microspore is a haploid (n) cell that gives a male gametophyte.
In conclusion, the description 'megaspore—2n' is INCORRECTLY paired with its chromosome count.
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Many species can reproduce either asexually or sexually. What might be the evolutionary significance of the switch from asexual to sexual reproduction that occurs in some organisms when the environment becomes unfavorable?
Answer:
When an organisms reproduces asexually its offspring is identical, this means it has the same DNA code.
For example viruses look for specific protein types at cell membranes to "anchor" or "stick" and then inject their genetic material. If all the individuals of a population have the same membrane proteins the whole population is vulnerable to be infected by the virus.
Sexual reproduction creates genetic variability due to mechanisms like crossing over and the assortment of chromosomes during meiosis. On top of it, different parents contain different information which leads to offspring with similar but different characteristics which can end up having different membrane proteins that can save them from the virus.
Final answer:
Sexual reproduction provides genetic diversity essential for adaptation and survival in changing environments, outweighing the rapid growth advantage of asexual reproduction in stable conditions.
Explanation:
The switch from asexual to sexual reproduction during unfavorable environmental conditions has significant evolutionary implications. Sexual reproduction introduces genetic variation through mechanisms such as crossover during prophase I and random assortment at metaphase I, which reshuffle mutations. This variation enhances the ability of a population to adapt and survive in unpredictable or changing environments, thereby potentially leaving more descendants than an otherwise similar asexually reproducing population.
On the other hand, while asexual reproduction allows rapid population growth and successful occupation of stable environments due to genetically identical offspring, it lacks the genetic diversity that can safeguard against drastic environmental changes. Therefore, in unstable conditions, sexual reproduction's advantage in generating genetic diversity becomes critical for the survival and continued evolution of species.
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You have just learned that in much of nature, the structure of a body part or organ is related to its function—the lesson provided you with some examples. For example, a maned wolf has very long legs (structure). You could deduce that those long legs will help the wolf travel more easily in its environment (it lives in grassland with tall grasses). Think about your own favorite animal, plant, or other organism. If you can, include a link to a website with photos or illustrations of this organism. Comment on how that organism’s structure relates to its function.
My favorite animal is the Arctic fox it has very small ears because with big ears very much heat is lost so the small ears cause the fox to lose less heat.
Answer:
Explanation:
Mangrove is a coastal ecosystem that occurs in the transition between land and sea in tropical and subtropical regions of the world, occupying environments flooded by tides such as estuaries, coastal lagoons, bays and deltas. These environments are not necessarily characterized by the mixture between fresh and salt water. The plants that make up the mangrove and dominate the landscape of this ecosystem are the mangroves.
There are adaptations that promote mangrove plants to survive in this very different place and they are: air roots so that they can breathe and exchange oxygen, because de soil has a less quantite about oxygen; there are glandules in its leaves to eliminate the large amount of salts and it is called halophytes; the seeds still germinate attached to the mother plant and are released at a stage of development called the propagule. The propagules accumulate a large amount of nutritional reserves, which allows their survival until they find a suitable place for their fixation.
These are some adaptations that we have on a mangrove ecosystem and it's an important ecosystem that there are so many species like crabs, fishes and a wide variety of animals and food for other animals. But unfortunately this paradise is disappearing and being destroyed by man.
Sex steroids are secreted by the __________ cells of the ovary and the ___________ cells of the testes.
Answer:
The sex steroids are the essential hormones for the proper function and development of the body, they monitor sexual differentiation, sexual behavior patterns, and the secondary sex characters. There are five prime categories of steroid hormones. These are estradiol (estrogen), testosterone (androgen), aldosterone, and cortisol.
Of these testosterone, estradiol, and progesterone comes under the category of sex-steroids. The progesterone and estradiol are produced by the ovarian granulosa cells of the ovary. On the other hand, the testicular Leydig cells are the location of testosterone synthesis.
Is there any reason that meiosis could not occur in an organism whose genome is always haploid?
Answer:
Meiosis is the process of cell division in which a single parent cell divides into four daughter cells. This type of division generally occurs in the gametic cells. This division is also known as reduction division.
The chromosome number is reduced to half in daughter cells in case of meiosis. The meiosis division does not occur in haploid cells because if the meiosis occur in haploid cell (X)there chromosome number reduces upto half (X/2). The half chromosome number in the haploid cell will be fatal for the organism and the organism will die.
Final answer:
Organisms with haploid-dominant life cycles, like fungi and certain algae, ensure genetic diversity without meiosis through mitosis and fertilization, as they are already haploid.
Explanation:
Organisms with a haploid-dominant life cycle, such as fungi and certain algae, do not use meiosis to produce gametes because their bodies are already haploid. Instead, these organisms ensure genetic diversification through the process of mitosis, followed by fertilization. Genetic variation is introduced through mutations and the recombination of genetic material during fertilization, where the fusion of haploid gametes from different organisms restores the diploid condition temporarily, before returning to a haploid state through other cellular processes. This strategy allows for the continuation of genetic diversity without the need for a meiotic reduction from a diploid to a haploid state.
Plant cell wall is mainly made up of
a. nuclic acid
b. colagen
c. starch
d. cellulose
Answer: Cellulose
Explanation:
In plant the cell wall is mainly composed of strong fiber made of carbohydrates polymer known as cellulose.
It is the major component of cotton fiber, wood and other plants. It also helps in paper production.
This substance helps in the protection of the plants from outer environment, also provides strength and rigidity to the plants.
Hence, the correct answer is cellulose.
In eukaryotic cells, transcription cannot begin until
a. the two DNA strands have completely separated and exposed the promoter.
b. several transcription factors have bound to the promoter.
c. the 5 ′ caps are removed from the mRNA.
d. the DNA introns are removed from the template.
Answer:
b. several transcription factors have bound to the promoter.
Explanation:
The initiation of RNA synthesis in eukaryotes includes assembly of RNA polymerase and several transcription factors at the promoter. The binding of TATA-binding proteins (TBP) to the TATA box serves to stabilize the TFIIB-TBP complex at the promoter.
The TFIIB is a transcription factor that is bound to both transcription factor binding proteins and DNA. This is followed by the binding of transcription factor TFIIF and the RNA Pol II enzyme to the TFIIB-TBP complex. Then the other transcription factors such as TFIIE, TFIIB and TFIIH also join the complex. The result is the formation of a closed complex.
Here, the function of TFIIA is to stabilize the TFIIB and TBP at promoter while TFIIB serves in the recruitment of RNA Pol II enzyme to the promoter.
TFIIE facilitates binding of TFIIH and has helicase activity to unwind the DNA duplex while TFIIF serves to prevent the binding of Pol II enzyme to the DNA sequences other than promoters.
Finally, the transcription factor TFIIH phosphorylates the enzyme RNA polymerase II and brings about a conformational change in the whole complex to facilitate the start of transcription.
In eukaryotic cells, the process of transcription, the creation of an RNA copy of a gene sequence, cannot begin until transcription factors have bound to the promoter region of the DNA.
Explanation:In eukaryotic cells, transcription, which is the process of making an RNA copy of a gene sequence, cannot start until several transcription factors have bound to the promoter. This process takes place in the cell nucleus. Preparatory steps include the DNA strands unwinding and the promoter region being exposed, but transcription only actually commences when transcription factors bind to the promoter, effectively marking the start site for RNA synthesis.
The options 'the 5 ′ caps are removed from the mRNA' and 'the DNA introns are removed from the template' pertain to post-transcriptional modifications and mRNA processing, not the start of transcription itself.
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Describe how the relative numbers of PDS and NPDs can be used to establish linkage.
Answer:
Linkage may be defined as the transmission of chromosome to the next generation that are located on same chromosome. Linkage results in the formation of recombination progeny.
PDS (parental ditypes) and non parental diypes can be used to find linkage in the Neurospora and other organism Parental ditypes has the genotype same as parental genotype whereas non parental ditypes contains recombinant progeny. If the ratio and arrangement of PDS and NPDs is 2:4:2 or 2:2:2:2. This means linkage is present between the gene and centromere of an organism.
What three factors does a species diversity index take into account?
a. Species richness, species diversity, and species productivity
b. Species richness, species abundance, and species evenness
c. Species evenness, species abundance, and species stability
d. Species abundance, species stability, and species richness
e. Species productivity, species stability, and species richness
Answer: b. Species richness, species abundance, and species evenness.
Explanation:
The species diversity index can be define as the measure of diversity of species. This takes into account for the number of species present in an area, also the relative abundance of species in an area. If the species richness increases, evenness of the species also increases and the diversity of species also increases.
Answer:
b. Species richness, species abundance, and species evenness
Explanation:
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The self-fertilization of an F1 pea plant produced from a parent plant homozygous for yellow and wrinkled seeds and a parent homozygous for green and round seeds resulted in a pod containing seven F2 peas. (Yellow and round are dominant.) What is the probability that all seven peas in the pod are yellow and round?
The combined probability that a pea will be both yellow and round is 9/16. When considering all 7 peas, this probability is raised to the power of 7, which produces a probability of approximately 0.008 that all 7 peas are yellow and round.
Explanation:The question is about Mendelian genetics, specifically focused on the probability of obtaining a specific phenotype in the F2 generation from a dihybrid cross involving two traits: seed color (yellow or green) and seed texture (round or wrinkled).
For each trait, yellow and round are the dominant phenotypes. According to Mendel, the probability of obtaining a yellow (Y) or round (R) seed in the F2 generation is 3/4. Because the assorting of alleles for color and texture are independent events, the combined probability that a seed will be both yellow and round is calculated by multiplying these probabilities (3/4) x (3/4) = 9/16.
If we want to find the probability of all 7 peas being yellow and round, we raise this probability to the power of 7 ('n' number of peas we are considering in this case). So, (9/16)^7 ≈ 0.008 is the probability that all seven peas in the pod are yellow and round.
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What is the basis for the difference in how the leading and lagging strands of DNA molecules are synthesized?
a. The origins of replication occur only at the 5′ end.
b. Helicases and single-strand binding proteins work at the 5′ end.
c. DNA polymerase can join new nucleotides only to the 3′ end of a pre-existing strand.
d. DNA ligase works only in the 3′ -> 5′ direction
Answer:c. DNA polymerase can join new nucleotides only to the 3′ end of a pre-existing strand.
Explanation:
The double stranded molecule of DNA having the pairing strands in the anti-parallel condition. One strand is remain in configuration as the 5’ to 3’ end, the complementary strand is positioned 3’ to 5’ end.
New DNA strand synthesis occurs from the 3’ to 5’ end as the enzyme DNA polymerase attach itself to the 3’ end of a DNA strand. It will add new nucleotides to the 3' end of the DNA.
Hence, the option (c) is the correct answer.
The primary reason for the difference in the synthesis of leading and lagging strands of DNA is that the DNA polymerase can only add new nucleotides to the 3′ ends of a pre-existing strand. DNA replication always occurs in the 5' to 3' direction, which influences the synthesis of both strands due to the anti-parallel structure of DNA.
Explanation:The basis for the difference in how the leading and lagging strands of DNA molecules are synthesized lies in the manner in which the DNA polymerase enzyme works. The key factor is that DNA polymerase can add new nucleotides only to the 3′ end of a pre-existing DNA strand.
This signifies that DNA replication always takes place in the 5′ -> 3′ direction. Considering the DNA double helix's anti-parallel structure, one strand (leading strand) is synthesized in a continuous fashion, as its orientation allows adding nucleotides at the 3′ end.
Conversely, the other strand (lagging strand) being oriented in 3′ -> 5′ direction, is synthesized discontinuously in small fragments called Okazaki fragments, as nucleotides can't be added to its 5' end. The enzyme DNA ligase then joins these pieces. Options a, b, and d may influence the process, but they're not the main reason for the leading and lagging strands' different synthesis.
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Proteins, which have diverse functions in a cell, are all polymers of the same kinds of monomers amino acids. Write a short essay (100–150 words) that discusses how the structure of amino acids allows this one type of polymer to perform so many functions.
Answer:
There is various kind of proteins found in the human body, which exhibits an array of functions and applications. All the proteins are formed of similar monomers, that is, amino acids, though one can witness different kinds of proteins.
This has been made possible because of the existence of different sequences of amino acids coded for in the DNA. Due to different sequences of amino acids, each protein will fold distinctly and therefore, will exhibit distinct functions inside the body.
Thus, the distinct sequences of amino acids permit for the great diversity of proteins and their functions in the human body even though each of the protein comprises similar kinds of monomers.
Lightly pigmented people are at lower risk of skin cancer that heavily pigmented individuals.
A. True
B. False
Answer:
False
Explanation:
There are a lot of studies corroborating that lightly pigmented people are at higher risk of develop skin cancer than heavily pigmented people.
Explain how the Lac operon is regulated, including all negative and positive components of regulation.
Explanation:
Three enzymes for lactose metabolism are grouped in the lac operon: lacZ (β-galactosidase), lacY (permease) and lacA (trans-acetylase). The transcription of this operon occurs only when lactose is available to digest, presumably to avoid wasting energy. Apart from these protein-coding genes we have P(promoter), O(operator and CBS(Cap-binding site), sequences that work as binding sites for transcriptional regulation.
Negative components: An important regulator is lacI when four of these molecules assemble they form a repressor, this will bind to the promoter, this won't allow the operon to be transcribed as long as the operator is occupied by a repressor. LacI can also prevent lactose to bind to the operator by binding with it and changing its shape.
Positive components: cAMP binding protein is another molecule that when is bound to CBS improves the chance of RNA polymerase to bind to the promoter to initiate transcription.
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Mendel's first law is the Law of Segregation. According to that law and according to other facts that Mendel discovered, what statement or statements are true in the following list?
a. genes behave like fluids rather than like separate particles
b. genes behave like separate particles rather than like fluids
c. the alleles of a given gene separate during the process of meiosis
d. both a. and c. are correct
e. both b. and c. are correct
Answer:
e. both b. and c. are correct
Explanation:
According to Mendel's law of segregation, each genetic trait is regulated by a pair of factors, alleles. These genetic factors are present as distinct particles.
The two alleles of a gene occupy the corresponding position on the homologous chromosomes. The two alleles of a gene are separated from each other during gamete formation due to the separation of homologous chromosomes during anaphase-I of meiosis-I.
This leads to the formation of gametes carrying one allele for each gene. This is the reason that the law of segregation is also called the law of purity of gametes.
Most CO2 from catabolism is released during
a. glycolysis.
b. the citric acid cycle.
c. lactate fermentation.
d. electron transport.
Answer:
The correct answer is option b. "the citric acid cycle".
Explanation:
The citric acid cycle is one of the most important pathways of the cellular metabolism because of its role in energy production and biosynthesis. The citric acid cycle is where most of the CO2 from catabolism is released. This happens by the release of two molecules of CO2 per turn in the cycle, using the two atoms of carbon from acetyl CoA.
The majority of CO2 produced during catabolism is released during the citric acid cycle. Neither glycolysis, lactate fermentation, nor the electron transport chain, emit CO2.
Explanation:The majority of carbon dioxide (CO2) produced during catabolism, the process by which the body breaks down molecules to produce energy, is released during the citric acid cycle, also known as the Krebs cycle or the tricarboxylic acid (TCA) cycle. Glycolysis, the first step of cellular respiration, does not release CO2. It converts glucose into pyruvate, without releasing any CO2. In lactate fermentation, no CO2 is produced either. The electron transport chain, the last step of cellular respiration, also doesn’t produce CO2. It utilizes the electrons and hydrogens gained from the Krebs cycle to create water and ATP, but it doesn’t generate CO2.
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Enzymes that break down DNA catalyze the hydrolysis of the covalent bonds that join nucleotides together. What would happen to DNA molecules treated with these enzymes?
a. The two strands of the double helix would separate.
b. The phosphodiester linkages of the polynucleotide backbone would be broken.
c. The pyrimidines would be separated from the deoxyribose sugars.
d. All bases would be separated from the deoxyribose sugars.
Answer:
The correct answer is option b. "The phosphodiester linkages of the polynucleotide backbone would be broken".
Explanation:
The phosphodiester linkages of the polynucleotide backbone is what binds each nucleotide to each other in the DNA molecules. These linkages are covalent bonds that take place between 3' carbon atom of one sugar molecule and the 5' carbon atom of another. The enzymes that break down DNA catalyze the hydrolysis of the phosphodiester linkage, which results in DNA cleavage within the backbone at specific or unspecific nucleotides.
Final answer:
Enzymes that break down DNA cleave the phosphodiester bonds in the DNA polynucleotide backbone, leading to the degradation of the DNA molecule into its constituent nucleotides. The correct choice is b - the phosphodiester linkages would be broken.
Explanation:
When enzymes that break down DNA are applied to the DNA molecule, they catalyze the hydrolysis of the covalent bonds that join nucleotides together. Specifically, these enzymes target the phosphodiester bonds in the polynucleotide backbone. Hydrolysis of these bonds would lead to a cleavage of the polynucleotide chain. This would result in the degradation of the DNA molecule into its individual nucleotides, where each phosphate group would be associated with a deoxyribose sugar and a nitrogenous base.
The correct answer to the student's question is: b. The phosphodiester linkages of the polynucleotide backbone would be broken. This process is different from denaturation, where hydrogen bonds between complementary bases break; breaking phosphodiester bonds involves a chemical reaction that cleaves the DNA backbone itself.
Enzymes such as nucleases perform this function, specifically endonucleases which cleave phosphodiester linkages within a DNA strand. Conversely, enzymes like DNA ligases can repair these breaks, forming a new phosphodiester linkage, though in this context, we are discussing the breaking of the DNA molecule by hydrolysis, not the joining of fragmented DNA.
Adenine in DNA is complementary to
a. uracil
b. adenine
c. guanine
d. cytosine
e. inosine
Answer:
DNA is also called deoxyribonucleic acid which is made up of two chains which wind around each other to form a double helix model. The 2 DNA strands are also called polynucleotides and they are made up of monomeric units known as nucleotides. These nucleotides are made up of one of four nitrogen-containing nucleobases: cytosine, guanine, adenine and thymine, a phosphate group, and sugar known as deoxyribose.
Nitrogen bases present on the two separate polynucleotides strands are bound together with the help of base pairing (such as adenine with Thymine) and with hydrogen bonds to form double-stranded DNA.
So, adenine in DNA is complementary to thymine.
Lysozyme is frequently found in such extracellular bodily fluids
assaliva and tears. It serves as a function in a first line
ofdefense against bacterial infections; however; it is not
effectiveagainst archaeal invaders. Explain.
Answer:
Lysozyme in the body fluid protects us from bacterial infection by damaging the cell wall of bacteria which ultimately kills the bacteria.
The bacterial cell wall is made up of NAG and NAM units which is joined by β (1-4) glycosidic bond. Lysozyme attacks on this bond and breaks it causing the damage in the cell wall of bacteria which leads to the death of bacteria.
Archaeal cell wall do not contain β (1-4) glycosidic linkage, it contains β(1-3) glycosidic linkage in its cell wall between NAT and NAG unit and lysozyme do not able to break β(1-3) glycosidic linkage between them so lysozyme do not cause any harm to the archaeal invaders.
Storage form of glucose in animal cells is:
a. starch
b. disaccharide
c. glycogen
d. glucose
Answer:
Glycogen. (Ans. C)
Explanation:
Glycogen is known as the storage form of glucose in animal cells. It a polysaccharide of glucose which is multi-functional and serves as an energy form storage in animals. Liver cells primarily stored glycogen, some glycogen also stored in muscle cells for immediate use if needed.
Glycogen molecule composed of many glucose molecules that are linked together with the help of the alpha acetal group. To create energy currency (adenosine triphosphate) glucose is the primary source used by every cell.
The storage form of glucose in animal cells is glycogen, a complex carbohydrate stored mainly in the liver and muscle cells. This is utilised as a source of energy when required.
Explanation:The storage form of glucose in animal cells is glycogen. This polysaccharide is a complex carbohydrate that animals and humans use to store glucose. It is found mainly in the liver and muscle cells. When your body requires energy, it transforms the stored glycogen into glucose which can be used as a source of energy. You can perceive it as animal's equivalent to plant's starch storage mechanism. So, the correct answer to your question is: c) Glycogen.
Learn more about Glucose Storage here:https://brainly.com/question/34701078
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The first stage of cellular respiration (Glycolysis) occurs in _________.
a. inner mitochondrial membrane
b. cytoplasm
c. mitochondrial matrix
d. nucleus
Answer: Option B. Cytoplasm
Explanation:
Cellular respiration is defined as the process in which sugar break down take place into as form and utilized by cell in the form of energy. In Cellular respiration food is utilized to create ATP and further used as energy source by using oxygen, and named aerobic respiration
There are four stages of cellular respiration includes first is glycolysis, second is pyruvate oxidation, third is citric acid cycle, and forth is oxidative phosphorylation.
Glycolysis is the first stage of cellular respiration which takes place in cytoplasm or cytosol.
In the process of Glycolysis six-carbon sugar undergo chemical changes and converted into two pyruvates, each containing three-carbon organic molecule. ATP and NADH are formed in this reaction.
Hence, the process take place in cytoplasm.