Answer:
b. they require more oxygen per unit of volume.
Explanation:
Large animals have a specialized organs for exhalation and inhalation of gases due to the requirement of large amount of oxygen for cellular respiration. Respiration is a process in which energy is released in the form of Adenine tri phosphate (ATP) in the mitochondria of the cell with the addition of oxygen. The waste product such as carbondioxide is produced which is removed from the body body through lungs.
Final answer:
Large animals require specialized organs for gas exchange because a. volume increases more rapidly than surface area as size increases as they grow. These organs, along with the circulatory system, ensure adequate oxygen intake and waste removal, essential for survival.
Explanation:
Large animals need specialized organs for gas exchange because a volume increases more rapidly than surface area as size increases. As a cell grows, its volume increases much more than its surface area. Since the surface of the cell is what allows the entry of oxygen, large cells cannot get as much oxygen as they need. This challenge is overcome in large animals that have specialized organs that effectively increase the surface area available for exchange processes, like the lungs, kidneys, and intestines. These organs are complex and have a high surface area relative to their volume, ensuring efficient gas exchange.
Additionally, the circulatory system plays a crucial role in moving materials and heat energy between the surface and the core of the organism, effectively supporting the gas exchange process. Thus, as animals become larger, the importance of having specialized organs becomes essential to maintain a functional surface area-to-volume ratio. This ratio is critical in many aspects of animal development, including the efficiency of muscle mass in supporting skeletons and in the regulation of body heat.
The promoters of positively controlled operons require activator proteins because The promoters of positively controlled operons require activator proteins because they are needed to bind to the allosteric site of RNA polymerase. RNA polymerase easily recognizes the consensus sequence. the promoters bind RNA polymerase weakly and utilize activator proteins to help RNA polymerase recognize the promoter. they are required to inactivate the repressor proteins.
Answer:
the promoters bind RNA polymerase weakly and utilize activator proteins to help RNA polymerase recognize the promoter.
Explanation:
The promoters of positively controlled operons require activator proteins because the promoters bind RNA polymerase weakly and utilize activator proteins to help RNA polymerase recognize the promoter.
In a forest, the nitrates contained in the nucleic acids of a bear can be made available again to primary producers after the bear dies, through the action of: A Vultures that eat fat and muscle from the carcass B Bacteria that decompose organic molecules back to their inorganic components C The toxic mushroom that the bear ate, which killed it D The salmon that are an important food source for the bear
Answer:
Option B
Explanation:
In any food cycle or ecosystem, the top most organisms are the decomposers. They are kept above the tertiary species because when the tertiary species die, these decomposers or micro-organisms feed on them and convert organic material with in them into reusable form/nutrients/chemicals.
These nutrients get mixed with the soil and are reused by the plants and trees (producers) for making food for all other organism in the food chain by entrapping solar energy.
Hence, option B is correct
four parts of a flowering plants and it's function
Answer:
Although all flowers are different, they have several things in common that make up their basic anatomy. The four main parts of a flower are the petals, sepals, stamen, and carpel (sometimes known as a pistil). If a flower has all four of these key parts, it is considered to be a complete flower.
Explanation:
(Hope you have a good day. Stay Safe!)
A man has extra digits (six fingers on each hand and six toes on each foot). His wife and their daughter have the normal number of digits (five fingers on each hand and five toes on each foot.) Having extra digits is a dominant trait. The couple's second child has extra digits. What is the probability that their next (third) child will have extra digits?
a. 3/4
b. 1/16
c. 1/2
d. 1/8
Answer:
c. 1/2
Explanation:
Polydactyly, the presence of extra digits on hands and feet, possesses an autosomal dominant mode of inheritance. To calculate the probability of the third child being polydactyl, we must first figure out the genotypes of the parents.Since polydactyly is a dominant trait, the mother must be homozygous recessive as she does not have extra digits.Since the first daughter is normal, the father is heterozygous for the allele. If he would have been homozygous dominant, all the children would be affected.Cross:
Parents [tex]M^{D} M^{d}[/tex] X [tex]W^{d}W^{d}[/tex]
Gametes [tex]M^{D}, M^{d}[/tex] X [tex]W^{d}, W^{d}[/tex]
Offspring 2 [tex]M^{D} W^{d}[/tex], 2 [tex]M^{d}W^{d}[/tex]
Probability of polydactyl child = 2/4 or 1/2 or 50%
Why is it important to classily living organisms?
The DNA of organisms may be more easily analyzed
Organisms may survive longer if they are classified
Scientists may study and discuss organisms more easily
Classifying is needed before scientists may make observations
Answer:
Scientists may study and discuss organisms more easily
Explanation:
Yes that's the point!
Answer:
Its C
Explanation:
I did the quiz!!
You decide to designate the alleles of the four loci as either additive (contributing to fruit segment number and represented by a superscript "+") or non-additive (not contributing to fruit segment number and represented by a superscript "0"). Using this convention, choose the correct genotype for the two pure lines and the F1, and indicate how many additive alleles each genotype has.
Answer:
The correct genotype of the two pure lines and the F1 is:
A⁺A⁺B⁰B⁰C⁰C⁰D⁰D⁰ and A⁰A⁰B⁺B⁺C⁺C⁺D⁺D⁺
The number of additive alleles on each genotype are two and six respectively.
Explanation:
Locus( plural form . loci) are fixed point on a chromosome in which genes are located. These genes are specific genetic material or genotype.
Now;
If we decide to designate the allele of the four loci into either additive (⁺) or non-additive(⁰); we have the following :
Let's the allele of the four loci to be
A⁺/A⁰, B⁺/B⁰, C⁺/C⁰ and D⁺/D⁰
However, from the diagram below; we deduce that the correct genotype for the two pure lines and the F1 is as follows:
A⁺A⁺B⁰B⁰C⁰C⁰D⁰D⁰ and A⁰A⁰B⁺B⁺C⁺C⁺D⁺D⁺ and the number of additive alleles on each genotype are two and six respectively.
The cross between both F1 traits will yield an heterozygous individual for the offspring. i.e A⁺A⁰B⁺B⁰C⁺C⁰D⁺D⁰ with only four additive allele
The two pure line genotypes are ++++ and 0000 respectively, with four and zero additive alleles. In the F1 generation, the genotype will be +0+0, indicating individuals possess two additive alleles.
Explanation:Let's consider two pure lines: one with additive alleles on all four loci (++++), and another with non-additive alleles on all four loci (0000). Thus, the first genotype (++++) has 4 additive alleles, and the second genotype (0000) has zero. If these two lines are crossed, in the F1 generation, every individual will inherit two additive alleles from the first parent and two non-additive alleles from the second, leading to a +0+0 genotype. Therefore, the F1 genotype will have 2 additive alleles.
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What must be added to the plate before examination of amylase production?
Gram's iodine reagent must be added to the plate before examination of amylase production.
Explanation:
The starch hydrolysis test is conducted to test the presence of the enzyme amylase in the test medium.
These are confirmatory lab tests done to detect and identify the presence of bacteria which can hydrolyze starch like amylose with the help of enzymes like amylase.
The test medium for this test is agar medium in a petri dish, where soluble starch is first added to initiate the microbial growth. Once incubation period of the microbes is over, dilute iodine solution is added to the petri dish in increased quantity. Iodine is a dye which helps to clearly identify the areas which are hydrolyzed by amylase and those which are not with the help of its color.
4. A population of water snakes is found on an island in Lake Erie. Some of the snakes are banded and some are unbanded; the banding phenotype is autosomal recessive. The frequency of banded snakes on the island is 0.4. There are 500 total snakes on the island. How many snakes are heterozygous for the banding allele
Answer:
233 snakes are heterozygous for the banding allele
Explanation:
According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the alleles. The genotypic frequencies after one generation are p² (Homozygous for allele p), 2pq (Heterozygous), q² (Homozygous for the allele q). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.
In the exposed example,
The banding phenotype is autosomal recessive, bbThe frequency of banded snakes on the island is 0.4There are 500 total snakes on the islandHow many snakes are heterozygous for the banding allele?
The frequency of banded snakes refers to the genotypic frequency for the trait, which is bb= q2= 0.4.
If q2= 0.4, then q = √0.4 = 0.63
The allelic frequency for b is 0.63.
This means that the allelic frequency for B or p is 0.37, which we deduce by clearing the equation p + q = 1
p + 0.63 = 1
p = 1 - 0.63
p = 0.37
The allelic frequency of B is 0,37, and the allelic frequency for b is 0,63. The population heterozygote frequency for this allele is 2 x p x q = 2 x 0,37 x 0,63 = 0.466. The percentage of the population that is heterozygous for this allele is 46%.
As the population size is 500 individuals, then we can calculate how many of these snakes are heterozygous. This is: 0.466 x 500 = 233
The number of heterozygous water snakes for the banding allele on an island in Lake Erie is determined using the Hardy-Weinberg principle, which calculates genotype frequencies based on allele frequencies.
Explanation:The question asks for the number of water snakes that are heterozygous for the banding allele on an island in Lake Erie, given that the phenotype of banding is autosomal recessive and its frequency is 0.4 in a population of 500 snakes. To determine the number of heterozygous individuals, we can use the Hardy-Weinberg principle which assumes that allele frequencies in a population will remain constant from one generation to the next in the absence of other evolutionary influences. The Hardy-Weinberg formula is expressed as p² + 2pq + q² = 1, where p is the frequency of the dominant allele, q is the frequency of the recessive allele, p² represents the frequency of homozygous dominant individuals, 2pq represents the frequency of heterozygous individuals, and q² represents the frequency of homozygous recessive individuals. Since q² is given as 0.4 (the frequency of banded snakes), we can calculate q as the square root of 0.4, which is approximately 0.63. To calculate p, we use p = 1 - q, which would be approximately 0.37. Subsequently, the number of heterozygous individuals (2pq) can be calculated as 2 * 0.37 * 0.63, then multiplied by the total number of snakes (500) to give the final count of heterozygous snakes.
What is a function if the backbone in animals?
A.it protects against predators
B. It regulated blood glucose
C. It surronds the spinal cord
D. It limits movements
Answer:
The answer is C hope this helps+
Answer:
The answer is C.
Explanation:
Don't forget that the human skeleton or backbone protects delicate organ such as the heart and lungs, which are protected by the ribs. It also protects structures such as the spinal cord, which is protected by the vertebral column.
A particular recessive genetic disorder is fatal before birth, so there are no homozygous recessive individuals. In a particular population, one in 15 individuals is a carrier for this disorder. What are the allele frequencies of the dominant (p) and recessive (q) alleles in this population?
Answer:
the allelic frequency for p is 0.967the allelic frequency for q is 0.033Explanation:
According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the alleles. The genotypic frequencies after one generation are p² (Homozygous for allele p), 2pq (Heterozygous), q² (Homozygous for the allele q). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.
In the exposed example,
A recessive genetic disorder is fatal before birth, so there are no homozygous recessive individualsIn a particular population, one in 15 individuals is a carrier for this disorder.What are the allele frequencies of the dominant (p) and recessive (q) alleles in this population?
If 1 of 15 individuals are carriers for this disorder, this means that 1/15 are heterozygous, 2pq. So, 2pq = 1/15 = 0.066
Now we must calculate the allelic frequencies.
We know that 1 in 15 individuals are heterozygous, and we also know that there are no recessive homozygous, q², because they can not survive, so of the 15 individuals only one is heterozygous and the rest 14 individuals must be dominant homozygous, p².
The dominant homozygous genotypic frequency is
p²= 14/15 = 0.933
And by clearing the next equation we can get the allelic frequency for p
p²= 0.933
p = √0.933
p = 0.967
So now we know that the allelic frequency for p is 0.967
This means that the allelic frequency for q or p is 0.033, which we deduce by clearing the equation p + q = 1
0.967 + q = 1
q = 1 - 0.967
q = 0.033
the allelic frequency for p is 0.967the allelic frequency for q is 0.033Termination of DNA synthesis in E. coli and humans differ significantly because of the genomic structures involved, namely a circular genome in bacteria and linear chromosomes in humans. Briefly describe structure and function of the ONE key E. coli protein and the ONE key human protein that are each needed for high fidelity termination of DNA replication.
Answer:
Main protein in ending high fidelity in E. Coli is the Tus protein that binds to Ter sequences in order to prevent replication forks from passing through the end region. In the Ter sequences, the Tus protein blocks replication by establishing a close association with a particular G-C base pair.
The main protein in human cells is telomerase, which contains an RNA primer and is required to extend the synthesis of lagging strands in linear chromosomal telomeres.
The key E. coli protein involved in the termination of DNA replication is the Tus protein, while in humans, it is the telomerase enzyme.
Explanation:
In Escherichia coli (E. coli), the termination of DNA replication is a highly regulated process that ensures the accurate duplication of the circular genome. The key protein involved in this process is the Tus (terminus utilization substance) protein. The Tus protein binds to specific sequences in the DNA called Ter sites (termination sites), which are located opposite the origin of replication on the circular chromosome. When the replication forks approach these Ter/Tus complexes, the Tus protein acts as a polar block to the progress of the replication machinery, causing the forks to stall and eventually terminate replication. This interaction is essential for the coordinated termination of replication and helps to prevent the formation of tangled or incompletely replicated DNA molecules.
In contrast, human cells have linear chromosomes, and the termination of DNA replication is not as straightforward as in bacteria. The ends of linear chromosomes are capped by telomeres, which are repetitive nucleotide sequences that protect the chromosome ends from degradation and fusion. The key protein involved in maintaining telomeres is the telomerase enzyme. Telomerase is a reverse transcriptase that adds telomeric repeats to the ends of chromosomes, compensating for the shortening that occurs during DNA replication due to the end-replication problem. This ensures that the ends of the chromosomes are maintained and that the cell can continue to divide without losing critical genetic information. The high fidelity of DNA replication termination in humans is thus maintained by telomerase activity, which is crucial for chromosomal stability and cell viability.
In summary, while the Tus protein in E. coli ensures the accurate termination of replication in a circular genome by blocking replication forks, the telomerase enzyme in humans maintains the integrity of linear chromosome ends, allowing for high-fidelity replication and preventing genomic instability. Both proteins are essential for the faithful duplication of their respective genomes."
Which of your cells have all of your DNA code?
suppose a male organism has a diploid chromosome number of 6. These chromosomes could be represented as A,a, B, b, C, and c where uppoer and lower cases (capital and small) of the same letter are homologous to one another. The karyotype (chromosome consititution) of this organism could then be designated as AaBbCc.
a. How would you universally describe a haploid set of chromosomes in this organism in terms of these letters (universal means one description fits all possible cases)?
b. List the eight different combinations of these chromosomes that could possibly be produced by spermatogenesis in this male.
Answer:
A. The haploid set of n is ABC
B. ABC, ABc, AbC, aBC, abc, abC, aBc, Abc.
Explanation:
A diploid cell is a cell that comprises of two complete sets of chromosomes. This is double the haploid chromosome number. Each pair of chromosomes in a diploid cell is noted to be a homologous chromosome set. A homologous chromosome pair consists of one chromosome donated from one parent and one from the other parent. This number is represented as 2n. It varies in different organisms. A diploid cell replicates via mitosis. It preserves its diploid chromosome number by making an identical copy of its chromosomes and distributing its DNA in an equal manner between two daughter cells.
Haploid refers to a cell that contains a single set of chromosomes. Gametes are made up of half the chromosomes contained in normal diploid cells of the body, these are also known as somatic cells. Haploid gametes are produced during meiosis. Meiosis is a type of cell division that reduces the number of chromosomes in a parent diploid cell by half.
Therefore,
Karyotype of this organism is designated as AaBbCc.
AaBbCc is the diploid set (2n).
Hence ABC or abc are the haploid set (n).
B. The eight different combinations of these chromosomes that could possibly be produced by spermatogenesis in this male are:
ABC, ABc, AbC, aBC, abc, abC, aBc, Abc.
How gas exchange occurs when the circulatory and respiratory systems work together.?
Answer:chase the engine
Because of the engine
Explanation:
Answer:The two systems work together in several ways. First, the respiratory system brings oxygen in the air to the alveoli in the lungs. The circulatory system then delivers the oxygen to the cells. Second, the circulatory system picks up carbon dioxide from the cells and carries it to the lungs, where it is released when we exhale.
Explanation:its the sample writing.
A lion eating a zebra is an example of which characteristics of life
Answer:
the answer is useing energy
Explanation:
the answer is useing energy bc a lion is a natural enemy of zebras. lions are very fast running animals so by running at extremely high speed,they preform hunting of zebra. by doing so,they utilize energy in getting their prey which is there meal. hope this helps!!:)
A lion eating a zebra exemplifies the characteristics of life, including nutrition, growth, response to stimuli, reproduction, homeostasis, and the role of evolution in adaptation to its environment.
The act of a lion eating a zebra exemplifies several key characteristics of life, primarily pertaining to the lion as an organism:
1. Nutrition: The lion obtains nourishment by consuming the zebra, demonstrating its ability to ingest and digest food.
2. Growth and Development: The lion's body will process the zebra's nutrients, contributing to its own growth and maintenance, a fundamental aspect of life.
3. Response to Stimuli: The lion's predatory behavior, hunting, and eating are responses to stimuli such as hunger and the presence of prey.
4. Reproduction: While not directly observed in this scenario, the lion's capacity to reproduce is a vital aspect of its life cycle.
5. Homeostasis: After eating, the lion must maintain internal balance (homeostasis) in terms of temperature, blood sugar, and other physiological factors.
6. Evolution: Over generations, the lion's predatory traits and behaviors may evolve in response to environmental pressures and changes in prey availability, highlighting the role of evolution in life.
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The complete question is:
A lion eating a zebra is an example of which characteristics of life what?
If someone nearby has a runny nose, can they spread their snot to you just by being close? Why or why not?
Answer:
Yes, if the touch their snot and touch you or an object you touch. They can also sneeze their existing snot at you if they didn't wipe it.
Yes, someone with a runny nose can spread their snot to you just by being close. The common cold, which often causes a runny nose, is typically caused by viruses like rhinoviruses, coronaviruses, and adenoviruses. These viruses can be transmitted through direct contact and droplet transmission, such as when an infected person coughs or sneezes and produces infectious aerosols.
Explanation:Yes, someone with a runny nose can spread their snot to you just by being close. The common cold, which often causes a runny nose, is typically caused by viruses like rhinoviruses, coronaviruses, and adenoviruses. These viruses can be transmitted through direct contact and droplet transmission, such as when an infected person coughs or sneezes and produces infectious aerosols.
When an infected person coughs or sneezes, mucus droplets are released into the air. These droplets can remain suspended in the air for some time and travel considerable distances. If you are close to the infected person, you can breathe in these droplets or they can land on surfaces that you touch, leading to potential transmission.
To minimize the risk of transmission, it is important to practice good respiratory hygiene, such as covering your mouth and nose with a tissue or your elbow when coughing or sneezing, and regularly washing your hands with soap and water or using hand sanitizer.
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Which of the following grows directly for a plants stem?
1. Overate leaves
2. Palmate leaves
3. Petiole leaves
4. Sessile leaves
Which of the following represents the correct sequence of events for sexual reproduction in most vertebrates? _____ Select one: a. mitosis → zygote → meiosis → gametes → embryo b. meiosis → zygote → mitosis → gametes → embryo c. mitosis → gametes → meiosis → zygote → embryo d. meiosis → gametes → mitosis → zygote → embryo e. meiosis → gametes → zygote → mitosis → embryo
Answer:
e. meiosis → gametes → zygote → mitosis → embryo
Explanation:
In the vertebrates, the life cycle alternates between the diploid and haploid phase. The vertebrate body is diploid and the haploid phase is only observed during the gamete formation.
The sex organs of the vertebrates produce gametes from the gamete mother cell through meiosis, which reduces the chromosome number to half and produce four haploid gametes.
The male and female gametes undergo fertilization event and form a diploid zygote. The zygote develops into the embryo through mitosis as mitosis produces the cells with an equal number of chromosomes.
Thus, Option-E is correct.
Suppose you do the test on a hypothetical Staphyloccocus species with the antibiotics penicillin (P 10) and chloramphenicol (C 30). You record zone diameters of 25mm the chloramphenicol and penicillin disks. Which antibiotic would be more effective against this organism?what does this tell you about comparing zone diameters to each other and the importance of the zone diameter interpretive chart?
Answer:
In an antibiogram, the larger the inhibition halo, the more sensitive bacteria will be to the drug that is exposed, that is why if the halo is 25 mm, it is considered quite wide, so these drugs such as chloramphenicol and amoxicillin affect the bacterial development in marine culture both bactericidal or bacteriostatic.
Both have a 25mm halo, so both would be SAME as specified.
Explication:
suppose that one of the two drugs has a smaller inhibition halo in the culture, that drug will be less effective in treating this bacterium, therefore more would allow its development, on the contrary, the larger halo would be the most effective drug (This is an example for you to understand that the greater the length of the inhibition halo, the greater the efficacy of the drug, and the lower the halo, the less efficacy)
What is an example of a key difference between DNA and RNA
Answer:
A key example is that DNA is double stranded where as RNA is single stranded.
Explanation:
22. An entomologist breeds 500 insects for research and
observes that 125 of the insects have red wings and 250 of
the insects have long antennae. In this insect population,
what is the probability of an insect having red wings and
long antennae?
Answer:
6
Explanation:
Leslie argues that an organism called a Euglena is a plant because it is eukaryotic, has a nucleus, makes its own food, and
is unicellular
Where is Leslie's error?
Plants do not make their own food
O Plants are multicellular.
O Plants are prokaryotic.
O Plants do not have a nucleus.
Answer:
Plants are mutiticelullar
Explanation:
All species of plants are mutiticelullar and just a few like mold,amoeba are unicellular.
Enzyme B helps in the process of breaking molecule A into molecules C and D Which of the following outcomes can be expected?
Answer:
A
Happy to Help!!
Enzyme B acts as a catalyst to convert molecule A into molecules C and D by reducing the activation energy needed for the reaction. The process could follow a sequential or ping-pong mechanism, where the enzyme binds substrates and facilitates their transformation into products.
Explanation:Enzyme B facilitates the breakdown of molecule A into molecules C and D. This process is an example of an enzymatic reaction where the enzyme lowers the activation energy, thus speeding up the reaction without being consumed in the process. The expected outcome would be the conversion of substrate A into products C and D, with enzyme B acting as a catalyst.
According to the provided information, enzymes work by reducing the activation energy (Option B), not by increasing it or altering the exergonic or endergonic nature of reactions. This reduction allows the enzymatic reaction depicted, A+B=C+D, to proceed more efficiently.
Enzymes may operate via several mechanisms, including sequential reactions where substrates bind in an ordered or random fashion. In the case of Enzyme B, if it follows a ping-pong mechanism, it would alternate between two states, without A and C bound simultaneously, to facilitate the reaction between molecules A and B to form C and D.
A beaker is divided by a membrane that is permeable to water and
impermeable to sodium ions. The sodium-water solution on side A is more concentrated than the sodium-water solution on side B. Which of the following statements will be true when equilibrium is reached across the membrane?
a) The absolute number of water molecules will be different on the two sides of the membrane, whereas the final solute concentration will be equal.
b) Equal numbers of water molecules will exist on each side of the membrane at equilibrium.
c) Since the solute can't cross the membrane, the solution will not reach equilibrium.
d) The solute will maintain the original concentration on both sides of the membrane. e) All answers are correct.
Answer:
a) The absolute number of water molecules will be different on the two sides of the membrane, whereas the final solute concentration will be equal.
Explanation:
According to the concept of osmosis, solvent molecules pass via a semi permeable membrane from area of lower solute concentration to higher solute concentration. Here, solution A and B have different concentration and they are separated by a semi permeable membrane which is only permeable to the water molecules.
Water will move from side B to side A since side B is less concentrated so it has more free water molecules. Eventually, the concentration will become equal on both sides and net movement of water molecules will become zero. However, absolute number of water molecules will be different in both the sides because originally there was more solute in side A so to make the concentration equal, more water molecules will now be present here than side B.
If we cross Surface fish (have eyes and are not albino) with fish from Cave Scarlet, the F1 progeny have eyes and are not albino. If we backcross these F1 Surface/Scarlet hybrids to Scarlet cavefish, 50% of the progeny are albino and 50% are not albino. In terms of the eye phenotype, there is a broad range of eye phenotypes extending from no eyes like the original Scarlet population to complete eyes like surface fish (with varying degrees of eye formation in between
Answer:
More than one gene is controlling the loss train of that eye. Thus, the Cross among the scarlet and grey species, F1 crossover has fractional eyes. Despite the fact that both sort of cavern fishes are visually impaired, yet cross between them has brought about offspring with fractional eyes. This shows multi-gene legacy for eye misfortune attribute. Blending of two genotypes has brought about incomplete characteristic in the descendants.
Meselson and Stahl were trying to determine if DNA replication was semiconservative, conservative, or dispersive by labeling E. coliDNA with a regimen of heavy nitrogen (H) for one round of replication and then transferring these cells to light nitrogen (L) for two more rounds of replication. Which of the following statements wouldnotbe true within the context of this experiment?a. If DNA replication were conservative, no DNA molecules of intermediate density (H-L) would have been seen.b. If DNA replication were dispersive, only DNA molecules that were of intermediate density (H-L) would have been seen.c. If DNA replication were semiconservative, the DNA molecules that were made would all continue to be heavy density (H-H).d. If DNA replication were semiconservative, the DNA molecules would consist of one parental strand basepaired to one newly replicated strand.e. If DNA replication were semiconserative, a higher proportion of DNA molecules from future divisions would have been low density (L-L)
Answer:
Option C
Explanation:
All of the following options are correct except option C. in the Meselson Stahl experiment, after the first round of replication, the DNA molecules made consisted mainly of one parental strand and one new strand each which is the DNA molecules of intermediate density (H-L). Then future replications will produce more of the low density L-L density. This form of replication is thus termed to be the semi conservative form of replication
Lipid bilayers formed between two aqueous phases have thisimportant property: they form two-dimensional sheets, the edges ofwhich close upon each other and undergo self-sealing to formliposomes.
a. What are the forces that drive bilayerformation? What are the consequences of this property for thestructure of biological membranes?
b. Explain why phosphoglycerides are capable of spontaneouslyassembling into the bilayer structure found in biological membranesbut triacylglycerols are not.
Answer:
a-The assembly process is driven by interactions between hydrophobic molecules (also called the hydrophobic effect), an increase in interactions between hydrophobic molecules (which causes clustering of hydrophobic regions) allows water molecules to bond more directly between Yes, the entropy of the system increased. This complex process includes non-covalent interactions, tales like Van der Waals, electrostatic bonding, and hydrogen bonding.
b-The fluidity of the membranes depends mainly on the length and the level of unsaturation of the phospholipids. Indeed, the shorter and more unsaturated the membrane lipids, the more fluid the membrane is. This is due to the fact that long chains show a greater association with each other. Specifically, for each -CH2- group added, the free energy of interaction between two chains decreases by 0.5 kcal mol-1, making the interaction stronger. When the chains are shorter, the interaction between the alkyl groups of the lipids is reduced.
Explanation:
The lipid bilayer is a thin polar membrane made up of two layers of lipid molecules, membranes are flat sheets that form a continuous barrier around cells and their structures
Kaylee wants to test her hypothesis that she performs better on tests after getting more sleep. In which way will she best be able to do this?
A. by analyzing her test scores when she gets the same amount of sleep
B. by asking her friends if sleep affects their test scores
C. by analyzing her average test scores after nights with different amounts of sleep
D. by reading online about the relationship between sleep and test scores
What is the source of energy in most ecosystems?
animals
plants
oxygen
sunlight
Its D. Sunlight. Since plants are producers they get their energy from the sun
Contrary to legend, camels do not store water in their humps, which actually consist of large fat deposits. How can these fat deposits serve as a source of water? Calculate the amount of water (in liters) that a camel can produce from 1.0 kg of fat.Assume for simplicity that the fat consists entirely of tripalmitoylglycerol.
Answer:
A) Oxidation of fats releases metabolic water
B) [tex]1.54[/tex] L of water per kg of tripalmitoylglycerol
Explanation:
A) During the oxidation of organic substances (fats) releases metabolic water along with release of energy
For example –
Oxidation of one mole of Palmitic acid (C16H32O7) acid releases 146 mole of H2O. In Tripalmitoylglycerol (fat C51H98O6) , there are 3 palmitic acid residues connected to glycerol molecule via ester bonds.
Oxidation of Palmitoyl CoA:
almitoyl-CoA + 23O2 + 108Pi + 108 ADP => 88nCoA + 108 ATP + 16CO2 + 23 H2O
B)
Molecular weight of tripalmitate [tex]= 807[/tex] g/mol
Number of moles in one Kg of fat
[tex]= \frac{1000}{807}\\[/tex]
[tex]= 1.24[/tex] mole
Mass of water
[tex]1.24 * 69 * 18.02[/tex] ([tex]69[/tex] moles of water and weight of water [tex]= 18[/tex] grams)
[tex]= 1541.8[/tex] grams
Density of water [tex]= 1[/tex] g/mL
Volume of water [tex]= 1.541[/tex] L
Camels do not store water in their humps, but the humps store fat which is metabolized to water and carbon dioxide through oxidation. This provides hydration to the camel. For instance, 1.0kg of fat can produce about 1.1 liters of water.
Explanation:When a camel's body uses these fat deposits for nutrition during periods without food, a reaction called oxidation happens. Oxidation of fat is a metabolic process in which the fat molecules are broken down into carbon dioxide and water using the oxygen that animals breathe in. If we consider that the complete oxidation of 1 gram of fat releases about 1.1 grams of water, then 1.0 kg (or 1000 grams) of fat can produce about 1100 grams of water. Because the density of water is 1 gram per milliliter, these 1100 grams are equivalent to approximately 1.1 liters of water.
Therefore, camels can stay hydrated for longer periods in the desert because of the water produced by the oxidation of their fat stores.
To summarize, though camels don't store water in their humps, the fat stored there can be metabolically converted to water in their bodies, helping them survive in harsh desert conditions.
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