Answer:
LASIK or laser-assisted in situ keratomileusis refers to a kind of refractive surgery. In the surgery, the surgeon initially cleaves a thin flap of tissue from the anterior part of the eye. After that, a laser removes away the tissue to reconfigure the cornea so that light aims better on the retina in the posterior part of the eye.
Cornea refers to a clear, safeguarding exterior covering of the eye. It generally offers protection against germs, dust, and other such deleterious particles. Apart from its protective function, the cornea also plays an essential part in amending the vision. The shape of the cornea is accountable for the bending of the light as it enters the eye.
The focusing of the objects relies hugely upon the curvature of the cornea. Thus, in the conditions where the cornea gets damaged, that is, its shape gets changed somehow, the vision gets affected and the manner in which the light moves within the eye gets distorted.
31.
Part 1. How is indeterminate cleavage different from determinate cleavage?
Part 2. Is there any benefit to the organism if they have determinate or indeterminate cleavage? i.e is one type of cleavage superior to the other?
Part 3. Why do you think that all animals do not display indeterminate cleavage?
Indeterminate cleavage results in identical cells capable of forming an embryo while determinate cells do not result in cells which are capable to develop embryo.
Indeterminate is superior to determinate cleavage.
Explanation:
Cleavage is the division of cells in the early embryonic stage. The two stages of cleavage described here are:
In indeterminate cleavage or regulative cleavage occurs when an embryo divides, each cell is capable of developing into complete embryo. eg: Deuterosomes
In determinate cleavage the resulting embryonic cells of blastomere cannot develop into embryos. It is also called as mosaic cleavage. The essential part of the cell might be missing which does not let the cell survive. eg: Protosomes
Indeterminate cleavage is of great importance as the cell grows and can produce new organism. The complete identical twin is formed. Its application can be seen in tomato plants.
3. All animals do not display intermediate cleavage because growth from intermediate cleavage is continuous and does not stop after adulthood which is not possible in animals.
Does the DNA support the hypothesis that the coelacanth is the closest living relative to amphibians, such as frogs?
Answer:
According to biologists the coelacanth is NOT the closest living relative to amphibians,
Explanation:
In this modern era of molecular biology there is evidence that the coelacanth and tetrapods are not closely related. While in the other hand, evidences indicated a close relationship between lungfishes and tetrapods. The molecular analysis was based on mitochondrial DNA sequences. I have attached picture of evolutionary relationship.
FIG. 1. Alternative hypotheses of sister group relationships between sarcopterygii and tetrapods.
(A) Lungfish as the sister group of
tetrapods.
(B) Coelacanth as the closest living relative of tetrapods.
(C) Coelacanth and lungfish equally closely related as sister groups of tetrapods.
Reference: Zardoya, R., & Meyer, A. (1996). Evolutionary relationships of the coelacanth, lungfishes, and tetrapods based on the 28S ribosomal RNA gene. Proceedings of the National Academy of Sciences, 93(11), 5449-5454.
Answer:
No, it doesn't support
Explanation:
The DNA based alternative hypothesis shows that the lung fish is actually the closest relative to amphibians rather than the coelacanth.
Geneticists sometimes use the following test for the nullness of an allele in a diploid organism: If the abnormal phenotype seen in a homozygote for the allele is identical to that seen in a heterozygote (where one chromosome carries the allele in question and the homologous chromosome is known to be completely deleted for the gene) then the allele is null. What is the underlying rationale for this test?
Answer:
There will be null activity for a gene within in the deletion, this means if a genotype has similar phenotype. This suggests that mutant allele and the allele with deletion has the same level of activity.
Explanation:
The limitation of such interpretation is that the phenotypes have the enzymatic activity according to a threshold level. This means that the mutant phenotype is visible till the enzymatic activity is lower than a threshold. If the level rises above the threshold the wild type phenotype is observed.
Homeotic genes contain a homeobox sequence that is highly conserved among very diversespecies. The homeobox is the code for that domain of a protein that binds to DNA in aregulatory developmental process. Which of the following would you then expect?A) That homeotic genes are selectively expressed over developmental time.B) That a homeobox containing gene has to be a developmental regulator.C) That homeoboxes cannot be expressed in non homeotic genes.D) That all organisms must have homeotic genes.E) That all organisms must have homeobox containing genes.
Answer:
wdawdsawdsaw
Explanation:
414w141432we
Suppose that rat liver expresses a protein called Yorfavase. This enzyme is composed of 192 amino acid residues, and thus the coding region of the yfg gene consists of 576 bp. However, the rat genome database indicates that the yfg gene consists of 1440 bp. Select which type of DNA does not contribute to the additional 864 bp found in the yfg gene.
O O O 5-end untranslated regulatory region centromeric DNA
DNA coding for a signal sequence noncoding intron O promoter sequence
Answer:
Centromeric DNA
Explanation:
Genes are responsible for formation of proteins. A codon produces a single amino acid and it comprises of three base pairs. The whole DNA of organism does not contribute to encode protein. There are sequences in DNA that does not contribute in coding known as non-coding DNA and perform different functions. These may include the O O O 5-end, DNA coding for a signal sequence, noncoding intron and O promoter sequence .
The centromere is the part of DNA that links the sister chromatids. The centromere is not converted to mRNA or proteins. It is also not involved in regulation of genes. So it did not contribute to 864 bp found in the yfg gene.
Measurements with small freshwater sponges such as Grantia have documented water flow rates around 3-4 ml/minute. Assuming that a sponge has a constant flow rate over the course of a day, how many ml of water could a single sponge such as Grantia filter in a 24-hour period?
A sponge with a flow rate of 3.5 ml/minute could potentially filter around 5040 ml of water in a 24-hour period.
Explanation:The process to find the total amount of water a single sponge such as Grantia filters in a 24-hour period involves simple multiplication. First, you need to know the flow rate of the sponge, which is given as 3-4 ml/minute. Let's take the average, which is 3.5 ml/minute. To find the total volume of water processed in a 24-hour period, we need to multiply the rate by the total number of minutes in a day, which is 1440 minutes. So, it's 3.5 ml/minute * 1440 minutes = 5040 ml/day. Therefore, this sponge could process around 5040 ml of water in a 24-hour period.
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A father with myotonic dystrophy has three daughters who are all carriers of the mutant allele and two sons who are unaffected noncarriers. the three daughters have six sons, of which four are affected and two are not, and four daughters, of which two are carriers and two are not. From this description, what type of mutation is probably responsible for myotonic dystrophy?
Answer:
Inherited Autosomal Dominant Mutations
Explanation:
The mutations in DMPK gene or ZNF9 gene contribute to the onset of Muscular Dystrophy. This is a type of inherited disorder that can run in families and due to it's autosomal transfer nature, it affects both sexes equally.
Assuming a typical monohybrid cross in which one allele is completely dominant to the other, what phenotypic ratio is expected if the F1's are crossed?
Answer:
Assuming the dominant allele is represented as A, and the genotype as AA and the recessive as a, and genotype as aa; then the F1 will produce the offispring. Aa, Aa, Aa,Aa in ratio: 1:1:1:1:1
This shows the dominance of allele A.
However if two of the F1 generation are crossed then:
The F2 is AA,Aa,Aa,aa the phenotypic ratio is
3 ; 1
While the Genotypic ratio is ratio;
1AA : 2Aa Aa; 1aa(1 ;2;1)
Explanation:
In a typical monohybrid cross in which one allele is dominant to the other, the expected phenotypic ratio when F1 generation is crossed is 3:1. This means that 3 out of 4 offsprings will express the dominant phenotype, while one will express the recessive phenotype.
Explanation:In a typical monohybrid cross where one allele is completely dominant to the other, the F1 generation would be heterozygous - meaning they carry one each of the dominant and recessive alleles. When these are crossed, the Mendelian genetics laws predict that there will be a phenotypic ratio of 3:1. This means that 3 out of 4 offsprings are expected to express the dominant phenotype, while 1 out of 4 is expected to express the recessive phenotype. For instance, in a cross between two purple-flowered pea plants (Pp) where purple color (P) is dominant over white (p), the expected phenotypic ratio in the F2 generation would be 3 purple-flowered to 1 white-flowered.
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Why did agriculture become the preferred way of supplying basic human needs?
Cross 1 Progeny:38 two-lobed, red18 two-lobed, yellow38 multilobed, red18 multilobed, yellowCross 2 Progeny:14 two-lobed, red14 two-lobed, yellow14 multilobed, red14 multilobed, yellowIn tomato plants, the production of red fruit color is under the control of an allele R. Yellow tomatoes are rr. The dominant phenotype for fruit shape is under the control of an allele T, which produces two lobes. Multilobed fruit, the recessive phenotype, have the genotype tt. Two different crosses are made between parental plants of unknown genotype and phenotype. Use the progeny phenotype ratios to determine the genotypes and phenotypes of each parent.PART A: For cross 1, determine two appropriate genotypes for both parents.a) Rrtt × Rrttb) Rrtt × rrTtc) RrTt×Rrttd) Rrtt × RrTt OR rrTt × RrTtPart B: For cross 1, determine two appropriate phenotypes for both parents.a) yellow fruit, two lobes AND red fruit, multiple lobesb) red fruit, two lobes AND red fruit, two lobesc) red fruit, two lobes AND red fruit, multiple lobesd) red fruit, two lobes AND yellow fruit, two lobesPART C: For cross 2, determine two appropriate genotypes for both parents.a) RrTt × Rrttb) RrTt × rrttc) RrTt × rrTtd) RrTt × rrtt OR Rrtt × rrTt
Question reformatted
In tomato plants, the production of red fruit color is under the control of an allele R. Yellow tomatoes are rr. The dominant phenotype for fruit shape is under the control of an allele T, which produces two lobes. Multilobed fruit, the recessive phenotype, have the genotype tt. Two different crosses are made between parental plants of unknown genotype and phenotype. Use the progeny phenotype ratios to determine the genotypes and phenotypes of each parent.
Cross 1 Progeny: 38 two-lobed red, 18 two-lobed yellow, 38 multilobed red, 18 multilobed yellow.
Cross 2 Progeny: 14 two-lobed, red 14 two-lobed, yellow 14 multilobed, red 14 multilobed, yellow
Part AFor cross 1, determine two appropriate genotypes for both parents.
a) Rrtt × Rrtt
b) Rrtt × rrTt
c) RrTt×Rrtt
d) Rrtt × RrTt OR rrTt × RrTt
Part BFor cross 1, determine two appropriate phenotypes for both parents.
a) yellow fruit, two lobes AND red fruit, multiple lobes
b) red fruit, two lobes AND red fruit, two lobes
c) red fruit, two lobes AND red fruit, multiple lobes
d) red fruit, two lobes AND yellow fruit, two lobes
Part CFor cross 2, determine two appropriate genotypes for both parents.
a) RrTt × Rrtt
b) RrTt × rrtt
c) RrTt × rrTt
d) RrTt × rrtt OR Rrtt × rrTt
Answer:
Part A - C
Part B - C
Part C - D
Explanation:
Part AThe genotype for two lobed red must be either RRTT, RrTT, RRTt or RrTt, as these are both dominant traits
The genotype for two lobed yellow must be rrTt or rrTT, as yellow is a recessive trait but two lobed is dominant.
The genotype for multilobed red must be Rrtt or RRtt, as multilobed is recessive but red is dominant
The genotype for multilobed yellow must be rrtt, as both these traits are recessive.
The answer cannot be a) as we have multilobed fruit which would require a T allele.
For a dihybrid cross between two heterozygous individuals (i.e. both RrTt) gives the expected ratio of 9:3:3:1, a classic Mendelian ratio for this type of cross. That is not the case for this cross, as there are double the amount of multilobed yellow than we would expected, and half the amount of two lobed red. This suggests one of the parents is homozygous for the t allele, but that they are both heterozygous for the R/r allele. Therefore, we can first check the ratios of RrTt x Rrtt (answer c) by a punnett square (see attached).
The possible gametes are RT, rT, Rt, rt for one parent (RrTt), and RT, Rt and rt for the other parent (Rrtt). This gives us the correct progeny.
If you ever struggle with these questions, just draw out all the possible punnett squares!
Part BWe have determined that the parental genotypes are RrTt x Rrtt. This means the parental phenotypes are Red two-lobed and red multi-lobed, corresponding to answer C.
Part CThe second cross has a ratio of 1:1:1:1. This gives the clue that the gametes are all present in equal numbers (i.e. each parent is heterozygous for one trait and homozygous for the other), OR that one parent is heterozygous for both and one is homozygous to both. This points to option d.
Capillary action is how plants draw moisture from the soil into their roots and stems. Explain the relationship between capillary action and intermolecular forces.
Answer: Intermolecular force is what holds molecules together and capillary action is moving of liquid towards upward into a tube that is capillary. So the Intermolecular forces which takes part in capillary action are; for keeping the molecule attached to tube is the adhesive force and the for keeping the liquid molecule together is cohesive force.
In the plants, mositure from soil is taken up from stem which is tube like and to which molecules have strong adhesion.
Capillary action relies on intermolecular forces, specifically cohesive forces among water molecules and adhesive forces between water and surfaces. These forces enable water to move against gravity in narrow spaces, aiding plant moisture uptake.
Capillary action is a phenomenon that occurs when a liquid, such as water, moves against the force of gravity within a narrow space, like a thin tube or a narrow pore in a solid material.
The relationship between capillary action and intermolecular forces lies in the cohesive and adhesive forces at play.
1. Cohesive Forces: Cohesive forces are the intermolecular forces that cause water molecules to stick to each other.
In a narrow tube or pore, these forces create a concave meniscus, where water molecules are attracted to each other along the walls of the space.
2. Adhesive Forces: Adhesive forces are the intermolecular forces that cause water molecules to be attracted to a solid surface, like the walls of a plant root or stem. These forces pull the water upward against gravity.
The combined effect of cohesive and adhesive forces allows water to climb or be drawn up through small openings, such as the tiny capillaries in plant roots.
This capillary action is essential for plants to transport water and nutrients from the soil to their stems and leaves, enabling their growth and survival.
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Rocks form in horizontal layers. However, as shown along mountainsides, rock layers can become twisted or bent.
Answer:
The earth is extremely old,and earth processes acted gradually to twist or bend rocks.
Explanation:
The transformation takes place through many years as the molten rock was pushed upward and sediment form into the rock that was lifted above sea level by the force generated by uprising rocks.The changes are very little and gradually takes place which gives rise to the formation of rocks.
1. Although generally not considered to be alive, a _______ is studied alongside other microbes such as bacteria2. The protein coat that surrounds the nucleic acid of a virus3. A viral life cycle that results in bursting of the host cell4. A viral life cycle in which the virus inserts its genome into the genome of its host, where it may remain dormant for long periods________.5. Viral genome that has inserted itself into the genome of its host________
Answer:
The correct terms are as - 1. virus, 2. capsid, 3. lytic cycle, 4. lysogenic cycle, 5. prophage.
Explanation:
A virus is considered as nonliving by various scientists as it required a living host to divide, however, it is studied with the microbes such as the bacteria and others. The virus is made up of nucleic acid surrounded by the protein coat called a capsid.
Viruses normally show two types of life cycles inside the host cell that are the lysogenic cycle and the lytic cycle.
The lytic cycle is characterized as the life cycle that ends with the destruction or bursting of the host cell. The lysogenic cycle includes being dormant inside the host genome until the favorable condition reestablished. Prophage is a virus that inserts itself into the host genome on its own.
Thus, the correct answer is - 1. virus, 2. capsid, 3. lytic cycle, 4. lysogenic cycle, 5. prophage.
The questions are about different aspects of virology including viral structure, and the lytic and lysogenic cycles. A virus, although not considered truly alive, is studied as a microbe. Its structure includes a protein coat (capsid) and its genetic content or nucleic acid. The lytic and lysogenic cycles describe how a virus interacts with its host.
Explanation:1. Although generally not considered to be alive, a virus is studied alongside other microbes such as bacteria. 2. The protein coat that surrounds the nucleic acid of a virus is known as a capsid. 3. A viral life cycle that results in bursting of the host cell is referred to as the lytic cycle. 4. A viral life cycle in which the virus inserts its genome into the genome of its host, where it may remain dormant for long periods is known as the lysogenic cycle. 5. A viral genome that has inserted itself into the genome of its host is referred to as a provirus or prophage.
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If you were to set up a PCR reaction (in vitro DNA synthesis) with a DNA template, primers,DNA polymerase, dATP, dGTP, dCTP, dTTP and a small amount of ddATP, what would be the result? If you were to set up a PCR reaction (in vitro DNA synthesis) with a DNA template, primers,DNA polymerase, dATP, dGTP, dCTP, dTTP and a small amount of ddATP, what would be the result? DNA synthesis would happen normally. All DNA molecules produced would be the same length as the template. DNA synthesis might be terminated after the addition of any adenine base (at random). DNA molecules of many different lengths would be produced. DNA synthesis would be terminated after the first adenine base is added. All DNA molecules produced would the same length
Answer:
DNA synthesis might be terminated after the addition of any adenine base (at random). DNA molecules of many different lengths would be produced.
Explanation:
Polymerase chain reaction may be defined as the molecular process of increasing the amount of DNA upto several times. Three important steps in PCR - denaturation, annealing and extension.
As given in the question, dd ATP ( dideoxy adenosine triphosphate) is added to the PCR reaction. ddATP lacks the hydroxyl group at 3' position of the sugar. This means the reaction will be terminated after this ddATP as no further nucleotide will be added. This will also leads to the genereation of the different length of DNA fragments as ddATP will randomly be inserted in the DNA molecule.
Thus, the correct answer is option (3) and (4).
In addition to providing yogurt with its unique flavor and texture, lactic acid-producing bacteria also provide which additional benefit during food production?a. Providing xenobiotics b. Lowering the pH to kill pathogenic bacteria c. Pasteurizing milk products d. Breaking down lactose for lactose-intolerant individuals
Answer:
Option b. Lowering the pH to kill pathogenic bacteria is correct answer.
Explanation:
bacterial motors are sensitive to pH. By decreasing the pH bacterial motors stops working. This was identified in a new research. But, with the weak acids and a lower internal pH they slow and ultimately stop moving (became dead).
Reference: Powell, K. Acid stops bacteria swimming. Nature (2003).
Plants are set up in a greenhouse. 1/3 of potted plants are grown under red light, 1/3 under green light, and 1/3 under natural light. Plant biomass is measured at the beginning of the experiment and after a 28 day period.
In this study, is the amount of plant growth the independent or dependent variable?
dependent
independen
Answer:
Plants are set up in a greenhouse. 1/3 of potted plants are grown under red light, 1/3 under green light, and 1/3 under natural light. Plant biomass is measured at the beginning of the experiment and after a 28 day period.
In this study, is the amount of plant growth the independent or dependent variable?
Dependent variable
Explanation:
Plant is the independent variable as the light varies in color which makes the plant to depend on them for growth
Each step in the flow of energy through an ecosystem is known as aA.food chain.B.trophic level.C.plateau of consumption.D.food web.
Answer: Option B) Trophic level
Explanation:
Trophic level refers to a feeding level, representing the number of links by which food energy is transferred from producers
to final consumers.
For example
Grass ------> Antelope ---------> Lion
This food chain begins with grass as producer, that is fed on by antelope which in turn is fed on by a carnivore, lion. Thus, there are three trophic levels by which energy is transferred in a step-by-step basis from grass to lion.
Thus, trophic level is the answer
Final answer:
Each step in the ecosystem's energy flow is known as a trophic level, representing the organism's position based on its role as a producer or consumer. The concept highlights the linear transfer of energy and nutrients through a food chain, emphasizing the energy losses at each trophic level.
Explanation:
Each step in the flow of energy through an ecosystem is known as a trophic level. In the context of ecology, a food chain provides a linear sequence of organisms through which nutrients and energy pass, including primary producers, primary consumers, and higher-level consumers. Each organism in a food chain occupies a specific trophic level, denoting its position based on its role as a producer or consumer. The concept of trophic levels is fundamental to understanding ecosystem structure and dynamics, emphasizing the flow of energy from one level to the next.
Trophic levels include:
Primary producers: Organisms that synthesize their own food from inorganic substances using light or chemical energy.
Primary consumers: Organisms that feed on primary producers.
Higher-level consumers: Organisms that feed on primary and other consumers.
Energy is transferred between trophic levels, but with significant losses at each step, usually around 10%, leading to the concept of an energy pyramid. This pyramid reflects the cumulative loss of usable energy at higher trophic levels, shaping the structure of food chains and ecosystems.
Phytoplankton 120,000 kJ/m2
Copepods 7,514 kJ/m2
Small fish 383 kJ/m2
Marine birds 14 kJ/m2
a. Calculate the efficiency of energy transfer to the copepods.
b. Calculate the efficiency of energy transfer from the secondary consumer to marine birds.
Answer:
a. 6.26%
b. 3.65%
Explanation:
A food chain is a series of event showing how organism feed on one another, it also shows the transfer of energy in an ecosystem.
From the question; we can have a food chain showing the transfer of energy right from the producer to the tertiary consumer
(Producer) Primary Sedcondary Tertiary
consumer (1°) consumer (2°) consumer (3°)
Phytoplankton ⇒ Copepods ⇒ small fish ⇒ Marine birds
120,000 kJ/m2 7,514 kJ/m2 383 kJ/m2 14 kJ/m2
W X Y Z
Efficiency of energy transfer is given as: [tex]\frac{EnergyAvaliableAfterTheTransfer}{EnergyAvailableBeforeTheTransfer}*100%[/tex]
The first question (a) says, Calculate the efficiency of energy transfer to the copepods.
i.e efficiency of energy transfer from phytoplankton to copepods =[tex]\frac{X}{W}*100%[/tex]
=[tex]\frac{7514}{120000}*100[/tex]
= 6.25%
b) Calculate the efficiency of energy transfer from the secondary consumer to marine birds.
i.e efficiency of energy transfer from small fish to marine birds = [tex]\frac{Z}{Y} *100[/tex]
=[tex]\frac{14}{383} *100[/tex]
=3.65%
In this environment, the color of Sea Lampreys is an example of: Select one: a. A consequence of non-random mating. b. Natural selection. c. A heterozygous advantage. d. An adaptation.
Answer:
Option d. An Adaption is correct.
Explanation:
Young Sea Lampreys usually feed on algae and other organisms that live on the bottom of sea. Due to this feeding activity it need to blend in the environment. This results in the adaption due to skin color (darker on top so that predator cannot see them and fading to a lighter colored belly because they move at the bottom).
The pKa for the side chain of histidine is 6.0. What is the ratio of the deprotonated imidazole side chain to the protonated side chain at pH 5.0 and at pH 7.5?
Answer:
When the pKa is 6.0, we can determine the fraction of protonated H is by:
pH = pKa + log [A]/[HA]
Where
A = Deprotonated imidazole side
HA = Protonated side
Given, pH = 5.0
5 = 6 + log [A]/[HA]
log [A]/[HA] = -1 (take antilog of both side)
[A]/[HA] = 0.1
The ratio of the deprotonated imidazole side chain to the protonated side chain at pH 5.0 = 0.1
Given, pH = 7.5
7.5 = 6 + log [A]/[HA]
log [A]/[HA] = 1.5 (take antilog of both sides)
[A]/[HA] = 31.62
The ratio of the deprotonated imidazole side chain to the protonated side chain at pH 5.0 = 31.62
Final answer:
The ratio of deprotonated to protonated histidine side chain at pH 5.0 is 0.1, and at pH 7.5 is 31.62, calculated using the Henderson-Hasselbalch equation.
Explanation:
The ratio of deprotonated to protonated histidine side chain can be calculated using the Henderson-Hasselbalch equation:
pH = pKₐ + log ([A⁻]/[HA])
where pKₐ is the acidity constant of the side chain, pH is the environment's pH, [A⁻] is the concentration of the deprotonated form, and [HA] is the concentration of the protonated form.
Calculation at pH 5.0:
pH = pKₐ + log ([A⁻]/[HA])
5.0 = 6.0 + log ([A⁻]/[HA])
log ([A⁻]/[HA]) = -1
[A⁻]/[HA] = 10⁻¹
[A⁻]/[HA] = 0.1
Calculation at pH 7.5:
pH = pKₐ + log ([A⁻]/[HA])
7.5 = 6.0 + log ([A⁻]/[HA])
log ([A⁻]/[HA]) = 1.5
[A⁻]/[HA] = 10^1.5
[A⁻]/[HA] = 31.62
At pH 5.0, the ratio of deprotonated to protonated histidine side chain is 0.1, meaning there is 10 times more protonated form present. At pH 7.5, the ratio is 31.62, indicating there is significantly more deprotonated form present.
Only one species of moth, Xanthopan morgani, is known to have a tongue long enough to reach the nectar in the Madagascar orchid Angraecum sesquipedale.
Would the orchid be more likely to reproduce if, through many generations, its nectaries became shorter so that other insect species were able to serve as pollinators? (Select all that apply.)a.Yes. If more insect species are able to use the orchid for food, then the orchid is more likely to reproduce.b.Yes. We can predict that the nectaries of these orchids will tend to shorten enough that other pollinators can also feed on them.c.No. If the nectaries were shorter, then the moth Xanthopan morgani would be able to feed without rubbing against the pollen.d.No. If the nectaries were shorter, then insect species that use other flowers as food sources would also drink its nectar, and it would be less likely these insects would visit another orchid of the same species while carrying its pollen.e.All of the above are plausible. It is impossible to predict.
Answer: e.All of the above are plausible. It is impossible to predict.
Explanation:
The long tongue of moth Xanthopan morgani is able to derive the nectar from the orchid. If through many generations the nectaries become shorter the moth may not be able to derive the nectar this may facilitate the other insect species to derive the nectar and pollination.
If more insect species will able to use the orchid for nectar then obviously the orchid is more likely to reproduce due to pollination. The nectaries may get even shorter that the nectar will readily available to the other species of insects. This may also interrupt the pollination of other species of orchids as Madagascar orchid is a source of nectar which will attract many pollinators also those were specific to the other orchid species.
All conditions are plausible. But several generations of evolution of the Madagascar orchid is required to be observed to predict the association of insects with the orchid.
Shorter nectaries in the orchid would increase the likelihood of other insect species serving as pollinators, but it wouldn't guarantee that the moth Xanthopan morgani can feed without rubbing against the pollen. This would not deter other insect species from visiting other orchids of the same species while carrying its pollen.
B and D are the correct answers.
Explanation:B and D are the correct answers.
If the nectaries of the orchid became shorter, it would allow other insect species to serve as pollinators. This would increase the likelihood of reproduction for the orchid as more insects would be able to access the nectar and potentially transfer pollen.
However, it is incorrect to select answer choices A and C. While more insect species being able to use the orchid for food increases the likelihood of reproduction, the reduction in nectary length would not guarantee that the moth Xanthopan morgani can feed without rubbing against the pollen. Furthermore, shorter nectaries would not necessarily deter other insect species from drinking the nectar and visiting other orchids of the same species while carrying its pollen.
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A dark-red strain and a white strain of wheat are crossed and produce an intermediate, medium-red F1. When the F1 plants are interbred, an F2 generation is produced in a ratio of 1 dark-red : 4 medium-dark-red : 6 medium-red : 4 light-red : 1 white Further crosses reveal that the dark-red and white F2 plants are true breeding.
a. Based on the ratios in the F2 population, how many genes are involved in the production of color?
b. How many additive alleles are needed to produce each possible phenotype?
c. Assign symbols to these alleles and list possible genotypes that give rise to the medium red and light red phenotypes.
d. Predict the outcome of the F1 and F2 generations in a cross between a true-breeding medium red plant and a white plant.
a). 2 genes are involved in production of colour in F2 generation.
b). 2 possible alleles for each phenotype
c).Rr (medium red), rr light red
d) F1 generation 1:2:1 ( 1 light red stain, 2 medium red) and F2 generation 1:1
1 medium red and 1 light red.
Explanation:
F1 generation:
dark red strain X white strain
RR X rr
r r
R rR Rr
R rR Rr
genotype ratio 4:1, all medium red
Cross between F1 generation of true breeding plants
R r
R RR Rr
r Rr rr 1 light red stain, 2 medium red, 1 white genotype ratio 1:2:1
In F2 Generation, F1 plants interbred ie true breeding
F2 generation:
R r
r Rr rr
r Rr rr
It produces 1 medium red and 1 light red plant. genotype ratio 1:1
Here alleles Rr is for medium red and rr is for light red
Additive allele depends on the allele concentration for a phenotype trait to appear.
Final answer:
The ratio of phenotypes in the F2 generation indicates that two genes are involved in the production of color in wheat, with a total of four additive alleles influencing the phenotype. True-breeding medium red (RaRa) crossed with white (rraa) would yield medium-red F1 offspring. F2 would show a ratio of 3 medium-red to 1 white.
Explanation:
The phenotypic ratio of the F2 generation, which is given as 1 dark-red : 4 medium-dark-red : 6 medium-red : 4 light-red : 1 white, suggests the involvement of two genes in color production due to the pattern not matching a simple 3:1 monohybrid cross. It fits a modified dihybrid cross ratio of 1:4:6:4:1, which is reminiscent of a 9:3:3:1 ratio with intermediate forms due to incomplete dominance.
To determine the number of additive alleles needed for each phenotype:
Dark-red: All four additive alleles are present (2 from each gene).
Medium-dark-red: Three additive alleles are present.
Medium-red: Two additive alleles are present.
Light-red: One additive allele is present.
White: No additive alleles are present.
We can assign the symbols R for the allele contributing to red color and r for its absence from one gene, and A for the allele contributing to red color and a for its absence from the second gene.
Thus, possible genotypes for medium red are RaRa or rrAA. For light red, the genotype could be Ra with the other gene being rr or aa, such as Rraa or Aarr.
For a cross between a true-breeding medium red plant (RaRa) and a white plant (rraa), the F1 would all be Rara (medium-red), and the F2 generation would show a 3:1 phenotypic ratio of medium-red (R-) to white (-rr) since the white parental plant can only contribute recessive alleles.
In a study of black guinea pigs and white guinea pigs, 100 black animals were crossed with 100 white animals and each cross was carried to an f2 generation. In 94 of the crosses, all the F1 offspring were black and an F2 ratio of 3:1 (black:white) was obtained.In the other 6 cases half of the F1 animals were black and the other half were white. Why? Predict the results of crossing the black and white guinea pigs from the 6 exceptional cases.
In genetics, the color of guinea pigs is determined by the presence or absence of a dominant allele for black coat color. The F2 ratio of 3:1 (black:white) suggests that black coat color is dominant. In the exceptional cases, the black guinea pigs were homozygous recessive and the white guinea pigs were heterozygous.
Explanation:In genetics, the color of guinea pigs is determined by the presence or absence of a dominant allele for black coat color. Since all the F1 offspring were black in 94 of the crosses, this suggests that black coat color is dominant. The F2 ratio of 3:1 (black:white) indicates that the black guinea pigs were heterozygous, and the white guinea pigs were homozygous recessive.
In the 6 exceptional cases where half of the F1 animals were black and the other half were white, it suggests that the black guinea pigs were homozygous recessive and the white guinea pigs were heterozygous. When these guinea pigs are crossed, the expected result would be a ratio of 1:1 (black:white) in the F2 generation.
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1.) Which processes produce carbon dioxide as a waste product?
a.) the Krebs cycle and the electron transport chain
b.) ethyl alcohol fermentation and the Krebs cycle
c.) ethyl alcohol fermentation and lactic acid fermentation
d.) lactic acid fermentation and the Krebs cycle
2.) The bonds between the phosphate groups in ATP have large amounts of chemical _____ energy.
a.) potential
b.) kinetic
c.) low
d.) high
3.) Where does the sour taste come from in foods such as cheese and yogurt?
a.) ethyl alcohol produced by fermentation
b.) lactic acid produced by fermentation
c.) ATP produced by the Krebs cycle
d.) carbon dioxide produced by cellular respiration
4.) What molecule is used by the enzyme ATP synthase to form ATP?
a.) FADH 2
b.) NADH
c.) ATP
d.) ADP
5.) What conditions cause cells to break down fat molecules?
a.) limiting calorie intake and increasing energy needs
b.) increasing energy needs and increasing calorie intake
c.) decreasing energy needs and increasing calorie intake
d.) limiting calorie intake and limiting oxygen exposure
6.) Why is ATP an example of chemical potential energy?
a.) It stores energy until a cell needs it.
b.) It drives reactions to make glucose.
c.) It creates and destroys energy.
d.) It assembles mitochondria.
7.) Which set of pairings correctly matches the process with its conditions?
a.) cellular respiration : aerobic : : fermentation : anaerobic
b.) cellular respiration : anaerobic : : fermentation : aerobic
c.) cellular respiration : anaerobic : : fermentation : anaerobic
d.) cellular respiration : aerobic : : fermentation : aerobic
8.) The purpose of cellular respiration is to enable cells to create and use _____.
a.) oxygen
b.) DNA
c.) carbon dioxide
d.) ATP
9.) How does ADP differ from ATP?
a.) ADP has one more adenine group than ATP.
b.) ADP has one more phosphate group than ATP.
c.) ATP has one more phosphate group than ADP.
d.) ATP has one more adenine group than ADP.
10.) What is involved in redox reactions?
a.) the addition of water to break down food macromolecules
b.) the bonding of ions to form chemical compounds
c.) the transfer of electrons between reactants
d.) the breaking down of water into hydrogen and oxygen atom
I set it for 64 points, so you will receive approximately 32 points.
Please answer as many as you can. Most answers out of the two answerers will become brainliest.
The ATP bonds have potential energy that is stored in it.
The bonds between the phosphate groups in ATP have large amounts of chemical potential energy but after breaking of bonds, this potential energy is converted into kinetic energy which is used to move things.
The sour taste come in foods such as cheese and yogurt from lactic acid produced by fermentation. Sour taste in food is mainly occurs due to the presence of citric, malic, oxalic, and tartaric acids in fruits and lactic acid in yogurt.
Limiting calorie intake and increasing energy needs are the conditions that cause cells to break down fat molecules.
ATP is an example of chemical potential energy because it stores energy until a cell needs it for performing different activities.
Cellular respiration is aerobic means needs oxygen for the generation of energy whereas fermentation is an anaerobic means it occurs in the absence of oxygen.
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Why do black redstarts exhibit migratory restlessness at night for fewer days than common redstarts?
Answer:
The statistics given has shown that night-migrating birds are driven by autonomous circadian clocks which is made possible by sunset cues. This timekeeping that is the autonomous circadian clocks system is probably the key factor in the overall control of nocturnal songbird migration.
Answer: Blact redstarts are associated with circadian clock. Seasonal melatonin and circadian clock lead to migratory nocturnal restlessness. The birds that exhibit migratory restlessness at night is due to high levels of melatonin at nights because melatonin are high in the night.
Explanation:
Migratory restlessness is the locomotive performance that birds engaged in before migration. It is common in birds with circadian rhythm.
The modern synthesis brought together Darwin's theory of evolution with Mendelian genetics. Why was this so important?
Answer:
Option C
Explanation:
Options for the question are
a. It demonstrated that nothing in biology makes sense except in the light of evolution.
b. It demonstrated that species arise from other species.
c. It provided a quantifiable mechanism to explain gradual evolutionary change.
d. It was the first time that change in biological organisms was observed.
e. It was not very important to evolutionary biology.
Solution -
The term modern synthesis by coined by Julian Huxley and the goal of this synthesis was to understand and determine how genetics bring evolutionary changes in a population and with this objective it combined the Darwin’s theory of evolution and Mendel’s work on heredity. Later on scientist Dobzhansky under this synthesis found that genetic mutation causes variation through the process of natural selection and hence lead to evolutionary changes in a population even if they are isolated for a long time.
Hence, option C is correct
Based on the following DNA sequence comparisons, which two unidentified microorganisms would you conclude are most closely related? Choose one: A. Both microbial genomes include a gene that encodes for streptokinase, an enzyme and virulence factor that activates human plasminogen. B. Both microbial genomes include genes that encode for protein components of the prokaryotic ribosome. C. Both microbial genomes include genes that encode for photosynthetic photoreceptor proteins. D. Both microbial genomes include a gene that encodes for nitrogenase, a nitrogen-fixing enzyme.
Answer:
A. Both microbial genomes include a gene that encodes for streptokinase, an enzyme and virulence factor that activates human plasminogen.
Explanation:
Streptokinase (Ska) is a single-chain 414 amino acid protein, produced by certain Streptococci species (group A, C, and G streptococci), which possess an activity closer to two host proteins (urokinase-type and tissue-type plasminogen activators), as it non-enzymatically process inactive plasminogen to proteolytically active plasmin.
Streptokinase is a species-specific virulence factor. If both microbial genomes produces streptokinase, then these microorganisms can both be grouped as Streptococci since they are closely related.
All other choices do not certainly ease the relationship beyond the domain level.
Answer:
A. Both microbial genomes include a gene that encodes for streptokinase, an enzyme and virulence factor that activates human plasminogen.
Explanation:
Streptokinase is an enzyme (also synthesize as a medication) produced specifically by streptococci spp. If the two unidentified microbes could produce this enzyme, then they should contain the gene that codes for streptokinase .Thus they share certain genome characteristics, and are related.
Other options related the similarities to the domain levels only
vanadium crystallizes in a body-centered cubic lattice and the density is 5.96g/cm3 what is the unit cell edge length in pm?
Final answer:
The unit cell edge length of vanadium in a body-centered cubic lattice is approximately 303.4 pm.
Explanation:
Vanadium crystallizes in a body-centered cubic (BCC) lattice. In a BCC structure, there are atoms at the eight corners of the unit cell and one atom in the center. The edge length (a) of the unit cell can be calculated using the formula:
a = (4 * radius) / √3
Given that the density of vanadium is 5.96 g/cm3, we can use its molar mass (50.9415 g/mol) and Avogadro's number (6.022 × 10^23 atoms/mol) to calculate the radius of a vanadium atom using the formula:
density = (molar mass * number of atoms) / (volume of unit cell)
By rearranging the formulas and substituting the given values, we can find the unit cell edge length:
a = (√[3 * molar mass * Avogadro's number / (4 * density)]) / 10
Substituting the values gives:
a = (√[3 * 50.9415 * 6.022 *10^23 / (4 * 5.96)]) / 10
Simplifying the expression gives:
a ≈ 3.034 Å
To convert Ångströms (Å) to picometers (pm), multiply by 100:
a ≈ 3.034 * 100 = 303.4 pm
Explain the significance of complementary base pairing in the conservation of the base sequence of DNA.
Which of the following statements about apoptosis is false? a. It is an important process in the development of human fingers and toes. b. Despite more than 600 million years of evolutionary divergence, humans and nematodes share similar apoptosis pathways. c. Because apoptosis genes kill cells, natural selection is seldom involved in apoptosis pathways. d. All of the above are true; none is false.
Answer:
C) Because apoptosis genes kill cells, natural selection is seldom involved in apoptosis pathways.
Explanation:
Apoptosis is a natural process taking in the cell which causes the physiological and biochemical changes in the cell which could cause the death of the cell.
The apoptosis is controlled at the genetic level therefore apoptosis is also known as the programmed cell death.
The studies have shown that the process is involved in the development of the finger and the toes in humans and the sequences controlling the process has been conserved during the evolution in a different organism.
This shows that humans and nematode have the same conserved sequence of apoptosis but the natural selection does not control the apoptosis.
Thus, the selected option is the correct answer.
Apoptosis is involved in human development and shared across species like humans and nematodes. It's incorrect to state that natural selection plays no role in apoptosis pathways, as these pathways have evolved to benefit the overall health of the organism.
Explanation:The question asks which statement about apoptosis is false. Let's go through each item:
a) Apoptosis does play a key role in the development of human fingers and toes. During embryological development, apoptosis is meant to eliminate the web-like tissues between individual fingers and toes to form fully separated digits.
b) Indeed, despite the vast evolutionary divergence, humans and nematodes share similar apoptosis pathways.
The false statement here is c) Even though apoptosis does involve killing off cells, natural selection plays a role here as well. It's involved in apoptosis pathways as these pathways have evolved to maintain the health of the organism by removing damaged or unneeded cells and preventing dangerous ones from proliferating.
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