Answer:
B - increases (deplolarization.)
Explanation:
the influx of sodium ion into the cell increases the positive charge of the axoplasm, cause charges reversal of the axon membrane and it is called depolarization.
Th influx of sodium ion is due combination of three factors
1, to the opening of voltage gated sodium channels,
2. increase in chemical gradient for sodium ions, and
3. high concentration of sodium ions outside compare to the axoplasm.
The combined effects of 2 and 3 above is called electrochemical gradients; and this pull sodium ions with the psotive charges through the voltage gated sodium channels into the axoplasm.
Few gated channels were open initially however as the intensity of the stimulus increases, more gated channels of sodium opens, with many Na+ diffused in.And if the voltage generated due to deoplarization is up to the threshold levels,Action potential occurs.
During an action potential, sodium (Na+) levels increase inside the cell due to the activity of the sodium-potassium pump in the cell membrane, which creates an electrochemical gradient. Option B
Explanation:During an action potential, sodium (Na+) levels increase inside the cell. This is due to the sodium-potassium pump in the cell membrane, which pumps three Na+ ions into the cell for every two K+ ions it pumps out.
This creates a short-term electrochemical gradient, with a higher concentration of Na+ ions inside the cell compared to outside. This change in concentration and the resulting electrical charge difference is what allows the signal to be transmitted down the neuron.
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Do all animals share a common ancestor?
Answer:
The most recent common ancestor of all currently living organisms is the last universal ancestor, which lived about 3.9 billion years ago. ... 6,331 groups of genes common to all living animals have been identified; these may have arisen from a single common ancestor that lived 650 million years ago in the Precambrian.
Which of the following is not true regarding the denaturation and reannealing of double-stranded DNA molecules? Which of the following is not true regarding the denaturation and reannealing of double-stranded DNA molecules? Decreasing the salt concentration of the solution lowers DNA's melting point (Tm). Increasing the G-C content of DNA raises its melting point (Tm). Single-stranded DNAs can only anneal to one another if they are 100% identical in nucleotide sequence. Melting point (Tm) is the temperature at which the DNA is one-half double-stranded and one-half single-stranded. DNA is more likely to denature when exposed to a high pH.
Answer: "Decreasing the salt concentration of the solution lowers DNA's melting point (Tm)" is not a true statement
Explanation:
Increasing salt concentration would lower the DNA's melting point (Tm), not otherwise.
For instance:
- In 8M urea (8M means 8 Moles per dm3), Tm is decreased by nearly 20°C.
- 95% formamide at room temperature would completely denature the double stranded DNA.
Thus, higher concentration of salts like urea or formamide lowers Tm, not otherwise
When giving a talk about the theories of Marshall McLuhan, Sandy first made sure to thoroughly explain the term "media ecology." This is an example of_________.
Answer:
Giving a definition.
Explanation:
Giving a definition deals with properly explaining the term(s) concerned to avoid further confusion and to make sure your audience ain't lost. It is not merely to copy what was printed in other dictionaries that gives a definition. Choose terms that your understanding or that personal observations has affected. By defining its roles, structure or existence, you can define a phrase.
A scientist discovers a new organism living deep beneath the earth surface this organism thrive despite intense pressure heat lack of water and sunlight this organism is probably a
Answer:
The organism that can live deep beneath the earth surface despite intense pressure heat lack of water and sunlight might be Nematodes.
Explanation:
Nematodes are able to cope extreme heat or extreme cold and dehydration. They have adopted by learning technique that allows them to survive. They can transform into a hardy form called the dauer stage. They can survive harsh conditions for longer durations at this stage. And again awaken themselves when conditions are favourable again. They can be found in hot springs, deserts, high up mountains and in the deepest oceans.What name is given to the rigid structure, found outside the plasma membrane, that surrounds and supports the bacterial cell?
The rigid structure that supports and envelopes the bacterial cell outside the plasma membrane is the cell wall, primarily composed of peptidoglycan. This provides the cell with strength, maintains its shape, and shields it from environmental threats.
Explanation:The rigid structure that surrounds and supports the bacterial cell, which is found outside of the plasma membrane, is called the cell wall. Bacterial cell walls are primarily composed of peptidoglycan, a large polymer of amino sugars. These sugars are linked together in a mesh-like structure providing the cell with strength and rigidity. The cell wall helps maintain the shape of the cell and also protects it from environmental threats such as changes in osmotic pressure.
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The rigid structure found outside the plasma membrane that surrounds and supports the bacterial cell is called the "cell wall."
The cell wall is a crucial component of bacterial cells, providing structural support and protection. It is a tough, rigid layer composed of peptidoglycan, a unique macromolecule that consists of sugar chains cross-linked by short peptides. The cell wall maintains the cell's shape and prevents it from bursting due to changes in osmotic pressure. It also plays a significant role in bacterial pathogenicity, as differences in cell wall structure are used to classify bacteria into two major groups: Gram-positive and Gram-negative.
Gram-positive bacteria have a thick layer of peptidoglycan, while Gram-negative bacteria have a thinner layer of peptidoglycan surrounded by an outer membrane. The cell wall is an essential target for antibiotics like penicillin, which disrupt its integrity and ultimately lead to bacterial cell death.
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If Meselson and Stahl had used CsCl gradient analysis and identified DNA molecules with two distinct densities after the first generation, which model of DNA replication would have been supported by this data?
O conservative replication model
O semiconservative replication model
O dispersive DNA replication model
Answer:
conservative replication model
Explanation:
In their experiment, Meselson and Stahl found that the DNA duplexes present after the first round of replication had intermediate density. This occurred since the semiconservative process of DNA replication of one DNA molecule (having both strands with 15N) forms two DNA duplexes each with one parental strand (15N) and one newly formed strand (14N).
If they had found two duplexes of two different density, this would have supported the conservative replication model. In this case, the DNA replication of the parent DNA molecule would have formed two DNA duplexes. One DNA molecule would have both the parental DNA strands with 15N while the other DNA duplex would have both newly formed strands with 14N.
If Meselson and Stahl had observed two distinct densities after the first generation using CsCl gradient analysis, it would have supported the conservative replication model. However, their actual findings supported the semiconservative replication model.
If Meselson and Stahl had used CsCl gradient analysis and identified DNA molecules with two distinct densities after the first generation, the model of DNA replication that would have been supported by this data is the conservative replication model. In this model, the parental DNA strands (blue) would remain associated in one DNA molecule while the new daughter strands (red) would remain associated in newly formed DNA molecules. However, Meselson and Stahl actually found a single band after one generation, which disproved the conservative model and supported the semiconservative replication model.
Cells of a normally rod-shaped bacterium (e.g., Bacillus subtilis) that have completely lost the ability to produce the MreB protein would mostly likely be Choose one: A. nonflagellated. B. stalked (like Caulobacter). C. coccoid-shaped. D. filamentous in form. E. unable to divide symmetrically.
Answer:
C. coccoid-shaped.
Explanation:
MreB protein serves as function as actin proteins do in the eukaryotic cells. MreB protein is involved in maintaining the cell shape. It is observed in many rod-shaped bacteria and archaea such as Bacillus subtilis. MreB polymerizes to form a spiral around the inside periphery of the cell and maintain the elongated shape of the cell. A bacterium that is not able to produce functional MreB protein cannot maintain the rod shape of its cell and is coccoid-shaped.
A dominant mutation in Drosophila called Delta causes changes in wing morphology in Delta/+ heterozygots. Homozygosity for this mutation (Delta / Delta) is lethal. In a population of 150 flies, it was determined that 60 bad normal wings and 90 had abnormal wings. a. What are the allele frequencies in this population? b. Using the allele frequencies calculated in part (a), how many total zygotes must be produced by this population in order for you to count 160 viable adults in the next generation? c. Given that there is random mating, no migration, and no mutation, and ignoring the effects of genetic drift, what are the expected numbers of the different genotypes in the next generation if 160 viable offspring of the population in part (a) are counted? d. Is this next generation at Hardy-Weinberg equilibrium? Why or why not?
What causes the myosin head to move into the 'cocked' position after it is released from actin? Group of answer choices ATP binding to myosin. ATP hydrolysis into ADP and Pi on the myosin head. ADP and Pi release from the myosin head Ca2+ binding to troponin.
Answer:
Explanation:
The molecular and cellular mechanisms and processes that explain muscle contraction in striated muscle occur in the myofibril sarcomere. Their understanding depends on the organization's understanding of the structure of the sarcomere. In an imaginary experiment we first assemble an ideal sarcomere.
Remember that the myofibril is a set of cylindrical compartments that are located next to each other, constituting an elongated cylinder. Each of these cylinders is a sarcomere and borders its neighbor on a line or band called, line or band z.
On each side of the z line, thin cylindrical filaments that are actin filaments are inserted. Each actin filament is formed by a double strand of actin molecules that are rolled over each other. In this organization, actin is called actin F.
The myosin head is 'cocked' into a high-energy state due to ATP hydrolysis into ADP and Pi, which occurs after ATP binds to myosin and it detaches from actin.
Explanation:The cause of the myosin head moving into the 'cocked' position after it is released from actin is ATP hydrolysis into ADP and Pi on the myosin head. When ATP binds to myosin, it results in the release of the myosin head from actin. Subsequently, the hydrolysis of ATP to ADP and inorganic phosphate (Pi) occurs due to the ATPase activity intrinsic to myosin. This process releases energy that changes the angle of the myosin head into the 'cocked' configuration. In this high-energy state, the myosin head has potential energy and is prepared for further movement upon binding to actin again, which is essential for muscle contraction.
An RFLP can be created:_________. a) only by a mutation creating a new restriction enzyme recognition site b) only by a basepair change within the restriction enzyme recognition site c) only by a deletion removing a restriction enzyme reognition site d) by any mutational event that cretes or deletes a restriction enzyme recognition site e) only in RNA
Answer:
Option B, only by a basepair change within the restriction enzyme recognition site
Explanation:
A restriction fragment length polymorphism (RFLP) consists of alternative alleles with varying sizes of restriction fragments. In the traditional RFLPs, base pairs were changes at the restriction sites. These restriction sites comprises of nucleotide sequences that are identified by restriction enzymes. In RFLP analysis, the restriction fragments are created when restriction enzyme divide DNA into fragments. These restriction fragments are then separated while gel electrophoresis.
Before the advent of PCR, RFLPs (which were predominant form of DNA variation) were used to analyze linkages.
Hence, option B is correct
Two autosomal genes control horn color in dragons. Pure-breedinggold-horned dragons were mated to pure-breeding silver-horneddragons. All of the F1 were gold. The F1 were intermated and the F2generation consisted of 147 gold, 17 silver and 92 bronze.
Fill in the blanks with whole numbers to indicate the geneticallybased phenotypic ratio that should be hypothesized to explain theF2 data.
gold: silver: bronze
Conduct a chi-square test to test the appropriate type ofepistasis.
In the chi-square test,
The expected number of gold-horned dragons is
The expected number of silver-horned dragons is
The expected number of bronze-horned dragons is
Answer:
The genetically based phenotypic ratio for of Gold: Silver: Bronze = 9:1:6
The expected number of gold-horned dragons is = 144
The expected number of silver-horned dragons is = 16
The expected number of bronze-horned dragons is = 96
Explanation:
Given that; F2 generation consisted of:
147 gold
17 silver
92 bronze.
Total autosomal genes control horn color in dragons = 256
When we conducted the punnet square for the Pure-breedinggold-horned dragons were mated with the pure-breeding silver-horneddragons, then intercrossed them for the F2 generation, It is seen that;
The genetically based phenotypic ratio for of Gold: Silver: Bronze = 9:1:6
∴
The expected number of gold-horned dragon = [tex]256*\frac{9}{16}[/tex]
= 144
The expected number of silver- horned dragon = [tex]256*\frac{1}{16}[/tex]
= 16
The expected number of Bronze-horned dragon = [tex]256*\frac{6}{16}[/tex]
= 96
Null Hypothesis = The observed ratio of Gold:Siver:Bronze
=144:16:96
=9:1:6
Observed Expected (OF - EF)² [tex]\frac{(OF-EF)^2}{EF}[/tex]
Frequency Frequency
(OF) (EF)
147 144 9 0.062500
17 16 1 0.062500
92 96 16 0.166667
Total: 0.291667
Since Degree of Freedom = 2
Our P-Value = 0.864
∴ Our P. Value is very high that we accept the null hypothesis
Which of the following is true of both starch and cellulose? a. They are both polymers of glucose. b. They are geometric isomers of each other. c. They can both be digested by humans. d. They are both used for energy storage in plants. e. They are both structural components of the plant cell wall.
Answer: Option A. They have both polymers of glucose
Explanation:
Starch and cellulose are both polymers of glucose. They have similar structure, Starch and cellulose are two similar polymers. They are both made from the same monomer, glucose, and have the same glucose-based repeat units. They consists of long chains of glucose molecules connected to (1-4)-glycosidic bonds. Glycosidic bonds are the standard way of attaching things to glucose. The (1-4) bit simply means that the glucose molecules in the chain are connected to each other to the 1st and 4th carbon in the glucose ring opposite each other.
The key similarity between glycogen and cellulose is that both are polymers of glucose.
Glycogen:
It is a polymer of glucose that is stored in animals as a storage of energy.
Cellulose:
It is also a polymer of glucose. It forms the cell wall of glucose.In plants starch work as the storage of energy.Therefore, the key similarity between glycogen and cellulose is that both are polymers of glucose.
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An organism is classified as a heterotroph or autotroph based on the type of ________ it utilizes.. A. respiration source (e.g., oxygen or other) B. nitrogen source C. carbon source D. none of the above
Answer: Option C) carbon source
Explanation:
The type of Carbon source is the distinction between heterotroph or autotroph. As autotrophs like green plants generate their food (glucose) from simple inorganic molecules like carbon dioxide (CO2) in the atmosphere, while heterotroph like man utilize carbon in carbohydrates such as starch that have been stored in the tissues of autotrophs.
Thus, the type of carbon source utilized distinguished heterotroph and autotrophs
If a cell contains a set of duplicated chromosomes, does it contain any more genetic information than the cell before the chromosomes were duplicated?
Answer:
No.
Explanation:
When the set of chromosomes are duplicated, it means the chromosomes will be double the number.
However the genetic information will not change. It will remain the same.
Duplication of chromosomes is usually occurs among the interphase stage of mitosis where chromosomes will duplicate to ensure the formation of two identical daughter cells.
In these cells there will be no addition or variation in genetic information.
Final answer:
A cell with duplicated chromosomes doesn't contain more genetic information than before; it simply has two identical copies of each chromosome, called sister chromatids. The total DNA in the cell is increased, but genetic content remains consistent. This replication ensures that each daughter cell gets a complete genetic set after cell division.
Explanation:
If a cell contains a set of duplicated chromosomes, it does not contain more genetic information than it did before the chromosomes were duplicated. During DNA replication, each chromosome in the cell creates an exact copy of itself, known as sister chromatids, which are attached at a region called the centromere. While the chromosome number remains the same, for instance in humans (n = 46), the physical amount of DNA within the cell is actually doubled. However, in terms of genetic content, the information stays the same, it's just that each chromosome now consists of two identical chromatids.
During the cell cycle, specifically in the S phase, the DNA replication occurs to prepare a cell for division into two cells. This process is crucial because it ensures that each daughter cell receives a complete set of the organism's genome. In human cells, after replication, there still are 46 chromosomes, but each chromosome is now made up of two sister chromatids. Therefore, after cell division during mitosis, each cell will have the same amount of genetic material, identical to the original parent cell, assuring the continuity of genetic information.
Which of the following statements is true regarding DNA replication? A. The 5' to 3' orientation of the new strand will depend upon where it is in the replication bubble relative to the position where replication first starts. B. The new strand is the opposite of the template strand. C. The new strand is an exact complement to the template strand. D. The new strand is exactly the same as the template strand.
Answer:
These options are true about DNA.
A. The 5' to 3' orientation of the new strand will depend upon where it is in the replication bubble relative to the position where replication first starts.
B. The new strand is the opposite of the template strand.
C. The new strand is an exact complement to the template strand.
Explanation:
Since the two strands of a DNA double helix are antiparallel, this 5′-to-3′ DNA synthesis can take place continuously on only one of the strands at a replication fork (the leading strand). On the lagging strand, short DNA fragments must be made by a “backstitching” process. The opposite strand with a base sequence directly corresponding to the mRNA sequence is refereed as coding strand. See images for clarification.
If you want to study more here is the reference:
Alberts B, Johnson A, Lewis J, et al. Molecular Biology of the Cell. 4th edition. New York: Garland Science; 2002. DNA Replication Mechanisms.
The new strand formed during DNA replication is an exact complement to the template strand.
Explanation:The correct statement regarding DNA replication is C. The new strand is an exact complement to the template strand. DNA replication is a semi-conservative process where each of the original parent DNA strands acts as the template for the synthesis of a new complementary strand. The parent DNA strands separate, and each strand serves as a template for the synthesis of a new strand. Each base on the parent strand pairs with its complementary base on the new strand, following the base-pairing rules (A with T and G with C).
Therefore, the new strand formed during DNA replication is an exact complement to the template strand. In other words, it has the opposite nucleotide sequence of the template strand, with the same specific adenine (A) to thymine (T) and guanine (G) to cytosine (C) ratios.
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An mRNA molecule has a sequence 5'- CAGAUCUAAUGCUUAUCGGAU-3'. When translated in a laboratory setting in which translation can be initiated anywhere along the molecule, how many reading frames are possible?
Answer:
Three reading frames
Explanation:
Translation of mRNA always happen in one direction from the 5' end to the 3' end of the RNA strand.
Reading frame refers to the grouping of three consecutive bases to form a codon that can constitute an amino acid.
There are six possible reading frames in any nucleotide sequence.
Three from the 3' to the 5' end and three possible reading frames from the 5' to the 3' end.
As mentioned earlier, translation in mRNA happens in one direction therefore the three possible reading frames are;
5'- C AGA UCU AAU GCU UAU CGG AU-3'. 5'- CA GAU CUA AUG CUU AUC GGA U-3'. 5'- CAG AUC UAA UGC UUA UCG GAU-3'.An mRNA sequence can be translated into a protein in three possible reading frames, therefore, for the given sequence that can be initiated anywhere, there are three possible reading frames.
Explanation:In the realm of molecular biology, an mRNA sequence can be translated into a protein in three possible reading frames. Each reading frame will read the sequence from a different starting point, thereby creating a unique series of codons, and potentially, a different protein sequence. Hence, for an mRNA sequence such as 5'- CAGAUCUAAUGCUUAUCGGAU-3' that can be translated anywhere in a laboratory setting, it means there are 3 possible reading frames. Each frame begins from a different nucleotide within the first three (frame 1 from C, frame 2 from A, frame 3 from G) and continues in triplets from there on.
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A woman with Turner syndrome is found to be color-blind (an X-linked recessive phenotype). Both her mother and her father have normal vision. a. Explain the simultaneous origin of Turner syndrome and color blindness by the abnormal behavior of chromosomes at meiosis. b. Can your explanation distinguish whether the abnormal chromosome behavior occurred in the father or the mother? c. Can your explanation distinguish whether the abnormal chromosome behavior occurred at the first or second division of meiosis? d. Now assume that a color-blind Klinefelter man has parents with normal vision, and answer parts a, b, and c
Answer:
A) The turner syndrome is characterized by lacking half chromosome numbers which are also termed as the monosomy as there is only half the number of X linked chromosomes that are responsible for the colorblindness will be greater. The chances are 50% of having this disorder in such a case.
B) The correct answer would be no, as the error will be shown in only one of the due to n number of chromosomes present in both parents.
C) Anaphase II of meiosis II it takes places as here only the strands are pulled to opposite poles of the cell
.
D) Assuming it as Klinefelter man with colorblindness has normal parents then,
a)double XX +Y , the colorblind gene is recessive in XX and Y. or recessive in XX but dominating Y only rules out the other recessive.
b) Explanation is similar as B)
c) Due to a nondisjunction event during meiosis, I or meiosis II in the female X chromosome is still present in it.
The type of synapse that occurs between the terminal end of the presynaptic cell and the terminal end of the postsynaptic cell is referred to as which kind of synapse?
Answer:
Synaptic cleft
Explanation:
Synaptic cleft is the gap between the terminal end of the presynaptic cell and the terminal end of the postsynaptic cell. It helps in transportation the neurotransmitters from a synapse to another.
Which of the following is/are true?
A. Redheaded females in a population will only breed with red headed males. The yellow-headed females will only breed with yellow-headed males. This population lives in the same geographic area. The red and yellow individuals could breed and produce fertile offspring, but they normally do not.
This would be an example of sympatric speciation.
B. Sympatric speciation is best described as a random event that disrupts the allele frequencies in a population
C. Redheaded females in a population will only breed with red-headed males. The yellow-headed females will only breed with yellow-headed males. This population lives on separate continents and rarely met in nature. The red and yellow individuals could breed and produce fertile offspring, but they normally do not.
This would be an example of sympatric speciation
D. Sympatric speciation is never due to sexual (mate) selection
E. Sympatric speciation requires geographic isolation
Answer: option B) Sympatric speciation is best described as a random event that disrupts the allele frequencies in a population
Explanation:
Sympatric speciation is an event/situation whereby organisms of the same species:
- live in the same territory or nearby territories ( i.e do not live in geographical isolation)
- DO NOT interbreed, but select a sexual mate from a much diverse territory to yield new species or offsprings.
This sexual selection then results in generations of offsprings that are genetically different from the rest of the same species due to uneven gene flow or disruption of alleles among the population of same species.
Thus, only option B is true.
If you are performing site-directed mutagenesis to test predictions about which residues are essential for a protein’s function. Which of each pair of amino acid substitutions listed below would you expect to disrupt protein structure the most? And why? how do you go abouts knowing this?
O Pro replaced by Gly or His,
O Gln replaced by Glu or Asn,
O Lys replaced by Asp or Arg
Answer:
O Gln replaced by Glu or Asn
Explanation:
Gln replaced by Glu or Asn, is a non-conservative substitution. Gln has a much shorter side chain or R-group than Glu or Asn. Additionally both Glu and Asn R-groups are polar and negatively charged while Gln is non-polar.In a protein, replacement with Glu or Asn would greatly destabilize the protein by introducing a charge and disrupting hydrophobic interactions formed between Gln and other non-polar residues. By comparing the R-groups of the residues, their length, hydrophobicity, charge and shape. One can predict if the substitution is disruptive or not, if residues are very similar then the interactions that they form are preserved. For example replacing Gln with Ala would be less disruptive.If the diastolic blood pressure at the heart level is 80.0 mm Hg, what is the diastolic pressure at the height of the head, which is 0.300 m above the heart? Ignore any pressure drop due to the viscosity of the blood. g
Answer:
56.6 mm of Hg
Explanation:
Given that;
the diastolic blood pressure at the heart level is 80.0 mm Hg = [tex](P_0)[/tex]
height = 0.300 m
the diastolic pressure at the height of the head, which is 0.300 m above the heart can be determined using the formula
P= [tex](P_0) - (pgh)[/tex]
where; [tex](pgh)[/tex] is the hydrostatic pressure applied by the column of the liquid (Blood) of height (h) and average density [tex]p[/tex], also the g = gravitational acceleration.
the average density [tex]p[/tex] of a human blood = 1060 kg/m³
gravitational acceleration. ( g ) = 9.81 m/s²
h = 0.300 m
∴ the [tex](pgh)[/tex] = 1060 kg/m³ × 9.81 m/s² × 0.300 m
= 3119.58 Pascal (Pa)
From the standard conversion rate, 1 mm of Hg(mercury) = 133.322 Pa
∴ the amount of mm of Hg(mercury) that can be gotten from 3119.58 Pascal (Pa) will be; [tex]\frac{3119.58}{133.322}[/tex]
= 23.40 mm of Hg(mercury)
P= [tex](P_0) - (pgh)[/tex]
P= (80.0 - 23.4) mm of Hg
P= 56.6 mm of Hg
Diastolic pressure at the height of the head, which is 0.300 m above the heart is 56.6 mmHg.
Given here,
Diastolic blood pressure at the heart level = 80.0 mm Hg
0.300 m above, diastolic pressure = ?
The Diastolic pressure can be calculated using the formula
[tex]\bold {P_d = h \times g \times \rho }[/tex]
Where,
[tex]\rho[/tex] = average density of a human blood = 1060 kg/m³
g - gravitational acceleration. = 9.81 m/s²
h = Height = 0.300 m
Diastolic pressure,
= 1060 kg/m³ × 9.81 m/s² × 0.300 m
= 3119.58 Pascal (Pa)
Since, 1 mmHg = 133.322 Pa
So, 3119.58 Pascal = 23.40 mmHg
Thus ,
Pd = (80.0 - 23.4) mmHg
Pd = 56.6 mmHg
Therefore, diastolic pressure at the height of the head, which is 0.300 m above the heart is 56.6 mmHg.
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A cross is done between two mutant T4 phage (a- b+ c+) and (a+ b-c)(no order implied by these genotypes). The genotypes of the progeny from this cross and their relative frequencies are given below: a-b+ c+ (30%) a+D c. (30%) a+ b+ c-(2%) a+ b+ c+ (8%) a- b-e+(2%) a-b-c-(8%) e+b-c+ (10%) a-b+ c-(10%) What is the distance between "a" and "b? Please give your answer in % recombination, but L number please! EAVE OUT the units (NO % sign), such as 20, or 3,
Since, the relative frequency is incorrectly mentioned in the question. The correct relative frequency table is attached below.
Answer:
20.
Explanation:
The genotype with the least recombinant frequency will represent the double cross overs.
The double cross overs progeny are a+ b+ c-(2%) and a- b-c+(2%).
The percentage of the recombinant frequency determined the distance between the genes. In the double crossovers, the b gene gas been exchanged and present in the middle.
The single recombinant crossovers as compared with the parents are a+ b+ c+ (8%) and a-b-c-(8%).
Distance between the gene a and b = Single cross overs + double cross overs
Distance between the gene a and b = 8 + 8 +2 +2 = 20.
Therefore, the distance distance between the gene a and b is 20 centi morgan.
Which statement is TRUE? a. All fatty acids in a triglyceride are unsaturated. b. All lipids contain fatty acids. c. All steroids contain a 4-ring region. d. All phospholipids have one saturated fatty acid tail and one unsaturated fatty acid tail.
Answer: Option C.
All steroids contain 4 ring region.
Explanation:
Steroids are drug like or biological compound that look like cortisol hormone. Steroids have four linked carbon ring and their rings are fused together. Steriod are hydrophobic and are not soluble in water. They have short tail. Steroids are important cell components and they are signaling molecules.
Snapdragons occur in nature as either green or yellow plants. A green snapdragon is homozygous and has the genotype CC. A yellow snapdragon is heterozygous and has the genotype Cc. Suppose that a gardener crosses two yellow snapdragons, and one-third of the offspring are green and two-thirds of the offspring are yellow.
What type of allele could be responsible for the 2:1 offspring ratio seen when two yellow snapdragons are crossed?
A. dominant allele
B. polymorphic allele
C. recessive lethal allele
D. codominant allele
Answer:
Snapdragons occur in nature as either green or yellow plants. A green snapdragon is homozygous and has the genotype CC. A yellow snapdragon is heterozygous and has the genotype Cc. Suppose that a gardener crosses two yellow snapdragons, and one-third of the offspring are green and two-thirds of the offspring are yellow.
What type of allele could be responsible for the 2:1 offspring ratio seen when two yellow snapdragons are crossed?
dominant allele
Explanation:
As a result of two yellow snapdragons crossed, the dominat yellow allele gives the reason why the reult reflects in such a way that there is yellow dominates over green and leads to why yellow has two-third while green has one-third
You treat cells with the ionic detergent sodium dodecyl-sulfate (SDS) to disrupt the membrane. Will integral membrane proteins be affected by this treatment and how?
Answer:
Yes. the Integral membrane will be affected.
Explanation:
Generally, Ionic detergents have hydrophilic head group which can be either negatively charged or positively charged. If treated with plasma membrane, it masks the native charge of the protein, colonizing the entire charges of the protein, converting the overall charge to its own.
To be specific SSD is anionic. It therefore adds overall negative charge(anion) to the entire protein charges, even if they have isoelectric points. ''just like the HIV-RNA genome which undergo retro-transcription of the entire human DNA genome to its own HIV- genome of the infected host!''.
This disrupts the hydrogen bonds, hydrophobic interactions and other non-covalent bonds of protein , thus the 3-Dimensional protein structure of tertiary protein and therefore of the integral membrane proteins . Denaturation of protein structure, and therefore the conformation protein function results.With resulting separation of the protein molecules based on their sizes.
A researcher identified a bacterial enzyme that is essential in the breakdown of glucose. The researcher wants to test a potential antibiotic that acts on the newly identified enzyme. She finds that glucose is indeed broken down at a slower rate when the potential antibiotic is present. Addition of a higher concentration of glucose does not have any impact on Vmax. Which of the following properly characterizes this finding?
a. The potential antibiotic may be a competitive or noncompetitive inhibitor of the enzyme. These possibilities could be tested by adding more enzyme.
b. The potential antibiotic is a competitive inhibitor of the glucose-converting enzyme and could be an effective treatment for bacterial infections.
c. The potential antibiotic is a noncompetitive inhibitor of the enzyme and likely changes the shape of the active site.
d. The potential antibiotic reduces the number of available enzymes and is, therefore, a competitive inhibitor.
Answer:
The potential antibiotic is a competitive inhibitor of the glucose converting enzyme and could be an effective solution for bacterial infections.
Explanation:
Enzyme kinetics mainly deals with the study of the rate of reaction of the enzyme. Different factors that can affects the enzyme rate are also studied in enzyme kinetics.
The competitive inhibitors of the enzyme do not show any effect on the Vmax value of the enzyme and binds to the enzyme only. These inhibitors do not have ability to binds with enzyme substrate complex. The increase in the enzyme and substrate concentration can remove the inhibitor affect.
Thus, the correct answer is option (b).
The potential antibiotic is acting as a noncompetitive inhibitor since higher glucose concentrations do not affect the Vmax, suggesting the antibiotic changes the enzyme's conformation rather than competing with glucose for the active site. So the correct option is c.
Explanation:The finding that the potential antibiotic slows down the breakdown of glucose by a bacterial enzyme but that a higher concentration of glucose does not affect the Vmax (maximum rate of enzyme activity) indicates the antibiotic is likely acting as a noncompetitive inhibitor. A competitive inhibitor would have a different effect: the Vmax could potentially still be reached if sufficient additional substrate (in this case, glucose) were added, as competitive inhibitors can be outcompeted by high concentrations of the substrate. Since adding more glucose did not change the Vmax, this suggests that the potential antibiotic is not a competitive inhibitor but rather implies that the antibiotic binds at a site other than the active site, thereby altering the enzyme's conformation and affecting its activity without directly blocking the substrate from binding. Therefore, the correct characterization of this finding would be option c: The potential antibiotic is a noncompetitive inhibitor of the enzyme and likely changes the shape of the active site.
what are the drawbacks to only regulating at the transcriptional level and not the translational level or protein level?
Answer:
Energy and Cellular Space
When gene Gene expression at( transnational level ) is regulated it ensures energy and cellular space conservation, If a gene were to be regulated at each time after transcription, then significant amount of ATPs beyond the synthetic capacity of the cells will be needed, the cells will need large surface to accommodate the mechanism s. Thus regulation saves more energy and space to regulate gene expression. by turning on genes when needed to expressed and off when not needed.
Sequence of Gene
In addition, for the timely synthesis of exact copies of protein needed by the cells for various cellular activities; it is important for cells to regulate and control how the DNA is translated to the required proteins. If this mechanisms were not regulated deletion , addition of nitrogenous bases in gene sequence during translation may lead to mutations and therefore wrong coding of the needed protein in the cells.
Protein Quantity
Furthermore, the need to know the quantity of protein to synthesize, when to stop the synthesis, necessitated regulation of the process.If required amount is not expressed wrong amino acids units will be synthesized leading to abnormalities in hormones and enzymes.
Explanation:
Which of the following specialized structures/inclusions would aquatic photoautotrophic bacteria likely possess? 1. Thylakoids 2. PHB Granules 3. Carboxysomes 4. Gas vacuoles 5. Chloroplasts
Answer:
1. Thylakoids 2. PHB Granules 3. Carboxysomes
Explanation:
1. Thylakoids are membrane bounded compartments present in bacteria for light dependednt reactions.
2. PHB Granules help in carbon fixation
3. Carboxysomes help to retain carbon when enough supply isn't available.
Answer: Option 1,2 and 3.
Thykaloids, PHB granules and carboxysomes.
Explanation:
Carboxysomes consist of polyhedral protein and enzymes Rubisco which is important to supply carbon and fix it.
Thykaloids are membrane bound part of chloroplast which is the site of light dependent reaction during photosynthesis.
PHB granules are important component which help to fix carbon .
What kind of evidence has recently made it necessary to assign the prokaryotes to either of two different domains, rather than assigning all prokaryotes to the same kingdomA) molecularB) behavioralC) ecologicalD) nutritionalE) anatomical
Answer:
A) molecular
Explanation:
Prokaryotes have 70S type of ribosomes. The small ribosomal subunit of ribosomes has some conserved sequences. This subunit is similar in the closely related species. On the other hand, the structure of the small ribosomal subunit is different in the organisms that are distantly related.
Microbiologist Carl Woese analyzed the small ribosomal subunits of some bacteria and Archaeans and found distinct differences. On the basis of the composition of this biomolecule, prokaryotes were divided into eubacteria and archaea.
Final answer:
Molecular evidence, especially differences in cell membrane structure and SSU rRNA sequences, has led to prokaryotes being classified into two separate domains: Bacteria and Archaea, which differ significantly from each other and from Eukarya.
Explanation:
The evidence that has recently made it necessary to assign the prokaryotes to either of two different domains, rather than assigning all prokaryotes to the same kingdom, is molecular evidence. Advances in genetic analysis have revealed significant genetic differences between two groups of prokaryotes, leading to the classification into two separate domains: Bacteria and Archaea. These domains are based on differences in the structure of cell membranes and in ribosomal RNA (rRNA), specifically the sequences of small-subunit ribosomal RNA (SSU rRNA). While both domains consist of organisms with prokaryotic cells, lacking a nucleus and true membrane-bound organelles, they are as different from each other as they are from the third domain, Eukarya, which includes all eukaryotic organisms with membrane-bound organelles and a nucleus.
In response to stress, cells in the adrenal cortex produce cortisol, a steroid hormone. Therefore, these cells would specifically exhibit large amounts of this organelle: Select one a. lysosomesb. rough ER C. smooth ER d. ribosomes e. vaults
Answer:C. smooth ER
Explanation:the endoplasmic reticulum is a membrane bound organelle found in the cytoplasm of the cell.it has a highly folded appearance.
Its function in the cell is the transportation of cellular products.
There are two types of endoplasmic reticulum;the rough endoplasmic reticulum and the smooth endoplasmic reticulum.
The rough endoplasmic reticulum have ribosomes attached to them ,giving them a granular appearance.the ribosomes synthesis proteins,which the rough endoplasmic reticulum transports outside or within the cell.
The smooth endoplasmic reticulum do not have ribosomes attached to it.its functions is to synthesis lipids.
It is also involved in the production of steroid hormones in the adrenal cortex and endocrine glands
Large amounts of this organelle C. Smooth ER Therefore , C. smooth ER is correct .
Smooth Endoplasmic Reticulum (Smooth ER):
The smooth ER is an organelle involved in various functions, one of which is the synthesis of lipids and steroids, including cortisol.
Steroid hormones like cortisol are lipids, and their synthesis primarily occurs in the smooth ER of cells.
In the adrenal cortex, cells are specialized to produce steroid hormones such as cortisol.
Cortisol is a crucial steroid hormone involved in the body's response to stress and helps regulate various physiological processes.
Cortisol Production:
The production of cortisol starts with cholesterol, which is transformed into cortisol through a series of enzymatic reactions, predominantly occurring in the smooth ER of the adrenal cortex cells.
The enzymes responsible for the synthesis of cortisol are predominantly located in the smooth ER membranes.
During stressful situations, the body signals the adrenal cortex to increase cortisol production.
The cells in the adrenal cortex respond by synthesizing more cortisol, leading to an increase in the smooth ER activity and volume to accommodate the higher demand for steroid hormone synthesis.
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