Answer:.
→The α‑helix is held together by hydrogen bonds between the amide N − H N−H and C = O C=O groups
→ In a β‑pleated sheet, the side chains are located between adjacent segments.
→ In an α‑helix, the side chains are located on the outside of the helix. .
→ The secondary level of protein structure refers to the spatial arrangements of short segments of the protein
Explanation:
This is the level of protein which results from spatial arrangement produced by the formation of hydrogen bonds between the oxygen atom of one carboxyl group(c=0) group, and hydrogen of the NH group of amino acids four places ahead of it .( The resulting structure is coiled and are therefore called alpha-helices)
AND
Hydrogen bonds between adjacent amino acids that join them side by side so that the bonds appear straight rather than coil, and the chains form upwards-downwards-upward- downwards format to form flat shaped structure called beta-pleated sheet.
The hydrogen bonding is due to strong polarities of the –NH- CO- groups of amino acids.
The two structures account for the spatial arrangement of secondary protein structure. Secondary structure is stabilized by the orientation and aggregation of these hydrogen bonds. . The outwards distributions of the side chains, the non-polar nature (hydrophobic) of alpha-helix makes some secondary proteins ideal as integral membrane proteins.
Note- peptide bond stabilizes primary protein structure.
The following statements are true for protein secondary structure:
The amide N-H and C=O groups form hydrogen bonds that hold the -helix together.In a helix, the side chains are found outside the helix.The secondary structure is stabilized by peptide bonds.Therefore, the correct options are A, C and E.
The polypeptide chain is regularly coiled to form a -helix, and hydrogen bonds between the amide NH and C=O groups help to stabilize the structure. Because they occur outside the helix and extend outward in a -helix, amino acid side chains enable interactions with the environment. The covalent bonds that link amino acids together, known as peptides, are essential for maintaining the secondary structure. Rotation is hindered by the planar structure of the peptide bond, which helps to form regular patterns such as the -helix.
Therefore, the correct options are A, C and E.
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You decide to designate the twist allele as FT to distinguish it from the forked allele F. Using the following allele symbols, identify the genotypes of the three F2 classes in Part C by dragging one label to each class. Labels can be used once, more than once, or not at all.
Image attached
Answer:
FTFT, F, FFT (in order left to right)
Explanation:
The twist allele is FT, the forked allele is F. We are told there are pure lines, so this means they are homozygous. That means the parents are FF x FTFT.
The F1 generation is both twisted and forked (as can be seen from the image), suggesting the alleles are codominant (both are expressed).
In the F2, there are three different types of flowers, 2 matching the parental and 1 matching the F1 twisted, forked, and both.
The order from left to right is twisted, forked both. We know twisted is the genotype FTFT, and forked is the genotype FF. The both phenotype would have a copy of each allele, so would be FFT
Each of two parents has the genotype red / blond, which consists of the pair of alleles that determine hair color, and each parent contributes one of those alleles to a child. Assume that if the child has at least one red allele, that color will dominate and the child's hair color will be red.
a. List the different possible outcomes. Assume that these outcomes are equally likely.
b. What is the probability that a child of these parents will have the blond/blond genotype?
c. What is the probability that the child will have red hair color?
The possible outcomes for the child's genotype are Rb, Rb, bR, and bb. The probability of the child having the Blond/Blond genotype is 25% while the probability of the child having red hair is 75%.
Explanation:In this genetics-based question, we're dealing with a common system of inheritance where the red hair allele is dominant over the blond hair allele. Let's denote the red hair allele as 'R' and the blond hair allele as 'b'. Since both parents have the genotype Red / Blond, their genotypes can be represented as 'Rb'.
a. The different possible outcomes for their children's genotypes (which determine hair color) can be determined via a Punnett square. The possibilities are: Rb (Red/Blond), Rb (Red/Blond), bR (Red/Blond), and bb (Blond/Blond).
b. The probability that a child of these parents will have the genotype Blond/Blond (bb) is 1 out of 4, or 25%.
c. If a child has at least one red allele (R), they will have red hair, owing to the dominance of the red allele. In this scenario, there are 3 out of 4 possibilities where the child has at least one red allele, so the probability that the child will have red hair is 75%.
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Which characteristics of DNA polymerase I raised doubts that its in vivo function is the synthesis of DNA leading to complete replication?
The question is incomplete. The complete question is as follows:
Which characteristics of DNA polymerase I raised doubts that its in vivo function is the synthesis of DNA leading to complete replication?
its composition of a single polypeptide chain .
deficiency of enzyme in some organisms that are still capable of DNA synthesis .
requirement of Mg2+ presence in order for the enzyme to work .
low stability under normal physiological conditions.
Answer:
Deficiency of enzyme in some organisms that are still capable of DNA synthesis
Explanation:
The DNA polymerase I may be defined as the important enzyme that play an important role in the DNA replication of prokaryotes. DNA pol I is the replicating enzyme, DNA repair enzyme and can also acts as the exonuclease.
DNA pol I has been studied invitro and Arthur Korenberg explain the discovery of the DNA pol I. This DNA pol I plays an important role in DNA repair rather than the replication process. This explained invivo by the fact that some in some organisms the deficiency of this enzyme do not halt the process of replication. If the DNA pol I acts as the main replaicating enzyme, the DNA synthesis must be stopped in the organisms that lack DNA synthesis.
Thus, the correct answer is option (2).
Energy:
a) recycles continuously through an ecosystem
b) is used over and over again
c) flows in only one direction through an ecosystem
d) tends to be concentrated by living organisms
Answer: Option C.
Energy flows in one direction in ecosystem.
Explanation:
In the ecosystem, energy flow in one direction. Energy is transfered from one trophic level to another.
Energy is transfered from producers, majorly plants,convert sunlight to chemical energy during photosynthesis and transfer it to herbivores( goat, sheep ) when they eat plants,the herbivores after using the energy transfer energy to the carnivores when the carnivores feed on the herbivores and to decomposer who feed on dead animals.
Energy flows in one direction through an ecosystem, from producers to consumers, and eventually dissipates as heat, while matter is recycled, such as through biogeochemical cycles.
Explanation:Energy in ecosystems behaves differently than matter. It flows in only one direction: from the sun or inorganic chemicals to producers, and then on to consumers. Producers, such as plants, are capable of capturing energy directly through photosynthesis and chemosynthesis. Consumers then acquire energy by feeding on these producers or on other consumers. This flow of energy is depicted using models like food chains and food webs. In contrast, matter is recycled within ecosystems, ensuring that elements such as carbon and nitrogen are continuously available in one form or another.
It is also pertinent to highlight the process of biogeochemical cycles, such as the water cycle, carbon cycle, and nitrogen cycle, whereby nutrients and water are constantly recycled, interacting with both biotic components and abiotic factors of an ecosystem. While matter recycles and maintains a certain balance, energy does not recycle but rather dissipates as heat with each transfer through the trophic levels, eventually exiting the ecosystem.
. Given that 3 of the 64 possible codons are stop codons, what is the chance of having a stop codon at any given position, assuming that the sequence is random?
Answer:
[tex]1 - (\frac{61}{64})^n[/tex]
Explanation:
Given
There are three stop codon
TAA, TAG, TGA.
Out of 64 available codons, 3 are stop codon thus the remaining are non-stop codon.
So the probability of choosing a non stop codon is
[tex]\frac{61}{64}[/tex]
Let us suppose the number of trials are "x" then the probability of choosing a non stop codon is
[tex](\frac{61}{64})^x\\[/tex]
Probability of choosing a stop codon is equal to
[tex]1 -[/tex] probability of choosing a non stop codon
[tex]1 - (\frac{61}{64})^n[/tex]
The chance of having a stop codon at any given position in a random sequence of codons can be calculated by dividing the number of stop codons by the total number of possible codons.
Explanation:In a given sequence of codons, the chance of having a stop codon at any given position can be calculated by dividing the number of stop codons by the total number of possible codons. The number of stop codons is given as 3, and the total number of possible codons is 64. Therefore, the chance of having a stop codon at any given position is 3/64, or about 0.046875, assuming that the sequence is random.
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pH has a major influence on protein structure by altering electrostatic interactions. In order to illustrate this point, let us think about polylysine (a polypeptide chain consisting of only L-lysine residues). At pH 10 and above, polylysine forms an ahelix. At a pH of 7 and below, however, this same polypeptide chain assumes an unfolded conformation (often referred to as ‘random coil’ or ‘random conformation’).
1) Can you explain why this transition occurs at pHs below the pKa of Lys?
2) What other residue(s) might you expect to show a similar pattern of pH-dependent folding and unfolding?
3) The residue(s) you might expect based on charge to be capable of forming α-helices do not do so in water when they alone make up a polypeptide chain. Can you come up with a reasonable explanation for why this might be?
4) Speculate on what you think might happen to the pH-dependence of α-helix formation if you had a polypeptide chain consisting of both glutamate and lysine residues.
Answer:
Explanation:
(1) Remember that this occurred in a polylysine which consists only of L-Lys, the pKa of Lys residues might be different in polylysine as compared with the free amino acid in the solution.
The transition can also occur if more than 50% of the lysine residues need to be charged in order to ‘break’ the helix. Note that there is a 50:50 ratio of the coordinated bond of protons and dissociated forms.
(2) Other residues that are positively charged at neutral pH. They are arginine and histidine.
(3) Both arginine and histidine are more voluminous than Lys. Steric interference can prohibit the formation of an α-helices.
Why do bacteria usually contain an even number of replisomes? Choose one: A. Bacteria need only one replisome; the other serves as a backup copy. B. Because replication is bidirectional, there are two replication forks emanating from the origin of replication. C. There are two copies of the gene that codes for DNA polymerase, an important enzymatic protein component of replisomes. D. DNA is double-stranded and each single strand requires its own replisome.
Answer: Option B.
Because replication is bidirectional, there are two replication forks emanating from the origin of replication.
Explanation:
Replisomes are complexes that involve in DNA replication. Replisomes consist of many proteins like ligase, helicase,topoisomerase,DNA polymerase III, e.t.c. The replisomes first unwind double stranded DNA to single strands DNA.
In prokaryotes,the nucleoid divides and requires two replisomes for bidirectional replication. Bacteria normally have even number of chromosome because the replication is bidirectional, and there are two forks which emanates from the origin of replication. The replication forks duplicate both the leading and the lagging strands.
The major function of replisomes is to carryout DNA replication.
Bacteria usually contain an even number of replisomes because DNA replication in bacteria is bidirectional, emanating from a single origin but forming two replication forks, each with its replisome for DNA synthesis (Option B),
Explanation:The correct answer to your question, 'Why do bacteria usually contain an even number of replisomes?' is B. Because replication is bidirectional, there are two replication forks emanating from the origin of replication. For a deeper understanding, let's explain it. In bacteria, the DNA replication starts at a single origin of replication but proceeds bidirectionally, creating two replication forks.
The enzyme helicase unzips the double helix at the origin of replication forming a replication fork. Each of these forks has a complex of proteins called replisome that is required for DNA synthesis. The replisome is responsible for copying the entire bacterial chromosome from the origin of replication to the terminus.
The synthesis is semiconservative where one strand—called the leading strand—is synthesized continuously in the direction of the replication fork, and the second strand, called the lagging strand, is synthesized discontinuously, away from the replication fork, creating Okazaki fragments. Thus, because of this bidirectional nature, there are always two replisomes in prokaryotic DNA Replication, hence an even number of replisomes.
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30. Select all accurate statements
A. All chordates have pharyngeal slits or clefts
B. All chordates have a dorsal, hollow nerve cord
C. All chordates are bilaterally symmetrical animals
D. All chordates will have a muscular post an*l tail
E. All chordates will have a notochord as adults.
Answer:
All the options A,B,C,D and E statements are correct.
Chordates are animals that posses a notochord, dorsal, hollow nerve chord, pharyngeal slits and they are bilaterally symmetrical.
Generally, membrane filters are a used to remove bacteria from liquids. b are made of microscopic pores c used to remove spoilage agents from alcoholic beverages. d all the above
Answer:All of the above
Explanation:
Membrane filters are very thin, highly porous media composed of foamed and/or stretched polymeric compounds. Because of their homogeneous structure they cannot contaminate the filtrate with fibres or particles from the membrane matrix.
Microfiltration membranes retain particles according to the size of the pore, and the affinity of the filter materials for the solute. While materials are held largely on the surface of the membrane, they can also retain within the matrix itself, as in depth filtration. Unlike depth filters, however, which bind solutes of nominal size ranges only, membrane filters retain particles with absolute accuracy due to their controlled, predetermined pore size. Some membranes also demonstrate retention of specific substances as for example in the case nitrocellulose membranes which bind proteins and nucleic acids in high concentration. This feature may be an advantage for some applications where exclusion based on chemical properties rather than porosity alone is required.
Membrane filters for microfiltration are used primarily for separating particulate materials or micro-organisms, larger than the rated pore size, from gases or liquids. On filtration of gases particles are also retained which are smaller than the pore size rated. Also they show a low particle capacity per unit area as compared to a depth filter. For some filtration applications it may be advisable to prefilter through a depth filter, eg, a glass fiber filter, prior to microporous filtration to prevent clogging. This may be necessary, because the solution is highly viscous, contain molecules which may precipitate in the membrane filter or are heavily contaminated by microorganisms and particulates.
This separation is carried out either for the purpose of cleaning to obtain highly pure or sterile products or recovering (concentrating) these materials in order to carry out further chemical, microscopic, microbiological or other analyses on the separated sample.
Typical applications include Sterile filtration
–Dialysis
–Fluid clarification/purification
–Gas filtration/particle control
–Microbiological investigations
–HPLC solvent filtration and sample preparation
Alleles of the gene that determines seed coat patterns in lentils can be organized in a dominance series:Marbled > Spotted = Dotted > ClearA lentil plant homozygous for the marbled seed coat pattern allele was crossed to one homozygous for the spotted pattern allele. In another cross, a homozygous dotted lentil plant was crossed to one homozygous for clear. An F1 plant from the first cross was then mated to an F1 plant from the second cross.What are the expected phenotypes of the F1 plants from the two original crosses?
Answer:
first original cross = 100 % marbled seed coat lentils
second original cross = 100% dotted seed coat lentils
Explanation:
Given,
Marbled > Spotted = Dotted > Clear ( Spotted and dotted alleles are codominant )
Let, marble = m
spotted = s
dotted = d
clear = c
First original cross: mm * ss :
m m
s ms ms
s ms ms
Entire progeny has ms genotype. Since m is dominant over s, result will be 100% marbled seed coat lentils.
Second original cross : dd * cc :
d d
c dc dc
c dc dc
Entire progeny has dc genotype. Since d is dominant over c, result will be 100 % dotted seed coat lentils.
How do you plant a lavender? What are the best conditions to grow it
Answer:
Lavender is best planted in the spring as the soil is warming up. If planted in the fall, use bigger plants to ensure survival over the winter
2. Plant lavender 2 to 3 feet apart
3.It thrives in any poor or moderately fertile soil.
4.Keep away from wet, moist areas.
Answer:
Lavender is grown in garden beds, pots and requires maximum sunlight and a soil that is well drained for good germination. The soil require for lavender growth should be moderately fertile. It can grow in both arid and humid lands but climate affects the its growth rate.
You are given a metaphase chromosome preparation (a slide) from an unknown organism that contains 12 chromosomes. Two that are clearly smaller than the rest appear identical in length and centromere placement.
What would most likely be true of these two chromosomes? Select all that apply.
They have similar banding patterns.
They contain identical genetic information.
They would replicate synchronously during the S phase of the cell cycle.
They are homologous chromosomes.
Answer:
True answers are as follow:
a. They have similar banding patterns.
c. They would replicate synchronously during the S phase of the cell cycle.
d. They are homologous chromosomes.
Explanation:
In Mitosis, chromosomes condense and align in the center before moving to each opposite pole is called meta-phase. (See attached picture)
In this stage you will observe similar banding pattern of homologous chromosomes that were replicated during S phase of cell cycle.
The two smaller, identical chromosomes most likely are homologous chromosomes, containing the same gene order and potentially identical genetic information. They would replicate together during the S phase of cell cycle.
Explanation:The two chromosomes that are smaller and identical in length and centromere placement are likely to be homologous chromosomes. As such, they would have similar banding patterns because they have the same order of genes. They would contain identical genetic information only if there hasn't been any genetic recombination during meiosis. The replication of these chromosomes would occur synchronously during the S phase of the cell cycle as DNA replication happens to all chromosomes of a cell at the same time.
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Which of the following statements is/are correct? (MORE THAN ONE ANSWER)
A. Nematodes are also known as 'segmented' worms.
B. Members of the Phylum Nematoda are all hermaphrodites
C. Given the following on the organisms:
I. An animal
II. Multicellular
III. has tissues
IV. Has a digestive tract
V. exhibits bilateral symmetry as an adult
This organism could be a member of the Phylum Nematoda
D. Nematodes possess a cuticle.
E. Some members of the phylum Nematoda are parasites
Answer:
E
Explanation:
some of the nematodes (non-segmented ) are parasites
Answer:
The statements which are correct about nematodes are options C, D and E:
C: Organisms from the phylum nematoda are animals, multicellular in nature, has tissues, have a complete digestive tract and exhibit bilateral symmetry.
D. Nematodes possess external structure called cuticle.
E. Some members of the phylum nematoda are parasitic, free living organisms that feed on other living materials.
What phenotypic ratio would you expect as a result of a test cross between two individuals where 9) one that is homozygous recessive for alleles at two independent loci?
What phenotypic ratio would you expect as a result of a test cross between a dihybrid organism and one that is homozygous recessive for alleles at two independent loci?
a. 3:1
b. 9:3:3:1
c. 1:1:1:1
d. 1:2:1
e. 9:4:2:1
Answer:
c. 1:1:1:1
Explanation:
When a heterozygous individual for two genes is test crossed with a double homozygous recessive individual, the progeny is obtained in 1:1:1:1 phenotypic ratio. This occurs as the heterozygous dominant individual forms four types of gametes in 1:1:1:1 ratio while the homozygous recessive individual would form only one type of gamete having one recessive allele for each gene.
For example, a test cross between TtRr (tall and red) and ttrr (short and white) would produce a progeny in following ratio=
1 tall, red: 1 tall, white: 1 short, red: 1 short, white
Here, T= tall, t= short, R= red, r= white
Final answer:
The genotypic ratio expected from the mating of two individuals heterozygous for a recessive lethal allele expressed in utero is 2:1 (homozygous dominant: heterozygous) among surviving offspring because the homozygous recessive genotype results in death before birth.
Explanation:
When two individuals that are heterozygous for a recessive lethal allele expressed in utero are mated, the expected genotypic ratio (homozygous dominant:heterozygous: homozygous recessive) among the offspring would be 2:1. This is because the homozygous recessive genotype, which would normally make up one-quarter of the offspring, results in death before birth due to the lethal allele. Therefore, the live offspring genotypic ratio becomes 1:2:0, as the homozygous recessive individuals do not survive. This contrasts with a typical Mendelian 3:1 phenotypic ratio observed in monohybrid crosses of non-lethal traits. The Punnett square for such a cross shows that out of the four possible genotypes, one (homozygous recessive) does not survive, leaving a genotypic ratio of homozygous dominant:heterozygous at 1:2 among surviving offspring.
which of the followings are true about V0 and Vmax? A. The unit for each of them is M B. Vmax is a special V0 when all enzymes bind substrate C. Vmax is independent of enzyme concentration D. V0 can be determined using a linear correlation of product and time in the beginning of a reaction
Answer:
The correct answer is option B.
Explanation:
The unit of Vmax and Vo is moles per second. Km refers to the concentration of the substrate, which is needed to attain the maximum reaction velocity. In case, when [S] is far greater in comparison to Km then Vo will be close to Vmax.
Vmax relies upon the concentration of the enzyme, Vmax enhances with the concentration of the enzyme and becomes steady when all the active sites of enzymes get occupied. Vo increases with the enhancement in the concentration of the substrate with time. Thus, the correct answer is option B, that is, Vmax is a special Vo when all the enzymes combine with the substrate.
In this exercise, you will research the effects of excess greenhouse gases in the atmosphere. You will apply the scienbific method and m ow greenhouse gases warm the planet by covering a jar with plastic wrap. You will then relate the findings of your experiment to global warming Procedure
In Exercise 1, we observed that gases have different properties In this section we will further investigate gases. How are greenhouse gases related to global warming?
Answer:
By heat trap mechanisms...
Explanation:
The main green house gases in our atmosphere are Carbon dioxide, Nitrous oxide, water vapors and ozone. All of these gases have a property of heat trap that trap the heat waves coming from the sun and heat up our atmosphere as a result. This is the reason why Venus is farther from sun but is more hotter than mercury because of it's dense Carbon Dioxide atmosphere.
Sickle cell anemia and albinism are both recessive traits in humans. Imagine that a couple, already pregnant with twins, has just learned that they are both heterozygous for both of these traits.
As the couple's genetic counselor, the couple asks you the following questions about how their carrier status will affect their offspring.
Part A
If the couple has fraternal twins, what is the probability that both children will be unaffected by both conditions?
Part B
If the couple has fraternal twins, what is the probability that both of the couple's children will have both sickle cell anemia and albinism?
Part C
What is the probability that one of the fraternal twins is a carrier of either, but not both, of the conditions?
(Hint: You will need to use both the product law and the sum law to answer this question.)
Part D
If the couple has fraternal twins, what is the probability of having two phenotypically normal children, one being a carrier of only the sickle cell anemia recessive allele, and the other being a carrier of only the recessive allele for albinism?
(Hint: You will need to use both the product law and the sum law to answer this question.)
Answer:
A. 81/256
B. 1/256
C. 4/9
D. 1/4
Explanation:
A. Given,
Both parents to be heterozygous.
Therefore, the chances for one child to inherit a defective gene from each parent
= Â ½ * ½ =Â ¼.
In the same vein, the probability of being affected by albinism
=Probability of possessing a defective gene from each parent
= Â ½ * Â ½
= Â ¼
Now, The probability of not being affected is represented using the formulae
= 1- the probability ofbeing affected
= 1-¼ Â
= Â ¾.
Thus, the probability of not being affected by both albinism and sickle cell anemia
= Â ¾ * ¾
= Â (3 * 3)/(4 * 4)
= Â 9/16.
So,
The probability of neither twins to be affected
= 9/16 * 9/16
= (9 * 9)/(16 * 16)
=81/256.
B. We obtain the chances for a child to possess one defective gene from each parent
= Â ½ * ½
= Â ¼.
In the same vein, the probability of being affected by albinism
=Probability of inheriting one defective gene from each parent
= Â ½ * Â ½
= Â ¼
Now, the probability of being affected by both albinism and sickle cellanemia
= Â ¼ * Â ¼
= 1/16
Therefore, the probability of both twins to be affected
= 1/16 * 1/16
= 1/256.
C. Take, the probability of twin 1 to be a carrier for albinism to be = 2/3 (note, the homozygous recessive is affected, not the carrier).
We can say that the chances of twin 1 to be a carrier for both the diseases
= 2/3 * 2/3
= 4/9.
In like vein, the chances of carrying neither recessive allele
= 1/3 * 1/3
= 1/9.
Therefore, the probability that one of the fraternal twins is a carrier of either, but not both,
= 1- (4/9 + 1/9)
= 1 - (5/9)
= 4/9
D. Given, the probability of one fraternal twin carrying a gene for either disease:
Take;
The provability of carrying the gene to be 3/4
= 3/4 * 3/4
= 9/16
Thw chances of non occurrence of gene = 1/3
And for fraternal twins
= 1/3 * 1/3
= 1/9
Therefore, the probability of having two phenotypically normal children, one being a carrier of only the sickle cell anemia recessive allele, and the other being a carrier of only the recessive allele for albinism
= 1 - (9/16+1/9)
= 1/4.
Slow release of bacteriophage progeny from a bacterial host cell A. does not kill the host cell B. is a feature particular to filamentous phage C. causes cell lysis D. B and C are correct. E. A and B are correct.
Answer:
C. causes cell lysis
Explanation:
The slow release of the bacteriophage progeny from a bacterial host cell causes bacterial cell lysis and further kills the host cell which is the bacteria, this results to the liberation of progeny viruses which then infect new bacterial cells.
Only a type of bacteriophage called the filamentous bacteriophage reproduces without killing the bacterial host cell. The filamentous phages infect majorly gram negative bacteria.
The correct statements regarding the slow release of bacteriophage progeny are that it is characteristic of filamentous phages and does not kill the host cell. Therefore, the correct answer is E. A and B are correct.
The slow release of bacteriophage progeny from a bacterial host cell is a feature particular to filamentous phages and does not result in cell lysis. Therefore, the correct answer to the question is E. A and B are correct. Filamentous phages can exit the host cell without lysing it through a process called extrusion, where new virions are released from the cell membrane without causing cell death. Unlike lytic phages, which replicate and cause lysis of the host cell, filamentous phages exhibit a different life cycle that allows the host cell to survive the release of new phage particles. Therefore, the statement that slow release of bacteriophage progeny causes cell lysis is incorrect, because this slow release is characteristic of filamentous phages and does not kill the host cell.
Samples of rejuvenated mitochondria are mutated (defective) in 3% of cases. Suppose 18 samples are studied, and they can be considered to be independent for mutation. Determine the following probabilities. (a) No samples are mutated. (b) At most one sample is mutated. (c) More than half the samples are mutated.
Answer:
a) [tex]0.58[/tex]
b) [tex]0.598[/tex]
c) [tex]0[/tex]
Explanation:
Given -
Total sample i.e n [tex]= 18[/tex]
Probability (p) [tex]= 3[/tex] % [tex]= 0.03[/tex]
We will use binomial distribution theory for determining the probability of mutated sample
Let X be the number of mutated sample
a) No samples are mutated i.e [tex]X = 0[/tex]
[tex]P(X=0) = 0.03^0 * 0.97^{18}\\= 0.5779 = 0.58\\[/tex]
[tex]0.58[/tex]
b) At most one sample is mutated
[tex]P(X=0) = 0.58 + 0.03^1 * 0.97^{17}\\= 0.598[/tex]
c) More than half the samples are mutated.
[tex]P(X = 10) + ........+ P(X = 18) = 0[/tex]
Explain what is happening to the whitebark pine in Yellowstone National Park as a result of climate changeResearchers often study ecosystems for a long period of time. Dr. Hadly has studied Yellowstone’s ecology for 30 years and the amphibians for 20 years. What is the value of long- term studies to advancing scientific understanding?
The given question is not complete. The complete question is:
Watch the film “Liz Hadly Tracks the Impact of Climate Change in Yellowstone (Links to an external site.)Links to an external site.” from the Scientists at Work series. Prior to watching the film, read the questions below and think about how you might answer them. You do not need to turn in your answers from before you watch the film.
Questions
Explain what is happening to the whitebark pine in Yellowstone National Park as a result of climate change
Researchers often study ecosystems for a long period of time. Dr. Hadly has studied Yellowstone’s ecology for 30 years and the amphibians for 20 years. What is the value of long- term studies to advancing scientific understanding?
Answer:
1. As a consequence of climate change, the whitebark pine is been targeted by pine beetles, which can thrive in winter conditions. An attack led by various beetles for one to two days has generated holes in the pine leading to its destruction.
2. The objective of long term studies is to find the connections between animals and plants and to witness how they are associating with each other in their micro-environment. Like in the documentary it was the association between the squirrel, pine trees and bear.
The beer needs to consume high content nutritious seeds of the pine tree, which safeguarded them in the winters due to the high amount of fat present in them that offered warmth. However, the beers were not able to reach the pine trees as they were far from their reach and thus, they took help from squirrels, which serve the objective for the bears.
Thus, it can be comprehended that all three creatures, that is, squirrel, tree and bear are associated with each other and developing a food web. All these studies need an ample amount of time and observation. Hence, Dr. Hardley took thirty years examining them and also took twenty years examining the influence of climate on amphibians living in the pond. Their species diversity declined substantially due to global warming that resulted in the drying up of the pond.
The gene frequency for a dominant allele is 0.68 and the recessive allele is 0.39. What percent of the population has a homozygous recessive genotype?
Answer:
15.21 %
Explanation:
If we recall the basic formula of Hardy-Weinberg's equilibrium ; we have the following below:
p + q = 1
p² + 2pq + q² = 1
where;
p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p² = percentage of homozygous dominant individuals
q² = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals
Given that p= 0.68 and q = 0.39
the percentage of the homozygous recessive genotype (q² ) will be
(0.39)² = 0.1521
= 0.1521 × 100
= 15.21 %
∴ the percentage of the population that has a homozygous recessive genotype = 15.21 %
Approximately 15.21% of the population has a homozygous recessive genotype.
Explanation:To calculate the percent of the population with a homozygous recessive genotype, we need to square the frequency of the recessive allele. In this case, the frequency of the recessive allele is 0.39, so squaring it gives us 0.39 * 0.39 = 0.1521. To convert this to a percentage, we multiply by 100, giving us 15.21%. Therefore, approximately 15.21% of the population has a homozygous recessive genotype.
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is rectus femoris NOT a muscle of the the medial forelimb of the fetal pig that you will be dissecting?
Answer:
Yes it in not a muscle of the medial forelimb of the fetal pig
Explanation:
Rectus femoris is not the part of forelimb of the fetal pig.
However, rectus femoris in fetal pig originates from the hip and that to through the ilium and extends towards the knee and insert into its tendon. These are responsible for flexing the hip and extension of lower knee and legs. This muscle is a part of muscles of thigh and lies at its top. In quadriceps these muscles are the most anterior muscles.
Answer: Rectus femoris is not a muscle of medial forelimb of the fetal pig.
Explanation:
Pigs are mammals like humans. Pigs have similar structures with humans but there are slight differences in these structures because pigs are quipedal and humans are bipedal.
Rectus femoris is not found in the medial forelimb of fetal pigs instead it is found in the medial hindlimbs of fetal pigs and is one of the quadriceps muscles. It starts from the hip through the ilium and extend to the knee.Rectus femoris function in knee extension, its extend the lower leg and it flexes the hip.
One method for separating polypeptides makes use of their different solubilities. The solubility of large polypeptides in water depends upon the relative polarity of their R groups, particularly on the number of ionized groups: the more ionized groups there are, the more soluble the polypeptide.
1. Which of each pair of the polypeptides that follow is more soluble at the indicated pH?
(a) (Gly)20 or (Lys-Ala)3 at pH 7.0:
O (Gly)20
O (Lys-Ala)3
(b) (Glu)20 or (Phe-Met)3 at pH 7.0
O (Glu)20
O (Phe-Met)3
Answer:
(a) (Glu)zo or(Phe-Met)3 at pH 7.0
O (Glu)zo ✔
O (Phe-Met)s ❌
(b) (Gly) zo or (Lys-Ala)3 at pH 7.0:
O (Gly12) ❌
O (Lys-Ala)✔
(c) (Ala-Asp-Gly)s or (Asn-Ser-His)s at pH 3.0:
O (Asn-Ser-His)s ✔
O (Ala-Asp-Gly)s ❌
(d) (Ala-Ser-Gly)s or (Asn-Ser-His)s at pH 6.0:
O(Ala-Ser-Gly)s ❌
O (Asn-Ser-Hish)s ✔
Explanation:
Polypeptides that has polar or charged side chains are more soluble than polypeptides with nonpolar side chains.
(a) At ph 7.0
(Glu)20 is negatively charged at pH 7 and more soluble
(Phe-Met)3 is observed to be less polar and less soluble
(b)At ph 7.0
(Lys-Ala)3 is positively charged (polar) and more soluble
(Gly)20 is uncharged as only the amino- and carboxyl-terminal groups are charged as its less polar and less soluble too.
(c) At pH 6.0
(Asn-Ser-His)5 has polar Asn side chains and partially protonated His side chains and it's more soluble unlike the (Ala-Asp-Gly)s at that pH.
(d) At pH 3.0
(Asn-Ser-His)s as partially protonated carboxylate groups of Asp residues and it is also neutral but the imidazole groups of His residues are fully protonated and positively charged. Hence it's more soluble than the (Ala-Ser-Gly)s at that particular pH.
Watch the animation, DNA Replication. After DNA replication, how many double-stranded DNA molecules exist?
Answer:
The DNA is a double helix or made up of two strands. The strands are separated during replication, each serving as a template to produce a complementary strand of each of the separated single strands. Therefore after replication, two double stranded DNA molecules will be present.
Final answer:
Two double-stranded DNA molecules are produced following DNA replication, with each new molecule containing one old and one new strand, reflecting the semiconservative nature of the replication process.
Explanation:
After DNA replication, two double-stranded DNA molecules exist. This is because DNA replication is semiconservative, meaning each new DNA molecule consists of one original and one newly synthesized strand. During DNA replication, the double helix of the parent DNA molecule is unwound by enzymes. Then, DNA polymerase synthesizes a new complementary strand for each of the two single strands from the parent molecule.
The entire replication process ensures that each daughter cell receives an exact copy of the DNA. Mechanisms such as DNA proofreading and the repair of mistakes, which involve enzymatic correction during replication, help maintain the accuracy of the genetic information passed on to each new cell.
The house fly, Musca domestica, has a haploid chromosome number of 6. How many chromatids should be present in a diploid, somatic, metaphase cell?
Answer:
12
12
24
Explanation:
Musca domestica is the scientific name of an animal popularly known as Housefly. From the question, it is known that this organism has a haploid chromosome number of 6.
In a diploid chromosomes, the number of chromatids that will be present will be twice of that in haploid.
Since haploid (n) = 6, diploid (2n) will be; 2 × 6 = 12
Somatic cells on the other hand also exhibit diploid number of chromosomes, again it means we will have (2n) = 2 × 6 = 12
In metaphase cell, cells do make sure they complete the S'phase of the cell cycle before cellular division.Therefore, the DNA present in the chromatids in the S'phase actively engage in chromatids doubling, as a result, 24 chromatids (i.e 12 × 2= 24) exists in the metaphase cell.
The chromatids should be present in a diploid, somatic, metaphase cell-
diploid - 12somatic - 12metaphase cell- 24Chromatids at different stageThe scientific name of the fly Housefly is Musca domestica. It is known that this organism has a haploid chromosome number of 6. In the diploid cells, the number of chromatids that will be present will be twice that in haploid.
Diploid genome (2n)
= [tex]2\times6=12[/tex].
Somatic cells also exhibit diploidy (2n)
= 12
Metaphase stage cells contain 24 chromatids as cells before entering cell division complete S-phase of the cell cycle in which DNA present in chromatids become double due to replication.
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Inversion of DNA sequences within chromosomes is a common process in evolution. The following gene arrangements in a particular chromosome are found in four different species.
1st sequence STUVWX
2nd sequence UVXTSW
3rd sequence UVWSTX
4th sequence SWVUTX
1. Assuming that the arrangement in 1 is the original arrangement, in what evolutionary order did the four species arise, such that there is only one inversion between each species. (Place 1st sequence first in your ordering.)
Answer:
2, 3, 4, 1
Explanation:
Inversion of DNA sequences refers to the rearrangement of the sequences where a segment is reversed end to end.
For instance in order for the gene of species to have the sequence STUVWX
The order of evolution will be
sequence UVXTSW will be the first sequenceWhen XTSW segment is inverted WSTX
sequence UVWSTX will be the result of the inversion.section UVWS is inverted to SWVU
sequence SWVUTX is formed.when WVUT undergoes inversion to TUVW
It becomes the original arrangement of STUVWX .Pellicle ___ Uniform fine turbidity QUESTIONS a. Evenly cloudy throughout b. Growth at top around the edge c. Growth on the bottom d. Membrane at the top e. Suspended chunks or pieces 2 What factors besides physical growth characteristics are important when recording data about an organism
Answer: (d) Membrane at the top
Explanation:
Pellicle: surface membrane produced by organisms floating on top of the medium. Sediment: organisms that sink to the bottom. Uniform fine turbidity: cloudy. Flocculent: Organisms that clump/suspended pieces or chunks.
(2) Other important factors apart from
physical growth too include would be the growth medium, incubation temperature and atmosphere (aerobic or anaerobic). Growth medium,incubation temperatureandatmosphere(aerobicoranaerobic).
A pellicle describing uniform fine turbidity means that the microorganisms are evenly dispersed throughout a medium, giving it a uniformly cloudy appearance. Apart from physical characteristics, environmental conditions, genetic predisposition, presence of competitors or predators, and availability of resources are crucial factors to consider when recording data about an organism.
Explanation:The term 'pellicle' in biology refers to a thin layer supporting the cell membrane in various protozoa. When it describes a uniform fine turbidity, it refers to how the organisms in a liquid medium are dispersed. In this case, 'a. Evenly cloudy throughout' pertains to a uniform fine turbidity, meaning that the microorganisms are evenly dispersed throughout the medium, giving it a uniformly cloudy appearance.
Beyond physical growth characteristics, other factors are important to consider when recording data about an organism. These may include the organism's environment, its genetic predisposition, the presence of predators or competitors, availability of resources such as food and space and potential changes in these conditions over time. All these elements contribute to the overall understanding of the organism and its behavior.
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during the course of development, the phenotype interacts with a environment to produce the genotype true of false
Answer:
True
Explanation:
Development in general involves the interaction between a genotype and its environment. For example, a organism born and raised in a cold climate likely will exhibit phenotipic features closest associated with such climatic conditions compared than other from the same species born and raised in a temperate region. It means that genes differentially expressed in different conditions can influence organismal development, although both organisms have potential to express the same genes
The statement is false. Genotype, a set of genes in an organism's DNA, interacts with the environment to influence phenotype, the physical expression of those genes. Environmental factors can modify phenotypic expression, demonstrating a reciprocal interaction between genotype and environment.
Explanation:The statement provided is false. During the course of development, the genotype - the set of genes in an organism's DNA - influences the organism's traits or phenotype. However, the environment can interact with the genotype, modifying the way these traits are expressed.
For instance, in Mendel's hybridization experiments, true-breeding plants with yellow pods and green pods produced hybrid offspring with yellow pods - the same color as one parent (phenotypically identical) but with a different genetic makeup (genotypically different).
Phenotypes and genotypes are interconnected, with the genotype serving as a blueprint, subject to environmental factors that can influence phenotypic expression. An example of environmental influence is sun exposure affecting skin color or temperature affecting sex determination in certain reptilian species. In essence, there is a reciprocal interaction between the genotype and the environment to determine the phenotype - not the other way around.
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A researcher conducts an experiment on the secretion of a particular hormone in mice. Scientists inject mice with a substance that stimulates the production of the hormone. The scientists then test the levels of hormones produced by the mice. The tool used to measure the hormones consistently detects the levels at 10 points lower than the actual hormone levels in the mice. This tool makes______ measurements, but the measurements aren’t _____.
first blank options:
1.)qualitative
2.)Reliable
3.)Valid
second blank options
1.)quantitative
2.)reliable
3.)valid
The first blank should be filled with Reliable, and the second blank should be filled with Valid. Therefore option 2 and 3 is correct.
The tool used to measure the hormones is reliable because it consistently produces consistent results, albeit with a consistent offset of 10 points lower. The reliability aspect pertains to the consistency and stability of the measurements.
However, the measurements from the tool are not valid because they don't accurately reflect the true hormone levels in the mice. Validity refers to the extent to which a measurement accurately measures what it's intended to measure.
In this case, the tool's measurements are consistently inaccurate by a fixed amount, which means they lack validity as they don't represent the actual hormone levels in the mice.
Therefore option 2 and 3 is correct.
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Which of the following statements about phospholipids is TRUE? A Option A: Each one has two fatty acid chains and the glycerol backbone is bonded to a small polar group. B Option B: Each one has three fatty acid chains. C Option C: The glycerol backbone is bonded to a small nonpolar group. D Option D: Their biological function remains unknown.
Answer:
Each one has two fatty acid chains and the glycerol backbone is bonded to a small polar group.
Explanation:
Phospholipid is a unique form of lipid. The bonding of the glycerol backbone to the polar phosphate group makes phospholipid to have dual solubility unlike general triglycerides.
The polar head is said to be hydrophillic that is water loving, while the two carbon chains that retained lipid features are hydrophobic water hating.
Therefore if a phopholipid is placed in water, in relation to its functions as component of cell membrane, it forms a bi-layer in which the water loving portion hydrophilic head points into the surrounding watery medium, while the hydrophobic layer points inwards far away from the watery medium into the internal cellular layer to form an impermeable barrier to hydrophilic (polar) substances.
This forms the basis of the phospholipd bilayer of the cell membrane. And it controls the permeability of the cell membrane to influx substances into the cells.
Phospholipids consist of a glycerol backbone bonded to two fatty acid chains and a small polar group, making it amphiphilic. They form the plasma membrane of cells, with the polar heads facing the water environment and the non-polar tails facing away from the water.
Explanation:The correct answer is Option A: Each one has two fatty acid chains and the glycerol backbone is bonded to a small polar group. Phospholipids consist of a three-carbon glycerol backbone with two fatty acid molecules attached to carbons 1 and 2, and a phosphate-containing group attached to the third carbon. This structure gives the phospholipid molecule an amphiphilic, or "dual-loving," characteristic with a polar (charged) head and non-polar (uncharged) tail. Phospholipids play a crucial role in forming the plasma membrane of cells, where the hydrophilic (water-loving) head faces towards the water environment and the hydrophobic (water-avoiding) tail faces away from the water. Examples of phospholipids include Phosphatidylcholine and Phosphatidylserine.
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