Answer:
The TSA or the tryptic soy agar is formed of casein and soybean meal, this formation helps in the appropriate growth of a huge array of non-fastidious and fastidious microbes. This combination of soy and casein provides organic nitrogen in the form of polypeptides and amino acids, which makes the medium more suitable for growth.
In the given case, if one permits the incubation of the dilution tubes for 24 hours prior to plating them on the TSA agar plates than there is a more probability of the result to get affected. As if unlimited resources are already present in the tubes, it will provide more favorable conditions for the formation of more colonies and thus will influence or change the colony-forming units per milliliter.
Even if the dilution is performed in a hood and in an autoclaved medium then also there will be an increase in the colonies of the microbes as in the time of 24 hours interval more microbes will get differentiated and will increase in number.
Final answer:
Allowing dilution tubes to incubate for 24 hours with unlimited resources before plating on TSA agar plates would impact the results by potentially causing confluent growth, making it difficult to distinguish individual colonies. TSA plates would typically have more colonies compared to EMB plates as TSA is non-selective and supports a broader range of bacterial growth.
Explanation:
If you allow dilution tubes to incubate for 24 hours before plating them on TSA agar plates, the results of the experiment would likely be impacted. Given unlimited resources in the tubes, this prolonged incubation could lead to an increase in bacterial growth, which may result in confluent growth (a mass of bacteria) in the test tube itself. Therefore, when you eventually do the plating, distinguishing individual colonies could be difficult or impossible, which would complicate or invalidate the results meant to measure bacterial concentration.
Regarding the comparison between TSA and EMB plates, due to the selective and differential properties of EMB agar, it inhibits the growth of gram-positive bacteria and highlights the growth of gram-negative bacteria, like Escherichia coli, by a color change. Since TSA agar is non-selective and supports a wider range of bacterial growth, you would generally expect more colonies on the TSA plates compared to the EMB plates, which would select only for specific types of bacteria.
Which of the following techniques is used sparingly because there is a slight but genuine risk of miscarriage or damage to the fetus while having a 99% accuracy in diagnosing genetic problems? amniocentesis CT scan ultrasound chorionic villus sampling
Answer: Amniocentesis
Explanation:
An amniocentesis is performed when a woman is between 14 and 16 weeks gestation. Women who choose to have this test are primarily those at increased risk for genetic and chromosomal problems, in part because the test is invasive and carries a small risk of miscarriage.
Amniocentesis is the technique used sparingly due to the slight risk of miscarriage or damage to the fetus. It offers 99% accuracy in diagnosing genetic problems. Other diagnostic techniques include chorionic villus sampling and prenatal genetic diagnosis.
Explanation:The technique referred to in the question is amniocentesis. This diagnostic procedure is usually done between weeks 15 and 20 of pregnancy, and is used to diagnose genetic disorders in the fetus. A small amount of amniotic fluid, which contains fetal tissues, is sampled from the amniotic sac surrounding the developing fetus. The fetal DNA is then examined for genetic abnormalities. The procedure has a high accuracy rate, but also carries a small risk of miscarriage or damage to the fetus, hence its use is usually reserved for cases wherein the potential benefits outweigh the risks.
Another technique is chorionic villus sampling (CVS) which involves sampling part of the placental tissue, usually performed between weeks 11 and 14 of pregnancy. This technique also carries risk and is used for similar reasons as amniocentesis.
Prenatal genetic diagnosis (PGD), ultrasound, and CT scans are other techniques used to detect conditions in the embryo or fetus, but they differ from amniocentesis and CVS in terms of their procedure, risk, and the type of information they provide. Note that, while ultrasound and CT scans provide images of the fetus, they do not provide the same genetic level of detail as amniocentesis or CVS.
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One factor that Saruman Enterprises considered in deciding whether to market a line of weight-loss supplements was the increasing rate of obesity in the United States. This trend is considered a(n) _______ factor.
Answer:
This trend is considered a(n) sociocultural factor.
Explanation:
In the field of business studies, sociocultural factors can be described as certain lifestyles or habits that characterize a particular society. The sociocultural environment of a particular area is used in business and marketing to make products which will attract the customers of a particular area. For example, with the increase in obesity of a particular place or city, people of that city will be more attracted to weight loss products.
Blue whales (Balaenoptera musculus) can weigh up to 170 tons and are the largest animals alive today. Their large size is thought to have protected them from large predators, such as Megalodon (Carcharocles megalodon), which would have preferred to eat smaller, less dangerous whales. What process best explains the evolution of the enormous blue whale body size
a. Natural selection.
b. Migration.
c. Genetic drift.
d. Non-random mating.
Answer:
Option c. Genetic Drift is the right answer
Explanation:
Due to random sampling of organisms results in the change of the frequency of an existing gene variant in a population is called as Genetic drift.
example: Baleen whales were moderately large. Baleen whales are the ancestors of the present-day blue whale. However, around 4.5 million year ago a massive evolution occurred in them which was resulted in blue whale origination.
Answer: A. Natural Selection.
Explanation: Blue whales (Balaenoptera musculus) can weigh up to 170 tons and are the largest animals alive today. Their large size is thought to have protected them from large predators, such as Megalodon (Carcharocles megalodon), which would have preferred to eat smaller, less dangerous whales. What process best explains the evolution of the enormous blue whale body size?
Which of the following statements is/are accurate?
A. One species of sea urchins can not fertilize another closely related species because the eggs do not have a receptor for the sperm. This is a good example of a hybrid breakdown.
B. Horses and Mules can be bred but their offspring is typically sterile. This could be an example of a hybrid breakdown.
C. Damselflies have sensory receptors that are sensitive to touch (tactile cues). Two related species of damselflies are unable to mate because their touch cues are not compatible. This is a good example of gametic isolation.
D. One species of sea turtle mates during the early spring and a closely related species mates during late spring. This is an example of temporal isolation.
E. One sponge species releases its gametes during the night and another species releases its gametes during the day. This is an example of ecological isolation.
Answer and explanation:
In nature there are different mechanisms that prevent the crossbreeding between different species, what in biology is called reproductive barriers. Some mechanisms that act by preventing hybridization between different species are:
Hybrid breakdown Gametic isolation. Mecanical isolation. Temporary isolation. Ethological aislaminet. Ecological insulation.These mechanisms are responsible for preserving the genetic integrity of each species by preventing hybridization between different species.
A. One species of sea urchins can not fertilize another closely related species because the eggs do not have a receptor for the sperm. This is a good example of a hybrid breakdown.This is not accurate. In the case of sea urchins, the encounter of gametes requires two chemical mechanisms:
The first mechanism is called chemotaxis, which consists of the presence of a chemical signal on the surface of the egg, for which only sperm has a receptor. Another mechanism is that - once the sperm and egg are found - the membrane of the egg releases substances that interact with receptors in the sperm, allowing the sperm to enter it.These two chemical mechanisms ensure that gametes of two different species cannot be joined and fertilized, which is an example of gametic isolation.
B. Horses and Mules can be bred but their offspring is typically sterile. This could be an example of a hybrid breakdown.This is accurate. Horses and donkeys belong to two different species, with a different chromosomal load:
Horses have 32 pairs of chromosomes. Donkeys have 31 pairs of chromosomes.Both species can be bred, but their descendant, mules (Equus africanus x ferus), have an odd number of chromosomes (63) and are infertile. This represents an exact example of hybrid breakdown.
C. Damselflies have sensory receptors that are sensitive to touch (tactile cues). Two related species of damselflies are unable to mate because their touch cues are not compatible. This is a good example of gametic isolation.This is not accurate. The sensitivity to the touch of damselflies is specific to individuals of the same species, preventing mating between male and female of different species.
This, like courtship, is a mechanism that prevents crossbreeding between different species, establishing an example of mechanical isolation.
D. One species of sea turtle mates during the early spring and a closely related species mates during late spring. This is an example of temporal isolation.This is accurate. When two related but different species - such as turtles - have their mating period at different times of the year, there is talk of temporal or seasonal isolation.
Temporal isolation is a reproductive barrier that prevents crossing between different species, due to their mating habits at different times.
E. One sponge species releases its gametes during the night and another species releases its gametes during the day. This is an example of ecological isolation.This is not accurate. In sponges, like some coral species, periods of release of gametes and fertilization vary throughout the day, with some synchrony between individuals of the same species.
The fact that some sponges release their gametes by day and other species do it at night is an example of the reproductive barrier called temporal isolation.
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Effect of pH on the Conformation of α-Helical Secondary Structures The unfolding of the α helix of a poly-peptide to a randomly coiled conformation is accompanied by a large decrease in a property called specific rotation, a meas-ure of a solution’s capacity to rotate circularly polarized light. Polyglutamate, a polypeptide made up of only L-Glu residues, has the α-helix conformation at pH 3. When the pH is raised to 7, there is a large decrease in the specific rotation of the solu-tion. Similarly, polylysine (L-Lys residues) is an α helix at pH 10, but when the pH is lowered to 7 the specific
This is because the pH of the solution changes the chemical and stereochemical properties of the solution.
Explanation:All the proteins or polypeptides are formed of polymers of amino acids. These amino acids comprise of an alpha carbon, where a hydrogen, a carboxyl group, an amino group and a variable group R is attached. All these four groups have their specific stereochemistry which gives the polypeptide a particular shape in their coiled form.
Here in case of polyglutamate, the polymer is formed of chains of glutamic acid. This glutamic acid has carboxyl group in the R group which remains free even in the polymerized state. In acidic pH, the carboxyl group has its normal structure - COOH, but as the pH increase to 7,the hydrogen ion dissociates making it - COO⁻. So the stereochemistry changes and the specific rotation also changes.
Similarly in poly lysine, there's amino group in the R group which remains stable in alkaline pH of 10,but in case of neutral or acidic pH, the structure becomes - NH3⁺. So, the specific rotation changes.
Final answer:
The secondary structure of proteins, like alpha-helix, is crucial for protein function and is stabilized by hydrogen bonds. The alpha-helix structure can be altered by pH changes, resulting in different physical properties, such as specific rotation, which indicates structural changes.
Explanation:
Effect of pH on Protein Secondary Structure
The secondary structure of proteins, like the alpha-helix (α-helix) and beta-pleated sheet (β-pleated), plays a critical role in the protein's overall structure and function. These structures are stabilized by hydrogen bonds between the amino acids in the protein chain. The α-helix is particularly stable due to hydrogen bonds that form between the carbonyl oxygen of one amino acid and the amide hydrogen four residues away. Changes in pH can lead to a disruption of these hydrogen bonds, causing the protein to unfold or alter its conformation. This can be observed through changes in physical properties, such as specific rotation, which is a measure of how a substance rotates polarized light. Polyglutamate and polylysine are examples of polypeptides that change from an α-helical structure to a randomly coiled conformation when the pH shifts away from their optimal range, thereby exhibiting a change in specific rotation.
Two selectively permeable sacs A and B were submerged in a 45% glucose solution that was contained in a beaker. Sac A contained a 15% glucose solution and Sac contained a 15% sucrose solution. In which direction did net movement of water molecules occur? a. Only from Sac A to the beaker b. Only from Sac B to the beaker c. From the beaker to both sacs d. From both sacs to the beaker
Answer:
Option a. Only from Sac A to the beaker is correct.
Explanation:
As beaker contains glucose which is a monosaccharide and Sac A also have glucose in it, So, therefore glucose from sac A will move into beaker through the process of OSMOSIS.
Sac A (15% glucose) is less concentrated as compared to beaker (45% glucose) therefore this phenomenon will occur. (See attached image for more detailed and graphical explanation)
The net movement of water molecules occurred from the beaker to both sacs due to the concentration gradient.
The net movement of water molecules occurred from the beaker to both sacs. This is because the 45% glucose solution in the beaker created a concentration gradient, with a higher concentration of solute outside the sacs compared to inside.
Water molecules will naturally move from an area of lower solute concentration to an area of higher solute concentration through a selectively permeable membrane, in order to balance the concentrations on both sides.
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In wheat plants, the feature of having colored kernels is dominant to having white kernels that lack pigment. A true-breeding plant with colored kernels is crossed to a true-breeding plant with white kernels, resulting in progeny that all harbor colored kernels. The F1 progeny are then crossed, and a few members of the F2 generation have white/colorless kernels. The modified ratio observed is 15 colored: 1 non-colored. Explain these results.
Answer: It occurred a dihybrid cross and epistasis.
Explanation: In dihybrid cross, two different genes controlled two different traits. When they interact with each other is called Epistasis. However, in wheat plants, the genes related to color kernels don't act opposedly to each other. In other words, the genes have the same role in producing protein, so they can substitute for each other.
In the color determination mechanism, a biochemical reaction is necessary to convert a precursor substance into a pigment and that reaction happens with the product of either genes. That's why having a dominant allele makes the wheat colorful. So, crossing colored kernels with white ones will produce a heterozygous F1 generation. Crossing this generation will produce a F2 generation with modified ratio of 15 colored: 1 non colored because, every individual who has dominant alleles will produce the substance and thus the biochemical reaction will happen. Only recessive homozygous ones won't have the substance and so won't have color.
According to the ideas of Thomas malhus what are the predicted changes to the human population after 2050
Answer:
The correct option is 'C' that is eventually the population stop increasing or would decreases due to lack of food and living space.
Explanation:
In 1798, he wrote an essay on Principle of population where he described that how the population will grow with economy.
He claimed that the population will grow until the food supply decreases and then the population will stop growing due to lack of food and space and those individuals that can fight against this condition and and could attain food, space and other essential necessities will survive.
Theoretical and experimental measurements show that in many cases, the contributions of ionic and hydrogen-bonding interactions to ΔH for protein folding are close to zero. Provide an explanation for this result. (Hint: Consider the environment in which protein folding occurs.)
The formation of favorable ______ ionic or ________ interactions in a _________ protein replace interactions between solvent (water) and the ionic species (or _________donors and acceptors) in the _________ state. The favorable ΔH obtained by formation of ____________ bonds in the ___________ protein is offset by the energy required to ___________ many interactions, with solvent going from the ________ to the ________ state.
Fill in the blanks above using the words listed below.
H-bonding
H-bond
intermolecular
restore
unfolded
folded
C-bond
break
intramolecular
C-bonding
Answer:
The formation of favorable C-bond ionic or C-bonding interactions in a folded protein replace interactions between solvent (water) and the ionic species (or restore donors and acceptors) in the unfolded state. The favorable ΔH obtained by formation of H-bond in the H-bonding protein is offset by the energy required to break many interactions, with solvent going from the inter-molecular to the intramolecular state.
Answer:
The answer is:
The formation of favorable intermolecular ionic or H-bonding interactions in a folded protein replace interactions between solvent (water) and the ionic species (or H-bond donors and acceptors) in the unfolded state. The favorable ΔH obtained by formation of intramolecular bonds in the folded protein is offset by the energy required to break many interactions, with solvent going from the unfolded to the folded state.
Explanation:
A population is made up of individuals where 149 have the A1A1 genotype, 18 have the A1A2 genotype, and 154 have the A2A2 genotype. What is the allele frequency of A1? Answer to 2 decimal places.
In population individuals where 149 have the A1A1 genotype, 18 have the A1A2 genotype, and 154 have the A2A2 genotype, so the allele frequency of A1 is 0.49.
What is the allele frequency?The incidence of a gene variant within a population is represented by the allele frequency. Alleles are different versions of a gene that share the same genetic locus on a chromosome.
To determine the allele frequency of A1
There is the formula:
frequency of A1 = (2 x number of A1A1) + Number of A1A2/ 2 x population
so, A1A1 = 149
A1A2 = 18
Population = 149 + 18 + 154 = 321
The frequency of A1 = (2 x 149) + 18/ 2 x 321
hence the frequency of A1 = 0.49
Therefore, the allele frequency of A1 is 0.49.
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Final answer:
The frequency of allele A1 in the given population is 0.49 when calculated by summing the number of A1 alleles and dividing by the total number of alleles in the population.
Explanation:
To calculate the allele frequency of A1 in the given population, we count the total number of alleles in the population and then determine how many of them are A1 alleles. Each individual has two copies of a gene, so we multiply the number of individuals by two to get the total number of alleles. The individuals with A1A1 genotype have two A1 alleles each, and the individuals with A1A2 genotype have one A1 allele each.
Total number of A1 alleles = (2 x 149) + (1 x 18) = 298 + 18 = 316
Total number of alleles in the population = 2 x (149 + 18 + 154) = 2 x 321 = 642
The frequency of A1 is thus = 316 / 642
= 0.49 (to two decimal places).
How would you calculate the percentage of individuals with a particular trait in a population knowing certain parameters?
Answer: Apply Hardy-Weinberg Equilibrium equation
Explanation:
Assuming the known parameters are the
frequency of two alleles 'B'and 'b' in a certain plant species, where 'B' is 0.19 and 'b' is 0.81
To calculate the percentage of heterozygous individuals in the population.
Apply Hardy-Weinberg Equilibrium equation, where heterozygotes are represented by the 2pq term.
Therefore, the number of heterozygous individuals (Bb) is equal to 2pq
2 × 0.19 × 0.81 = 0.31 or 31%
(b) To calculate the percentage of homozygous recessives in the population.
The homozygous recessive individuals (bb) are represented by the q^2 term in the H-W equilibrium
0.81 × 0.81 = 0.66 or 66%
Final answer:
Calculating the percentage of individuals with a particular trait in a population can be achieved through methods such as the Hardy-Weinberg formula for allele frequencies or capture-recapture for population size estimation. QTL simulations indicate the contribution of specific loci to genotypic variability. Genetic counseling makes use of genotype probability calculations from family pedigrees for inherited diseases.
Explanation:
To calculate the percentage of individuals with a particular trait in a population, we can use different methods depending on the information available. The Hardy-Weinberg formula is a tool that can be employed to estimate the allele frequencies in a population. When we know the frequency of individuals affected by a genetic condition, assuming genotype aa, and that all affected individuals have this genotype, we can derive allele frequencies using the equation q² = frequency of aa. For instance, if 0.04% of the population has the aa genotype (q² = 0.0004), we calculate the frequency of allele q by taking the square root of 0.0004, yielding q = 0.02. This allows us to then calculate allele p (since p + q = 1), and subsequently the frequency of the other genotypes (AA and Aa) within the population.
Another method for understanding population dynamics involves simulating quantitative trait loci (QTL). If we wanted one QTL to contribute 75% to the total genotypic variability, a second QTL to contribute 20%, and a third to contribute 5%, we would allocate their effects according to these proportions in the simulation model.
For estimating population size using marked individuals, the capture-recapture method is commonly utilized. If 80 deer are tagged and released, and then from a later capture of 100 deer, 20 are found to be marked, the population size (N) can be estimated using the ratio of marked to unmarked individuals in the second capture.
Finally, in the context of genetics, if we are advised of the mode of inheritance and genotypes present within a pedigree, we can calculate the probabilities of these genotypes being passed down to offspring. This is incredibly useful for genetic counseling and assessing risks of inherited conditions.
Some mutations result in proteins that function well at one temperature but are nonfunctional at a different (usually higher) temperature. Siamese cats have such a "temperature-sensitive" mutation in a gene encoding an enzyme that makes dark pigment in the fur. The mutation results in the breed's distinctive point markings and lighter body color.
Let's complete the question by adding the missing piece of information
The mutation results in the breed's distinctive point markings (ears, mask, tail and legs) and lighter body color. Use this information to explain the pattern of the cat's fur pigmentation.
Answer:
The mutation of the TYR gene results in the enzyme tyrosinase to be heat susceptible. Tyrosinase takes part in the production of melanin to give darker fur in colder areas. The areas like the tail, legs, ears, and face do lack as much body heat and so will get darker.
Explanation:
A unique protein (enzyme), known as tyrosinase, is the major workhorse in the development of the melanin. A research team from the University of California, USA, led by L. A. Lyons, discovered that Siamese cats have tyrosinase that went through mutation due to the changes in the DNA helix and is temperature-sensitive as it's activity reduces with a rise in temperature. This explains why cat’s warm parts of the body are coated with white, melanin-lacking hair since Tyrosinase is deactivated in these regions and melanin is not developed – hair is white-colored. On the other hand, in cooler boundary the enzyme is active and the melanin is formed – hair has dark color.
Which of the following statements about the evolution of development is false? a. Evolution by natural selection "works" like a tinker, assembling new structures by combining and modifying available materials. b. Nearly all evolutionary innovations are the result of modifications of previously existing mechanisms and structures. c. The genes that control development are highly conserved. d. All of the above are true; none is false
All of the above are true; none is false
Explanation:
Evolution is the phenomena by which an organism gets modified or develops more advanced characters which makes it more fit for the survival in nature.
The first statement is true because evolution refers to the modification of existing characters nothing new develops spontaneously to overcome natural selection. During natural selection organisms that survive gradually surpass the selection pressure. And the evolved character of it gets fixed within the species.
Evolution occurs within the existing mechanisms and structure which find out new ways of surviving in the changing environment.
The genes that control the development are highly conserved and this is the reason we can trace back the similarities with common ancestors. And make a comparitive study of structural anatomy.
Final answer:
The false statement is: c. The genes that control development are highly conserved.
Explanation:
The correct answer is: c. The genes that control development are highly conserved.
Genes that control development can be highly variable between species, allowing for diverse developmental processes and structures to evolve. While certain core developmental genes may be conserved across different species, there is also significant variation that allows for evolutionary innovation.
For example, the Hox genes, which play a crucial role in determining body structure during development, can vary greatly between species, leading to diverse body plans across different organisms.
You just bought two black guinea pigs from the pet store that are known to be heterozygous (Bb). You also know that black fur (BB) is dominant over white fur (bb), and that a lethal recessive allele is located only one cM away from the recessive b allele. You decide to start raising your own guinea pigs, but after mating these animals several times, you discover they produce only black progeny. a) How would you explain this result? b) If the original black guinea pigs produce an average of 10 offspring per mating, how many matings would you have to make with these same parents before you'd expect to see a white guinea pig? c) Indicate the most likely genotype of the white offspring.
Answer:
21 Mating
Explanation:
Ans A)The recessive lethal allele is tightly linked to, and thus co-segregates with, the recessive B allele, which is lethal in the homozygous state (bb)
Ans B)
1% of the gametes will be recombinant (Bl or bL)
0.5% of the gametes will be bL
0.5% of the gametes will be Bl
white animals will have the following genotype
genotype probability
bL/bL(0.005)(0.005) = 0.000025
bL/bl(0.005)(0.495) = 0.002475
bl/bL(0.495)(0.005) = 0.002475
Probability of white animal =0.004975
1/0.004975 = 201 therefore on average you'd have to look at 201 progeny before seeing a white animal so on average you'd have to do 21 mating
The idea that the bacterial genome is ""loose"" in the cytoplasm is incorrect because Choose one: A. the DNA is attached to the cell envelope and organized into domains through supercoiling and DNA-binding proteins. BY. attached ribosomes prevent tangling of the DNA. C. bacterial cells have a nucleus. D. the DNA is usually condensed into a chromosome.
Answer:D. the DNA is usually condensed into a chromosome
Explanation:
The bacterial genome is found loose in the cytoplasm but with a form. This unique feature is called the chromosomal DNA and it is not contained within a nucleus in the bacterial cell.
Answer: Option D.
DNA is usually condensed into a chromosome.
Explanation:
Bacterial genome refers to the complete set of DNA or complete set of genetic material of bacteria. Bacteria chromosomes are circular DNA. The chromosomes are packed by histone proteins into a condensed structure called chromatin. The condensed DNA is supercooled.
The body tubes are long, semi-rigid tubes, usually made of _______________________chloride or some other type of plastic. A typical ET tube has nine basic parts. The proximal end (sticking out of mouth) has a standard ________ mm adaptor.
Answer:
The body tubes are long, semi-rigid tubes, usually made of _polyvinyl_ chloride or some other type of plastic. A typical ET tube has nine basic parts. The proximal end (sticking out of mouth) has a standard _15_ mm adaptor.
Explanation:
An endotracheal tube (ET) is a flexible plastic tube that is inserted into the trachea through the mouth to help a patient breathe. Then it is connected to a ventilator for oxygen supply to patients's lungs. This process of insertion of tube is refereed as endotracheal intubation.
The body tubes in question are usually part of endotracheal tubes, made of polyvinyl chloride or a similar plastic. The proximal end of these tubes that sticks out of the mouth has a standard 15 mm adaptor.
Explanation:The body tubes referred to in the question are typically part of medical devices known as endotracheal (ET) tubes, used to secure a patient’s airway during operations or critical care. These tubes are long, semi-rigid tubes, typically made of polyvinyl chloride or some other type of plastic. A typical ET tube has nine basic parts. The proximal end, which sticks out of the patient's mouth, has a standard 15 mm adaptor. This universal size allows it to attach to a variety of equipment like ventilators or bag valve masks.
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In Drosophila, vestigial (partially formed) wings (vg) are recessive to normal long wings (vg+) and the gene for this trait is autosomal. The gene for the white-eye trait is on the X chromosome. Suppose a homozygous white-eyed, long-winged female fly is crossed with a homozygous red-eyed, vestigial winged male.
a. What will be the genotypes and phenotypes of the F1 flies?
b. what will be the genotypes and phenotypes of the F2 flies?
c. what will be the genotypes and phenotypes of the offspring of a cross of the F1 flies back to each parent?
Answer/ Explanation:
a. The genotype of a homozygous white eyed long winged female would be Vg+Vg+XrXr. We denote the white allele as recessive (r) because the XY male only has one copy and yet has red eyes, so the red eye trait (R) must be dominant. A homozygous red eyed vestigial winged male would have be VgVgXRY. The possible gametes for the female are Vg+Xr only. For the male, the possible gametes are VgXR or VgY
The attached punnett square shows the results of the cross. The females will all be Vg+VgXRXr. The males will all be Vg+VgXRY (must inherit Y from father). That means they will all have normal length wings, the males will have white eyes and the females will have red eyes.
b. The F2 flies arise from intercrossing the F1, so the cross will be Vg+VgXRXr x Vg+VgXRY. The possible gametes for the mother are: Vg+XR, Vg+Xr, VgXR or VgXr. The possible gametes for the father are Vg+Xr
, Vg+Y
, VgXr
, VgY
. The attached punnet square shows this cross. The ratio of the phenotypes will be 6:6:2:2, or 3:3:1:1 (long-winged red eye: long-winged white eye: vestigial wing red eye: vestigial wing white eye), genotypes shown in the attachment.
c. F1 cross back to the mother would be Vg+VgXRY x Vg+Vg+XrXr. The genotypes are shown in the attached punnet square. The offspring will all be long-winged with white eyes. The F1 to the father would be Vg+VgXRXr x VgVgXRY. The ratio would be 3:3:1:1 long-winged red eye: long-winged white eye: vestigial wing red eye: vestigial wing white eye
Final answer:
In crossing a homozygous white-eyed, long-winged female Drosophila with a homozygous red-eyed, vestigial-winged male, the F1 generation will have all females with red eyes and long wings and all males with white eyes and vestigial wings. The F2 generation shows a mix of possible phenotypes depending on the sex and inherited genes. Backcrossing the F1 with the original parents yields specific outcomes based on the traits of the parents and the mechanisms of inheritance at play.
Explanation:
Crossing Flies: Drosophila Melanogaster Genetics
In Drosophila melanogaster, the genetics of eye color and wing size are well-studied, with the white-eye trait being X-linked and vestigial wings being autosomal recessive. Considering a cross between a homozygous white-eyed, long-winged female and a homozygous red-eyed, vestigial-winged male, we can predict the outcomes for the F1 and F2 generations, as well as the results of backcrossing F1 individuals with each parent.
a. F1 Generation
b. F2 Generation
c. Backcross to Parents
Cross with white-eyed, long-winged female parent:Produces red-eyed, long-winged females and white-eyed, vestigial-winged males.Cross with red-eyed, vestigial-winged male parent:Results in a similar mix of phenotypes as the F2 generation, depending on sex and inheritance patterns.Because the researcher had access to twenty years of data on the same participants in an education program, the researcher was able to perform a(n) _________ study.
Answer:
Descriptive study
Explanation:
Any research study can be classified into two groups
a) Descriptive and
b) Analytical study
A Descriptive study is also referred as non-analytic study as it involves understanding of something happening in a population as a whole since time frame being referred by doing through literature review, studying case reports, conducting surveys, doing comparisons and referring previous researches. It helps to understand prevalence, incidence, or experience of a group
While analytic study is based on establishing relation between two factors through experiment
Because the researcher had access to twenty years of data on the same participants in an education program, the researcher was able to perform a longitudinal study. This type of research is particularly powerful for tracking changes and developments over time.
Longitudinal studies follow the same group of individuals and assess them repeatedly across a protracted timespan. For instance, psychologist Rich Lucas was able to utilize longitudinal data from over 20,000 Germans across two decades to explore the implications of marriage on happiness. Such studies can reveal impactful insights into the long-term effectiveness of educational programs and behavioral trends but can also be resource-intensive to conduct.
Longitudinal research contrasts with cross-sectional research, which analyzes different segments of a population at a single point in time. While cross-sectional studies can be less expensive and quicker to execute, they do not provide the sequential data over time that longitudinal research can offer.
A bacteria culture starts with 200 bacteria and in 1 hour contains 400 bacteria. How many hours does it take to reach 2000 bacteria?
Answer:
At 0hr it was 200 bacteria
At 1 hr it doubled and was 400
Means: at every 1 hr the population of the bacteria doubles
400 bacteria = 1hr
2000 bacteria = X
X = 2000 × 1 / 400
X = 5 hrs
LASIK, or laser-assisted in situ keratomileusis, is a surgical procedure on the eye that millions of people have undergone to improve visual acuity. It involves the use of a laser is to reshape the cornea in individuals suffering from astigmatism, near-sightedness (myopia), or far-sightedness (hyperopia).
1. How could changing the shape of the cornea affect one’s visual acuity?
Answer:
LASIK or laser-assisted in situ keratomileusis refers to a kind of refractive surgery. In the surgery, the surgeon initially cleaves a thin flap of tissue from the anterior part of the eye. After that, a laser removes away the tissue to reconfigure the cornea so that light aims better on the retina in the posterior part of the eye.
Cornea refers to a clear, safeguarding exterior covering of the eye. It generally offers protection against germs, dust, and other such deleterious particles. Apart from its protective function, the cornea also plays an essential part in amending the vision. The shape of the cornea is accountable for the bending of the light as it enters the eye.
The focusing of the objects relies hugely upon the curvature of the cornea. Thus, in the conditions where the cornea gets damaged, that is, its shape gets changed somehow, the vision gets affected and the manner in which the light moves within the eye gets distorted.
How does oceanic lithosphere form?
The system that enables 4 nucleotides to dictate the sequences of millions of proteins is called the __________.
Answer:
Genetic code
Explanation:
Genetic code refers to the set of rules by which the information encoded in the genetic material is translated into the proteins. The genetic code is present on the genes in the form of triplets of nucleotides called codons which specifies the specific amino acid.
The codon system of the genetic code rules allows the 4 different nucleotides to form specific amino acids by the combination of the four different nucleotide in the triplets.
Thus, Genetic code is the correct answer.
would cephalization be an advantage to jellyfish? why or why not?
Answer:
Cephalization does not confer an advantage to free-floating or sea organisms. Many aquatic species display radial symmetry like the jelly fish for instance.
Aquatic animals are not cephalized because they must be able to find food, take caution and defend themselves from harm from all directions.
What are the emergent properties of the nervous system that cannot be predicted by studying individual neurons?
Answer:
Consciousness, human emotion etc, are emergent properties
Explanation:
The properties which are characteristics of an entire system and not its constituting members are called as emergent properties.
Consciousness with in an individual human being can be termed as an emergent property of nervous system. A single neuron cannot generate or holds the sense of consciousness, self-awareness, pride or honor etc. The entire nervous system can also generate complex human emotions such as fear, joy, pride etc. Neuro-biologists have not been able to depict the expression of these functions at micro level such as a single neuron and thus these properties are termed as emergent properties.
Dendrites are A. the conduction zone of a nerve cell. B. the input zone of a nerve cell. C. a type of glial cell. D. small interneurons.
Answer:
B. the input zone of a nerve cell.
Explanation:
Dendrites are the small extensions that come out of the soma or cell body of a neuron. The function of dendrites is to receive the nerve signals or information from the axons of the presynaptic neurons and carry them towards the cell body or soma. In a synapse, the signals from the axons of the presynaptic neurons are revived by dendrites of postsynaptic neurons. The plasma membrane of dendrites have receptors to which the chemical messengers from other cells bind. In this way, dendrites serve as input zone of a nerve cell.
Dendrites are the input zone of a nerve cell. They receive information from other neurons and deliver it to the body of the neuron. They are not glial cells, small interneurons, or the conduction zone of a nerve cell.
Explanation:Dendrites are best described as B. the input zone of a nerve cell. These are branch-like structures that protrude from the neuron. They play a crucial role in receiving information from other neurons and transmitting it to the body of the neuron. Unlike the axon, which is the output zone of the nerve cell, the dendrites play an essential role in receiving neural inputs or signals, acting as the 'input zone'. They are not a type of glial cell (which are non-neuronal cells in the nervous system), nor are they small interneurons or the conduction zone of a nerve cell.
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An enzyme-linked immunosorbent assay requires:
A) a radioactive substrate.
B) a radioactive standard for binding to the antibody.
C) aromatic amino acids.
D) a catalytic antibody.
E) an antibody that binds the protein of interest.
Answer: Option E.
Eliza require an antibody that bind to protein of interest.
Explanation:
Enzyme linked immunosorbent assay is a plate like or immunological technique or assay that Is use to detect and measure peptides, proteins,hormones and antibodies.
Enzyme linked immunosorbent assay require an antibody that can bind to a specific protein of interest.
Would penicillin be expected to have an equal ability to kill both Gram-positive and Gram-negative bacteria? Why or why not?
Answer:
No, due to the difference in Peptidoglycan layer
Explanation:
Since the mode of operation of the drug penicillin is to prevent the proper build-up of peptidoglycan in the cell wall of bacteria and bursting up its cell wall, Penicillin would have greater effect in killing Gram-negative bacteria with thin layer of peptidoglycan in their cell wall, unlike Gram-positive bacteria with a thick layer of peptidoglycan in their cell wall.
Suppose that the narrow-sense heritability of wool length in a breed of sheep is 0.92, and the narrow-sense heritability of body size is 0.87. The genetic correlation between wool length and body size is –0.84. If a breeder selects for sheep with longer wool, what will be the most likely effects on wool length and body size?
Answer:
There is an increase in wool length and a little decrease in body size
Explanation:
Given that:
the narrow-sense heritability of wool length in a breed of sheep is 0.92; &
the narrow-sense heritability of body size is 0.87.
As these traits approaches 1, the higher the rate that helps them to provide a positive feedback for selection.
The question proceeds by stating that " The genetic correlation between wool length and body size is –0.84."
You see, the functionality of this negative (-0.84) genetic correlation tends to create an imbalance between these heritable traits. As such, there is a shift in the paradigm in which one traits tends to increase and the other trait to decrease.
Now, if a breeder selects a sheep with longer wool, there is an increase in wool length since he selects it although there will be a little decrease in body size due to the negative genetic correlation effect between the two heritable traits.
"In 1958, Meselson and Stahl conducted an experiment to determine which of the three proposed methods of DNA replication was correct. Identify the three proposed models for DNA replication."
Answer:
Three proposed models: 1) Conservative
2) Semiconservative
3) Dispersive
Correct one: 2) Semiconservative
Explanation:
Meselson and Stahl proposed that the process of DNA replication could be conservative, semiconservative or dispersive. Conservative DNA replication will form one DNA duplex with both newly formed strands and the other duplex with both parental strands. The semi conservative mechanism will form two DNA helices each of which would have one parental strand and one newly formed strand (hybrid). The dispersive mechanism will form DNA double helices with patches of the new and parental strand.
However, they found that DNA replication follows a semi-conservative mechanism as after two rounds of replication of heavy chain DNA in the light-medium formed two molecules of hybrid DNA and two DNA molecules with light chains only. There was no DNA molecule with both the strands having the heavy chains.
The three proposed models for DNA replication individually tested by Meselson and Stahl in 1958 were the conservative, semi-conservative, and dispersive models. Their experiment proved the semi-conservative model, where each new DNA molecule consists of one old and one new strand, to be the accurate depiction.
Explanation:In 1958, Matthew Meselson and Franklin Stahl conducted an experiment with the goal of determining the accurate model for DNA replication. The three proposed models that they evaluated were the conservative model, the semi-conservative model, and the dispersive model.
In the conservative model, the parent DNA molecule remains intact and an all-new molecule is formed based on its template. In contrast, the semi-conservative model suggests that each of the two new DNA molecules consists of one original and one new strand, replicating its structure from the parent molecule. Lastly, the dispersive model proposes that each strand of both daughter molecules contains a mixture of old and newly synthesized DNA.
As a result of the experiment, they found that the semi-conservative model was the correct depiction of DNA replication.
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The protein of the bicoid gene in Drosophila determines the _____ of the embryo. Group of answer choices anterior-lateral axis posterior-dorsal axis anterior-posterior axis posterior-ventral axis
Answer:
Option c. anterior-posterior axis is the right answer.
Explanation:
Bicoid is a maternal effect gene whose protein concentration gradient patterns the anterior-posterior (A-P) axis during Drosophila embryogenesis. it was the first protein demonstrated to act as a morphogen. See image having more explanation.
Final answer:
The protein of the bicoid gene in Drosophila determines the anterior-posterior axis of the embryo.
Explanation:
The protein of the bicoid gene in Drosophila determines the anterior-posterior axis of the embryo. The bicoid protein is essential for the establishment of the head and thorax regions in the developing embryo. It is distributed in a concentration gradient along the anterior-posterior axis and helps specify different cell fates in this axis. Higher concentrations of bicoid protein at the anterior end lead to the formation of head structures, while lower concentrations at the posterior end result in the development of abdominal segments.
The bicoid gene in Drosophila is essential for determining the anterior-posterior axis of the embryo. Axis formation is a pivotal step in embryogenesis, with genes like bicoid playing a crucial role in defining the future head and tail regions of the organism. Research involving axial development has highlighted the importance of such genes in ensuring the proper symmetry and structure of the developing organism. Mutations in the genes responsible for axis formation, including Hox genes, can result in drastic changes to body symmetry, leading to developmental abnormalities. In vertebrates and invertebrates alike, Hox genes are pivotal in determining body structure and the organization of various body segments.