Final answer:
The mean relative fitness of the population is 0.85. The expected allele frequency change for allele A after one generation with selection is a decrease to approximately 0.62 due to the relative fitness differences among the genotypes.
Explanation:
The question at hand requires an understanding of genetic structure, allele frequency, and the Hardy-Weinberg principle to calculate both the mean relative fitness within a population and the expected allele frequency change for allele A after one generation with selection.
Mean Relative Fitness Calculation:
To calculate the mean relative fitness (w), we multiply the relative fitness of each genotype by its proportional representation in the population and sum the results:
w(AA) = 1.00 × (500/1000) = 0.50,w(Aa) = 0.80 × (250/1000) = 0.20,w(aa) = 0.60 × (250/1000) = 0.15.The sum is the mean relative fitness: wμ = 0.50 + 0.20 + 0.15 = 0.85.
Expected Allele Frequency Change:
The next step is to calculate the expected change in allele frequency using the Hardy-Weinberg equation. First, determine the frequency of the alleles (p for A and q for a). The current frequency of A (p) = (2×500 + 250) / (2×1000) = 0.625. The frequency of a (q) = 1 - p = 0.375. To calculate the expected frequency of allele A after one generation of selection, we must account for the relative fitness of each genotype:
p' = (p² × w(AA) + pq × w(Aa)) / wμ,p' = (0.625² × 1.00 + 0.625 × 0.375 × 0.80) / 0.85 ≈ 0.62.Hence, the expected frequency of A after one generation is approximately 0.62, indicating a slight decrease due to selection.
The mean relative fitness within the population is 0.85, rounded to two decimal places.
To find the mean relative fitness within the population, we need to calculate the weighted average of the relative fitnesses of each genotype, considering their frequencies.
Given:
- AA genotype count [tex](n_AA)[/tex] = 500
- Aa genotype count [tex](n_Aa)[/tex] = 250
- aa genotype count [tex](n_aa)[/tex] = 250
- Relative fitness of AA genotype [tex](w_AA)[/tex] = 1.00
- Relative fitness of Aa genotype [tex](w_Aa)[/tex] = 0.80
- Relative fitness of aa genotype [tex](w_aa)[/tex] = 0.60
First, calculate the total number of individuals in the population (N):
[tex]\[ N = n_{AA} + n_{Aa} + n_{aa} \][/tex]
[tex]\[ N = 500 + 250 + 250 \][/tex]
[tex]\[ N = 1000 \][/tex]
Next, calculate the contribution of each genotype to the total fitness:
[tex]\[ Contribution_{AA} = n_{AA} \times w_{AA} = 500 \times 1.00 = 500 \][/tex]
[tex]\[ Contribution_{Aa} = n_{Aa} \times w_{Aa} = 250 \times 0.80 = 200 \][/tex]
[tex]\[ Contribution_{aa} = n_{aa} \times w_{aa} = 250 \times 0.60 = 150 \][/tex]
Now, sum the contributions of all genotypes:
[tex]\[ Total \, Contribution = Contribution_{AA} + Contribution_{Aa} + Contribution_{aa} \][/tex]
[tex]\[ Total \, Contribution = 500 + 200 + 150 = 850 \][/tex]
Finally, calculate the mean relative fitness (w_mean):
[tex]\[ w_{mean} = \frac{Total \, Contribution}{N} \][/tex]
[tex]\[ w_{mean} = \frac{850}{1000} \][/tex]
[tex]\[ w_{mean} = 0.85 \][/tex]
So, the mean relative fitness within this population is 0.85, rounded to two decimal places.
The blood of a normal healthy adult rat contains 5 million red blood cells (RBCs) per microliter of blood. Each deciliter of blood contains 15 grams of hemoglobin. Each RBC contains 250 million molecules of hemoglobin (a fairly close approximation for a healthy adult retired breeder female rat). Suppose you have 8.56 mL of blood drawn from a healthy female retired breeder rat, then how many RBCs are in this blood sample?
Answer:
There are 4.28×10^10 RBCs in the blood sample.
Explanation:
Concentration of blood = 5×10^6 RBC/10^-6 L = 5×10^12 RBC/L of blood
Volume of blood sample = 8.56 mL = 8.56/1000 = 0.00856 L
Number of RBCs = concentration of blood × volume of blood sample = 5×10^12 RBC/L × 0.00856 L = 4.28×10^10 RBCs
"What would Avery, Macleod, and McCarty have concluded if their results had been that only RNAse treatment of the heat-killed bacteria prevented transformation of genetic virulence?
Answer:
RNA was the genetic material
Explanation:
If avary, Macleod, and McCarty would have seen that only RNAse treatment of the heat-killed bacteria prevented the transformation of genetic virulence than might have concluded that RNA is the genetic material as transformation does not occur because the RNAase degraded the RNA.
But if the genetic material is DNA then RNAase will not work because RNAase can not degrade DNA and then DNA will pass from virulent heat killed strain to nonvirulent strain and will cause transformation.
Fill in the blanks, your answers should be lower case. The process of makes an RNA copy from a DNA template. The RNA is made by the enzyme which makes the RNA from (3 or 5)' to (3 or 5)'. The process of makes a protein from an mRNA. The protein is made by the which catalyzes the peptide bond between each amino acid. In bacterial cells these two processes occur in the cytoplasm. However, in eukaryotes the process occurs in the nucleus and the process occurs in the cytoplasm
The process of transcription makes an RNA copy from a DNA template.
The RNA is made by the enzyme RNA polymerase which makes the RNA from (3 or 5)' to (3 or 5)'.
The process of Translation makes a protein from an mRNA.
The protein is made by the peptidyl transferase in Ribosome which catalyzes the peptide bond between each amino acid.
In bacterial cells these two processes of transcription and translation occur in the cytoplasm.
However, in eukaryotes the process of transcription occurs in the nucleus and the process of translation occurs in the cytoplasm
Explanation:
The process of translation and transcription comprises the central dogma of molecular biology. By these two process the information stored in the genes flows into the proteins.
The process of creating an RNA copy from a DNA template is known as transcription, conducted by the enzyme RNA polymerase. The translation of an mRNA into a protein is facilitated by the ribosome. While both processes occur in the cytoplasm in bacteria, in eukaryotes, transcription occurs in the nucleus and translation in the cytoplasm.
Explanation:The process of transcription makes an RNA copy from a DNA template. The RNA is made by the enzyme RNA polymerase which makes the RNA from 3' to 5'. The process of translation makes a protein from an mRNA. The protein is made by the ribosome which catalyzes the peptide bond between each amino acid. In bacterial cells, these two processes occur in the cytoplasm. However, in eukaryotes, the transcription process occurs in the nucleus and the translation occurs in the cytoplasm.
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"Suppose that two parents are both heterozygous for sickle cell anemia, which is an autosomal recessive disease. They have eight children. Use the binomial theorem to determine the probability that three of the children have sickle cell anemia and five of the children are healthy. Round your answer to the nearest tenth."
If two parents are both heterozygous for sickle cell anemia, the probability that three of the children have sickle cell anemia and five of the children are healthy is approximately 0.1 or 11.0%.
What is the probability of sickle cell anemia?The probability of having 3 children with sickle cell anemia and 5 healthy children out of 8 total children is explained below.
The probability of any individual child having sickle cell anemia with heterozygous parents is 1/4; the probability of any individual child being healthy is 0.75.
Using the binomial theorem,
P(3 children with sickle cell anemia and 5 healthy children)
= [tex]P^8[/tex]₃ ×[tex](1/4)^3[/tex] ×[tex](3/4)^5[/tex]
= P(3 children with sickle cell anemia and 5 healthy children) = 0.110 that is nearly 11%.
Hence, if two parents are both heterozygous for sickle cell anemia, the probability that three of the children have sickle cell anemia and five of the children are healthy is approximately 0.1 or 11.0%.
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Final answer:
The probability that three out of eight children of two heterozygous parents for sickle cell anemia have the disease and five are healthy is approximately 20.7%.
Explanation:
The question posed involves the application of the binomial theorem to a genetic problem involving sickle cell anemia, an autosomal recessive disease. To find the probability that three of the children have sickle cell anemia and five of the children are healthy when both parents are heterozygous for the trait, we can use the binomial probability formula:
P(X=x) = ⁿCₓ × (p)ˣ × (q)ⁿ⁻ˣ
Where 'n' is the total number of children (8), 'x' is the number of children affected with sickle cell anemia (3), 'p' is the probability of having sickle cell anemia (1/4), and 'q' is the probability of being healthy (3/4). The term 'ⁿCₓ' represents the binomial coefficient, which can be calculated using combinations.
P(3 have sickle cell) = ⁸C₃ × (1/4)³ × (3/4)⁵
Calculating further:
P(3 have sickle cell) = 56 × (1/64) × (243/1024)
P(3 have sickle cell) = 56 × (243/65536)
P(3 have sickle cell) = 13608/65536
Converting to a percentage and rounding to the nearest tenth gives us approximately:
P(3 have sickle cell) = 20.7%
Restriction enzymes, which are extensively used in molecular biology, are products of bacterial "immune" system. Since bacterial genomes span several million base pairs (E. coli > 4 million bps), and the presence of restriction site within a genome is more than likely, how does a bacteria manage to protect itself from innate REs?
Explanation:
In the term of disease by intracellular microscopic organisms, they have the ability and repeated inside phagocytic cells, which causes the circulating antibodies to be distant to intracellular bacteria The natural safe reaction against these microscopic organisms is intervened basically by phagocytes and NK cells Innate immunity additionally arrives in a protein substance structure, called innate humoral immunity. for example, the body's supplement system and substances called interleukin-1 (which causes fever) and interferon25] Discuss what your analysis above indicates about the applicability of the Hardy-Weinberg criteria to this population. Which assumptions, if any, of the Hardy-Weinberg criteria are violated
Answer: The Hardy–Weinberg principle, also known as the Hardy–Weinberg equilibrium, model, theorem, or law, states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences.
Explanation: the population is in Hardy-Weinberg equilibrium for a gene, it is not evolving, and allele frequencies will stay the same across generations. There are five basic Hardy-Weinberg assumptions: no mutation, random mating, no gene flow, infinite population size, and no selection.
what color would the ecoli cell appear under the microscope following a gram stain? explain why with a clear explanation of the gram staining process and the peritnent cellular componests that cause cells to stain
Answer:E. Coli will appear pink in color under the microscope following gram staining
Explanation:
The Gram stain is a differential technique that is commonly used for the purposes of classifying bacteria. The staining technique distinguishes between two main types of bacteria (gram positive and gram negative) by imparting color on the cells.
Being Gram-negative bacteria, E. coli have an additional outer membrane that is composed of phospholipids and lipopolysaccharides. The presence lipopolysaccharides on the outer membrane of bacteria gives it an overall negative charge to the cell wall. Because of these properties, E. coli does not retain crystal violet during the Gram staining process.
In organisms with large genomes, inversions are more likely to be tolerated if the breakpoints occur in: reciprocal translocations. noncoding DNA. open reading frames. coding DNA. closed reading frames.
Answer:
Non-coding DNA.
Explanation:
Inversion is a type of chromosomal abnormality in which the sequence of a segment of a chromosome is inverted or rotated at an angle of 180 degrees. This type of abnormality can change the reading frame of the gene and can cause mutations.
But if the genome sequence is non-coding that is not involved in the formation of protein synthesis than even if the reading frame is inverted will not affect the phenotype of the cell. Also, the non-coding sequences are removed by the splicing mechanisms.
Thus, Non-coding DNA is correct.
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Linda wrote a few steps to describe how carbon circulates between the atmosphere and living organisms. Step 3 is missing.
Step 1: Carbon enters the atmosphere as carbon dioxide from respiration and combustion
Step 2: Plants use carbon dioxide to make glucose
Step 3:
Step 4: Animals exhale carbon dioxide
Which of the following best describes step 3?
a. Dead animals are buried underground and form fossil fuels.
b. Carbon present in glucose is converted into oxygen by animals.
c. Carbon present in glucose is transferred into plant-eating animals.
d. Dead animals decompose and recycle the carbon in the form of carbon dioxide.
Answer:D
Explanation:decomposition of dead plants and animals.
Answer:
c
Explanation:
edit: yup its c i just did the test and got it right
. The Vmax of a glucose transport into a certain preparation of red blood cells is determined to be 1206nmol glucose/s without ATP present in the buffer. The Vmax of a glucose transport into the same preparation of red blood cells is determined to be 1158nmol/s when ATP is present in the buffer. However, when cells that do not express any glucose transporter are probed for rate of glucose transport, Vmax is infinite. Explain these data.
Answer:
During a process of facilitated diffusion, presence ATP increase the rate of glucose transport through a protein membrane of a red blood cells. Vmax will then be on increased.
Explanation:
The rate of glucose transport in to the certain sample of blood is a facilitated diffusion process. It is understandable to be on increase through a protein membrane with the presence of enzyme to speed up the rate of the biological reaction. This is reflected when it was initially observed that with the enzyme ATP presence in the buffer, the Vmax of a glucose transport into the certain preparation of red blood cells is determined to be 1206nmol glucose/s(considerably low).
When ATP is added, the Vmax is on increase. That is, the Vmax of a glucose transport into the same preparation of red blood cells is determined to be 1158nmol/s. Then Vmax rises to infinity, when cells that do not express any glucose transporter are probed for rate of glucose transport.
It can be further explained that glucose moves into the blood through the permease in the membrane between the cell and the blood.Thus, ATP is used as an energy source to drive Na+ out of the cell, resulting in glucose transport from the intestine to the blood.
Membrane proteins must have an asynchronous distribution on the cell membrane for the system to function. This is an example of the membrane synthetic apparatus determining where in a membrane a protein should be localized. The Na+K+ ATPase must be localized to the membrane between the cell and our blood.
The data reflects that glucose transport via GLUT1 is highly efficient and typically passive, but slightly inhibited by ATP. Without GLUT1, glucose is minimally transported across the membrane, emphasizing the necessity of transport proteins. The minor reduction in Vmax with ATP indicates possible regulatory mechanisms.
The data indicates different behaviors for glucose transport under various conditions:
Vmax of glucose transport without ATP is 1206 nmol/s, showing a high rate of passive, protein-mediated transport via GLUT1.When ATP is present, Vmax decreases slightly to 1158 nmol/s, possibly due to competition or inhibition of the transport process.Without glucose transporters, Vmax is infinite, suggesting that glucose does not pass through the membrane efficiently without these transporters, highlighting the necessity of proteins like GLUT1 for effective glucose transportation.The data helps illustrate passive transport via GLUT1, a glucose transporter protein, and how ATP impacts the transport rate.
It can be desirable to produce eukaryotic proteins in prokaryotes such as E. coli. To do this several DNA sequences must be joined or cloned together. Place the labels in the appropriate position to allow for the expression of the insulin protein in a prokaryotic cell. cDNA is DNA that is synthesized from the mature mRNA from eukaryotic cells using the enzyme reverse transcriptase.
5' end_____ _____ _____ _____3' end
1. Poly A signal sequence
2. Genomic clone of insulin
3. Rho terminator
4. -95 CAT box
5. -10 TATA
6. -25 TATA box
7. cDNA of insulin
8. -35 Sequence
Final answer:
To express human insulin protein in E. coli, assemble a DNA sequence that contains a -35 Sequence, a -10 TATA box for bacterial promoter activity, followed by the cDNA of insulin, and ends with a Rho terminator for transcription termination.
Explanation:
To generate a recombinant version of the human insulin protein in a prokaryotic cell, such as E. coli, we need to assemble a proper expression cassette. The cassette needs to have a eukaryotic gene sequence converted to cDNA, combined with prokaryotic promoter and terminator sequences to allow expression in the host cell.
The correct order for cloning the insulin cDNA for expression is:
-35 Sequence-10 TATA box (also known as Pribnow box)cDNA of insulinRho terminator or a similar prokaryotic terminatorThe -35 Sequence (-35 Sequence) and the -10 TATA box (-10 TATA) are part of the bacterial promoter necessary for initiating transcription. The cDNA of insulin (cDNA of insulin) is inserted downstream of the promoter, allowing it to be transcribed. Finally, a termination sequence such as the Rho terminator is necessary to stop transcription.
Why these specific elements? The -35 and -10 sequences are binding sites for the RNA polymerase in prokaryotes, which begins the process of transcription. The cDNA of insulin, obtained through reverse transcription of the mature mRNA, excludes any introns that would be present in the genomic clone (thus, we do not use the Genomic clone of insulin or the Poly A signal sequence which are found in eukaryotic expression systems). The Rho terminator is a sequence that facilitates the termination of transcription in prokaryotes.
Imagine that zika virus has a 1% incidence in the population. A test for the virus has a 3% false positive rate and no false negative rate. If someone takes the test and gets a positive result, what is the chance that they are infected?
Answer:
There is 97% Chance that the person is infected.
Explanation:
According to the question, the test for the virus has 3% false positive and there is no false negative rate.
Therefore if someone takes the test and gets a positive result, the chances of the person to be infected ( True positive ) is
100 - 3 = 97%
Note: False positive result is a result that indicates that a given condition is present when it is not actually present.
An individual suffers a blood clot in an artery that delivers blood to his leg. The leg begins to take on a blue hue, becomes colder than the rest of his body and he experiences numbness in the leg.
He is most likely experiencing:
a) anemic hypoxia
b) ischemic hypoxia
c) hypoxic hypoxia
d) histotoxic hypoxia
e) allergic hypoxia
Answer: Anemic hypoxia.
Explanation:
Anemic hypoxia is a medical condition that result when few haemoglobin are present in the blood leading to decreased ability of the blood to carry oxygen.
Oxygen is essential for human proper body functioning.
This is normally cause by lack of red blood cells ( haemoglobin) that carry oxygen.
The symptoms of anemic hypoxia are;
Skin color change to blue or cherry red.
Numbness
Cold hands and feet.
Headache
Weakness.
From the questions, the symptoms are similar to anemic hypoxia.
in general terms, what two general factors in a medium affect small speed of sound?
Please explain! I will give brainliest!
Answer:
Density of the medium, temperature of the medium, and stiffness of the medium.
Explanation:
Medium
Medium has a huge effect of the speed of sound. When most people discuss the “speed of sound” they are talking about the propagation of sound waves through the medium of “Air”. For anyone who has gone underwater and listen to people talking above it is likely that one would notice the muted an “odd” way that voices sound underwater. This is because the “medium” of water greatly bends, distorts and changes the speed of sound wave.
There is a whole aspect of science that measure and defines the effect of different mediums (gaseous and liquid) on the speed of sound. This is called Fluid Dynamics. Underwater communication is possible if you understand how this wave propagation as well as another important factor (pressure).
Because of elasticity of materials sound will, as a rule of thumb, generally travel faster in solids than in liquids and faster in liquids than in gases.
♦ Temperature
Temperature has a large effect on the speed of sound. Not as much as the “Medium” does, but far more than anything else. Temperature affects the speed of sound because temperature can affect the “elastic” qualities of different mediums. At the very basics lower temperatures will decrease the speed of sound while higher temperatures will increase the speed of sound, all other factors being equal.
♦ Pressure
Pressure is the final factor that has a significant impact on the speed of sound. The effect of pressure on the speed of sound is due to the materials inertial properties. In short, the more pressure that is applied to the material or medium the denser it becomes and the greater the “inertia” becomes. This makes any interactions between particles slower. Therefore the speed of sound throughout the medium is slowed due to the greater pressure.
Answer:
Elasticity and density
Explanation:
The speed of sound depends on elasticity and density of the medium through which it is travelling. The sound travels faster in liquids than gases. The sound travels faster in solids than in liquids. The lesser the elasticity and higher the density , the sound travels slower in a medium.
Speed = elasticity / density
Why does it make biochemical sense that chaperones recognize hydrophobic surface area? What catastrophic event are chaperones meant to prevent in cells?
Answer:
Molecular chaperons in the cells helps in protein folding. These are the group of proteins that have functional similarity and they also assist protein folding.
They have the ability to prevent the non specific binding and aggregation by the binding of the non native proteins.
Molecular chaperons helps in recognizing the hydrophobic surfaces of the unfolded proteins because they themselves are hydrophobic in nature and will combine to the hydrophic binding and bonding.
This helps in guiding the protein to folding.
Chaperones recognize hydrophobic surface areas to prevent protein aggregation during folding, which could lead to cellular dysfunction. They help proteins fold correctly and prevent catastrophic events such as the formation of protein aggregates.
Explanation:Chaperones recognize hydrophobic surface areas because many proteins require assistance in the folding process to prevent them from aggregating during folding. Hydrophobic regions on the protein's surface tend to be exposed and can interact with other hydrophobic regions, leading to protein aggregation. Chaperones bind to these hydrophobic regions and help the protein fold correctly by preventing aggregation. This prevents catastrophic events such as the formation of protein aggregates, which can lead to cellular dysfunction and disease.
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Which of the following processes directly require ATP? Choose all that apply.
a.Diffusion of calcium into the motor neuron at the neuromuscular junction.
b.Release of cross-bridge (interaction) between actin and myosin.
c.Movement of calcium ions back into the sarcoplasmic reticulum after contraction.
d.Conformation change of troponin resulting in the movement of tropomyosin off the actin active sites.
Answer:the following processes directly require ATP includes:
B (Release of cross-bridge (interaction) between actin and myosin.)
C (Movement of calcium ions back into the sarcoplasmic reticulum after contraction.)
Explanation:
Muscle contraction occurs in various parts of the body to ensure proper body functioning. This process requires the release of calcium ion and the use of ATP ( Adenosine Triphosphate) as source of energy at various levels for the process to take place. The distinct role of ATP in muscle contraction includes:
-ATP is directly required as it causes detachment from actin after power stroke when it binds at one of the reactive sites of myosin. This explains option B (Release of cross-bridge (interaction) between actin and myosin.)
-it powers the pump that transports calcium ions back into the sarcoplasmic reticulum after contraction. This explains option C (Movement of calcium ions back into the sarcoplasmic reticulum after contraction)
- it activates the myosin head so it can bind to actin and rotate by the action of ATpase enzyme.
The following processes which directly require ATP include
Release of cross-bridge (interaction) between actin and myosin.Movement of calcium ions back into the sarcoplasmic reticulum after contraction.The myosin head has to move in order for the release of cross-bridge
between them to be achieved. The myosin head movement brings about
movement of the actin also.
The movement of calcium also requires the use of ATP as it helps in the
pumping of the calcium ions into the sarcoplasmic reticulum.
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Two strains of true-breeding maize that both produce ears of corn with white kernels are crossed and found to produce F1 plants that all make ears of corn with red kernels. If these F1 plants are backcrossed to one of the parents, what proportion of the offspring should have white kernels
Answer:
50% or 1/2
Explanation:
Since the two parental strains are true-breeding, that is, homozygous; it means that the red kerneled ears of corn offspring produced are heterozygous and one of the parental strains must have been homozygous recessive and the other homozygous dominant.
Assuming the ears of corn's colour is coded for by the allele A, it means that one of the parents has the genotype AA while the other has aa with the offspring having genotype Aa.
AA and aa genotypes produce white ears of corn while Aa genotype produce red ears of corn.
Thus, if Aa is backrossed to either of AA and aa:
Aa x AA = AA (white), AA (white), Aa (red) and Aa (red)
Aa x aa = Aa (red), Aa (red), aa (white) and aa (white)
Hence, the proportion of the offspring with white kernels will be 50% in each case.
Case Study: Parents of a 3-year-old boy noticed that their son had a waddling gait, fell frequently and had difficulty getting up again, and was not able to run. By age five, there was progressive muscular weakness and muscle wasting, and weakness of the trunk muscles led to abnormal posture. By age 9 he was confined to a wheelchair. Contractures appeared, first in the feet, as the gastrocnemius muscles tightened.1. This hereditary X-linked recessive disease characterized by progressive muscular weakness is ______.2. What does dystrophy mean? Why is this term used to describe this case?
Answer:
Muscular dystrophy
Explanation:
Muscular dystrophy is an inheritable genetic condition that involves mutations in genes that encodes the production of muscle proteins to build and to preserve healthy proteins. The disease is characterised by progressive weakness and loss of muscle mass.
Dystrophy is referred to as a disorder in which the a tissue or an orgasm progressively wastes away. Thé dystrophy term is used because the disease involves the gradual and progressive wasting away of the muscle which is a type of tissue causing muscular weakness allowing the child to experience these symptoms.
Have your partner focus on an object on the far side of the room (e.g., the chalkboard or a chart) for 1 minute. Then have your partner switch focus to an object in your hand (e.g., a pencil). Watch your partner's pupils carefully as the point of focus is changed from far to near.
Answer:
Eye changes or adjusts itself according to the focus of our eye.Explanation:
Pupil changes its shape when the eye focuses on near or far object. When we observe something far away, the pupil gets widen. But when we look at something near to us, pupil get smaller. So, pupil adjust itself according to the focus of our eye. If the partner switches from far to near, the pupil changes from wide to small. Pupil changes its shape according to bright as well as dim light also.
Many chemotherapy drugs target rapidly dividing cells, such as cancer cells. Why might cancer cells divide, and therefore evolve, more quickly than other cells
Answer:
The faster that cancer cells divide, the more likely it is that chemotherapy will kill the cells, causing the tumor to shrink. They also induce cell self-death or apoptosis. Chemotherapy drugs that kill cancer cells only when they are dividing are called cell-cycle specific.
Explanation:
In cattle, coats may be solid white, solid black, or black-and-white spotted. When true-breeding solid whites are mated with true-breeding solid blacks, the F1 generation consists of all solid white individuals. After many F1×F1 matings, the following ratio was observed in the F2 generation: 12/16 solid white 3/16 black-and-white spotted 1/16 solid black Part A How many gene pairs are involved in the inheritance of cattle coat color?
Answer:
Two locus
Explanation:
Let assume the gene for white (1st locus) be W i.e. ww = recessive
here, Let the alleles for the 2nd locus be B and b.
White: W_B_, W_bb
Black&White: wwB_
Black: wwbb
This is dominant epistasis. In dominant epistasis, where the dominant allele of the 1st locus (W) masks or hide the expression of the 2nd locus. When the two alleles in the 1st locus are recessive (ww), the alleles in the 2nd locus can be expressed(B and b).
Two locus gene pairs are involved in the inheritance of cattle coat color
Assumption of Generation:Let us assume the gene for white (1st locus) be W i.e. ww = recessive here, Let the alleles for the 2nd locus be B and b.
White: W_B_, W_bbBlack&
White: wwB_Black: wwbb
This represents dominant epistasis. In this, the dominant allele of the 1st locus (W) should mask or hide the expression of the 2nd locus. At the time when the two alleles in the 1st locus are recessive (ww), so the alleles in the 2nd locus could be expressed(B and b).
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a) There are abiotic and biotic factors that keep population size of species under control Provide two examples that affect (or have affected) population size of native species in WA (one example for density dependent and one for density independent). You need to pick two real examples, provide specific species names and habitats/ecosystems where those species occur. You need to use different examples from the ones I used in lectures
b) Explain why species that overlap a great deal in their fundamental niches have a high probability of competing. Now explain why species that overlap a great deal in their realized niches and live in the same area, probably do not compete significantly.
c) What are two important pieces of ecological knowledge that you could apply to the decisions you make about your own life? Mainly from an ecological point of view, but it is good that you make a connection with other aspects of your life (personal or professional).
Answer:
Answer A: A key example of biotic and abiotic protagonists is the presence of rocky structures that exist in the mountains of southern Argentina (Patagonia) that thanks to them exist the ideal temperatures for the procreation of animals such as the Patagonian condor and the possibility of form nests in these rocky structures that secure their young.
Answer B:
The explanation for this is that they have populations that increase in number and have their niches realized and constant resources necessary to develop the species in this way, considering the dominant species in the area and the one that predominates, in the case of fundamental niches. , they are necessary and even obligatory niches for many populations, which if populations do not go to them, they die, extinguish, or decrease their number, that is why in the face of this reality made up of a fundamental niche since it is fundamental to persist life of the population.
Answer c:
Two key principles are: one, maintain the equilibrium, that is why one seeks to have an environmentalist position preserving nature, and two considering that we are part of that balance but that does not give us the right to break it with environmental contamination, intervention of natural food systems and many other factors.
Explanation:
Both crystal violet and safranin are basic stains and may be used to do simple stains on Gram-positive and Gram-negative cells. This being the case, explain how they end up staining Gram-positive and Gram-negative cells differently in the Gram stain.
Answer:
Gram positive cells retains purple stains of crystal violet, gram-negative cells takes up red or pink stain of safranin
Explanation:
Gram staining is a simple method of distinguishing between Gram-positive cells and Gram-negative cells. The method make use of the following steps
Heat fix cells on a glass slide and then:
1. Stain cell with crystal violet for 1 min
2. Rinse with water
3. Decolourize with alcohol for 30 seconds and wash with water
4. flood cells with iodine as mordant
5. Flush with water
6. Counterstain with safranin for 1 min and wash with water
7. View under the microscope
Gram-positive cell possesses a thick peptidoglycan in its cell wall which helps retain the purple colour of the crystal violet, While Gram-negative cells will loose out this colour when flushed with alcohol but takes up the red or pink colour when stained with safranin.
Crystal violet stains Gram-positive cells purple, while safranin stains Gram-negative cells pink in the Gram stain.
Explanation:Crystal violet and safranin are used in the Gram stain to differentiate between Gram-positive and Gram-negative cells. Crystal violet is a primary stain that binds to the thick peptidoglycan layer of Gram-positive cells, making them appear purple.
Safranin, a secondary counterstain, is applied after decolorization and stains the decolorized Gram-negative cells pink. The differences in cell wall structure and the interaction of the stains with the cell walls result in the differential staining of Gram-positive and Gram-negative cells.
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g Imagine two different sigma factors with different promoter recognition sequences. What would happen to the overall gene expression profile in the cell if one sigma factor were artificially overexpressed?
Answer:
Transcription is affected via the modulation of the concentrations of the different types of holoenzymes, so saturated promoters are only weakly affected by sigma factor competition. However, in case of overlapping promoters or promoters recognized by two types of sigma factors, we find that even saturated promoters are strongly affected. Active transcription effectively lowers the affinity between the sigma factor driving it and the core RNAP, resulting in complex cross-talk effects. Sigma factor competition is not strongly affected by non-specific binding of core RNAPs, sigma factors and holoenzymes to DNA. Finally, we analyze the role of increased core RNAP availability upon the shut-down of ribosomal RNA transcription during the stringent response. We find that passive up-regulation of alternative sigma-dependent transcription is not only possible, but also displays hypersensitivity based on the sigma factor competition.
You purchase two identical houseplants and place them side by side on your windowsill. You water both plants equally. One plant, plant a, you leave alone. On the other plant, plant b, you inject florigen into the cells of the apical meristem. Which of the following would you expect to occur? (Select all answer options that apply.)
a. While flowering, the leaves will stop growing on plant b because florigen inhibits leaf meristems.
b. In subsequent years, plant b will only generate flowers in the same place.
c. Plant b will produce a flower at its apical meristem.
d. Once the flower is gone, plant b will grow taller, but only from growth of lateral meristems.
e. Both plants will produce flowers at the same time, but plant b will have more of them.
f. Once the flower is gone, the apical meristem will develop again, and plant b will continue to grow from that stem.
Answer: Option E.
Both plants will produce flower at the same time but plant B will have more of them.
Explanation:
Florigen is a protein or hormone like molecule that help to control or boost flower production in plants. It is a flowering hormone. It is produced in plants leaves and acts on apical meristem or growing tips. The hormone is normally injected at the growing tips or apical meristem where flowers growth will be activated. Since plant B is injected with florigen,it will further trigger flower production on the plant than plant A .
Answer: C
Explanation:
Mechanical energy is the energy of _____.
Answer:
Mechanical energy is the energy of potential and kinetic energy i.e sum of potential and kinetic energy.
Explanation:
Mechanical energy is the energy posses by an object which enable it to work due to it's potion or movement.
Mechanical energy can be inform of potential energy( energy stored or due to object position) or kinetic energy( object in motion).
Mechanical energy is the sum of potential energy and kinetic energy. This energy is associated with object's position or motion.
Inosine monophosphate is a branch point for the synthesis of a variety of purine nucleotides. Match the appropriate reaction for the synthesis of either AMP or GMP:
1) AMP A) condensation with Asp
2) GMP B) oxidative hydration
C) transamination with Gln
D) release of fumarate
1. --> 1:A, C; 2: B, D
2. --> 1:C, D; 2: A, B
3. --> 1:A, B; 2: C, D
4. --> 1:A, D; 2: B, C
Answer:
1:A, D; 2: B, CExplanation:
1.Purine Bases Can Be Synthesized by two pathways: i) de Novo and ii) Salvage Pathways.
2.Purine nucleotides synthesis starts with Phosphoribosyl pyrophosphate (PRPP), and it leads to the synthesis of nucleotide, inosine 5'-monophosphate (IMP). Inosine monophosphate is a branch point for the synthesis of many purine nucleotides amd leads to the synthesis of a variety of purine nucleotides.
3.The first reaction of purine synthesis is catalyzed by the enzyme glutamine phosphoribosylpyrophosphate amidotransferase.
20. During filtration, which of the following does NOT enter the bowman's capsule from the bloodstream?
A. glucose
B. water
C. ions
D. plasma proteins
E. amino acids
Answer: option D.
Plasma proteins.
Explanation:
Filtration is the transfer of water and solutes like glucose, ions, amino acids from the plasma to the renal tubules in the renal corpsicule plasma . Renal corpsicule help to filter blood. The fluid present in the blood in the glomerulus is pass to the Bowman's capsule to form glomerular filtrate , plasma protein do not enter Bowman's capsule. The filtrate is processed to form urine in the nephron.
What do you think might be the evolutionary benefit of the milk production regulation mechanism in breastfeeding?
Answer: The benefit of evolutionary benefit of milk production regulation mechanism is that the mother will not waste her energy or nutrients required for her growth and development to produce milk at that time rather it will be during her late pregnancy stage where the mammary gland will be activated by oxytocin hormone to produce milk automatically to breast feed infants at birth.
Explanation:
Evolution of lactation or mammary glands in mammals has been very important to feed the infants.
Lactation refers to production of milk from the mammary glands. Lactation normally occur at post pregnancy stage. At this stage the infants are born and milk is been secreted by the mammary glands to breast feed the young ones. Lactation doesn't occur all the time but it occur during late to post pregnancy stage. This is activated by oxytocin which help to implant embryo in the uterus and help in lactation.
The milk production regulation mechanism in breastfeeding is a complex process that is controlled by a variety of hormones, including prolactin, oxytocin, and estrogen. This mechanism ensures that the mother produces enough milk to meet the needs of her baby, but not so much that it becomes a burden.
There are several possible evolutionary benefits of this mechanism.
First, it helps to ensure that the baby is always fed, even if the mother is not constantly producing milk. This is important for the survival of the baby, as it needs to be fed frequently in order to grow and develop.
Second, the mechanism helps to prevent the mother from producing too much milk, which can lead to mastitis, an infection of the breast tissue. Mastitis can be painful and can interfere with breastfeeding.
Third, the mechanism helps to ensure that the milk is of the right composition for the baby. The composition of milk changes over time to meet the changing needs of the baby. For example, the milk produced in the early days after birth is high in colostrum, which is a nutrient-rich liquid that helps to protect the baby from infection.
Hence, the milk production regulation mechanism in breastfeeding is a complex and efficient process that has evolved to ensure the survival and health of the baby.
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A 35-year-old man presents with anemia, neutropenia, thrombocytopenia, myeloblasts with the presence of Auer rods, and one or two distinct nucleoli and promyelocytes. Cytochemistry examination demonstrates peroxidase and Sudan black B (SBB) positive and TdT terminal eoxynucleotidyl transferase (TdT) negative. This hematologic picture is consistent with: A) Acute lymphoblastic leukemia (ALL) B) Acute myeloblastic leukemia (AML) C) Chronic myelocytic leukemia (CML) D) Chronic lymphocytic leukemia (CLL) E) None of the above
Answer:
B) Acute myeloblastic leukemia (AML
Explanation:
It is a type of cancer that affects the blood as well as the bone marrow. It causes a situation whereby there is excessive immature white blood cells thus affecting the production of matured or normal white and red blood cells, blood platelets through the interference of the myeloid cells.
It is treatable medical condition that requires medical diagnosis. It requires laboratory tests with imaging always needed. It is also known as Acute myelogenous leukemia, acute granulocytic leukemia and acute nonlymphocytic leukemia.
Answer: The hematologic picture is consistent with Acute myeloblastic leukemia (AML). Therefore the correct option is B.
Explanation:
Leukaemia is a type of cancer that affects the blood forming tissues such as the bone marrow, leading to excessive or over production of abnormal blood cells ( usually the white blood cells). They occur in two categories; it can be Acute leukaemia or Chronic leukaemia.
There are different forms of acute leukaemia which is classified according to the lineage ( myeloid or lymphoid). They include:
- Acute Myeloblastic Leukaemia (AML)and
- Acute Lymphoblastic Leukaemia ( ALL).
In AML, the following Laboratory diagnosis differentiates it from other leukaemia, these includes:
- myeloblasts with the presence of Auer rods, and one or two distinct nucleoli and promyelocytes, and
-peroxidase and Sudan black B (SBB) positive and TdT terminal eoxynucleotidyl transferase (TdT) negative. Therefore the correct option is B