Answer: D) Degradation of defective proteins activates mRNA repair pathways
Explanation:
A recent publication (J Phys Condens Matter. 2013. 18;25(37):374104.) provided the information on this matter. According to researcher RNA polymerase have ability to backtrack and translocate the error on both RNA transcript and DNA template. By doing so RNA polymerase cleave the defective part on RNA transcriptome by the process called proofreading. While proofreading contribute in accuracy of transcription but on contrary it also slower the process of transcription.
DNA polymerase possesses robust error correction mechanisms while RNA polymerase's error correction is limited. However, while errors from both can result in defective proteins, the DNA mutation is permanent and inheritable unlike the temporary and non-inheritable effects of RNA polymerase errors.
Explanation:True, DNA polymerase possesses multiple mechanisms for editing and error correction, ensuring the high accuracy of DNA replication. Mistakes can occur, but repair mechanisms often promptly correct these errors. If an error is not corrected, it leads to a mutation which is serious because DNA mutations are permanent and transmitted to successive generations. This is different from RNA polymerase, which has a limited error correction capacity.
Unlike a DNA mutation, if an error in transcription by RNA polymerase occurs, it can lead to the production of a defective protein. However, these defective proteins are often short-lived as the mRNA transcripts have short half-lives and degrade quickly. Additionally, each mRNA molecule is usually translated only once before degradation. Therefore, the effects of RNA polymerase errors are temporary and not inherited across generations, unlike DNA mutations.
Lastly, while both DNA and RNA polymerase mechanisms can lead to the production of defective proteins, the extent and duration of their effects differ significantly because of the inherent differences in DNA and RNA stability and the mechanisms of their polymerases.
Learn more about DNA and RNA Polymerases here:https://brainly.com/question/1037413
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A sample ( 885 mg ) of an oligomeric protein of M r 155,000 was treated with an excess of 1‑fluoro‑2,4‑dinitrobenzene (Sanger's reagent) under slightly alkaline conditions until the chemical reaction was complete. The peptide bonds of the protein were then completely hydrolyzed by heating it with concentrated HCl . The hydrolysate was found to contain 4.85 mg of DNP‑Val. 2,4‑Dinitrophenyl derivatives of the α‑amino groups of other amino acids could not be found. Calculate the number of polypeptide chains in this protein. Give the answer as a whole number.
Answer:
In 1945, Frederick Sanger described its use for determining the N-terminal amino acid in polypeptide chains, in particular insulin.[4] Sanger's initial results suggested that insulin was a smaller molecule than previously estimated (molecular weight 12,000), and that it consisted of four chains (two ending in glycine and two ending in phenylalanine), with the chains cross-linked by disulfide bonds. Sanger continued work on insulin, using dinitrofluorobenzene in combination with other techniques, eventually resulted in the complete sequence of insulin (consisting of only two chains, with a molecular weight of 6,000).[5]
Following Sanger's initial report of the reagent, the dinitrofluorobenzene method was widely adopted for studying proteins, until it was superseded by other reagents for terminal analysis (e.g., dansyl chloride and later aminopeptidases and carboxypeptidases) and other general methods for sequence determination (e.g., Edman degradation).[5]
Dinitrofluorobenzene reacts with the amine group in amino acids to produce dinitrophenyl-amino acids. These DNP-amino acids are moderately stable under acid hydrolysis conditions that break peptide bonds. The DNP-amino acids can then be recovered, and the identity of those amino acids can be discovered through chromatography. More recently, Sanger's reagent has also been used for the rather difficult analysis of distinguishing between the reduced and oxidized forms of glutathione and cysteine in biological systems in conjunction with HPLC. This method is so rugged that it can be performed in such complex matrices as blood or cell lysate.[6][7]
Explanation:
Example: A sample (525 mg) of an oligomeric protein of Mr 117,000 was treated COOH with an excess of1-fluoro-2,4-dinitrobenzene (Sanger's reagent) under slightly alkaline conditions until thechemical reaction was complete. The peptide bonds of the protein were then completelyhydrolyzed by heating it with concentrated HCI. The hydrolysate was found to contain 3.37 mgof DNP-Val (shown at the right), 2,4-Dinitrophenyl derivatives of the α- amino groups of otheramino acids could not be found H3C Calculate the number of polypeptide chains in this protein.Give the answer as a whole number Number A second oligomeric protein of M 230,000 wasshown by a similar endgroup analysis to consist of five polypeptide chains. SDS polyacrylamidegel electrophoresis in the presence of a reducing agent shows three bands: α (M, 30,000), β (M40,000) and γ(M-60,000), indicating three distinct polypeptides. SDS electrophoresis withoutreducing agent also yields three bands, with Mr of 30,000, 40,000, and 120,000 Which of the
Which of the following describes a situation of scarcity?
A. Someone offers free advice about getting into college.
B. A person lets the kids in the neighborhood use his pool.
C. Someone distributes free bottles of water at the beach.
D. A child charges friends for a ride on his new bike.
Answer:
A child charges friends for a ride on his new bike
Explanation:
Answer:
D.
Explanation:
A child charges friends for a ride on his new bike.
Day-to-day choices can help reduce the risk of heart disease. One of the major risk factors for development of heart disease is elevated LDL, which can be affected by the types and amounts of dietary fat consumed as well as other dietary factors. Read the statements below and select all of the correct statements regarding how various dietary fats affect LDL cholesterol levels. Select all that apply. a. Typically, the higher your consumption of unsaturated fats, the higher the LDL cholesterol levels in your blood. b. Trans fats are worse for heart health than saturated fats because they raise LDL cholesterol and lower HDL cholesterol. c. Increasing intake of plant foods may be one of the easiest ways to decrease LDL cholesterol. d. Typically, the higher your consumption of saturated fats, the higher the LDL cholesterol levels in your blood. e. Dietary cholesterol does not affect blood cholesterol levels.
Answer:
b. Trans fats are worse for heart health than saturated fats because they raise LDL cholesterol and lower HDL cholesterol.
c. Increasing intake of plant foods may be one of the easiest ways to decrease LDL cholesterol.
d. Typically, the higher your consumption of saturated fats, the higher the LDL cholesterol levels in your blood.
Explanation:
Generally, the consumption of saturated fat increases the amount of LDL cholesterol in the blood system while the consumption of unsaturated fat can reduce the amount of LDL cholesterol in the blood system. In addition, the consumption of plant food can reduce the amount of LDL in the blood system. Trans fats are more harmful to one's heart than saturated fat because they can increase the amount of cholesterol in the blood system.
Answer:b,c,d
Explanation:a. Unsaturated fats are mostly beneficial to health, e.g- omega 3 fatty acids, and are not known causes of LDL increase.
b. Trans fats are known to increase LDL cholesterol while also decreasing HDL cholesterol levels. We do not find this in saturated fats which are known to only increase LDL cholesterol.
c. Plants possess "soluble fiber" that drastically help to reduce cholesterol levels. They are known to be very low in saturated fats, and also free of cholesterol.
d. Saturated fats are common causes of elevated LDL in blood stream.
e. Dietary cholesterol as opposed to serum cholesterol are fats found in our diets(food) that may contain high levels of cholesterol that increase LDL levels in blood.
Why do all Krebs cycle reactions occur twice for each molecule of glucose that undergoes respiration?
Answer: This is because 2 pyruvate molecules were produced in Glycolysis,and each must be metabolize to extract one ATP molecule each contained.
In addition to produced required number of reduced Co-enzymes for Electron Transport Chain (ETC) reactions to take place for ATP production
It must also be twice/glucose molecule to generate required volume of oxygen for decaboxylation,because the oxygen in the reaction came from.from . redox and not from the atmosphere.
Therefore the both the link reaction and the kreb.cycle
must occur twice to produce mandatory 2 ATPs ,and generate the required amount of reduced Co Enzymes NADH and FADH2 for ETC(electron transport Chain ) to occur.
Explanation:
Answer:
The Krebs cycle reaction occurs twice because glucose produces two molecules of Pyruvate and then two acetyl CoA.
Explanation:
The citric acid cycle goes around twice for each molecule of glucose that enters cellular respiration because there are two pyruvates from the single glucose molecule and thus, two acetyl CoA is produced per glucose which is the starting molecule in the Krebs cycle.
Red-flowering snapdragons are homozygous for allele R1. White-flowering snapdragons are homozygous for allele R2. Heterozygous plants (R1R2) bear pink flowers. What phenotypes should appear among F1 offspring of the crosses listed below? What are the expected probabilities for each phenotype?
Answer:
a) 1/2 red, 1/2 pink; b) all pink; c) 1/4 red, 1/2 pink, 1/4 white; d) 1/2 white, 1/2 pink
Explanation:
The flower color trait in snapdragons shows incomplete dominance: the heterozygous genotype produces an intermediate phenotype between the two different homozygous genoytpes.
The possible genotypes and phenotypes are:
R1R1 : redR1R2: pinkR2R2: whitea. R1R1 X R1R2The R1R1 individual only produces R1 gametes. The R1R2 parent produces 1/2 R1 gametes and 1/2 R2 gametes.
For that reason, the F1 will be:
1/2 R1R1 (red)1/2 R1R2 (pink)b. R1R1 X R2R2The R1R1 individual only produces R1 gametes. The R2R2 parent only produces R2 gametes.
For that reason, the F1 will be :
100% R1R2 (pink)c. R1R2 X R1R2Both parents are heterozygous. This is a monohybrid cross, and from Mendel's Laws we expect the following offspring:
1/4 R1R1 (red)2/4 R1R2 (pink)1/4 R2R2 (white)d. R1R2 X R2R2The R1R2 parent produces 1/2 R1 gametes and 1/2 R2 gametes. The R2R2 individual only produces R2 gametes.
For that reason, the F1 will be:
1/2 R2R2 (white)1/2 R1R2 (pink)2. A likely explanation for an abnormal human phenotype associated with a trisomy is: A) the presence of multiple recessive mutant alleles. B) the extra chromosome has typically undergone significant rearrangements. C) the absence of genes necessary for certain cellular processes. D) altered gene dosage (also known as genetic balance). E) the random inactivation of a complete autosome.
Answer:
The correct answer is B the extra chromosome has typically undergone significant rearrangements.
Explanation:
Trisomy is a type of aneuploidy which is characterized by the presence or occurance of an extra copy of a particular chromosome.As a result the chromosomes in which the trisomy is occurring contain 3 arms instead of 2. Trisomy occur due to non disjunction of a particular chromosome during meiosis.
As a result a second copy of chromosome is present inside the cell containing that abnormal chromosome.If the same thing occur in the cell of gamet undergoing fertilization then there is a high chance that the embryo will carry that extra chromosome.
In case of humans Trisomy result in the rearrangement extra chromosome which ultimately give rise to
1 Trisomy 18 occur due to extra copy of chromosome no 18
2 Trisomy 13 occur due to extra copy of chromosome no 13
3 Trsomy 21 occur due to extra copy of chromosome no 21 which is also called down syndrome.
Final answer:
An abnormal human phenotype associated with a trisomy is likely due to altered gene dosage.
Explanation:
Trisomy is a state where humans have an extra autosome. The most common trisomy is of chromosome 21, which leads to Down Syndrome. Trisomic individuals suffer from an excess in gene dose, which disrupts the normal balance of gene dosage. This can lead to functional challenges and developmental delays. The most likely explanation for an abnormal human phenotype associated with a trisomy is altered gene dosage.
Indicate at which step of the replication-transcription-translation process each type of RNA first plays a role.During which step of the replication-transcription-translation process does each type of RNA first play a role?replication?translation?transcription/ rna processing?
Answer:
Ribosomal RNA (rRNA) is used in the translation process
Messenger RNA (mRNA) is produced during the transcription process and is used in the translation process
Transport RNA (tRNA) is used in the translation process
and if you count the RNA produced by RNA primase than that is used in the replication process.
Answer:
that answer is correct
Explanation:
What structure is used to seal the DNA into an opening created by the restriction enzyme during recombinant DNA technology
Answer:
DNA Ligase
Explanation:
DNA ligase is a specific type of enzyme, that facilitates the joining of DNA strands together by catalazing the formation of phophodiester bond.
Is used both DNA repair and DNA replicaation. It has extended use in molecular biology for recombinant experiments.
Answer:
DNA ligase
Explanation:
DNA ligase is an enzyme responsible for the joining of DNA strand ends. If two pieces of DNA have matching ends, ligase can link them to form a single, unbroken molecule of DNA.
What forms blood clots in an artery that supplies blood to the myocardium courseherp?
Answer: Blood clot in the coronary arteries( arteries of the heart), called Coronary thrombosis is caused mostly by deposits and build up of low density cholesterol, fibrous tissues, and inflammatory cells. on the walls of the artery. Smoking, sedentary life style(lack of exercise),and hypertension are other causes..
Obesity is another causative condition for coronary thrombosis.
These deposits and build up coalesces in different parts of the coronary arteries; to form structures called plaques. The latter may break down at this arterial segments which leads top formation of blood clot. (thrombosis)
The clot results in the blockage of the arteries that supplies the affected site, with blood containing oxygen and nutrients. This leads to tissue hypoxia(lack of oxygen) to the cardiac muscles cells of the myocardium and therefore myocardial infraction.(HEART ATTACK)
In some cases these clots are pumped out of the heart to the cerebral arteries, due to proximity of the brain to the heart. In this case this can lead to stroke,
Explanation:
Blood clots in coronary arteries form when plaque ruptures, causing platelets to clump together. This can narrow or block the arteries, leading to angina or heart attacks. Atherosclerosis is the primary cause of this condition.
Plaque begins the process of forming blood clots in the coronary arteries, which carry blood to the myocardium. Atherosclerosis, a condition that causes the coronary arteries to narrow over time, is caused by plaque, which is composed of cholesterol deposits. At the point when an area of plaque bursts, it causes platelet sections known as platelets to adhere to the injury site and bunch together, shaping blood clumps. These blood clots may cause severe conditions like angina (chest pain) or a heart attack by further narrowing the coronary arteries and significantly reducing blood flow. Now and again, the blood coagulation may totally hinder the conduit, altogether influencing the oxygen supply to the heart muscle and causing myocardial localized necrosis (cardiovascular failure).During DNA replication
A.each strand can act as a template for the replication of the moleculeB.the two strands of the original DNA molecule must be separatedC.enzymes facilitate all the steps involvedD.each of the original two strands of DNA will wind up in a different chromatid
Answer:
The answer is A: Each strand can act as a template for the replication of the molecule
Explanation:
During DNA replication, each of the two strands that make up the double helix serves as a template from which new strands are copied. The new strand will be complementary to the template or parent strand.
7. A bioengineer is trying to understand the biomechanics of a hole created in the skin for a transcutaneous implant. The engineer made a hole using a circular biopsy drill in the dorsal skin of a dog. The diameter of the drill is 5 mm. If the hole becomes an ellipse with a minor and major axis of 3 and 7 mm, answer the following questions. a. In which direction is the internal stress in the skin greater? b. In which direction are the collagen fibers more oriented? c. How can the bioengineer obtain a circular rather than an elliptical hole for the implant? d. Assuming the implant is non-deformable compared to the skin, what problems will arise between skin and implant when a load or force is applied to the skin or implant by handling accidentally?
The internal stress and collagen fiber orientation in the skin in response to a transcutaneous implant are greater and more oriented respectively along the major axis of the ellipse. To achieve a circular hole, drilling technique adjustments or material considerations may be necessary. Load or force application can cause complications due to deformability mismatch between the skin and a rigid implant.
A bioengineer analyzing the biomechanics of a transcutaneous implant in the skin of a dog is faced with a situation where a circular hole drilled by a biopsy drill has deformed into an elliptical shape with minor and major axes of 3 mm and 7 mm respectively. Such analysis is crucial in understanding the mechanical behavior of the skin in response to implants and the stress distribution around the implant site.
a. The internal stress in the skin is greater in the direction of the major axis, which is 7 mm. This is due to the fact that the skin stretches more along the major axis, causing increased tension and stress in that direction.
b. The collagen fibers within the skin are more likely to be oriented along the direction of least stress, which would be along the minor axis of the ellipse (3 mm).
c. To obtain a circular hole rather than an elliptical one, the bioengineer could consider altering the drilling technique, such as optimizing the speed and pressure, using different drill materials, or modifying the post-drilling processes to ensure the skin maintains its shape. Additionally, the mechanical properties of the skin and the drill's cutting edge sharpness should be considered to minimize deformation.
d. If a load or force is applied to the skin or implant, especially if the implant is non-deformable compared to the skin, there could be issues with alignment, pressure sores, and stress concentration at the interface. This mismatch in deformability can lead to inadequate load transfer, potential implant loosening or migration, and increased risk of skin breakdown or irritation.
What can possibly kill sperm in the Great Sperm Race?
Answer:
Answer:
THE GREAT SPERM RACE:
A sperm's race to fertilize an egg is not so easy. Out of about 250 million sperm ejaculated into the human vagina during intercourse, not more than one in a hundred will survive the Great race to the end due to the hurdles it has to face in the hostile, ACIDIC CHAMBER to the cervix (it has hundreds of tiny branching tunnels that can trap, crush and slowly kill sperm).
If ovulation is not occurring soon the sperm will "drown in a thick flow of cervical mucus
"The correct answer is that a variety of factors can kill sperm in the Great Sperm Race, including an acidic pH, the presence of white blood cells, and the harsh conditions within the female reproductive tract.
In the Great Sperm Race, which is a metaphor for the journey sperm must undertake to fertilize an egg, sperm face numerous challenges that can impede their progress or lead to their demise. Here are some of the key factors that can kill sperm:
1. Acidic pH: The vagina is naturally acidic, which helps protect against infections. However, this acidic environment can be hostile to sperm. Sperm prefer a more alkaline pH, which is typically found in the cervical mucus during ovulation. If the pH is too acidic, it can damage or kill sperm.
2. White Blood Cells (WBCs): The presence of WBCs, or leukocytes, in the reproductive tract can be indicative of an infection or inflammation. WBCs can attack and kill sperm as part of their immune response to foreign bodies
3. Harsh Conditions: The female reproductive tract can be a challenging environment for sperm due to its complex structure and varying conditions. Sperm may encounter barriers such as thick cervical mucus, which can be difficult to penetrate, or they may be exposed to unfavorable temperatures or toxic substances.
4. Immunological Responses: Sometimes, the woman's immune system may mistakenly identify sperm as foreign invaders and mount an immune attack against them, leading to their destruction.
5. Sperm Defects: Not all sperm are capable of reaching the egg. Some may have defects such as abnormal morphology (shape) or poor motility (movement), which can prevent them from successfully navigating the reproductive tract.
6. Anti-Sperm Antibodies: Both men and women can produce antibodies that target sperm, leading to sperm agglutination (clumping) or immobilization, which can prevent sperm from reaching the egg.
7. Environmental Factors: External factors such as exposure to certain chemicals, radiation, or high temperatures can also damage or kill sperm.
Understanding these factors is crucial for diagnosing and treating infertility, as well as for developing effective contraceptive methods. Interventions may include adjusting the timing of intercourse to coincide with the more sperm-friendly environment during ovulation, treating infections, or using assisted reproductive technologies when necessary."
How Light Bleaches Rhodopsin
For this exercise, you will take a scenario of bleaching occurring within the rod cells and rank the steps in the process from beginning to end.
Rank the order of the bleaching process from beginning to end.
1 - Vitamin A from the bloodstream replenishes what is lost. The vitamin A returns to its original shape and gets reincorporated into rhodopsin.
2 - Rhodopsin absorbs light, and the vitamin A changes shape.
3 - Vitamin A detaches from the rhodopsin, and some vitamin A is lost.
4 - Light travels to the macula within the retina.
5 - Light enters the eye through the cornea.
Light contacts the eye.--------------- Light-absorbing capabilities of proteins are regenerated
Answer:
Salut!
Light enters the eye through the cornea.Light travels to the macula within the retina.Rhodopsin absorbs light, and the Vitamin A changes shape.Vitamin A detaches from the rhodopsin, and some vitamin A is lost.Vitamin A from the bloodstream replenishes what is lost. The vitamin A returns to its original shape.Explanation:
Retina is the part of the eye that contains photosensitive cells that capture light and produce the electrical signals that the brain perceives as images. These photosensitive cells are of two kinds:
RodsConesRods contain the photosensitive pigment, rhodopsin that is needed for vision at night or in dim light. Cones function in bright light.
Rhodopsin Bleaching:
Vision in bright or excessive light requires a process called rhodopsin bleaching which is the degradation of rhodopsin upon exposure to light. Upon contact with light, rhodopsin goes through structural changes characterized by the conversion of a pigment derived from Vitamin A, 11-cis retinal to all trans retinal. This chemical conversion initiates a photo-transduction reaction (reaction in which a photon of light is converted into electrical signals) that produces the electrical signals that travel to the brain via the optic nerve. The brain converts the electrical signals to images. This is followed by rhodopsin regeneration in the dark in which all trans retinal is converted back into 11 cis retinal.
Which of the following represents a sensory input that is not part of both the somatic and autonomic systems?(A) Vision(B) taste(C) Baroreception(D) Proprioception
Answer:
the correct answer is (C) Baroreception
Bacteria and other microbes can be used to "clean up" an oil spill by breaking down oil into carbon dioxide and water. Two samples isolated from the Deepwater Horizon leak in the Gulf of Mexico were labeled A and B. The DNA of each was isolated and the percent thymine measured in each sample. Sample A contains 20.7 % thymine and sample B contains 30.9 % thymine. Assume the organisms contain normal double‑stranded DNA and predict the composition of the other bases.
Answer:
Sample A: Thymine= 20.7% ; Adenine= 20.7%; Guanine = 29.3% ; Cytosine= 29.3%
Sample B: Thymine= 30.9% ; Adenine= 30.9% ; Guanine= 19.1% ; Cytosine base= 19.1%
Explanation:
In a double-stranded DNA molecule, the percentage of adenine base is equal to that of the thymine base while the percentage of guanine base is equal to that of the cytosine base. It is called Chargaff's rule and is based on the complementary base pairing of purine and pyrimidine bases.
Sample A: Percentage of thymine= 20.7%
This means that the sample has= 20.7% adenine base
Total thymine and adenine bases in sample A= 20.7 + 20.7 = 41.4%
Therefore, proportion of guanine + cytosine bases in the sample= 100-41.4= 58.6%
Percentage of guanine base= 58.6/2 = 29.3%
Percentage of cytosine base= 29.3%
Sample B: Percentage of thymine= 30.9%
This means that the sample has= 30.9% adenine base
Total thymine and adenine bases in sample A= 30.9 + 30.9 = 61.8%
Therefore, proportion of guanine + cytosine bases in the sample= 100-61.8= 38.2%
Percentage of guanine base= 38.2/2 = 19.1%
Percentage of cytosine base= 19.1%
Most vaccines currently administered are delivered by intramuscular injection. Yet the pathogenic organisms these vaccines aim to protect against usually enter the body by a different route. For instance, many viruses infect us via the respiratory tract, and many bacterial pathogens infect via the gastrointestinal tract. One major advantage of delivering vaccines against these organisms via their normal route of infection would be______________.
Answer: It is essential for influencing immune responses at the initial site where pathogen first invaded the organism thus stimulating rapid immune response against the pathogen.
Explanation:
This is because the route of vaccination is essential for influencing immune responses at the initial site where pathogen first invaded the organism. It is known as mucosal vaccine strategies. This is the most effective point of protection against the pathogens and the resolution of infection (state such as reduction of inflammation) at the initial point of pathogen entry. In addition; it also activate local innate immune response.
Answer:
It is essential for influencing immune responses at the initial site where pathogen first invaded the organism
Explanation:
mark ne brainy plz!
A) Prof. Robo is studying skin color of Banana Slugs. He has isolated 8 recessive mutants that are white instead of yellow. He performs the following complementation analysis. (+) indicates yellow color (-) indicates white color. Based on complementation analysis, how many genes affect skin color? Place the various mutants in the different complementation groups.
Answer:378
Explanation: 378
have been cloned and
The other 207 loci have been mapped out, the true gene identities have yet to be determined.
Two large populations of horses are being systematically crossed (mares from one population bred to stallions of the other and vice versa). Coat color is not a factor in determining which animals are selected and which individual matings are made (random matings). Frequencies of coat color genes at the C locus for population 1 are.85 for Cand.15 for c. Frequencies for Care.6 and care.4 for population 2. Given these values, what are the gene and genotypic frequencies of the F1?
a. p=0.725, q = 0.275: P=0.51, H=0.43, Q=0.06
b. p=0.725.q = 0.275; P=0.06. H=0.56, Q=0.51
c. p=0.4.q = 0.6: P=0.12. H=0.56, Q=0.32
d. p=0.725.q = 0.275: P=0.34. H=0.57. Q=0.09
Answer:
a. p=0.725, q = 0.275: P=0.51, H=0.43, Q=0.06
Explanation:
Let state our given parameters from the question:
Frequencies of coat color genes at the C locus for population 1 are .85 for C
This implies that the Allelic frequency C for population p1 =0.85
Frequencies of coat color genes at the c locus for population 1 are .15 for c
This implies that the Allelic frequency c for population q1 = 0.15
Frequencies for Care .6 i.e p2= 0.6
Frequencies for care .4 i.e, let that be q2= 0.4
The table below shows a diagrammatic representation of the above expression:
Alllelic Frequency C c
Population 1 (p1) 0.85 (q1) 0.15
Population 2 (p2) 0.6 (q2) 0.4
Now, from above: let think of the table as a punnet square and then cross it together;
(p1) = 0.85 (q1) = 0.15
p2 = 0.6 p1p2 p2q1
= 0.6 × 0.85 = 0.15 × 0.6
= 0.51 (P) = 0.09 (H)
q2 = 0.4 p1q2 q1p2
= 0.85 × 0.4 = 0.4 × 0.15
=0.34 (H) = 0.06 (Q)
From the above table, the heterozygous are represented by (H)
∴ Frequency of heterozygous can be calculated as:
= 0.09 + 0.34
= 0.43
Thus, we can conclude that the progeny F1 genotypic frequencies are:
P= 0.51
H= 0.43
Q= 0.06
Now, let us calculate the allelic frequencies, p and q in F1
p = P + 1/2 × (H)
= 0.51 + (1/2 × 0.43)
= 0.51 + 0.215
= 0.725
q = Q + 1/2 × (H)
= 0.06 + (1/2 × 0.43)
= 0.06 × 0.215
= 0.275
Hence, p=0.725, q = 0.275: P=0.51, H=0.43, Q=0.06 , This makes option a the correct answer.
The major advantage of using artificial chromosomes such as YACs and BACs for cloning genes is that
plasmids are unable to replicate in cells.
only one copy of a plasmid can be present in any given cell, whereas many copies of a YAC or BAC can coexist in a single cell.
YACs and BACs can carry much larger DNA fragments than ordinary plasmids can.
YACs and BACs can be used to express proteins encoded by inserted genes, but plasmids cannot.
all of the above
H. Simply inserting an entire eukaryotic gene into a prokaryotic expression system will most likely not work for the following reason.
Prokaryotes lack promotors
Prokaryotes lack introns
Prokaryotes lack polymerase
Prokaryotes lack repressors
Question is a multiple choice question.
Answer:
Question 1. The major advantage of using artificial chromosomes such as YACs and BACs for cloning genes is that:
Answer:
(C)
Explanation:
YACs and BACs can carry much larger DNA fragments than ordinary plasmids can.
Question 2. Simply inserting an entire eukaryotic gene into a prokaryotic expression system will most likely not work for the following reason.
Answer:
(B)
Explanation:
Prokaryotes lack introns
In this lesson we discussed the difference between strong and weak bases. Which of these bases would be classified as weak?
A. sodium fluoride
B. ammonia
C. potassium hydroxide
D. calcium hydroxide
E. none of the above
Answer:
E. none of the above i think
Explanation:
You are interested in two traits in rabbits, each of which is controlled by a separate gene with two alleles:
a) coat color (brown, B, is completely dominant to white,
b) and tail (tailed, T, is completely dominant to tail-less,
c) You cross a brown, tailed rabbit that is heterozygous at both loci with a white, tail-less rabbit and produce a large number of offspring.
Among the offspring you find only two phenotypes in equal proportions: brown, tailed and white, tail-less.
Answer:the scenario here with the brown tailed rabbit RrTt
The white tailess rabbit is rt.
Crossing both yields the below offsprings possibilities;
For colour: Rr or rr
For tail: Tt or tt
That means they would produce equal numbers of Brown tailed & white tailess rabbits.
Explanation:
Farmers use a variety of approaches in an effort to produce more successful yields. These may include hormones, antibiotics, herbicides, pesticides, and forms of biotechnology like plant breeding or genetic engineering. Sort each statement into the bin with the corresponding agricultural approach. Drag the appropriate items into their respective bins.
Answer:
Statements are missing. These are,
1-Consumers may reduce exposure to these by washing and peeling fruits and vegetables.
2-These may be used to increase weight gain and meat production in cattle.
3-These include organophosphates and sex pheromones.
4-Golden rice is yellow because it contains beta-carotene, which the plant can now manufacture because genes for certain enzymes were inserted in the rice genome.
5-Their use may be avoided in favor of integrated pest management (IPM).
6-Scientists inserted the Bt gene into corn to make it resistant to the corn borer.
7-These include recombinant bovine somatotropin (rbS
Hormones:
These include organophosphates and sex pheromonesThese include recombinant bovine somatotropin (rbS)These may be used to increase weight gain and meat production in cattle.Antibiotics:
NA.
Pesticides/Herbicides:
Consumers may reduce exposure to these by washing and peeling fruits and vegetables. Their use may be avoided in favor of integrated pest management (IPM)Biotechnology/Genetic Engineering:
Golden rice is yellow because it contains beta-carotene, which the plant can now manufacture because genes for certain enzymes were inserted in the rice genome.Scientists inserted the Bt gene into corn to make it resistant to the corn borer.Consider seven recessive genes, a, e, d, j, l, y and r. The seven loci are closely linked in the centromeric region of an autosome but their order is unknown. Six deletions are examined in an individual that is heterozygous for all seven genes. All six deletions resulted in pseudodominance of at least two genes, and two deletions spanned the region of the centromere. Deletion Centromere included in deletion Genes exhibiting pseudodominance del 1 No e and j del 2 Yes a, l and r del 3 No a and d del 4 Yes e, l and r del 5 No d and y del 6 No e and l Determine the gene order and the location of the centromere.
Answer:
Gene order: y,d,a,(r,l),e,j
Centromere: between l and r
Explanation:
If you see your data you can conclude that:
- deletions that include the centromere include r and l genes.
- At one side of the centromere is gene a and at the other side is gene e
- when you see mutation 3, the gene that is away from the centromere is d, then in mutation 5 the gene that is further is y
- On the other side, the mutation 6 tell us that e gene is by l gene side
- In mutation one you see that the gene further the centromere by that side is j
An outbreak of salmonellosis occurred after an epidemiology department luncheon, which was attended by 485 faculty and staff. Assume everyone ate the same food items. Sixty-five people had fever and diarrhea, five of these people were severely affected. Subsequent laboratory tests on everyone who attended the luncheon revealed an additional 72 cases.
The ratio of severe cases to other clinically apparent cases was:
A. 65/485
B. 5/60
C. 72/485
D. 65/72
E. 5/65
Answer:
72/485
Explanation:
this question was just on my bio test i took
How can the female orgasm potentially aid conception?
Answer:
Two main hypotheses:
• "poleaxe" hypothesis.
It says orgasm in women makes them feel relaxed as it allow resting after sex inorder for sperm to reach its destination quickly
Its Shortcoming is that it isn't sure whether or not lying down after sex can help conceive. It's critics say women who remained horizontal after insemination have chances of conceiveing.
•The Upsuck Theory.
says contractions of the uterus help "suck up" the semen that gets deposited in the vagina, near the cervix during sex. As orgasm moves the sperm through the uterus and fallopian tubes.
Sperm retention is greater when female organism occurs a minute or more after sex.
The ________ extends through the hindbrain, midbrain, and forebrain.
a. Reticular formation
b. Medulla
c. Pons
d. Cerebellum
Answer: option C) Pons
Explanation:
Pons, or better still, the pons Varolii is a band of nerve fibers located within the brain stem. While, the brain stem is known to connect the spinal cord(Hind brain) to the fore brain
Thus, Pons extends through the hindbrain, midbrain, and forebrain.
Answer:
Reticular formation
It's correct
The calcium carbonate (CaCO3) stones located on the maculae are called:
a. ampullae
b. stereocilia
c. ossicles
d. ottoliths
Answer:
d. ottoliths
Explanation:
An otolith is a calcium carbonate structure in the saccule or utricle of the inner ear, majorly in the vestibular system of vertebrates.
The otolith organs is chiefly made up of the saccule and utricle
Human hair color is a classic, if oversimplified, example of recessive epistasis. Red hair is caused by a recessive allele r. However, if an individual has a dominant R allele, they may have either brown or blonde hair depending on whether they have a dominant B allele (which causes brown hair), or are homozygous for a recessive b allele (which causes blonde hair). The B and R loci are on located on different chromosomes. If a couple with genotypes Bb; Rr and bb; rr have children, what hair colors and in what proportions are expected among their children?
Answer:
1 Brown: 1 Blonde: 2 Red
Explanation:
According to the given information, the recessive allele "r" gives red color to hair but is epistatic to alleles B and b. Therefore, the genotype with two copies of the "r" allele would have red-colored hair. The genotypes with at least one copy of "B" and “R" alleles each would have brown hair while the "R" allele would give blond hair in presence of allele "b".
Therefore, a cross between BbRr and bbrr would produce progeny in following phenotype ratio= 1 Brown: 1 Blonde: 2 Red
Human hair color is a polygenic trait influenced by multiple genes. For a couple with genotypes Bb; Rr and bb; rr, a Punnett square predicts their children will have an equal chance of brown, blonde, or red hair in a 1:1:1:1 ratio.
Explanation:Hair color in humans is an example of a polygenic trait influenced by multiple genes. Individuals with certain combinations of alleles for these genes will display various hair colors such as black, brown, blonde, or red. In the scenario provided, where one parent has the genotype Bb; Rr (capable of brown or blonde hair with a possibility of red) and the other parent has the genotype bb; rr (red hair), we can predict the possible hair colors of their offspring using a Punnett square to combine their genotypes.
Since red hair requires two recessive r alleles and the second parent provides only recessive r alleles, all the offspring must carry at least one r allele, ensuring the possibility of red hair. Also, as brown hair is dominant over blonde and requires at least one B allele, which only one parent carries, we can expect some children to inherit brown or blonde hair depending on whether they inherit the B allele from the first parent.
Combining these alleles from both parents would yield a 1:1:1:1 ratio of the following phenotypes: Brown (Bb; Rr), Blonde (bb; Rr), Red (Bb; rr), and Red (bb; rr).
The presence of circular dna in mitochondria and chloroplasts suggests that these organelles evolved from ___.
Answer:
Bacteria.
Explanation:
These are DNA that formed closes loops.
They have no ends.
A typical example is plasmids of bacteria., circular bacteria chromosomes,
In viruses example are ccc DNA (circular closed circular DNAf) formed by viruses.
Mitochondria and chloroplasts evolved from prokaryotes through endosymbiosis.
Explanation:The presence of circular DNA in mitochondria and chloroplasts suggests that these organelles evolved from prokaryotes. Circular DNA is a characteristic feature of prokaryotic cells, such as bacteria, and is not typically found in the nucleus of eukaryotic cells. This indicates that mitochondria and chloroplasts were likely once free-living prokaryotic organisms that were engulfed by a larger host cell through a process called endosymbiosis. Over time, these organelles became integrated within the host cell and established a mutually beneficial relationship.
Learn more about Endosymbiosis here:https://brainly.com/question/33444517
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Integrated pest management requires Select one:
a. a complete genetic make-up of pests.
b. a complete knowledge of the pest's life history.
c. extensive application of pesticides.
d. All of these are correct.
Answer:
Option B
Explanation:
Integrated pest management is a method to control the pest through some common sense practices.
In this method, in depth study of pest’s life cycle along with interaction of pest with the environment must be done. This knowledge is then combined with the available knowledge of pest control methodologies in order to deal with pests in a most economically and environment friendly way.
Hence, option B is correct