Answer:
1. Deoxyribose sugar
2. Phosphate molecules
Explanation:
The back bone of DNA is made up pf two molecules.
1. Deoxyribose sugar
2. Phosphate molecules
deoxyribose is a 5 carbon sugar molecule while phosphate is inorganic salt .
Final answer:
The backbone of the DNA molecule is composed of a sugar-phosphate chain, formed by the alternating units of phosphate groups and a five-carbon sugar called deoxyribose.
Explanation:
The sides or backbone of the DNA molecule are made up of alternating units of phosphate groups and a five-carbon sugar known as deoxyribose. These two components are bonded together through a type of chemical reaction known as dehydration synthesis, which forms a covalent bond between the phosphate group of one nucleotide and the deoxyribose sugar of the next, resulting in a long polymer chain. This sugar-phosphate backbone forms the structural framework of DNA to which the nitrogen-containing bases are attached. The structure of DNA is a double helix, where these backbones twist around each other with the complementary nucleotide bases pointing inwards, held together by hydrogen bonds.
Part B - Diet and Lifestyle Choices During Pregnancy
Pregnancy comes with a unique set of diet and lifestyle guidelines. Some substances are particularly beneficial for the mother or baby, others are harmful, and some should be used with caution.
Drag the appropriate items to their respective bins.
1- green leafy vegetables
2- sashimi and sushi made with
raw fish
3-alcoholic beverages
4-unpasteurized milk
5-cooked fish and seafood
6-sodas and sports drinks
7-caffeinated beverages
8-low-fat sources of calcium
and vitamin D
Avoid During Pregnancy
Consume During Pregnancy
Consume with Caution During Pregnancy
Answer:
Explanation:
Avoid During Pregnancy
3-alcoholic beverages
7-caffeinated beverages
4-unpasteurized milk
2.sashimi and sushi made with
raw fish
Consume During Pregnancy
1- green leafy vegetables
8. low-fat sources of calcium
and vitamin D
Consume with Caution During Pregnancy
6-sodas and sports drinks
5-cooked fish and seafood
soda and sports drink should be used with Doctor recommendation during pregnancy. A high amount of cooked fish is also not advisable due to its adverse effects on the pregnant women but a daily dose recommended is good for body protein needs.
green leafy vegetables and low-fat sources of calcium and vitamin D are highly recommended during pregnancy
alcohol and caffeine are not good
while raw fish is also not good during pregnancy because they have high mercury, bacteria and viruses which are more harmful during pregnancy.
Tasting involves many different cell-signaling processes that ultimately generate nerve signals transduced by membrane depolarization. Sweet tastes result in PIP2 hydrolysis, while salty tastes allow sodium ions to directly alter the membrane potential. What can you deduce about the signaling mechanisms for sweet and salty?
Answer:
We can deduce that:
1. Sodium ions directly enters the cells, indicating the signal is transduced by an ion channel.
2. Sweet utilizes the GRCP signaling pathway, activating phospholipase C.
Explanation:
IN SALTY, the Sodium ion passes through dynamic various chambers that can be located on the apical membrane of the taste buds.
Sweet utilizes GRCP signaling pathway va activation of phospholipase C in order to produce IP3 and DAG.
If a membrane protein in an animal cell is involved in the cotransport of glucose and sodium ions into the cell, which of the following is most likely true?
A) The sodium ions are moving down their electrochemical gradient while glucose is moving up.
B) Glucose is entering the cell along its concentration gradient.
C) Sodium ions can move down their electrochemical gradient through the cotransporter whether or not glucose is present outside the cell.
D) Potassium ions move across the same gradient as sodium ions.
E) A substance that blocked sodium ions from binding to the cotransport protein would also block the transport of glucose.
Answer:
The correct answer is E) A substance that blocked sodium ions from binding to the cotransport protein would also block the transport of glucose.
Explanation:
Cotransport means that one molecule is being transported through the membrane while another molecule is transported with that energy by the same transporter. Therefore if one substance is blocking the sodium ions from binding to the protein, therefore the glucose couldn't be transported either.
Two nutrient solutions are maintained at the same pH. Actively respiring mitochondria are isolated and placed into each of the two solutions. Oxygen gas is bubbled into one solution. The other solution is depleted of available oxygen. Which of the following best explains why ATP production is greater in the tube with oxygen than in the tube without oxygen?
A. The rate of proton pumping across the inner mitochondrial membrane is lower in the sample without oxygen. B. Electron transport is reduced in the absence of a plasma membrane. C. In the absence of oxygen, oxidative phosphorylation produces more ATP than does fermentation. D. In the presence of oxygen, glycolysis produces more ATP than in the absence of oxygen.
Was told the answer is A. but i need help justifying that and why it is not D
Answer:
Glycolysis is not an oxygen dependent process and the rate of glycolysis is not affected by the amount of oxygen present. The amount of oxygen present affects the rate of oxidative phosphorylation in the mitochondria. Hence with more oxygen present, there is increased oxidative phosphorylation in the presence of high amounts of final electron acceptor O2 which pumps more protons across the membrane and results in a steeper proton gradient across the mitochondrial membrane thus increasing the rate of ATP production. Thus the answer is A as the rate of proton pumping is increased with higher rate of oxidative phosphorylation due to the abundance of final electron acceptor O2.
Oxygen doesn't affect glycolysis' rate. Oxygen affects oxidative phosphorylation. With more oxygen, oxidative phosphorylation rises, pumping more protons across the mitochondrial membrane and increasing ATP synthesis.
What is oxidative phosphorylation?During the process of oxidative phosphorylation, electrons that were obtained from NADH and FADH2 mix with oxygen (O2). The energy that is released as a result of these oxidation and reduction events is then used to fuel the synthesis of ATP from ADP.
Glycolysis does not require oxygen, and the presence of oxygen has no effect on the rate at which it occurs. Oxygen has an effect on the oxidative phosphorylation that occurs in mitochondria. In the presence of high amounts of final electron acceptor O2, the process of oxidative phosphorylation speeds up when there is more oxygen present. This results in the transfer of more protons across the mitochondrial membrane, which boosts the production of ATP.
Therefore, option (A) is correct and (D) can't be true.
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A 47-year old man presents with severe pain in his right great toe. He is unable to bear weight on his right foot and there was no apparent trauma. He reports the pain started last night and continued to increase in severity. Based on his physical exam and laboratory findings he is diagnosed with gout. Gout can be caused by several different mechanisms and blood samples are drawn and urinalysis is performed. The results from this individual are listed below.Patient Plasma Urate(mg/100ml) Urinary Uric Acid(mg/24hr)Normal <7.0 413±78Patient 10.5 215 What is the best explanation for the presentation of gout in this individual and please explain why the other choices are not consistent with the data or what additional information you may need to decide the root cause?1. overproduction of purines2. decreased salvage of purines3. decreased urinary excretion of uric acidAfter further discussion with the patient, he tells you that he recently switched from diet soda to regular soda because he read that the ‘fake sugar’ diet soda was bad for him; he tells you he drinks between 4 and 5 L of soda a day. (NOTE: The primary sugar in most sodas is fructose or high fructose corn syrup.)What is the potential impact of this dietary modification on the presentation of gout?
Answer:
the best explanation is 1. overproduction of purines
Explanation:
Purines are nitrogenous base that are broken down to uric acid which is excreted through the kidney as a constituent of urine. Different conditions that lead to an impaired removal of plasma uric acid cause gout. Soda drinks and many soft drinks contain low levels of purines and high level of fructose. Fructose is the only carbohydrate known to increase uric acid levels by increase the rate of degradration and synthesis of purine
Individual cells do not survive for the entire lifetime of an organism because a. the enzyme telomerase is readily destroyed by the environment, resulting in cell death. b. DNA replication is subject to errors that cause cell death. c. Okazaki fragments disrupt protein synthesis, resulting in cell death. d. the repeating telomeric sequence of TTAGGG interferes with normal DNA replication and leads to cell death.
Answer:
b. DNA replication is subject to errors that cause cell death.
Explanation:
Answer : none
Explanation:the removal of the RNA primer following dna replication leads to a shortening of the chromosome and eventual cell death
Both prokaryotes and eukaryotes use the information encoded on the genes in their DNA to synthesize proteins. How does the process of translation in prokaryotes differ from translation in eukaryotes?
Translation in prokaryotes differs from eukaryotes in that it occurs simultaneously with transcription and may involve polycistronic mRNA.
Explanation:The process of translation in prokaryotes differs from translation in eukaryotes in several ways. In prokaryotes, translation occurs simultaneously with transcription, meaning the mRNA is translated as it is being transcribed. This is because prokaryotes lack a nucleus and do not have separate compartments for transcription and translation. In contrast, in eukaryotes, transcription occurs in the nucleus and the mRNA must undergo several processing steps before it can be transported to the cytoplasm for translation. Additionally, prokaryotes often have polycistronic mRNA, meaning multiple genes are translated from the same mRNA molecule, whereas eukaryotes generally have monocistronic mRNA, with each mRNA molecule encoding only one gene.
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The immune system Choose one:
A. is nonspecific in its responses to invaders.
B. is composed entirely of cells, rather than tissues and organs.
C. kills all nonself cells.
D. is a system that differentiates self from nonself to neutralize harmful invaders.
Answer:
The correct answer is D. is a system that differentiates self from nonself to neutralize harmful invaders.
Explanation:
A defense system that protects the host body from pathogen infection is called the immune system. This immune system is made up of many different barriers which helps in preventing illness from pathogen attack.
Physical and chemical barriers provide the first line of defense to the body for example skin, mucous membrane, lysozyme in saliva provide first line of defence. Second line of defense is provided by non-specific immune cells like mast cells, macrophages, dendritic cells, etc.
The third line of defense is provided by pathogen-specific immune cells like T lymphocytes and B lymphocytes. These cells have the ability to differentiate self and non-self cell and kills the non-self cell that come from outside the body.
The authors of a cross-sectional study hypothesized that lack of regular exercise is associated with obesity in children. Their study of 12 children in Michigan, however, failed to show a statistically significant association between exercise habits and obesity (OR = 1.9, p = 0.11). Is this A) Selection Bias, B) Confounding, C) Information Bias, D) Random Error
Answer:
A) Selection Bias
Explanation:as it depends on sampling method of data collection
The Punnett square predicts the ratio of genotypes in the offspring, based on the genotypes of the parents. The Punnett square below examines the chance of offspring having freckles, which is a dominant trait. A Punnett square is shown. The columns are labeled Upper F and f. The rows are labeled Upper F and f. Clockwise from upper left the boxes contain: Upper F Upper F, Upper F f, Upper F f, f f. Based on the Punnett square, what is the probability that the offspring will have freckles?
Answer:
According to the description, the punnet square can be drawn as follows:
F f
F FF Ff
f Ff ff
The results of the punnet squares show that there is a 75% chance for the children to have freckles. As freckles is a trait which can arise from just a single allele, hence we can say that is is a dominant trait.
The children from this cross have 25% chance of being homozygous for the freckles trait. There is a 50% chance that the children will be heterozygous for freckles trait and there is a 25% chance that the children will not have freckles.
Answer:
C. 75%
Explanation:
Which of the following statements is correct concerning the spinal cord?A. The white matter contains cell bodies for spinal nuclei.B. Damage to sensory tracts in the spinal cord leads to paralysis.C. Just like the cerebrum, the gray matter is found on the superficial surfaces.D. Spinal nerves have mixed motor and sensory function.
Answer:
D. Spinal nerves have mixed motor and sensory function.
Explanation:
spinal's nerves are mixed nerves, which means that they carry sensory and motor information. They carry sensory information to the central nervous system to give a motor answer that will travel through the spinal nerves to specific muscles. The White matter, which is peripheral in the spinal cord, contains axons while the gray matter, which is central in the spinal cord, contains cell bodies.
Spinal nerves have mixed motor and sensory function is the statements is correct concerning the spinal cord. Thus, option (d) is correct.
The spinal cord is essential for the movement of nerve impulses from the brain to the rest of the body. It transmits impulses for pressure, temperature, movement, feeling, and pain.
Gray matter and white matter both make up the spinal cord, with the gray matter being central and the white matter being on either side. While nerve fibres that carry messages are found in the white matter, the gray matter houses the neuronal cell bodies.
As a result, the significance of the concerning the spinal cord are the aforementioned. Therefore, option (d) is correct.
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Many promoter regions in genes transcribed by eukaryotic RNA polymerase II do not contain a TATA box sequence. However, all three eukaryotic polymerases require TBP for transcription initiation regardless of the presence of a TATA box in the promoter region. What hypothesis might explain why TBP is necessary for transcription from TATA-less promoters
A. The DNA-bending function of TBP cannot be performed by another protein in the general transcription factor complex.
B. TBP is still required for initial recognition of promoter sequences. This protein has an equal affinity for DNA sequences with and without a TATA box.
C. TBP is required for stabilizing the complex as it assembles on the promoter.
D. TBP is required for the catalytic activity of the RNA polymerase.
Answer:
A. The DNA-bending function of TBP cannot be performed by another protein in the general transcription factor complex.
Explanation:
TATA box sequence binding protein binds to aspecific region on DNA known as TATA Sequence. TATA binding protein binding to the DNA Sequence is necessary because it generates bending of DNA till 80° angle.
BINDING of all transcription factors depends solely upon the binding of TBP directly, that is why TATA binding proteins bonding to DNA is very important for initiation of transcription. Any other protein does not have the capacity of band formation with DNA. Hence, A. the DNA-bending function of TBP cannot be performed by another protein in the general transcription factor complex.
According to the graph, which of these statements is true?
a. This enzyme works best at very low temperatures.
b. This enzyme works best at very high temperatures.
c. This enzyme works equally well at all temperatures.
d. This enzyme works best at middle temperatures.
Answer:
The correct answer is D
Explanation:
That's why in the graph the best reaction of the enzyme is on 40° C.
On low temperatures the enzyme activity is too low.
On very high temperatures it fall, so the enzyme works better in the middle temperatures.
Riparian zones are important to the natural environment because they _______. a. contain pollution from runoff b. reduce soil erosion c. support diverse organisms d. all of the above Please select the best answer from the choices provided A B C D
Answer:
The correct answer will be option-D
Explanation:
A riparian zone is the zone formed at the meeting area of land with water mostly from the lakes, ponds and the river. The plants growing in the riparian zones are known as the riparian vegetation.
The riparian zones prove useful to natural environment and humans as the growing vegetation help bind the soil which reduces soil erosion. It also reduce pollution level in the water bodies as it contains the pollution from the run off.
The riparian zones provides habitat to a variety of plants mainly from bryophytes and pteridophytes and animals which can live in the riparian zones.
Thus, option-D is the correct answer.
Answer:
the correct answer is D ALL OF THE ABOVE
What is the energy requirement in kJ/mol to transport a proton across the mitochondrial inner membrane in plant cells at night when the outside temperature is 15ºC, the pH differential across this membrane is 1.4 pH units, and the membrane potential of 160 mV. Choose the ONE correct answer.
A. +21.8 kJ/mol
B. +75.4 kJ/mol
C. +23.1 kJ/mol
D. +46.2 kJ/mol
E. -21.8 kJ/mol
F. -23.1 kJ/mol
G. +19.0 kJ/mol
H. +44.6 kJ/mol
Answer:
C. +23.1 kJ/mol
Explanation:
the formula to use to calculate the energy requirement in kJ/mol to transport a proton across the mitochondrial inner membrane in plant cells is:
ΔGt = RTIn [tex]\frac{C_{2} }{C_1}[/tex] + ZFΔV
let's list the values of the data we are being given in the question to make it easier when solving it.
Z= 1
F= 96500C (faraday's constant)
ΔV= 160mV = 0.160V
R= 8.314( constant)
T= 15ºC ( converting our degree Celsius into kelvin, we will have 273.15k+ 15 = 288.15K)
∴ T= 288.15K
Putting it all together in the formula, we have:
ΔGt = 8.314 × 288.15 × 2.303 log [tex]\frac{C_{2} }{C_1}[/tex] + 1 × 96500 × 0.160
ΔGt = 5517.25 [tex][ -(log(H^{+}_{out} ) + log(H^{+}_{in} )_{in}[/tex] +15440
ΔGt = 5517.25 [tex](pH_{out} - pH_{in}[/tex] +15440
Given that the pH differential gradient across the membrane is 1.4pH units. It implies that;
ΔGt = 5517.25 × 1.4 + 15440
= 7724.15 +15440
= 23164.15 Joules/moles
= +23.1 KJ/mole
Fill in the blanks.
Osteichthyans" commonly referred to as "fishes" have a ________________ endoskeleton.
Fishes control their buoyancy with an air sac known as a ________________.
Fishes breathe by drawing water over _______ in chambers covered by a bony flap called the ________________.
Answer:
bony
swim bladder
mouth
operculum
Explanation:
Fill in the blanks.
Osteichthyans" commonly referred to as "fishes" have a _____Bony___________ endoskeleton.
Fishes control their buoyancy with an air sac known as a _____swim bladder_
Fishes breathe by drawing water over ___mouth ____ in chambers covered by a bony flap called the __operculum.______________.
A mutation occurs in an operon that prevents the transcription factor from binding to its recognition site on the DNA. In which type(s) of gene regulation would this mutation result in constitutive expression of the structural gene?
Answer:Negative inducible and Negative Repressible
Explanation:
Negative control of operon includes Negative inducible and Negative repressible both prevent transcription, but are different in there mode of operation.
Negative inducible operon: This is usually a process by which an active repressor regulatory protein binds to the DNA thereby inhibiting RNA polymerase from transcribing gene.
Negative repressible operons: This is also a process when an negative inducer binds to the operator to which prevent DNA transcription
Answer:
Explanation:
In the negative inducible and negative repressible regulations of gene.
Which of the following statements about valves in the heart is/are true? A. A valve enables blood to only flow in one direction. B. A valve keeps blood from flowing in a reverse direction. C. All of the valves open when the blood pressure on one side of the valve is higher than on the other side of the valve. D. Statements A and B are correct. Statement C is incorrect. E. Statements A, B, and C are correct.
Answer: Option D is the correct answer.
The correct statements about valves are;The heart valves enable blood to flow in one direction.The valves keep blood in flowing in reverse direction.
Explanation:
Heart valves are thin tissue paper membrane that is attached to the heart. The heart valves and chambers are lined with endocardium.The heart valves is divided into four which are;
1. The two atrioventricular valves , the biscupid and triscupid valves which are between the upper chambers of the atria and the lower chambers of the ventricles
2. The two semilunar valves, the aortic and the pulmonary valves which are in the arteries leaving the heart. The biscupid and the aortic valves are in the left heart while the pulmonary and triscupid valves are in the right heart. During ventriculsr systole when pressure rises in the left ventricle and it is greater in the aorta, the aortic valves then open and allow blood to leave in through the left ventricle in the aorta.
The statement which is true about valves in the heart is: Choice D.
A valve enables blood to only flow in one directionA valve keeps blood from flowing in a reverse direction.Discussion:
Heart valves by definition are thin tissue membranes attached to the heart. The heart valves and chambers are lined with endocardium.
The heart valves are divided into four which are;
The two atrioventricular valves , the biscupid and triscupid valves.The two semilunar valves, the aortic and the pulmonary valves located in the arteries leaving the heart.All of which are characterized by enabling blood flow only in one direction and keeping blood from flowing in the reverse direction.
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Should any of Marjorie's family members whom she comes into contact with be considered at risk of pertussis? If so, who?A) Yes, the five-year-old cousin should be considered at risk.B) Yes, the aunt should be considered at risk.C) Yes, the six-week-old cousin should be considered at risk.D) No, no one should be considered at risk.
Answer:
C) Yes, the six-week-old cousin should be considered at risk.
Explanation:
young are at higher risk of pertussis as compared to older children and adult because young infants are not fully immunized. At the young age of 6 week, the cousin may not have started the DPT vaccination. so babies at this age gets easily affected when they are in contact with the infected people.
Suppose that you are in charge of designing a fracking job site. Which of these locations would be most ideal for the site?
Answer:
on a flat, vacant plot on the outskirts of the city
Explanation:
Took the test
plants make their own food. The chemical for this process is shown by the following equation CO2 + H2O+energy>C6H12+O2 what are the products in this chemical reaction
In peas, tall plants (T) are completely dominant over short plants (t). You cross two heterozygous pea plants Use a Punnett square to determine the expected genotypic and phenotypic ratios of the offspring.What is the male pea plant's genotype?What is the female pea plant's genotype?What are the male pea plant's gametes?What are the female pea plant's gametes?What is the expected genotypic ratio of the offspring?What is the expected phenotypic ratio of the offspring?
Answers and Explanation:
Find enclosed the Punnet square.
T= dominant allele
t= recessive allele
Tt= heterozygous genotype
TT and tt= homozygous genotype
- The male pea plant's genotype is Tt (heterozygous)
- The female pea plant's genotype is Tt (heterozygous)
- Male gametes: T and t
- Female gametes: T and t
- The expected genotypic ratio of the offpring will be:
TT: Tt: tt ⇒ 1: 2: 1
1/4 of the offpring will be TT genotype (dominant homozygous), 1/2 of the offpring will be Tt (heterozygous) genotype and 1/4 of the offpring will be tt genotype (recessive homozygous).
- The expected phenotypic ratio of the offpring will be:
tall : short ⇒ 3 : 1
3/4 of the offpring will be tall (TT and Tt) and the restant 1/4 will be short (tt).
Irwin’s sells a particular model of fan, with most of the sales being made in the summer months. Irwin’s makes a one-time purchase of the fans prior to each summer season at a cost of $40 each and sells each fan for $60. Any fans unsold at the end of the summer season are marked down to $29 and sold in a special fall sale. Virtually all marked-down fans are sold. The following is the number of sales of fans during the past 10 summers: 30, 50, 30, 60, 10, 40, 30, 30, 20, 40.
Estimate themean and variance of the demad for fans each summer.
Answer:
answer is in the image below
Explanation:
In the scheme of feeding bacteria to worms to induce an RNAi response in worms, explain the purpose of the following components: lac promoter incorporated into the genome of E. coli, T7 gene incorporated into the genome of E. coli, and T7 promoter on the plasmid introduced into E. coli.
Answer:
lac promoter incorporated into the genome of E. coli, = for the repression of the gene
T7 gene incorporated into the genome of E. coli,+ for the formation of T7 gene
and T7 promoter on the plasmid introduced into E. coli= for the attachment of Polymerase to transcriped RNA from 5 to 3 direction
Explanation:
Answer:
For repression of gene the lac parameters must be incorporated into genome of E.coli
Where for the formation T7 gene,T7 gene must be incorporated into genome of E.coli
For attachment of polymerase to transcript RNA from 5 to 3 direction the T7 on the placimid must be introduced into E.coli
___________ arise from two eggs fertilized by two different sperm cells
Answer:
the answer is identical twins
Answer:
Identical twins
Explanation:
In a population of flowers growing in a meadow, C1 and C2 are autosomal codominant alleles that control flower color. The alleles are polymorphic in the population, with f (C1) = 0.9 and f (C2) = 0.1. Flowers that are C1C1 are yellow, orange flowers are C1C2, and C2C2 flowers are red. A storm blows a new species of hungry insects into the meadow, and they begin to eat yellow and orange flowers but not red flowers. The predation exerts strong natural selection on the flower population, resulting in relative fitness values of C1C1 = 0.30, C1C2 = 0.60, and C2C2 = 1.0.a.Assuming the population begins in H?W equilibrium, what is C1 allele frequency after one generation of natural selection?b.Assuming the population begins in H?W equilibrium, what is C2 allele frequency after one generation of natural selection?c.Assuming random mating takes place among survivors, what is the genotype C1C1 frequency in the second generation?d.Assuming random mating takes place among survivors, what is the genotype C1C2 frequency in the second generation?e.Assuming random mating takes place among survivors, what is the genotype C2C2 frequency in the second generation?
Final answer:
The allele frequencies of C1 and C2 and the genotype frequencies in a population of flowers are influenced by autosomal codominant alleles and natural selection. Calculations for changes in allele and genotype frequencies involve the initial frequencies, relative fitness values, and the Hardy-Weinberg principle, but exact values require additional data.
Explanation:
In a population of flowers with autosomal codominant alleles influencing flower color, the C1 and C2 alleles contribute to different phenotypes. Using the given allele frequencies and relative fitness values, we can calculate the impact of natural selection on this population.
a. C1 allele frequency after one generation of natural selection
To calculate the C1 allele frequency after one generation of selection, we must multiply the initial frequency by the relative fitness and then normalize it. However, without the actual population numbers or total fitness value, we can't provide a specific numerical answer.
b. C2 allele frequency after one generation of natural selection
Similarly, the C2 allele frequency is calculated by paying attention to the initial frequency and the impact of the relative fitness. As the C2C2 individuals have a relative fitness of 1.0 and are not being eaten, the C2 allele is likely to increase in frequency.
c. Genotype C1C1 frequency in the second generation
Assuming random mating among survivors of the first generation, the C1C1 frequency can be calculated using the Hardy-Weinberg principle. However, the exact frequency would need further calculation with the revised allele frequencies after the first selection event.
d. Genotype C1C2 frequency in the second generation
The frequency of the C1C2 genotype would also be calculated according to the Hardy-Weinberg equilibrium equation as 2pq, reflecting the genotype's expected proportion in the population.
e. Genotype C2C2 frequency in the second generation
The C2C2 genotype frequency would increase in the second generation, assuming no other evolutionary forces come into play and random mating occurs, we predict this using the Hardy-Weinberg principle (q²).
Why is conserving biodiversity important? Choose all that apply.
protecting endangered species
protecting habitat diversity
maintaining genetic diversity
preserving heirloom varieties of food crops
Answer:
The correct options are :
protecting endangered species protecting habitat diversitymaintaining genetic diversityExplanation:
Biodiversity can be described as the different varieties of plants and animals which are present in a habitat. If biodiversity wasn't conserved then survival of most plants and animals species would become difficult whenever changes occurred. Without biodiversity, degradation of a habitat would occur. Hence, the conservation of biodiversity is very important for protecting the wild-life animals and plants. Biodiversity boosts up an ecosystem. No matter how small an organism is, their conservation is important for the maintenance of biodiversity.
Conserving biodiversity is important for protecting endangered species, maintaining ecosystems, ensuring the adaptability of species through genetic diversity, and ensuring food security through preservation of heirloom crop varieties.
Conserving biodiversity is critical for several intertwined reasons. Protecting endangered species ensures the survival of organisms that may have unique ecological roles or genetic traits beneficial to the ecosystem or human society. Protecting habitat diversity helps maintain the health and integrity of ecosystems, which in turn supports a variety of life forms. Maintaining genetic diversity within species is crucial for their adaptability and survival in the face of environmental changes, including climate change. Lastly, preserving heirloom varieties of food crops is important for food security and agricultural sustainability, as these varieties can be more resilient to pests, diseases, and changing climate conditions.
The concentration of agarose gel affects the resolution of DNA separation. For a standard agarose gel electrophoresis, a 0.8% gives good separation or resolution of large 5–10kb DNA fragments, while 2% gel gives good resolution for small 0.2-1kb fragments. 1% gels is often used for a standard electrophoresis. If you have very similar sized DNA molecules that are running too close together on an agarose gel, what solution would you apply to resolve this issue?
Answer:
Use a higher % agarose gel.
Explanation:
Agarose gels have a porous matrix. The higher the concentration of agarose, the smaller the pores, so larger DNA molecules will have more difficulty moving through the gel and they will run slower than small DNA molecules.
The higher % agarose gel has thus a better resolving power (the measurable interval between two entities -the DNA bands- is smaller). For that reason, a 2% agarose gel will allow you to differentiate better between two bands of close molecular weight, if you let the DNA fragments run long enough.
James Marcia developed a system by which adolescents can be classified into one of (our identity statuses, according to whether they have (1) made a thorough exploration (and/or experienced a crisis") or (2) made a "commitment" in life. The identity status which would best be described as "no conflict, no crisis no clue" would be the identity foreclosure status.-permissive the identity diffusion status.-neglectful the identity achievement status.-Authoritative the identity foreclosure status.-coercive, domineering
Answer:1 premiussie 2 neglectful
Explanation:
Sorry
The first step in the catabolism of amino acids is the removal of the nitrogen as ammonia, forming a keto acid that can enter one of the carbon catabolic pathways. The alpha-keto acid pyruvate can be formed from the amino acids alanine, cysteine, glycine, serine and threonine. Consider the route for alanine catabolism. Anine reacts with to produce pyruvate and. This reaction is catalyzed by aspartate aminotransferase. alanine aminotransferase. alanine dehydrogenase. The substrate for the first step can be regenerated by reacting with NAD^+ This reaction is catalyzed by glutamate dehydrogenase. alpha-ketoglutarate dehydrogenase. alanine dehydrogenase. aspartate dehydrogenase. The coenzyme/prosthetic group required in the first reaction is thiamine pyrophosphate. biotin. pyridoxal phosphate. lipoic acid.
Coenzymes are organic compounds that many enzymes require for catalytic activity. They are frequently vitamins or vitamin derivatives.
The coenzyme prosthetic group required in the first reaction is thiamine pyrophosphate.
What is thiamine pyrophosphate?Thiamine pyrophosphate is a vitamin B1 derivative produced by the enzyme thiamine diphosphokinase.
Thiamine pyrophosphate is a cofactor found in all living systems that catalyzes a variety of biochemical reactions.
Thus, we can conclude that option A is correct.
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