Answers
Explanation:
Step 1. The two quality loci, A and B, assort autonomously. The alleles A and B are predominant over the alleles an and b. In this way, when a cross happens between AaBb X AaBb, the subsequent gametes would be AB, Ab, aB, and ab.Step 2.The offsprings which have in any event one A and B allele, will show AB phenotype. Along these lines, AABB, AaBb, AABb, AaBB, will all have AB phenotype.
If two gene loci, A and B, assort independently, the probability of producing an AB gamete from AaBb would be 1/4 or 25%
Since both A and B are independently inherited, Aa and Bb can be crossed just we would have it in a monohybrid cross following Mendelian pattern.
Thus:
Aa x Bb
AB Ab aB ab
The gametes and their respective probabilities of appearing would be:
AB - 1/4Ab - 1/4aB - 1/4ab - 1/4In other words, all 4 gametes have an equal chance of being produced during gametogenesis.
More on independent assortment of genes can be found here: https://brainly.com/question/13041062
Related Questions
Based on your results in Part A, which of the following statements most appropriately reflects the relative effectiveness of the vancomycin treatment, teixobactin treatment, and the control?
Teixobactin reduced the number of MRSA colonies about as effectively as did vancomycin relative to the control.
Neither teixobactin or vancomycin were effective at reducing the number of MRSA colonies relative to the control.
Teixobactin was significantly more effective than the vancomycin at reducing the number of MRSA colonies relative to the control.
Teixobactin was significantly less effective than the vancomycin at reducing the number of MRSA colonies relative to the control.
Answers
Answer:
Option C is correct,which states that Teixobactin was significantly more effective than the vancomycin at reducing the number of MRSA colonies relative to the control.
Teixobactin reduced the number of MRSA colonies about as effectively as vancomycin relative to the control. Statement 1
Explanation:The most appropriate statement reflecting the relative effectiveness of vancomycin treatment, teixobactin treatment, and the control is:
Teixobactin reduced the number of MRSA colonies about as effectively as did vancomycin relative to the control.
This means that both vancomycin and teixobactin were similarly effective in reducing the number of MRSA colonies compared to the control group. Therefore, teixobactin was not significantly more or less effective than vancomycin.
Learn more about Teixobactin Against MRSA here:https://brainly.com/question/36741462
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5. Which of the following statements are accurate?
A. Red blood cells are known as erythrocytes.
B. Hemoglobin would be found in white blood cells and functions in carrying oxygen in body tissues.
C. The liquid portion of the blood is known as the hematocrit.
D. Platelets are the only thing in the blood that contributes to blood clotting.
E. Lymphocytes are the only type of white blood cells and are immune cells
Answers
Answer: Option A ,Band C
Explanation:
Red blood cells are called erythrocytes. They are produced from the bone marrow and deposited into the blood stream when they are matured. The red blood cells carry oxygen through out the body.
Lymphocytes are white blood cells that are produced from the bone marrow and help in body defence, they are found in lymph tissues. They are immune cells.
Platelets are tiny cells that contribute to blood clots
In one experiment, scientists raised mice in germ-free conditions so the mice lacked intestinal microbes. The mice were fed a low-fat diet rich in complex plant polysaccharides, such as cellulose, that are often called fiber.
When the mice were 12 weeks old, the scientists transplanted the microbial community from the intestine of a single "donor" mouse into all of the germ-free mice. Then they divided the mice randomly into two groups and fed each group a different diet.
Group 1 (the control group) continued to eat a low-fat, high-fiber diet.
Group 2 (the experimental group) ate a high-fat, high-sugar diet.
Identify the components of this experiment by matching the terms with the appropriate category below:
Factors being tested Factors controlled Factors to be measured
(independent variable) (Kept consistent) (dependent variable)
1. initial composition of the microbial community
2. age of the mice
3. change in body fat
4. diet
5. weight gain
6. final composition of the microbial community
Answers
Answer:
The answers are:
1. initial composition of microbial community (controlled variable)
2. age of the mice (controlled variable)
3. change in body fat (dependent variable)
4. diet (independent variable)
5. weight gain (dependent variable)
6. final composition of the microbial community (controlled variable)
Explanation:
Before I go ahead to explain each of the choices of classification, let us define each of the different types of variable.
1. Independent variable: this is a variable that is under the control of the experimenter. It is made up of the factors that are being tested. In our example, the independent variables is; diet. The aim of the experiment is to study the effect of the type of diet on the mice. The diet can also be varied and assigned at will by the experiment.
2. Dependent variable: a dependent variable is one whose outcome is reliant on the effect of the changes in the independent variable. It is the factor to be measured. In our example, the factors that obey these definitions are; change in body fat and weight gain. these two factors depend on the types of diet taken by the mice, hence they are the factors to be measured.
3. Controlled variable: these are variables that are kept constant throughout the experiment to rule out biases, because they produce the same effect on the dependent variable as the independent variable and must be keep constant or uniform in the test groups. in our example, the factors that are controlled include; initial and final compositions of the microbial communities and the age of the mice, because the microbial composition if not uniform among the groups can cause biases in how the diets are processed, and the age of the mice also must be uniform to avoid age-related differences in the processing of the diets
You are studying the source of new virus that has recently infected humans. You suspect that the virus was transferred from other primates (they exhibit a similar infection), specifically chimpanzees, gorillas, or orangutans. You sample blood from several infected humans and sequence some viral genes. You then build a phylogenetic tree with the human sequences and all the known strains from each primate. Draw a hypothetical phylogenetic tree that would suggest that the virus came from gorillas, and this transfer occurred twice independently. Label chimp sequences (c), gorilla (g), orangutans (o), and humans (h).
Answers
Answer:
it doesnt make sense can you try to put it some other way
Explanation:
Carlita was hired today for a job as a laboratory assistant. She will work for professor of molecular biology who wants to make new discoveries about genes, DNA , RNA, and their functions. Carlitas work will most likely contribute to which of these specific research goals? PLEASE HELPPP HURRYYY
Answers
Answer: Option C) Determining the link between genes and the traits they influence.
Explanation:
First, the field of molecular biology seeks to understand the relationship between genes and proteins. Now, since genes are specific sequences on the DNA that codes for functional proteins, and these proteins make up the various traits physically expressed in living organisms such as skin color, blood group etc.
Then, the specific research goal of Carlita would be determining how genes control the expression of individual traits at the molecular level
Answer: C
Explanation:
Fungi and plants form mutually beneficial symbiotic relationships called mycorrhizae. Classify the resources as a. supplied by the fungus or the plant. b. Supplied by fungus. c. Supplied by fungus d. Supplied by plant.
Answers
Answer:
Mycorrhiza is a symbiotic association in between the green plants and fungus.
The micorrhiza are located in the roots of the vascular plants. The Green plants performs the process of photosynthesis and the sugars produced are supplied to the Micorrhizas.
The fungus supplies the nutrients, water such as phosphorus that is absorbed from the soil to the plant body.
Answer:Answer:
Mycorrhiza is a symbiotic association in between the green plants and fungus.
The micorrhiza are located in the roots of the vascular plants. The Green plants performs the process of photosynthesis and the sugars produced are supplied to the Micorrhizas.
The fungus supplies the nutrients, water such as phosphorus that is absorbed from the soil to the plant body.
Explanation:
Answer:
Mycorrhiza is a symbiotic association in between the green plants and fungus.
The micorrhiza are located in the roots of the vascular plants. The Green plants performs the process of photosynthesis and the sugars produced are supplied to the Micorrhizas.
The fungus supplies the nutrients, water such as phosphorus that is absorbed from the soil to the plant body.
The medial deltoid attached to the humerus at an angle of 15 deg. What are the size of the rotary and stabilizing components of muscle force when the total muscle force is 500 N
Answers
Answer:
The size of the rotary and stabilizing components of muscle are 129.41 N and 482.96 N
Explanation:
According to the angle produced between the deltoid attached and the humerus
The Rotary Formula (or Vertical Component) is
FR = 500 N · sin 15° = 129.41 N
and the Stabilizing Formula (or Horizontal Component) is
FS = 500 N · cos 15° = 482.96 N
Final answer:
For a muscle force of 500 N attached at a 15-degree angle, the rotary and stabilizing components are calculated using trigonometry, resulting in a rotary component of approximately 129.4 N and a stabilizing component of approximately 482.95 N.
Explanation:
The question pertains to calculating the rotary and stabilizing components of a muscle force when the total muscle force is given and the muscle is attached at a specific angle. To solve this, we apply knowledge of trigonometry and vector decomposition. Given that the total muscle force is 500 N and it's attached at a 15-degree angle to the humerus, the force can be decomposed into two components: the rotary (perpendicular) component and the stabilizing (parallel) component.
Calculations:
Rotary component (Fr): This is the component of the force acting perpendicular to the lever arm. It is calculated using the sine function (Fr = Total Force × sin(θ)). For a 15-degree angle, Fr = 500 N × sin(15°) = 500 × 0.2588 ≈ 129.4 N.
Stabilizing component (Fs): This is the component of the force acting parallel to the lever arm. It is calculated using the cosine function (Fs = Total Force × cos(θ)). For a 15-degree angle, Fs = 500 N × cos(15°) = 500 × 0.9659 ≈ 482.95 N.
In conclusion, for a total muscle force of 500 N attached to the humerus at a 15-degree angle, the rotary component is approximately 129.4 N and the stabilizing component is approximately 482.95 N.
In poodles, black fur is dominant to white fur. A black poodle is crossed with a white poodle. In a litter of four, all of the puppies are black. What is the BEST conclusion?
Answers
Answer:
The black poodle is true breeding (homozygous) for the black fur colour.
Explanation:
Let the allele for fur colour be represented by B.
Black fur (B) is dominant over white fur (b).
Black poodle (B _) is crossed with white poodle (bb):
B _ x bb
Progeny (all black) = Bb, Bb, _b and _b.
Since all the progeny are black, _b = Bb (remember, B is dominant over b)
Hence, B_ is also BB.
The best conclusion is that the black poodle is true breeding (homozygous) for the black fur colour.
Answer:
Explanation: all the puppies in the F1 generation are black because homozygous black (BB) is crossed with a homozygous white(bb) and since the black fur is dormant all the puppies will be black ie heterozygous black (Bb).
Starting with a population of genetically identical mice, you discover two new independent mutant strains in which all of the animals have epileptic seizures. In both strains, you know that the epileptic seizures are due to a single DNA mutation. You cross a mutant mouse from one strain to a mutant mouse from the second strain and find that none of their offspring undergo spontaneous seizures. From this experiment you would conclude that the two mutant strains of mice most likely have mutations in:
a-the same DNA base position within a particular gene
b-the same gene, but not necessarily the same DNA base position.
c-two different genes.
Answers
Answer:
B - the same gene, but not necessarily the same DNA base position. They do not show seizures when they are crossed, but show seizures in independent populations.
When they are not in the same position and hence they do not come together to express the disease.
Thus, the answer is B Explanation
a. If they are in the same DNA base positions,, their offspring might have expressed epileptic seizures.
c. If they are from different genes it would have interfered with the trait.
Final answer:
The two mutant strains of mice most likely have mutations in two different genes, as inferred from the observation that their offspring did not exhibit epileptic seizures, suggesting compensatory heterozygosity.
Explanation:
When a mutant mouse from one strain is crossed with a mutant mouse from another strain and the offspring do not exhibit epileptic seizures, it suggests that the mutations are likely in two different genes. This conclusion comes from the understanding that if the mutations were in the same gene (whether at the same DNA base position or different ones), the offspring would more likely show the dominant phenotype, in this case, seizures - if the seizure trait was a result of a dominant allele.
Since none of the offspring have seizures, this indicates that the mice are likely heterozygous for a recessive seizure allele from each parent, and thus each mutation is compensating for the other in a complementary fashion, which is consistent with the mutations being in two different genes.
Which of the actions below is preformed by all cells to maintain homeostasis
A.They divide rapidly into many specialized cells
B.They use light energy to produce food molecules
C.They obtain energy from nutrients and remove waste products
D.They combine with another cell to increase in size
Answers
Answer: Option C.They obtain energy from nutrients and remove waste products
Explanation:
Homeostasis is the maintenance of a stable internal environment. Hence, cells allow processes within its cytoplasm that produces energy by degrading macromolecules like carbohydrates or lipids while expelling by-products such as gases, salts etc out of its cell membrane.
Answer:C
Explanation: they maintain a balance between the nutrients obtained from food and waste product generated during digestion of food.
You can make a cell extract that is able to perform glycolysis in vitro (in a test tube) if glucose is added. Arsenate is a potent inhibitor of triose phosphate dehydrogenase, the enzyme required for the 6th step in glycolysis. If both arsenate and glucose are added to the cell extract, what happens?
a. ATP levels decrease.
b. Both ATP and pyruvate levels decrease.
c. Both ATP and pyruvate levels increase.
d. ATP levels increase.
Answers
Answer: B
Explanation:
If glucose and arsenate are both added to the cell extract, at first glycolysis will start.
In step one of glycolysis, glucose is phosphorylated to glucose-6-phosphate catalyzed by hexokinase which splits the ATP into ADP, and the Pi is added on to the glucose.
In step 3 of glycolysis, fructose-6-phosphate is further phosphorylated to fructose 1,6-bisphosphate. The enzyme is phosphofructokinase. This again involves hydrolysis of another ATP molecule.
A total of two ATP is used.
Step 6 in glycolysis reaction which involves generation of 2 ATP's molecules is inhibited by arsenate. Hence all other glycolytic reaction would not take place. Therefore no ATP is produced and pyruvate is not produced also.
ATP level decreases because ATP is only used up but no ATP is gained from the inhibited pathway. Also the inhibition of the step 6 enzyme cut short the pathway and pyruvate the end product of the pathway is not formed.
Biennials produce only a small leaf crown in the first year of growth.
True
False
Answers
The answer is true because biennialsare mostly two.
True, biennials like carrots and beets produce a small crown of leaves in the first year of growth as they focus on vegetative development and storage of resources, preparing for flowering and seeding in their second year.
True. Biennials are plants that have a two-year lifecycle. In the first year, they focus on vegetative growth, developing a small crown of leaves and storing resources in their roots and other structures. This vegetative phase includes crops such as carrots, beets, and cabbage, where typically only a rosette of basal leaves is produced. During the second year, these plants utilize the stored resources to flower, produce fruit, and set seeds, followed by the end of their lifecycle. The cold period, or winter, plays a crucial role in vernalizing these plants, preparing them for the reproductive phase. Due to the significance of this vegetative growth in the first year, these plants do not need to be re-planted annually like annuals, but they do not survive beyond the second year or season like perennials.
n corn long ears (L) is dominant to short ears (l); glossy kernels (G) is dominant to opaque kernels (g) and high starch (S) is dominant to normal starch (s). A heterozygous plant with long ears, glossy kernels and high starch is crossed with a short eared plant with opaque kernels and normal starch. The following phenotypes are observed. Which genes are linked? (Gene name is based on the dominant trait.)
Answers
The question is incomplete. I have attached the complete question
Answer:
Ears and starch are linked, glossy isn't linked to either
Explanation:
The deviation from expected frequencies indicates that ears and starch genes are linked
A maternal effect can cause the offspring phenotype ratio to depart from that of classic Mendelian inheritance. In a species of snail, the dominant allele N codes for right-handed shell coiling and recessive allele n codes for left-handed shell coiling. If an Nn female with right-handed shell coiling males with an Nn male, what is the shell coiling phenotypic ratio of their offspring?
A. 4.right-handed coil:0, left-handed coil
B. 3. Right-handed coil: 1.left-handed coil
C. 3.bidirectional coil: 1.left-handed coil
D. 0.right-handed coil: 4.left-handed coil
Answers
Answer:
B. 3. Right-handed coil: 1.left-handed coil
Explanation:
Phenotype is what you see - the visible or observable expression of the results of genes, combined with the environmental influence on an organism's appearance or behavior.
When Nn is crossed with Nn, they will produce offspring with NN, Nn, Nn and nn genotype.
N - Dominant allele
n - recessive allele
The phenotypic ratio of this offspring is 3 right-handed coil and 1 left-handed coil.
Answer:
B. 3 right-handed coil: 1 left-handed coil
Explanation:
Female genotype : Nn
Male genotype: Nn
N: dominant allele (right handed coil)
n: recessive allele (left handed coil)
When we cross the Male (Nn) x Female (Nn) the phenotypic ratio will be right handed coil 3 : left handed coil 1.
Solution:
Nn x Nn
Four possible genotypes = NN, Nn, Nn, and nn
So, NN = right handed coil
Nn (2) = right handed coil
nn= left handed coil
Which of the following is a NOT mechanism by which the biofuels are made? a. by reacting animal fat or vegetable oil with waterb. by fermenting sugar components of starchy cropsc. by reacting animal fat or vegetable oil with alcohold. by treating animal fat or vegetable oil with hydrogen
Answers
Answer:
A. by reacting animal fat or vegetable oil with water.
D. by treating animal fat or vegetable oil with hydrogen
Explanation:
Water is insoluble in animal fat or vegetable oil so it makes a layer on the surface of oil. When animal fat or vegetable oil react with hydrogen, it produces solid oil called ghee not biofuels. Biofuel is produced when animal fat or vegetable oil react with alcohol.
Answer: Option A and D.
Explanation:
Biofuel is fuel that produced from either plants, algae or animals. There are four main types;
Ethanol that is made from corn.
Biodiesel made from animal fats and vegetable oil.
Green diesel made from algae and plant sources.
Biogas e.g methane made from animal manure and digested organic material.
Ethanol a type of biodiesel is made from fermentation of sugar components in starch crops.
Biodiesel is made from reaction between animals fats or vegetable oil with alcohol.
Describe the differences in the circulation pattern in the thoracic cavity for the right side v. the left side after the vessels leave the aortic arch in humans versus cats.
Answers
Explanation:
In the cat, the left normal carotid supply route falls off the brachiocephalic arteryIn the human, the left normal carotid supply route is an immediate branch off the aortic curveGeneral way of circulation through the cat:
Blood streams from the head through the foremost vena cava into the right chamber Blood from the back finish of the body moves through the back vena cava into the right chamber The right atrium, blood courses and contracts through the tricuspid valve into the right ventricleIt can be by the pulmonary artery, and into the lungsIt can be through the pneumonic veins into the left atrium and Through the mitral valve into the left ventricleFinal answer:
The circulation pattern in the thoracic cavity involves the aortic arch, which leads to various arteries and returns blood to the right side of the heart via the superior and inferior vena cava. Both humans and cats have a diaphragm that aids respiration crucial for systemic circulation. Differences between species may include the branching pattern and vessel size from the aortic arch.
Explanation:
The question asks us to describe the differences in the circulation pattern in the thoracic cavity for the right side versus the left side after the vessels leave the aortic arch in humans compared to cats. Humans, as well as cats, have a four-chambered heart with the systemic and pulmonary circuits. However, the branching pattern of the aorta and the arrangement of the major arteries can vary between species. In humans, the aorta arcs to the left and gives rise to the brachiocephalic artery, the left common carotid artery, and the left subclavian artery. After the blood flows through the systemic circuit, it returns to the right side of the heart via the superior vena cava and inferior vena cava. The diaphragm, present in all mammals, plays a crucial role in separating the thoracic and abdominal cavities and assisting with respiration which is pivotal for the systemic circulation. In contrast, cats might display variations in the branching pattern from the aortic arch and the relative size of the blood vessels, which may affect the blood flow dynamics on either side of the heart after leaving the aortic arch. For both humans and cats, the heart pumps deoxygenated blood from the right side through the pulmonary circuit to the lungs and oxygenated blood from the left side through the systemic circuit to the body.
In a population of 3000 fruit flies, 270 of them contain white eyes. White eye color is a recessive trait. What are the allelic frequencies for the red eye allele and white eye allele
Answers
p + q = 1
So p + 0.09 = 1
p = 0.91
Red eye: 0.91
White eye: 0.09
Answer:Red is 91%, white is 9%
Explanation:3000-270 =2730 red in percentage is 2730/3000 *100/1 =91%
White is 270/3000*100/1= 9%.
Allele frequency or gene frequency is the relative frequency of an allele expressed as a fraction or percentage.
ose the item in column 2 that best matches each item in column 1. axonemal microtubules nucleation treadmilling desmin EB1 Arp2/3 complex myosin subfragment 1 (S1) tropomodulin gelsolin microvilli lamin phosphoinositides MAPs plectin lamin A. anchors spectrin filaments to red blood cell membrane B. hemidesmosome C. crosslinks microtubules to intermediate filaments D. inhibits microtubule assembly E. forms scaffold underlying nuclear envelope F. stabilizes and organizes microtubules G. is in cilia and flagella H. formation of tubulin oligomers I. binds to microtubule plus ends J. prevents loss of subunits from filament minus end K. connects crisscrossing actin filaments into 3D networks L. binds to profilin and CapZ M. curves bacterial cells N. increase cell surface area O. decorates actin microfilaments P. nucleates microtubule branches Q. nucleates actin filament branches R. assembly and disassembly on opposite ends of the same filament S. breaks and caps actin filaments T. keeps muscle myofibrils in register U. desmosome
Answers
Answer:
axonemal microtubules: (G) is in cilia and flagellanucleation: (H) formation of tubulin oligomerstreadmilling: (R) assembly and disassembly on opposite ends of the same filamentdesmin: (T) keeps muscle myofibrils in registerEB1: (I) binds to microtubule plus endsARP2/3 complexes: (P) nucleates actin filament branches myosin subfragment 1 (S1): (O) decorates actin filamentstropomodulin: (J) prevents loss of subunits from filament minus ends gelsolin: (S) breaks and caps actin filamentsmicrovilli: (N) increase cell surfacefilamin: (K) connects crisscrossing actin filaments into 3D networksphosphoinositides: (L) binds to profilin and CapZMAPs: (F) stabilizes and organizes microtublesplectin: (C) crosslinks microtubles to intermediate filamentslamin: (E) forms scaffold underlying nuclear envelope.Explanation:
Axonemal microtuble: an axoneme is a microbule based cytoskeleton structure of a eukaryotic cilium or flagellum. It provides motility, and also gives support to the structure.nucleation: it is a process where interaction of several tubulin molecules lead to the formation of microtubules seed. This process occurs spontaneously in purified tubulin solutions.treadmilling: is a phenomenon, that especially occurs in actin filaments and microtubules,where continuous removal of actin monomers from pointed ends of filament occur and their reincorporation at the other endsdesmin: it is a specific protein found in muscle that integrates sarcolemma, Z disk and nuclear membrane in sarcomeres and regulates sarcomere architectureEB1: its a plus-end tracking protein that binds to plus end of microtules and modulates their dynamics and interactions with intercellular organelles.ARP2/3 complex: its a seven-subunit central actin nucleator that binds to the side of filament to promote new filament growth as a branch thus forming a complex actin network.myosin subfragmnet (S1): S1 is the head fragment of myosin globular heads, that contains the ATPase and actin-binding activities of myosin.tropomodulin: it is a protein that binds and caps the pointed ends(or minus ends) of actin filaments thereby regulating its length in muscle cells as well as non muscle cell.gelsolin: is an actin binding protein that binds and caps actin filament(responsible for assembling and disassembling of filaments)microvilli: found in the apical surface of epithelial cells or the stereocillia, spine like protrusions, that increase the cell surface area for diffusion upto 100 times.filamin: is an actin binding protein that holds two filaments at large angles.phosphoinositides: it regulates the activities of many actin binding proteins and have the tendency to bind with profilin and CapZ(its a capping protein that caps the barbed ends of actin filament)MAPs: microtubules associated protein or MAPs binds to and stabilizes microtubules lattice. MAPs have repeating domains that allow eash MAP molecule to associate with more than one tubulin dimer.plectin: a structural protein of the cytoskeleton that maintains tissue integritylamin: are fibrous protein of type V intermediate filaments that provide structural function and transcriptional regulation in the cell nucleus.1. True/False: DNA replication is considered "semi-conservative" because only half of the chromosomes are copied during S phase of interphase.
Group of answer choices
True
False
2.The monomers of DNA consist of
Group of answer choices
a. deoxyribose, phosphate, and a nitrogenous base
b. deoxyribose, phosphate, and hydrogen bonds
c. ribose, phosphate, and nitrogenous bases
d. ribose, phosphate, and hydrogen bonds
e. none of the above are correct
Answers
Answer:
1) The statement is false
2) Option A) deoxyribose, phosphate, and a nitrogenous base
Explanation:
1) In semi-conservative DNA replication:
- a parent double-stranded DNA splits in two.
- Each strand is then read by the enzyme, DNA polymerase, to ensure accurate synthesis of a new daughter strand
- the newly synthesized strand contains nucleotides that are complimentary to free nucleotides present in the parent strand.
Thus, because the parent strand is retained in the newly synthesized DNA, DNA replication is described as semi-conservative
2) DNA consists of several repeating units of polydeoxynucleotides where each is made up of a nitrogenous base (Adenine, thymine, cytosine or guanine) linked to
deoxyribose sugar by an N-glycosidic linkage, and then the sugar linked to a phosphate group by phosphoester bond.
Answer: 1. True.
Option A. deoxyribose, phosphate and nitrogenous base.
Explanation:
DNA replicates in semi conservative because when a parent helix replicates, it produce two daughter helixes which contain one of the two helical strand of the parent. During S phase in interphase, DNA is unwind by enzyme helicase and it is duplicated. The two single strands is use as a template to form identical strands.
2. The monomers of DNA are also called nucleotides. They include 5 carbon sugar(deoxyribose, Nitogenous base and phosphate group.
If Sylvia is described as trustworthy, altruistic and tender-minded, then she probably is fairly high on the Big Five superfactor known as extraversion neuroticism openness agreeableness conscientiousness
Answers
Answer:
Agreeableness
Explanation: cares about others and contributes to the happiness of others.
Answer:
non argumentive ad will most likely agree
Explanation:
Bacteria are grown in N15 over time, then switched to growth in N14 medium. After three divisions (60 min) what is the percentage of bacterial cells that have N15 labeled DNA? (A) 100% (B) 50% (C) 25% (D) 10% (E) 5%
Answers
Answer:
C. 25%
Explanation:
A bacterial cell gives rise to two daughter cells by one division. If a single bacterial cell enters into cell division, it will form a total of 2^3= 8 cells after three rounds of cell division. The two DNA strands of the parent DNA duplex having N15 will be present in two bacterial cells out of the total 8 cells. This would occur since the process of DNA replication forms a new DNA duplex with one parental strand and one new strand. So, after three round of cell division 25% cells will have DNA with N15 (2/8 x 100= 25%)
Final answer:
After three divisions in 14N medium starting with 15N-labeled DNA, due to semi-conservative replication, 25% of bacterial cells will contain DNA with 15N label per Meselson and Stahl's experiment.
Explanation:
The experiment by Meselson and Stahl with E. coli growing in heavy nitrogen (15N) and then switching to light nitrogen (14N) demonstrates the semi-conservative model of DNA replication. After the first division in 14N medium, each DNA molecule contains one strand with 15N and one with 14N, making it fifty percent 14N. By the principle of semi-conservative replication, after three divisions, there will be one more division than there are strains that have any 15N isotopes. Thus, one quarter of the bacterial cells should contain 15N since those are the ones that did not undergo the last division and still contain one original heavy strand. This makes the answer (C) 25%.
The autonomic nervous system can change the rate of the heart by: Group of answer choices beta1 adrenergic receptor activation. Increases in cAMP lead to increased amounts of Na influx (though If channels) and Ca2 influx of the pacemaker cells. This increases the frequency of APs of the pacemaker, and increases the rate of contraction. beta2 adrenergic receptor activation. Increases in cAMP lead to increased amounts of Na influx (though If channels) and Ca2 influx of the pacemaker cells. This increases the frequency of APs of the pacemaker, and increases the rate of contraction. muscarinic ACh receptors activation. Activation leads to reduced activity of Ca2 channels, and increasing activation of K channels, hyperpolarizing cells and reducing the rate of contraction. alpha1 adrenergic receptor activation.
Answers
Answer:
The correct answer is: β1 adrenergic receptor activation. Increased in cAMP lead to increased amounts of Na influx (though if channels) and calcium influx of pacemaker cells. This increases the frequency of APs of the pacemaker and increases the rate of contraction.
Explanation:
cAMP generated in response to β1 adrenergic receptors,results in excitation contraction coupling by initiating PKA and causing phosphorylation of L type Ca2+ channels and ryanodine receptors. This increases the concentration of intracellular Ca2+ ions in atrial cells, ventricular cells and AV pacemaker cells. This increase in firing rate. As a result rate of contraction increases.
1. List the cellular structures over which an action potential travels, starting at the dendrites and traveling to where neurotransmitter molecules are released.
Answers
Answer:
A neuron or a nerve cell is a unique cell, which performs an essential function of transmitting the nerve impulse in the form of action potential from one nerve cell to another. The point of communication between two neurons is known as a synaptic junction from where the transmission of a signal between presynaptic axon end and the postsynaptic dendrite occurs.
The synapse present in the dendrite receives a signal in the form of neurotransmitter from the presynaptic axon. This results in the formation of an action potential that gets transmitted towards the cell body of a neuron. From the cell body, the conduction of impulse takes place via a long tubule composition known as an axon, which constitutes the nodes of Ranvier and reaches the nerve endings or the axon terminals.
In the terminals, the change in potential results in the opening of the ions channels that discharge neurotransmitters in the form of acetylcholine, which further combines with the ligand-receptor situated on the next dendrite and thus repeats the process. Thus, the action potential travels from the dendrite to the cell body, and from there it travels to the terminals of the axon and eventually towards the ion channels and the ligand-receptor.
"As you examine the specimens (slides, whole specimens, etc.) in lab, determine where each species belongs on the phylogenetic tree based on the traits provided. List 5 additional traits you can add to the phylogeny. "
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To determine where each species belongs on the phylogenetic tree, examine their shared traits and make inferences about their evolutionary history. Five additional traits can be added to the phylogeny, such as anatomical structures, biochemical reactions, reproductive strategies, behaviors, and genetic sequences.
Explanation:When examining specimens in the lab, you can determine where each species belongs on the phylogenetic tree based on their shared traits. By comparing the traits of different species, you can identify common characteristics and make inferences about their evolutionary history.
For example, if two species share a trait that is not found in any other species, they are likely more closely related and would appear closer on the phylogenetic tree.
To add five additional traits to the phylogeny, you can consider various characteristics such as anatomical structures, biochemical reactions, reproductive strategies, behaviors, and genetic sequences. By including these traits, you can further refine the phylogenetic tree and understand the relationships between different species more comprehensively.
If Harry's negligent act injures Sally, and Susan, while attempting to come to Sally's aid breaks her arm in the process then, Harry is liable for the harm to __.
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Answer: Sally
If Harry's negligent act injures Sally, and Susan, while attempting to come to Sally's aid breaks her arm in the process then, Harry is liable for the harm to Sally.
Explanation:
As a mathematical expression, we can say the action of Harry H, is directly proportional to the injury to Sally S; while the mistake of Susan N is directly proportional to injury to Sally S.
So, if H = S and N = S, it is safe to say
H = N = S.
Thus, Harry negligence create room for Susan mistake, which eventually harmed Sally.
Explain why the underproduction of albumin by a cirrhotic liver contributes to excessive filtration, which leads to ascites. Refer again to the forces that determine net filtration pressure in the hepatic capillaries, and to the function of albumins
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Answer:
Albumin is produced mainly by the liver. Albumin is the major protein that is acts as the carrier protein for steroids, thyroid hormones and fatty acids in the blood.
Albumin stays in blood vessels as they are too large to move across the capillary walls, in blood vessels by contributing to osmotic pressure or colloid osmotic pressure that maintain water volume. So, the albumin is the protein that is involve primarily on the water reabsorption.
Your BFF is a red-headed, left-handed, blue-eyed girl. She finds out that you are taking Human Genetics and wants you to explain to her the genetic basis of her three traits. So, your task is to provide her with the most up-to-date explanation, to the extent that is possible. You have to do so in 400-500 words.
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Answer:
Red hair, left handed, and blue eyes are recessive traits.
Any trait has its two allele, dominant and recessive. Dominant trait allele if present with the recessive allele, then the phenotype will be of dominant allele.
Suppose A = dominant allele for brown color hair, a = recessive allele for red color hair.
As mentioned above red color hair is recessive in nature. for the recessive traits to show, the alleles should be recessive only.
Aa (brown hair) (parent 1) is crossed with Aa (parent 2) (brown hair). They both have gametes A and a.
Their children will have genotype AA, aa, and Aa. AA = brown color, aa = red color hair, and Aa= brown color hair.
Same case goes with the left hand and blue eyes traits.
So, the girl parents must be having recessive alleles for all the traits, and she must have inherited all the recessive allele only.
Explanation:
First of all, you should ask your friend if any of her parents also have these characteristics. In this you must explain the concept of heredity, which is the genetic phenomenon that allows us to be able to inherit genetic information from our parents, through DNA. That's because each of us has half the DNA of each of our parents, so we have characteristics in common with them.
After explaining heredity, you should talk about your friend's blue eyes. Blue eyes are characterized by a small amount of melanin, which is the brown pigment that colors our eyes, skin and hair. The more melanin, the darker the eyes. What determines the amount of melanin in the eyes is DNA. If your friend's parents have light eyes, it means they have little melanin and that trait was passed to her through DNA.
Red hair, on the other hand, is the result of a genetic mutation that is passed from parents to children through DNA. People with red hair have a mutation in the gene responsible for the production of melanin. This mutation causes the gene to produce very low levels of melanin for the hair, but it produces large amounts of pheomelanin which is a reddish pigment.
However being left-handed is not related to heredity. Many scientists believe that what determines whether a person is left-handed or not is a set of genes that choose the left or right side of the body as dominant. However, other scientists believe that this is actually linked to a matter of skill and custom.
Alleles of the gene that determines seed coat patterns in lentils can be organized in a dominance series: marbled > spotted = dotted (codominant alleles) > clear. A lentil plant homozygous for the marbled seed coat pattern allele was crossed to one homozygous for the spotted pattern allele. In another cross, a homozygous dotted lentil plant was crossed to one homozygous for clear. An F1 plant from the first cross was then mated to an F1 plant from the second cross. a. What phenotypes in what proportions are expected from this mating between the two F1 types? b. What are the expected phenotypes of the F1 plants from the two original parental crosses?
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Answer:
two types of cross are given,marbled and spottedhere, marbled is dominant (MM) ; spotted is co-dominant (Ss)
in F1 generation → MS (marbled spotted coat)
2. dotted and clear
here, dotted is co-dominant (Dd) ; clear is recessive (dd)
in F1 generation → Dd (clear background with dotted coat)
if MS × Dd , then, it will give,
25% spotted and dotted ( as both are co-dominant {Ss=Dd}) : 50% marbled (as marbled is dominant{ MM}) : 25% spotted ( as spotted {Dd} is co-dominant and clear {dd} is recessive)
a. 25% spotted dotted: 50% marbled: 25% spotted
expected phenotypes from F1 is →MS (marbled)- as marbled is the most dominant among all, it won't let spotted to be expressed.
and Dd - as clear is recessive only dotted will be expressed.
b.marbled and dotted.
Final answer:
The expected phenotypes from the mating of two F1 lentil plants from different crosses are 1:1 marbled to spotted/dotted, as marbled is dominant while spotted and dotted show codominance, and clear is recessive. The F1 offspring from the first cross will have a marbled phenotype, and from the second cross, a dotted phenotype due to codominance.
Explanation:
We are addressing complex inheritance patterns in lentils involving multiple alleles with a dominance series. To predict the offspring's phenotypes, we first need to understand Mendel's laws and how dominance affects the genotype-phenotype relationship. We also employ Punnett squares to visualize genetic crosses.
a. Expected Phenotypes in Offspring from F1 Mating
The F1 plants from the first cross (marbled × spotted) will all exhibit the marbled phenotype, as marbled is dominant over spotted. The F1 plants from the second cross (dotted × clear) will all show the dotted phenotype due to codominance between dotted and spotted alleles, and dominance over clear.
When these two F1 plants are crossed (marbled/dotted × dotted), the expected offspring phenotypes are as follows, assuming each allele has equal chance of passing on:
Marbled: Represents the dominant allele and is expected to show up in half the offspring if the marbled allele is present.
Spotted/Dotted (Codominance): Should appear in the other half of the offspring, as these are the alleles present in both F1 parents. The spotted and dotted patterns are codominant and will be expressed equally if one of each is inherited.
Clear: Will not be seen in the offspring, as it is recessive to all other alleles.
Thus, the phenotypic ratio of the offspring from the cross between the two F1 plants is expected to be 1:1 marbled to spotted/dotted with no clear phenotypes.
b. Expected F1 Phenotypes from Original Parental Crosses
The F1 phenotype resulting from a homozygous marbled crossed with homozygous spotted will be all marbled, following Mendelian inheritance patterns. In the cross of homozygous dotted with homozygous clear, all offspring will display a dotted pattern, indicative of codominance interaction over the clear allele.
Explain why macromolecules (food) and water are essential to life at the cellular level. Consider the equation Food + Water + x = Life, what additional factor (x) would you add? Explain why you consider that this factor is essential to life.
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Answer:
x = oxygen
Explanation:
Food and water is essential to life at cellular level because this water is responsible for making food with the help of CO2 during photosynthesis in plant and this food provides energy to the cell and support the life of plant cell. In an animal cell, H2O helps in releasing waste product from cell and provide a medium for reactions to occur.
Apart from food and water oxygen is also required to support life because oxygen is required by cells to oxidize food and release energy from food so this energy is used to perform metabolic function of cells that support cell life.
Final answer:
Macromolecules and water provide essential nutrients vital for cellular functions. Biological macromolecules are carbohydrates, lipids, proteins, and nucleic acids, synthesized through dehydration synthesis. The additional factor 'x' needed for life is oxygen, essential for cellular respiration and ATP production.
Explanation:
Macromolecules (food) and water are essential to life at the cellular level because they provide an organism with critical nutrients. These nutrients include four major classes of biological macromolecules: carbohydrates, lipids, proteins, and nucleic acids, each serving vital roles such as energy storage, structural support, cellular communication, and genetic information storage. Carbohydrates are primarily used for energy, lipids for storing energy and building cellular structures, proteins for numerous functions including tissue repair and enzyme catalysts, and nucleic acids for storing and transmitting genetic information. The synthesis of these macromolecules occurs through dehydration synthesis, a process where monomers link together by losing water molecules.To the equation Food + Water + x = Life, I would add the additional factor (x) as oxygen. Oxygen is crucial for cellular respiration, the process by which cells derive energy. Without oxygen, cells would not be able to produce ATP, the energy currency of the cell, which is vital for many cellular processes.The recent hominin fossil finds from Ileret, Kenya, negate the conventional view held since 1960 that H. habilisand H. erectusevolved one after the other. Instead, they lived side by side in eastern Africa for perhaps half a million years.A. TrueB. False
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Explanation:
The recent finding of the fossils which showed that Homo erectus and Homo habilis lived side by side in eastern Africa for perhaps half a million years challenged the conventional way that these two species evolved one after the other(H.habilis 1.44 million years old and H.erectus 1.55 million years old)
The fossils were found in Kenya and took years to prepare the specimens for study and to be sure of the identification of the species, the scientists saidUniversity of Utah geologists determined the dates of the fossils from volcanic ash depositsThe most recent Homo habilis that had been known was about the same age as the earliest Homo erectus, said Daniel Lieberman, a professor of biological anthropology at Harvard University, “Now we have extended the duration of the habilis species, and there’s no doubt that it overlaps considerably with erectus”The fact that the two hominid species lived together in the same lake basin for so long and remained separate species, Meave Leakey said in a statement from Nairobi, “suggests that they had their own ecological niche, thus avoiding direct competition”