Answer:
Both type A and B blood are Dominate types and O is recessive. So the mother would Have to be oo and couldn't pass on an A and The man cant pass on an A because Type B blood can only be found as BB or Bo. This is due to A and B both being Dominate the would end up as AB not just A/B
Answer:
Both type A and B blood are Dominate types and O is recessive.
Explanation:
If a person’s blood cortisol concentration is 10 µg/mL, and 10 mg of cortisol can be excreted from the body in a 24-hour period, what volume of blood plasma in mL was “cleared” during this amount of time?
Answer:
[tex]1000 ml[/tex] of blood plasma was “cleared”
Explanation:
Given,
Concentration of a person’s blood cortisol [tex]= 10[/tex]µg/mL
The amount of cortisol excreted from the body in a period of 24 hour [tex]= 10[/tex]mg
The volume of blood plasma cleared during this time period is equal to
amount of cortisol excreted from the body divided by Concentration of a person’s blood cortisol
The above definition could be represented as following-
volume of blood plasma cleared [tex]= \frac{A}{B} \\[/tex]
Where A represents amount of cortisol excreted from the body
while B represents the concentration of a person’s blood cortisol
Substituting the given values in above equation, we get -
[tex]= \frac{10 * 10^{-3}g}{10*10^{-6}\frac{g}{ml} } \\= 1000 ml[/tex]
[tex]1000 ml[/tex] of blood plasma was “cleared”
The volume of blood plasma cleared of cortisol in 24 hours is calculated to be approximately 1000 mL. Clearance rate is found to be roughly 0.694 mL/min.
To find the volume of blood plasma that was "cleared" of cortisol in a 24-hour period, we must use the given values.
The blood cortisol concentration is 10 µg/mL, and 10 mg (which is equal to 10,000 µg) of cortisol is excreted in 24 hours.
Calculate the total amount of cortisol excreted in 24 hours: 10 mg = 10,000 µg.Use the formula for clearance, where clearance (C) is the volume of plasma cleared per minute (mL/min):C = (Amount of Substance Excreted in 24 Hours) / (Concentration in Blood Plasma * Time in Minutes)
Substitute the known values into the formula:
C = 10,000 µg / (10 µg/mL x 1440 minutes)
C = 10,000 µg / 14,400 µg/mL
C ≈ 0.694 mL/min
Total Volume Cleared in 24 hours = Clearance Rate x Time
Total Volume Cleared in 24 hours = 0.694 mL/min x 1440 minutes
Total Volume Cleared in 24 hours ≈ 1000 mL
Hence, approximately 1000 mL of blood plasma was cleared of cortisol in this 24-hour period.
In bones, a layer of osteoblasts and osteoclasts line the interior of the bone (this cellular layer covers all trabeculae, is adjacent to the medullary cavity and lines the inside of osteons). This layer is called the endosteum. What is its role in bone?
Answer:
Endosteum plays an important role in bone repair, bone remodelling and appositional bone growth.
Explanation:
Endosteum consists of a soft and thin connective tissue that lines the cavity of femur and humerus. Some of the major functions of endosteum are as follows:
Bone remodelling: Endosteum can stimulate the bone resorption that leads to the formation of new bone from the outside.
Bone repair: Hematoma, at the time of bone injury causes the division of endosteal cells and helps in bone repair.
Appositional bone growth: Endosteum that line the osteoblast cell can secrete bone matrix and increases the bone diameter.
Protection from infection known as species resistance is a result of
Answer:
Protection from infection known as species resistance is a result of both the absence of necessary receptors and lack of suitable environment in the body.
What are the three types of body membranes?
The body contains two main types of tissue membranes: connective tissue and epithelial membranes. The three types of epithelial membranes are mucous membranes, serous membranes, and the cutaneous membrane (skin), each serving unique protective and functional roles in the body's operation.
Explanation:There are two broad categories of tissue membranes in the human body: connective tissue membranes and epithelial membranes. Among these, three specific types of epithelial membranes exist, playing critical roles in the body's function and structure. These are:
Mucous membranes: These are found lining the body cavities and hollow passageways that open to the external environment. They secrete mucus, which is essential for keeping these surfaces moist and protected.Serous membranes: These line body cavities that are closed to the exterior, such as the peritoneal, pleural, and pericardial cavities. Serous membranes secrete a lubricating fluid that reduces friction between the organs they envelop.Cutaneous membrane: More commonly known as the skin, it covers the body's exterior. The skin is a complex organ vital for protection, temperature regulation, and sensation.Lastly, connective tissue membranes include synovial membranes, which line the moveable joint cavities, facilitating smooth joint movements. Understanding these membranes is essential for comprehending how the body protects, separates, and ensures the smooth operation of its internal mechanisms.
Other nutrients will require a transporter in the membrane, as well as a concentration gradient. This process of ____________ is the process by which some monosaccharides are absorbed.
Answer:
Facilitated diffusion
Explanation:
Facilitated diffusion is the process of diffusion in which ion or some substances are transported across the membrane by a specific carrier membrane protein. The process does not require the input of energy and is driven by a concentration gradient.
The absorption of simple sugars in small intestine occurs through facilitated diffusion. The polar sugar molecules can not cross the nonpolar inner core of the membrane. Hence, the membrane proteins transport the sugars down the concentration gradients without any input of energy.
Birds, reptiles, and insects dispose of excess nitrogen atoms by excreting uric acid, rather than urea, sparing these animals the need for large quantities of water to dissolve the urea. Humans and other primates also make and excrete uric acid as the catabolic degradation product of purines. Uric acid has a very low water solubility. High concentrations in the blood can form insoluble crystals of sodium urate, forming painful stones in the kidneys and ureters. High concentrations in the blood can also lead to painful arthritic joint inflammation, a condition known as gout. The final step in the synthesis of uric acid is the oxidation of xanthine, catalyzed by the enzyme xanthine oxidase. This enzyme is a target for some drugs used to treat gout and a more serious condition of uric acid overproduction called Lesch-Nyhan syndrome. Draw the structures of xanthine and uric acid, the substrate and product of xanthine oxidase.
Answer:
Xanthine oxidase enzyme converts xanthine (substrate) to uric acid (product). Xanthine is known by its chemical or IUPAC name 3,7-dihydropurine-2,6-dione and chemical formula C₅H₄N₄O₂. It is a purine base present in a maximum number of human tissues and fluids and in other living organisms. It is converted from hypoxanthine by the action of the xanthine oxidoreductase enzyme. (see structure in the figure)
Uric acid is the compound created by the action of xanthine oxidoreductase. It is a heterocyclic compound of C, N, and O and H. The chemical formula of uric acid C₅H₄N₄O₃. It produced by the metabolic breaking of purine.
C₅H₄N₄O₂ (xanthine) + O₂ + H₂O → C₅H₄N₄O₃ (uric acid) + H₂O
Fred has chronic emphysema. Blood tests show that his pH is low but almost normal but his bicarbonate levels are elevated significantly. How can this be? What would urinalysis show?
Final answer:
Fred's chronic emphysema has led to respiratory acidosis with compensatory metabolic alkalosis, indicated by low but almost normal pH and high bicarbonate levels. His kidneys are attempting to balance the blood's pH, likely leading to alkaline urine shown in urinalysis.
Explanation:
Fred has chronic emphysema, a condition often associated with respiratory acidosis due to hypercapnia, which is an accumulation of CO₂ in the bloodstream. The elevation in bicarbonate levels suggests that his body is attempting to compensate for the acidosis. In a compensatory response, the kidneys retain bicarbonate to neutralize the excess acid. A urinalysis in Fred's case might show an elevated pH due to the body trying to excrete excess hydrogen ions (acidic ions) and retain more bicarbonate.
Chronic respiratory conditions like emphysema impair the lungs' ability to adequately expel CO₂, leading to an increase in the partial pressure of CO₂ (pCO₂) and a decrease in blood pH, initiating acidosis. However, over time, the kidneys attempt to regenerate and reabsorb bicarbonate ions as a compensatory mechanism, which is why Fred's bicarbonate levels are significantly elevated. Meanwhile, the urinalysis might further confirm this compensation by showing increased bicarbonate or alkaline urine, indicating the kidneys are working to correct the pH imbalance.
2. All of the following are true of crossing over, EXCEPT
crossing over occurs immediately after an egg is fertilized.
homologous chromosomes exchange genetic material.
crossing over results in unique combinations of alleles.
crossing over occurs during meiosis I.
crossing over occurs only in cells destined to become sperm or egg.
Answer:
Option (1)
Explanation:
Meiosis is also known as reduction division. Meiosis is the process of cell division by which a parent cell divides to give four daughter cells each with half number of chromosomes.
Crossing over may occur during meiosis I of the cell cycle. Crossing over will not occur in the cells immediately after the fertilization of egg and zygote restores the original number of chromosomes.
Thus, the correct answer is option (1).
The statement 'crossing over occurs immediately after an egg is fertilized' is NOT true.
Crossing over, also called recombination, occurs in meiosis, a type of reductional cell division by which one parental cell divides to produce four gametes (germinal cells) having half of the genetic material.Crossing over specifically occurs during the first phase of Meiosis I i.e., during Prophase I.This process (crossing over) increases the genetic variation of the resulting germinal cells (either spermatic or egg cells).In conclusion, the statement 'crossing over occurs immediately after an egg is fertilized' is NOT true.
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Can the genotype for a gray-bodied fly be determined? Why or why not? Describe all of the possible genotypes for a fly with that phenotype
Answer:
The genotype of a fly can be determined gray bodied is dominant over black body flies. The possible genotype of a fly is either YY or Yy.
Explanation:
The genotype of a gray bodied fly can be determined with the help of punnett square. The gray bodied fly is crossed with black bodied fly, the original genotype of the fly can be determined. This type of cross is known as test cross.
If the cross results in the formation of all gray bodied progeny, the fly is homozygous dominant with the genotype YY. The cross results in the mixture of progeny with gray and black body , then the genotype of a fly is heterozygous dominant Yy.
Thus, the genotype of a fly can be determined by the test cross. The fly may have genotype either YY or Yy.
A fellow student in the class is looking at a slide of onion root tip cells and is excited that he can see the chromosomes undergoing anaphase 1. How do you know that he is incorrect The chromosomes would not be coiled up during Anaphase 1 and therefore he would not be able to distinguish them Onion root tip cells do not undergo anaphase 1 He would need to use an electron microscope to see chromosomes Plants do not carry out mitosis
Answer:
The correct answer should be the second option, that is, Onion root tip cells do not undergo anaphase 1.
Anaphase I is a part of meiosis which occurs only in cells involved in reproduction. For example, gametes.
Onion root tip cells undergo mitosis.
The chromosome can be observed using compound microscope.
Thus, all other options are incorrect except second option.
Answer:
Explanation:
In the body of a multicellular organisms such as onion plant, there are two types of cells, namely somatic cells and reproductive cells. Somatic cells undergo mitosis and involved in vegetative growth whereas reproductive cells undergo meiosis that form gametes during sexual reproduction. Root tip of opinion consists of meristematic cells which are somatic cells which do not divide by meiosis. A fellow student recognises anaphase I in root tip cells is incorrect because anaphase I is a stage of meiosis I. Root tip cells only divide by mitosis. However, during anaphase I, a pair of chromosome separate which cannot be observed during mitosis in root tip cells.
List the parts of the Central Nervous System and the parts of the Peripheral Nervous System.
Answer:
Central nervous system includes brain and spinal cord. Peripheral nervous system includes autonomic nervous system and somatic nervous system.
Explanation:
Nervous system consists of electrically excitable cell called neurons and collection of nerves that transmit information in the body.
Nervous system is mainly divided into two parts - PNS (peripheral nervous system) and Central nervous system (CNS) . The CNS of the body consists of spinal cord and brain that regulates and coordinates all the functions of the body.
PNS consist of cranial and spinal nerves. This system is divided into Autonomic nervous system ( controls involuntary actions of body) and somatic nervous system ( control the voluntary actions of the body).
The Central Nervous System, composed of the brain and spinal cord, controls most functions of the body. The Peripheral Nervous System, made up of nerves and ganglia beyond the CNS, serves as a communication line connecting all parts of the body to the CNS. The PNS has two subdivisions: the autonomic and somatic nervous systems.
Explanation:The Central Nervous System (CNS), is composed of the brain and spinal cord, serving as the main data center within the body. It is essentially contained within the cranial and vertebral cavities of the body. In contrast, the Peripheral Nervous System (PNS) encompasses all bodily structures beyond the brain and spinal cord. It consists of nerves and ganglia, carrying messages from the CNS to the muscles, organs, and senses in the periphery of the body. It can be further divided into two major subdivisions: the autonomic and somatic nervous systems.
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Secrete a hormone that controls the blood calcium level
Answer:
The parathyroid glands secretes PTH hormone that controls the blood calcium level.
Explanation:
The PTH is the parathyroid hormone which is secreted by parathyroid gland. It maintains the blood calcium level whenever required in the body.
The hormone increases the amount of calcium by preventing calcium excretion from the urine. The PTH also stimulates the release of calcitriol. The calcitriol allows absorption of calcium from the diet.
3. Which of the following situations is least likely to lead to geographic isolation and allopatric speciation?
a. An eight-lane highway divides a population of snails with limited mobility.
b. A deep, narrow river divides a population of hummingbirds.
c. A deep, narrow river divides a population of white-footed mice (not known for their swimming ability).
d. A patch of land bisects a river and divides a population of minnows.
e. A mountain range divides a population of moths.
Answer:
b. A deep, narrow river divides a population of hummingbirds.
Explanation:
Hummingbirds can fly across the river. The deep and narrow river can not separate a population of hummingbirds into two patches since hummingbirds from one side of the river can fly and reach to the other side of the river.
Without any geographical isolation, hummingbirds can continue to interbreed among themselves. The continued interbreeding would not allow their reproductive isolation and thereby would prevent the allopatric speciation.
Identify the major weight bearing part of a vertebra
Answer:
Vertebral body
Explanation:
The vertebral body serves as weight-bearing part of a vertebra. The vertebral body is the thick and disc-shaped structure and forms the anterior portion of the vertebrae. A vertebral body has rough inferior and superior surfaces. Being disc-shaped, the vertebral bodies of subsequent vertebrae are stacked upon each other and thereby, serve to carry the weight.
The major weight bearing part of a vertebra is the vertebral body, which progressively increases in size and thickness going down the vertebral column. Lumbar vertebrae carry the greatest amount of body weight and have specific characteristics to support this.
Explanation:The major weight bearing part of a vertebra is the vertebral body. The vertebral bodies progressively increase in size and thickness going down the vertebral column, with the lumbar vertebrae supporting the greatest amount of body weight due to their large size and thickness. They have short transverse processes and a short, blunt spinous process that projects posteriorly.
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A period of sustained skeletal muscle contraction in which individual twitches cannot be detected is called
Complete tetanus represents a period of sustained skeletal muscle contraction without detectable individual twitches, occurring at high frequencies of motor neuron signals that fuse contractions into a continuous state.
A period of sustained skeletal muscle contraction in which individual twitches cannot be detected is called complete tetanus. This occurs when the frequency of motor neuron signaling is so high that muscle fibers respond with continuous contractions. A single twitch doesn't produce significant muscle activity in a living body, rather a series of action potentials is necessary for a graded muscle response. During incomplete tetanus, the muscle experiences quick cycles of contraction with short relaxation phases. However, in complete tetanus, the stimulus frequency is high enough that the relaxation phase is eliminated, and the contractions fuse into one sustained contraction.
Saddle joints have concave and convex surfaces. Name the two bones of the hand that articulate to form a saddle joint. A. The trapezium of the carpal bone and the thumb's metacarpal. B. The scaphoid of the index finger and the triquetrum of the middle finger. C. The trapezium of the ring finger and the capitate of the fourth finger. D. The scaphoid of the middle finger and lunate of the index finger.
Answer:
Option (A).
Explanation:
Saddle joint is a type of synovial joint and also known as sellar joint. These joints are present in thumb, shoulder and inner ear.
The bone has one part concave inward and other part is convex outward. This bone can articulate with thumb's metacarpal and trapezium of the carpal bone. Saddle joints show abduction and adduction movements.
Thus, the correct answer is option (A).
The saddle joint in the hand is the first carpometacarpal joint, located between the trapezium carpal bone and the first metacarpal bone of the thumb, allowing for the distinct opposability of the thumb.
Explanation:The saddle joints in the human body are a type of synovial joint where the articulating bone surfaces each resemble a saddle, which is concave in one direction and convex in the other. This design allows for a wide range of movements.
Specifically, in the hand, the saddle joint is found at the first carpometacarpal joint, which is between the trapezium, one of the carpal bones, and the first metacarpal bone that forms the base of the thumb. This versatile joint enables the thumb to move across two planes, allowing for the distinctive human ability to oppose the thumb against the other fingers.
Based on this description, the correct answer to which two bones articulate to form a saddle joint in the hand is: A. The trapezium of the carpal bone and the thumb's metacarpal.
3. During meiosis,
cells divide once.
cells double in size but do not divide.
cells do not divide.
cells divide twice.
cells do not divide but lose half their chromosomes.
Answer:
Cells divide twice
Explanation:
Meiosis may be defined as the process of cell division in which a single cell divides to give four daughter cells and each daughter cells have half number of chromosomes.
Meiosis is also known as reduction division as the chromosome number reduces upto half in the progeny cells as compared with the parent cell. The cells divide twice in meiosis and give rise to four daughter cells.
Thus, the correct answer is option (4).
Meiosis is a cellular division that results in four daughter cells with half the number of chromosomes of the parent cell. Specifically, cells in meiosis divide twice across two phases: Meiosis I and Meiosis II.
Explanation:During meiosis, cells divide twice. Meiosis is a type of cellular division that results in four daughter cells, each with half the number of chromosomes of the parent cell. It consists of two successive divisions: Meiosis I and Meiosis II. During Meiosis I, homologous chromosomes pair up and then separate, resulting in two cells. In Meiosis II, these cells divide again, resulting in four cells with half the number of chromosomes as the original cell.
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In the small intestines, a process called ____________ aids in mixing. During this process the contents ____________ move along the tract.
Answer: Segmentation;do not
Explanation:
The food that we ingest through mouth and the food moves from mouth to esophagus and then from esophagus to intestine followed by stomach.
Segmentation can be defined as the contraction that occurs in large and small intestine. In this process contraction moves the chyme in both directions which allows a better of the secretions with the chyme. This process do not takes place along the tract.
The correct answers are segmentation and do not.
How does each of the following molecule pass through the plasma membrane in the described scenarios?
A. Both simple diffusion and facilitated diffusion
B. facilitated diffusion
C. active transport
D. simple diffusion
selectABCD 1.
Molecular oxygen (O2) and carbon dioxide (CO2) traveling down their concentration gradient.
selectABCD 2.
inorganic ions traveling down their concentration gradient
selectABCD 3.
glucose traveling against its concentration gradient
selectABCD 4.
water molecules (H2O) traveling down their concentration gradient
Answer:
1. simple diffusion.
2. facilitated diffusion.
3. active transport.
4. both simple and facilitated diffusion.
Explanation:
1. Simple diffusion can be defined as transport of small, non-polar, uncharged molecules, such as urea, gases, water through the lipid bilayer without the help of transport proteins.
As molecular oxygen and carbon dioxide are gases, these molecules would be transported by simple diffusion.
Thus, statement 1 correctly matches with option (D).
2. Facilitated diffusion involves transport of polar, charged, and large molecules, such as ions, glucose, and amino acids with the help of membrane proteins, known as transport proteins.
Passive transport is an example of of facilitated diffusion, which includes transport of molecules from higher to lower concentration (along their concentration gradient) without use of energy.
As ions are travelling down their concentration gradient, it is an example of facilitated transport.
Thus, statement 2 correctly matches with option (B).
3. Active transport is a type of facilitated diffusion that involves transport of molecules against their concentration (from lower to higher) with the help of energy. As glucose is travelling against its concentration, it is an example of active transport.
Thus, statement 3 correctly matches with option (C).
4. Water molecules are small, charged molecules that can be transported by both simple diffusion through lipid bilayer or facilitated diffusion through transport proteins, aquaporins.
Thus, statement 4 correctly matches with option (A).
Final answer:
The transport methods for various molecules across the plasma membrane are specific to their nature and include simple diffusion for O2 and CO2 (D), facilitated diffusion for inorganic ions (B), active transport for glucose against its concentration gradient (C), and osmosis/simple diffusion or through aquaporins for water (A).
Explanation:
Transport across the plasma membrane includes various mechanisms such as simple diffusion, facilitated diffusion, active transport, and osmosis. These mechanisms depend on the nature of the molecule being transported and the direction relative to the concentration gradient.
Molecular oxygen (O2) and carbon dioxide (CO2) traveling down their concentration gradient can move through the plasma membrane by simple diffusion (Choice D).
Inorganic ions traveling down their concentration gradient typically require facilitated diffusion through channel or carrier proteins (Choice B).
Glucose, when traveling against its concentration gradient, is transported by active transport mechanisms that require energy (Choice C).
Water molecules (H2O) traveling down their concentration gradient generally pass through the membrane by means of osmosis, which can occur via simple diffusion or through aquaporins (Choice A).
Frutose’s molecular weight is 180 g/mole. A can of coke cola (330ml) contains 35 g fructose. The molar concentration fructose in coke cola is
Answer:
The molar concentration of [tex]35[/tex] g fructose in [tex]330 ml[/tex] solution is [tex]0.589[/tex]moles/litre
Explanation:
Given -
Molecular weight of fructose [tex]= 180 \frac{g}{mole}[/tex]
Can of coka coal has fructose [tex]= 35[/tex] gram
Volume of can of coke [tex]= 330[/tex] ml
Number of moles in [tex]35[/tex] grams of Fructose
[tex] = \frac{35}{180} \\= 0.194[/tex]
Molar concentration
[tex] = \frac{0.194}{0.33l} \\= 0.589\\[/tex] moles/litre
Hence, the molar concentration of [tex]35[/tex] g fructose in [tex]330 ml[/tex] solution is [tex]0.589[/tex]moles/litre
Final answer:
The molar concentration of fructose in a can of Coca-Cola is determined by dividing the mass of fructose by its molecular weight to find the moles, and then dividing by the volume in liters to find the concentration, which is 0.589 mol/L.
Explanation:
The molar concentration of fructose in coke cola can be calculated using the formula for molarity, which is moles of solute per liter of solution (mol/L). The moles of fructose are found by dividing the mass of fructose by its molecular weight. In this case, 35 g of fructose is divided by 180 g/mole to find the moles of fructose. To find the volume in liters, we convert 330 mL to liters by dividing by 1000, since there are 1000 mL in a liter.
We calculate the moles of fructose:
moles of fructose = 35 g / 180 g/mole = 0.1944 moles.
Then we convert the volume of Coca-Cola to liters:
volume in liters = 330 mL / 1000 = 0.33 L.
Finally, the molar concentration of fructose in Coca-Cola is:
molar concentration = 0.1944 moles / 0.33 L = 0.589 mol/L.
What are the possible reason for no colonies to grow on agarose plate?
How does inefficient ligation result in no colony growth on agarose plate?
Answer: Possible Reasons- Low concentration of DNA fragments, reaction volume more than 10μl, unequal concentration of control and experiment for transformation.
Incorrect ligation and poor ligase activity result in no colony growth on agarose plate.
Explanation:
Low concentration of DNA will not allow colonies to grow on agarose plate. Reaction volume must be less than 10μl. Different concentration of control and experiment will also affect growth of colonies.
Inefficient ligation result in no colony growth on agarose plate. This could be due to incorrect ligation between blunt and sticky ends. T4 DNA ligase joins the ends of DNA. Ligase poor activity will not allow colonies to grow on agarose plate.
A cell in the testes of a male bald eagle is getting ready to begin meiosis. The diploid number for this bird species is 66. How many replicated chromosomes are there in this testicular cell at metaphase I? How many sister chromatids are there in metaphase I? In each daughter cell, how many sister chromatids are there in prophase II? How many chromosomes are there in each sperm cell?
Final answer:
At metaphase I in eagle testicular cell, there are 66 replicated chromosomes and 132 sister chromatids. During prophase II, each daughter cell will have 33 sister chromatids. Upon completion of meiosis II, each sperm cell will have 33 chromosomes.
Explanation:
When a cell in the testes of a male bald eagle, which has a diploid number of 66, is preparing to undergo meiosis, it will first replicate its DNA during interphase. At metaphase I, there would be 66 replicated chromosomes, because the chromosomes have paired up but not yet separated into daughter cells. Since each replicated chromosome consists of two sister chromatids joined at the centromere, there would be a total of 132 sister chromatids at metaphase I.
In meiosis II, specifically prophase II, each daughter cell from meiosis I will have 33 sister chromatids, since the cells are now haploid. Finally, when meiosis II is completed, each sperm cell, which is a gamete, will have 33 chromosomes (haploid).
Defecation
a. is initiated by distension of the wall of the colon.
b. depends on voluntary control of the external sphincter of the anus.
c. depends on the motor impulses from sympathetic pathways.
d. decreases pressure in the rectum.
Answer:
A. is initiated by distension of the wall of the colon
Explanation:
not sure
SOME ONE PLS HELP ME THIS IS URGENT
According to the image below, which season is about to begin in the southern hemisphere?
Image shows Earth tilted 23.5 degrees away from the sun.
Summer
Autumn
Winter
Spring
Answer:
Pretty sure it's Winter.
Explanation:
Tilted away from the Sun.
Answer:
Autumn
Explanation:
Select the correct statement. a) Some drugs readily pass through the skin. b) Drugs taken by inhalation are usually slower acting and less potent than drugs taken orally. c) Usually more drug is needed when given intravenously than when administered orally to achieve an effect. d) The more times a drug is injected in the same vein, the easier the injection becomes.
Answer:
for 1 drugs taken by inhalation act just as fast as drugs taken intravenously
your answer is a some drugs readily pass throught the skin brainliest?
Explanation:
Answer:
a) Some drugs readily pass through the skin.
Explanation:
The skin has the ability to absorb various substances. Some penetrate the most superficial layers and others the deepest of the organ and may even enter the bloodstream. Penetration of mediations, for example, is not always possible because of natural barriers - such as the stratum corneum (the outermost layer of the skin, composed of keratin). However, some medications pass quickly through the skin.
Parasympathetic stimulation
a. causes chief cells to stop the production of H+ and Cl-.
b. causes the gastric glands to increase production of gastric juice.
c. leads to a decrease in the secretion of gastrin.
d. causes a decrease in salivation.
Answer: b. causes the gastric glands to increase production of gastric juice.
Explanation:
The parasympathetic nerves are part of the autonomous nervous system. The parasympathetic predominates mainly during "passive responses" of the organism, such as satiety, rest and digestion. Its functions are: Promotes energy conservation, Reduces heart rate, Promotes glandular secretion, Protects the retina from excess light, Promotes emptying of cavities, Promotes rest and repair, Physiologically antagonizes sympathetic nervous system.
Answer:
b. causes the gastric glands to increase production of gastric juice
Explanation:
Parasympathetic stimulation is a function that is carried out by the Parasympathetic Nervous System which is a part of the Autonomic Nervous System.
Parasympathetic stimulation results in the process of digestion, production of saliva, production of urine, Production of tears, decrease in the heart rate and blood pressure.
Parasympathetic stimulation is involves in the process of digestion because after the injection of food , when the body is at rest, Parasympathetic stimulation increase the rate of the digestion process by, stimulating the gastric glands so that they can produce more gastric juice. This gastric juice contains enzymes and some other substances such as
a. Pepsin for the digestion of protein.
b. Renin
c. Hydrochloric acid (HCl) in other to deactivate any bacteria found food present in the stomach.
d. Mucus which helps to protect the lining of the digestive system.
Does anyone know how asymmetric liposome works? asymmetric liposome
Answer:
Asymmetric liposomes have different lipids in outer and inner leaflets, which would greatly increase the flexibility of vesicle in drug delivery systems. It has been well known that the phospholipid distribution in natural membranes is asymmetric. For example, phosphatidyl tcholine and sphingomyelin concentrate at the outer leaflet whereas phosphatidylethanolamine, phosphatidylinositol and phosphatidylserine are mainly localized in the inner leaflet. Typically, Lipids are self-assembled symmetrically in artificial liposomes regardless of the preparation methods. As drug delivery carriers, asymmetric liposomes with advanced functions are appealing candidates for targeted accumulation and controlled drug release. Their outer and inner leaflet could be manipulated depending on the nature of encapsulated drug molecules. For example, asymmetric liposomes help deliver negatively charged siRNA to target organs by having positively charged inner layer that encapsulates siRNA with high efficiency, and negatively charged outer surface prevents nonspecific uptake of the asymmetric liposomes. The unique tunability of asymmetric liposomes opens a wide door for multi-site functionalization, resulting in highly engineered liposomes as advanced drug delivery vesicles
34. Which of the following types of nucleic acid is NOT a pyrimidine, a type of biomolecule, which makes up an organism’s genome?
a. Adenine
b. Uracil
c. Cytosine
d. Thyomine
Answer: a. Adenine
Explanation:
Two types of chemically similar nucleic acids, DNA (deoxyribonucleic acid) and RNA (ribonucleic acid), are the principal information-carrying molecules of the cell. The bases adenine, guanine, and cytosine are detected in both DNA and RNA while thymine is observed only in DNA, and uracil is located only in RNA. When it comes to defining the nucleic acid-based in the nature of its nitrogen- and carbon-containing ring structure we have the variation that answers the question. While b. Uracil, c. Cytosine, and d. Thymine only contains a single ring, which we call pyrimidines, the biomolecule a. Adenine contains a pair of fused rings that we call purines, the right answer for the question.
A phenotypically normal prospective couple seeks genetic counseling because the man knows that he has a translocation of a portion of his chromosome 4, which has been exchanged with a portion of his chromosome 12. Although his translocation is balanced, he and his wife want to know the probability that his sperm will be abnormal. What is your prognosis regarding his sperm?
Final answer:
A translocation between chromosome 4 and chromosome 12 in the man's genetic makeup can affect his sperm by increasing the likelihood of producing abnormal sperm. Consulting with a genetic counselor is recommended to assess the specific risks and make informed decisions.
Explanation:
A translocation occurs when a portion of one chromosome breaks off and attaches to a different chromosome. In the case of the prospective couple, the man has a balanced translocation between chromosome 4 and chromosome 12. This means that the genetic material is exchanged between these two chromosomes without any material being lost or gained.
However, translocations can still affect fertility because they can cause problems during meiosis, the process of cell division that produces sperm and eggs. During meiosis, chromosomes need to pair up and separate properly to produce healthy gametes. In the case of a translocation, the rearrangement of genetic material can disrupt the pairing of chromosomes, leading to a higher likelihood of producing abnormal sperm or eggs.
Therefore, while the translocation itself may not cause any noticeable physical abnormalities in the man, there is a higher chance that his sperm might be abnormal due to the rearrangement of chromosomes. It is important for the couple to consult with a genetic counselor to assess the specific risks and make informed decisions about their reproductive options.
How many ATP and NADPH molecules are required in the C3 pathway to make one six-carbon sugar? If the synthesis of a molecule of ATP were to require four protons, would you expect that these relative requirements for ATP and NADPH could be met by noncyclic photophosphorylation in the absence of cyclic photophosphorylation?
Answer:
The correct answer is 18 ATP and 12 NADPH and noncyclic photophosphorylation can produce ATP molecule in the absence of cyclic photophosphorylation.
Explanation:
ATP and NADPH molecules are synthesized during light reaction of the photosynthesis which is utilized during the reactions of the dark phases. Dark reaction or Calvin cycle (C3 cycle) is the cyclic pathway of producing glucose triose phosphates (3C) from carbon dioxide and water. This reaction proceeds into 3 phases: carboxylation, reduction and regeneration.
First ATP and NADPH are utilized during the reduction step in the reduction of 3-phosphoglycerate to glucose-3-phosphate by the transfer of phosphate group from 6 ATP to 3-phosphoglycerate and 6 NADPH reduction as it donates an electron. Regeneration step also uses 3 ATP in conversion of G3P to RUBP molecules.
A total of 9 ATP and 6 NADPH are utilized in producing 3C G3P molecule, So, to produce 6C sugar molecule 18 ATP AND 12 NADPH are used.
During chemiosmosis synthesis of ATP, 4 protons produce 1 molecule of ATP which could be able to generate 3 molecules of ATP for each pair of NADPH. so, noncyclic photophosphorylation or Z-scheme will be able to produce ATP in the absence of cyclic photophosphorylation.
Thus, 18 ATP and 12 NADPH and noncyclic photophosphorylation can produce ATP molecule in the absence of cyclic photophosphorylation are the correct answer.