Work in groups of 4. At Jo, preparation of pots will occur in the potting shed and will be done section-by-section with your demonstrator. 1. Form a group and give yourselves a name - it has to be unique so that you can locate your plants at all times. 2. Collect two pot labels. Write your group name and the species you are growing on the labels. Leave enough space to write in the treatments next week. 3. Collect two pots. Fill each one with white sand and stick in the label. Add water to the pots until there is a trickle from the base: at this point the sand is holding as the largest volume of water it can under natural circumstances, which is called its 'field capacity 4. Collect 10 seeds. Check the seed packet the depth at which the seeds should be sown, and then sow five seeds in each pot. Cover them with sand and water lightly. 5. Place your pots on the bench in the glasshouse, with their labels in them. The seeds will germinate over the coming week. Step 2, you will thin the seedlings down to 3 per pot and then apply nutrients to them. One pot will receive nitrogen (N), phosphorus (P), potassium (K) and micronutrients, and the other will receive only N, K and micronutrients i.e. no phosphorus. Stage 2 Application of treatments (30 min) Check that the seeds in your pots have germinated. Make notes about each plant in your lab journal, and take photos. Look carefully for signs of fungal disease. Thin the plants out to two per pot, and apply the fertilizer treatments as follows 1. Identify the two largest and healthiest seedlings; these will remain in the pot. Gently remove all the other seedlings. 2. Weigh out the required amount of each fertilizer using a balance. Remember you will need two lots of N, K and micronutrients and only one lot of P. The micronutrients may be supplied as a liquid, so follow the instructions available for these. 3. Choose one pot to be the control; the other will be the treatment' pot. Label the pots accordingly. The control will receive P, N, K and micronutrients, while the treatment plants will receive only N, K and micronutrients i.e. no P. 4. Water the pots until they are at field capacity before you add the nutrients. Wait for water to stop running out of the pots before proceeding. 5. Sprinkle the nutrients as evenly as you can across the surface of the pot, and then water gently. 6. Return your pots to the glasshouse. Stage 3 Observations of growth (15 min) Observe your plants to see how they are progressing. Record your observations notes on features that might be symptoms of disease or nutrient deficiency - like leaf colour change, differences in size and texture. Take photographs to use in your lab report (How will you include a scale bar?). 6.48 .45 1 Stage 4 Experiment Harvest and Data Analysis (60 min) Collect your group's pots and observe your plants carefully. Record detailed observations of leaf colour and size. Take photographs. 7. Collect and label two paper bags: include your group's name, species and whether its contents are the control and or the treatment. 8. Following the instructions of your demonstrator, gently turn the plants and soil in the control pot out onto a mesh grid. Do not separate the shoot and root systems. This is important - we want to keep the plants intact and have as much of the root system as possible. Gently wash as much sand from the roots as possible. When done, wrap the whole plants loosely in paper towel and place them in the correct paper bag. 9. Repeat step 2 with the treatment plant. 10. Bring the plants in their bags to the lab for weighing. 11. Determine the fresh weights for the whole plants from the control pot. (Total Fresh Weight TFW). Blot as much water as possible from the plants. Place weigh boat on the balance, and use the Tare button to reset to zero. Then weigh each plant on the balance + tfw It 0. 3.76 Record your results.

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Answer 1

Leave enough space to write in the treatments next week. Collect two pots and fill each one with white sand and stick in the label.

Collect two pot labels and write your group name and the species you are growing on the labels.  Add water to the pots until there is a trickle from the base. At this point, the sand is holding as the largest volume of water it can under natural circumstances, which is called its 'field capacity.

Blot as much water as possible from the plants. Place weigh boat on the balance and use the Tare button to reset to zero. Then weigh each plant on the balance. The result should be recorded.

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Related Questions

Please answer all of the following True or False Questions
• the number of chromosomes does not vary during mitosis
• Poly A-directed cleavage and polyadenylation do not constitute a way to produce different mRNA isoforms
• Balancer chromosome in flies are useful because they prevent the production of recombinant progeny
• Recombination can only occur in cells undergoing meiosis

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The answers to the given True/False questions are:

True: The number of chromosomes does not vary during mitosis.

False: Poly A-directed cleavage and polyadenylation constitute a way to produce different mRNA isoforms.

True: Balancer chromosome in flies are useful because they prevent the production of recombinant progeny.

False: Recombination can occur in cells undergoing both meiosis and mitosis.

The number of chromosomes does not vary during mitosis. The number of chromosomes remains the same during mitosis. Each daughter cell will contain the same number of chromosomes as the parent cell.

Poly A-directed cleavage and polyadenylation constitute a way to produce different mRNA isoforms .Poly A-directed cleavage and polyadenylation do constitute a way to produce different mRNA isoforms. The poly(A) tail of an mRNA molecule plays an important role in mRNA stability, export from the nucleus, and translation.

Balancer chromosomes in flies are useful because they prevent the production of recombinant progeny. Balancer chromosomes are useful in flies as they prevent the production of recombinant progeny and help maintain specific mutations within a population of flies.

Recombination can occur in cells undergoing both meiosis and mitosis. Recombination can occur in both meiosis and mitosis. It can result in a new combination of genes on a chromosome. In meiosis, recombination between homologous chromosomes is a source of genetic diversity, and in mitosis, it can lead to cancer.

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14. Explain how Snyder agar is both a selective and differential medium: 15. a. What is one way bacteria use sugar to produce dental caries? b. What type of growth environment do bacteria need to produce acid? What type of metabolism are they doing to produce acid?

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14. Snyder agar is both a selective and differential medium because it has a low pH level which selects for the growth of oral bacteria like streptococci that thrive in this environment.

15a. Bacteria use sugar to produce dental caries through a process called glycolysis, which involves the breakdown of sugar molecules into pyruvate.

15b. The type of growth environment bacteria need to produce acid in an acidic growth environment.

15c. The type of metabolism to produce acid is known as anaerobic metabolism.

Snyder agar also contains a pH indicator which enables the differentiation of lactate fermenters (which produce acids that lower the pH and change the agar from green to yellow) from non-lactate fermenters that do not change the color of the agar.

Bacteria use sugar to produce dental caries through a process called glycolysis, which involves the breakdown of sugar molecules into pyruvate. This metabolic pathway yields ATP, which is an energy source for the bacteria and also produces acid as a by-product. The acid produced lowers the pH of the surrounding environment, which leads to the demineralization of tooth enamel and the formation of cavities.

Bacteria need an acidic growth environment to produce acid. They use the sugar from their surroundings and metabolize it through the process of fermentation to produce acid. This type of metabolism is known as anaerobic metabolism since it does not require oxygen to produce energy. The acid produced by bacteria can also create an acidic environment in which the bacteria can grow and thrive.

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1. This slide shows tissue from the urinary system. What structure is this tissue taken from? 2. What tissue type is found at the arrow? 1. What is the name of the cells found at the tip of the arro

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The slide represents the tissue from the urinary system. The structure from where the tissue is taken can be any of the organs of the urinary system such as the kidneys, ureters, bladder, or urethra. The urinary system works to eliminate waste from the body, maintain electrolyte and fluid balance, and regulate blood pressure.

1. The slide represents the tissue from the urinary system. The structure from where the tissue is taken can be any of the organs of the urinary system such as the kidneys, ureters, bladder, or urethra. The urinary system works to eliminate waste from the body, maintain electrolyte and fluid balance, and regulate blood pressure.

2. The tissue type found at the arrow is transitional epithelial tissue. Transitional epithelium is a type of tissue found in the urinary system. It is made up of layers of cells that can expand and contract as needed to accommodate changes in the volume of urine within the urinary system.

The tissue has a unique appearance due to the way the cells are shaped. They are rounded when the bladder is empty, and flattened when it is full. This tissue lines the ureters, bladder, and urethra.1. The name of the cells found at the tip of the arrow is transitional epithelial cells. They are specialized cells that make up the transitional epithelial tissue found in the urinary system. The cells are able to stretch and contract as the bladder fills and empties. They have a unique shape, which is why they are named "transitional." The shape of the cells changes depending on the degree of stretch of the organ they are lining. When the bladder is empty, the cells are rounded, and when it is full, they are flattened.

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____ of S. aureus binds to host cell IgG via Fc receptors.
a. Protein A b. Leukocidin c. Enterotoxin d. T-cell superantigen

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Protein A of S. aureus binds to host cell IgG via Fc receptors. The correct answer is a.  

Protein A is a virulence factor produced by Staphylococcus aureus, a bacterium responsible for various infections in humans. Protein A has a unique ability to bind to the Fc region of immunoglobulin G (IgG) antibodies. IgG antibodies play a crucial role in the immune response by binding to pathogens and marking them for destruction by immune cells.

By binding to host cell IgG, Protein A interferes with the normal immune response. It can inhibit opsonization, which is the process of coating pathogens with antibodies to enhance their recognition and elimination by immune cells in Human Immunodeficiency Virus. Instead, Protein A binds to the Fc region of IgG, preventing its interaction with Fc receptors on immune cells.

This binding allows S. aureus to evade immune detection and phagocytosis, which is the engulfment and destruction of pathogens by immune cells. By interacting with IgG via Fc receptors, Protein A contributes to the pathogenicity and persistence of S. aureus infections in the host.

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In Mendel's peas, yellow seeds are dominant to green. A pure-breeding yellow plant is crossed with a pure-breeding green plant. All of the offspring are yellow. If one of these yellow offspring is self-fertilized, what will be the expected proportion of plants with green seeds in the next generation?

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All the offspring in the next generation will possess yellow seeds, and there will be no plants with green seeds. All plants will have yellow seeds due to the dominance of the yellow allele.

Mendel's pea experiments led to the formulation of the theory of inheritance, which states that each parent contributes one allele to their offspring. An allele represents a version of a gene, such as "green" or "yellow." In this context, yellow seeds are dominant, while green seeds are recessive.

When a pure-breeding yellow pea plant is crossed with a pure-breeding green pea plant, all offspring will exhibit yellow seeds in accordance with Mendel's laws of inheritance.

If one of the offspring from the aforementioned cross is self-fertilized, the next generation will inherit two alleles for seed color, one from each parent. However, since the yellow allele is dominant and the green allele is recessive, the presence of just one yellow allele will result in the expression of the yellow seed phenotype. Therefore, all the offspring in the next generation will possess yellow seeds, and there will be no plants with green seeds.

In conclusion, the expected proportion of plants with green seeds in the subsequent generation is zero.

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Question 6 O pts Why do you think COVID is more severe in the elderly with respect to the respiratory system and lymphatic system? Look at sections 24.11 and 23.7 in the text book and use the informat

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Overall, COVID-19 is more severe in the elderly due to age-related changes in the respiratory and immune systems that can exacerbate symptoms and increase the risk of complications.

As COVID-19 enters the body, it can infect various cells, including respiratory and immune cells, by using ACE2 receptors that are present on their surface. These cells become damaged or die, leading to inflammation and other symptoms. The elderly are at a higher risk of developing severe COVID-19 infections due to age-related changes that occur in their respiratory and immune systems.

The respiratory system is responsible for the exchange of gases between the body and the atmosphere, and it consists of the nose, throat, bronchi, and lungs. In the elderly, the respiratory system undergoes changes that can make it harder to breathe. For example, the airways may become narrower, and the lungs may lose their elasticity. Additionally, the elderly are more likely to have pre-existing conditions such as chronic obstructive pulmonary disease (COPD) or asthma that can exacerbate COVID-19 symptoms.

The lymphatic system is responsible for fighting infections and maintaining fluid balance in the body. It consists of lymph nodes, lymphatic vessels, and lymphoid organs such as the spleen and thymus. As the immune system responds to COVID-19, the lymphatic system may become overwhelmed, leading to a buildup of fluid in the lungs and other organs. This can cause severe respiratory distress in the elderly, especially those with weakened immune systems due to age or other health conditions.

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How is the start codon aligned with the P-site in the prokaryotic initiation complex? O a. The Shine-Dalgarno sequence in the mRNA binds to the 16S rRNA of the 30S ribosomal complex, with the start codon aligning under the P- site. O b. IF-2 binds a GTP and an fMet-tRNA, with the tRNA anticodon base pairing with the start codon in the mRNA. O c. The mRNA is bound by a complex of initiation factors; one that binds the 5' cap, an ATPase/helicase, and a protein that binds to the poly(A)- binding proteins. O d. The 48S complex scans through the mRNA, starting at the 5' cap and reading through until the start codon aligns with the tRNA in the P-site. e. The second codon aligns base-pairs with IF-1 in the A-site. Which of the following is TRUE regarding translation in prokaryotes? O a. Which charged tRNA enters the ribosome complex depends upon the mRNA codon positioned at the base of the A-site. O b. Both RF1 and RF2 recognise all three stop codons. O c. The formation of the peptide bond is catalysed by an enzyme within the 50S subunit. d. Elongation factor G (EF-G) delivers an aminoacyl-tRNA to the A-site. e. The binding of elongation factor Tu (EF-Tu) to the A site displaces the peptidyl-tRNA and stimulates translocation. Clear my choice

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The start codon is aligned with the P-site in the prokaryotic initiation complex through the process of IF-2 binding a GTP and an fMet-tRNA, with the tRNA anticodon base pairing with the start codon in the mRNA. This is the true statement regarding the prokaryotic translation.

Thus, the correct answer is option b, "IF-2 binds a GTP and an fMet-tRNA, with the tRNA anticodon base pairing with the start codon in the mRNA. "During the translation process in prokaryotes, IF-1 binds to the A site of the small ribosomal subunit.

Whereas the initiation factor IF-2 binds a GTP molecule and recruits the formylated initiator methionine tRNA (fMet-tRNA) to the small subunit of the ribosome. Following this, IF-2 hydrolyses the GTP to GDP, and the 50S subunit binds to the 30S subunit, completing the 70S ribosome complex.

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5. Ion channels are pore-forming membrane proteins that allow ions to pass through. Describe the basic features and biological roles of three classes of gated ion channels. (10 marks)

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Voltage-gated ion channels are involved in generating and transmitting electrical signals, ligand-gated ion channels mediate responses to specific chemical signals, and mechanosensitive ion channels enable cells to respond to mechanical forces in their environment.

Voltage-gated ion channels are a class of ion channels that open or close in response to changes in the voltage across the cell membrane. They play a crucial role in generating and propagating electrical signals in excitable cells, such as neurons and muscle cells. Voltage-gated ion channels allow the selective flow of ions (e.g., sodium, potassium, calcium) across the cell membrane, enabling the generation of action potentials and the transmission of nerve impulses.

Ligand-gated ion channels, also known as receptor-operated channels, are ion channels that open or close in response to the binding of specific molecules, called ligands, to their receptors. Ligands can be neurotransmitters, hormones, or other signaling molecules. When a ligand binds to the receptor, it induces conformational changes in the ion channel, leading to its opening or closing. Ligand-gated ion channels are involved in various physiological processes, including synaptic transmission, muscle contraction, and sensory perception.

Mechanosensitive ion channels are ion channels that respond to mechanical forces, such as tension, pressure, or stretch. They are found in various tissues and cell types, including sensory neurons, epithelial cells, and cardiovascular cells. Mechanosensitive ion channels participate in diverse biological functions, including touch sensation, hearing, regulation of blood pressure, and osmoregulation. When mechanical forces act on the ion channels, they undergo structural changes that modulate ion permeability, allowing ions to enter or exit the cell and thereby transducing mechanical stimuli into electrical signals.

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A eukaryotic cell gets infected with a bacterium and begins to die. The researchers find that the bacteria produce a protein that sequesters free protons from the cytosol (removes H+ from the cytosol). Given the function of the bacterial protein, what explains why the host cell dies?
a. The bacterial protein increases the pH of the cytosol, causing host proteins to denature, fold improperly and lose function.
b. The lack of protons causes the lipid bilayer to become too fluid, killing the muscle cells.
c. The protons are no longer able to act as a cofactor and transcription is inhibited.
d. The bacterial protein inhibits ATP synthesis by substrate level phosphorylation in the cytosol.

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The statement that explains why the host cell dies in a eukaryotic cell infected with a bacterium that produces a protein that sequesters free protons from the cytosol is that The bacterial protein increases the pH of the cytosol, causing host proteins to denature, fold improperly and lose function. Correct option is A.

What are eukaryotic cells?

Eukaryotic cells are cells that have a true nucleus with a nuclear membrane, genetic material that is organized into chromosomes, and membrane-bound organelles such as mitochondria, endoplasmic reticulum, Golgi bodies, lysosomes, and peroxisomes. Eukaryotic cells are found in animals, plants, fungi, and protists.

Denaturing of host proteins: When the bacterial protein increases the pH of the cytosol, it affects the host cell in many ways. One of the significant ways is that it causes host proteins to denature, fold improperly and lose function.

Host proteins lose their 3D structure due to the altered pH of the cytosol, causing them to no longer be able to perform their designated functions. When the host proteins fail to perform their functions, the cell can no longer maintain homeostasis, leading to cell death.

Therefore, the correct option is A. The bacterial protein increases the pH of the cytosol, causing host proteins to denature, fold improperly, and lose function.

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Identify whether the structure is part of the conducting division or the respiratory division. conducting division respiratory division trachea larynx nasal cavity primary bronchi respiratory bronchioles pharynx alveolar sacs tertiary bronchi

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The conducting division and respiratory division are the two parts of the respiratory system. The structure that belongs to the conducting division or the respiratory division can be identified as follows:

Conducting Division The conducting division includes the nasal cavity, pharynx, larynx, trachea, bronchi, bronchioles, and terminal bronchioles.

The main purpose of this division is to transfer air from the external environment into the respiratory tract.Respiratory DivisionThe respiratory division is made up of respiratory bronchioles, alveolar ducts, and alveoli.

This division is responsible for facilitating gas exchange between the respiratory system and the bloodstream. It is important to note that respiratory bronchioles are located at the junction of the conducting and respiratory divisions of the respiratory tract.

The following structures belong to the conducting or respiratory division:

Nasal cavity: Conducting divisionPharynx: Conducting divisionLarynx: Conducting divisionTrachea: Conducting divisionPrimary bronchi: Conducting divisionTertiary bronchi: Conducting divisionRespiratory bronchioles: Respiratory divisionAlveolar sacs: Respiratory division.

The conducting division includes the nasal cavity, pharynx, larynx, trachea, bronchi, bronchioles, and terminal bronchioles. On the other hand, the respiratory division is made up of respiratory bronchioles, alveolar ducts, and alveoli. The respiratory bronchioles are located at the junction of the conducting and respiratory divisions of the respiratory tract.

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Place the steps of the molecular process of muscle contraction in correct order. Myosin head groups form cross bridges Action potential arrives at sarcolemma ATP binds to myosin head groups Electrical

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The correct order of the molecular process of muscle contraction is as follows:Action potential arrives at sarcolemma. Electrical stimulation (an action potential) arrives at a motor neuron, travels down the motor neuron to its end, and causes the release of acetylcholine (ACh).

ATP binds to myosin head groups. The binding of ATP causes the cross-bridge between actin and myosin to weaken. It prepares myosin for the next cycle of contraction. Myosin head groups form cross bridges. The myosin heads interact with active sites on actin to form cross-bridges. The process of muscle contraction occurs when the myosin head group binds to actin on the thin filament and generates tension by forming a cross-bridge. To generate this tension, a series of steps occur in a cycle that repeats as long as there is a stimulus.The steps in the molecular process of muscle contraction in the correct order are as follows.

Electrical stimulation (an action potential) arrives at a motor neuron, travels down the motor neuron to its end, and causes the release of acetylcholine (ACh).2. ACh binds to receptors on the sarcolemma, initiating an action potential. The action potential travels along the sarcolemma and down T-tubules, causing the release of Ca2+ from the sarcoplasmic reticulum.3. Ca2+ binds to troponin, causing tropomyosin to move and expose the active sites on actin.4. Myosin heads bind to active sites on actin, forming cross-bridges.5. ADP and P release from the myosin head, causing the head to rotate and generate tension on the actin filament (the power stroke).6. ATP binds to the myosin head, causing the cross-bridge between actin and myosin to weaken.

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7. Which neurons of the autonomic nervous system will slow the heart rate when they fire onto the heart? If input from those neurons is removed, how will the heart rate respond? (2 mark)

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The neurons of the autonomic nervous system that slow down the heart rate are the parasympathetic neurons, specifically the vagus nerve (cranial nerve X). When these neurons fire onto the heart, they release the neurotransmitter acetylcholine, which binds to receptors in the heart and decreases the rate of firing of the heart's pacemaker cells, thus slowing down the heart rate.

If input from these parasympathetic neurons is removed or inhibited, such as through the administration of certain drugs or in certain pathological conditions, the heart rate will increase. This is because the parasympathetic input normally provides a balancing effect to the sympathetic nervous system, which tends to increase the heart rate. With the removal of parasympathetic input, the heart will be under the influence of the unopposed sympathetic stimulation, leading to an increase in heart rate.

The parasympathetic neurons that slow down the heart rate are part of the vagus nerve (cranial nerve X), specifically the cardiac branches of the vagus nerve. These neurons innervate the sinoatrial (SA) node, the natural pacemaker of the heart.

When these parasympathetic neurons are activated, they release acetylcholine, which binds to muscarinic receptors on the SA node. This binding leads to a decrease in the rate of depolarization of the SA node cells, slowing down the generation and conduction of electrical impulses in the heart. As a result, the heart rate decreases.

If the input from the parasympathetic neurons is removed or inhibited, such as in conditions where the vagus nerve is damaged or in the absence of parasympathetic stimulation, the heart rate will be influenced primarily by sympathetic stimulation. The sympathetic nervous system is responsible for increasing the heart rate and enhancing cardiac output in response to various stressors and demands.

Therefore, in the absence of parasympathetic input, the heart rate will increase as the sympathetic influence becomes dominant. This can lead to a higher heart rate, increased contractility, and overall increased cardiovascular activity.

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1. According to the Cell Theory, cells are viewed as the minimal functional units of organisms. True/ False 2. The region of a eukaryotic cell that is enclosed by the plasma membrane but not enclosed by any internal membrane is termed the _______________.
A. extracellular environment
B. cytoplasm
C. lumen
D. cytosol

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According to the Cell Theory, cells are viewed as the minimal functional units of organisms. True The cytosol, also known as the cytoplasmic matrix or groundplasm, is the liquid component of the cytoplasm in eukaryotic cells. So correct answer is D

The Cell Theory is a biological theory that states that cells are the basic building blocks of all living organisms, and that all organisms are made up of one or more cells. The theory further suggests that cells are the functional and structural units of life, and that cells are responsible for carrying out all of the functions necessary for the survival of an organism. This includes processes such as metabolism, reproduction, and responding to stimuli.

2. The region of a eukaryotic cell that is enclosed by the plasma membrane but not enclosed by any internal membrane is termed the _______________.  It is the region of the cell that is enclosed by the plasma membrane but not enclosed by any internal membrane. The cytosol contains various organelles, including the mitochondria, ribosomes, and the cytoskeleton. It also contains various dissolved molecules, such as enzymes, nucleic acids, and ions. The cytosol plays a vital role in various cellular processes, such as protein synthesis, cell division, and cell signaling.

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Using named examples of genetic conditions explain the inheritance patterns of:
i. a recessive autosomal condition
ii. a dominant autosomal condition
iii. a sex-linked condition
You should use genetic inheritance diagrams. The diagrams should give the genotypes and phenotypes of the parents and F1 zygotes, the gametes produced and the way that the gametes could combine during a monohybrid cross.

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Genetic conditions are determined by the presence of gene abnormalities that can either be inherited or developed later in life. The following is a detailed explanation of the inheritance patterns of genetic conditions.

1. A recessive autosomal condition: An example of a recessive autosomal genetic condition is cystic fibrosis. The pattern of inheritance is represented by parents who are carriers of the cystic fibrosis gene but do not have the condition.

2. A dominant autosomal condition: An example of a dominant autosomal genetic condition is Huntington's disease. The pattern of inheritance is demonstrated by parents where at least one of them has the dominant gene.

3. A sex-linked condition: An example of a sex-linked genetic condition is hemophilia. The pattern of inheritance is represented by parents, with males being more likely to inherit the condition than females.

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How is a polynucleotide chain read in a nucleic acid structure?
From the 5'-end to the 3'-end.
From the 3'-end to the 5'-tail.
From the poly(U) head to the poly(A) tail.
From the poly-p head to the 5'-end.

Answers

In a nucleic acid structure, a polynucleotide chain is read from the 5'-end to the 3'-end. (Option A)

A polynucleotide chain is an extended chain of nucleotides, which includes both DNA and RNA. DNA has a double-stranded helix structure, while RNA has a single-stranded structure.

The nucleotides in a polynucleotide chain are linked together by phosphodiester bonds. The phosphodiester bonds create a backbone for the polynucleotide chain, which alternates between a phosphate group and a sugar molecule. A nucleotide is a molecule that consists of a nitrogenous base, a pentose sugar, and a phosphate group. The nitrogenous base can be either a purine (adenine or guanine) or a pyrimidine (cytosine or thymine in DNA or uracil in RNA).

In a polynucleotide chain, the nitrogenous bases pair up through hydrogen bonds. Adenine pairs with thymine (DNA) or uracil (RNA) through two hydrogen bonds, while guanine pairs with cytosine through three hydrogen bonds. This base pairing allows DNA to replicate and RNA to transcribe genetic information.

Thus, the correct option is A.

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The topic is hiochemistry however i could not find it. May i ask how many types of enzyme regulation seen here and may i ask what types are there i know there is covalent modication as there is phosphorylation. According to my tracher there is allosteric inhinition and activation but may i ask where is it ? Also she mentioned there is proteinprotein interaction can anyone olease point out where and is there other types of regualtion seen here ? thank you

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There are four types of enzyme regulation (i) Covalent modification (ii) Allosteric regulation (iii) Protein-protein interactions (iv) Gene regulation.

Enzymes are proteins that catalyze biochemical reactions, increasing reaction rates by decreasing activation energy. The rate of enzyme-catalyzed reactions can be regulated by numerous mechanisms, which are generally classified into four types: covalent modification, allosteric regulation, protein-protein interactions, and gene regulation.

What are the types of Enzyme Regulation ?

Covalent modification: It is a type of enzyme regulation that involves the covalent attachment of a molecule, usually a phosphate, to an enzyme protein to alter its activity. Enzyme phosphorylation is the most common form of covalent modification and is frequently involved in signal transduction pathways. It can also include other types of covalent modifications, such as methylation, acetylation, and ubiquitination.

Allosteric regulation: It is a type of enzyme regulation that involves the binding of a regulatory molecule to a site on an enzyme that is distinct from the active site. This binding induces a conformational change in the enzyme that alters its activity. Allosteric regulation can be either positive (activating) or negative (inhibiting).

Protein-protein interactions: It is a type of enzyme regulation that involves the interaction of two or more proteins that affect enzyme activity. This interaction may involve the formation of protein complexes that modify enzyme activity.

Gene regulation: It is a type of enzyme regulation that involves the regulation of the expression of genes that encode enzymes. This regulation can occur at many levels, including transcriptional, translational, and post-translational regulation.

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Which of the following "edge effects" is/are often associated with forest fragmentation of the Eastern Deciduous Forešt? None of these are associated with this fragmentation. All of these are associated with this fragmentation. Reduction in population sizes of year-round residents that are attracted to habitat edges and nest in cavities due to competition with migrants. Mesopredator release and increased predation (e.g., on ground nests of birds) near forest edges.
Increases in most ground-nesting birds that breed in the interior of forest fragments. A reduction in the population size of the Brown-headed Cowbird.

Answers

The "edge effects" often associated with forest fragmentation of the Eastern Deciduous Forest include:

1. Reduction in population sizes of year-round residents that are attracted to habitat edges and nest in cavities due to competition with migrants.
2. Mesopredator release and increased predation (e.g., on ground nests of birds) near forest edges.

Therefore, the correct option would be: "Reduction in population sizes of year-round residents that are attracted to habitat edges and nest in cavities due to competition with migrants" and "Mesopredator release and increased predation (e.g., on ground nests of birds) near forest edges."

While the mechanisms of vocal production are similar across primates, there are important differences between the production of human speech and nonhuman primate vocalizations. Some of these differences can be directly attributed to anatomical changes during evolution. What do anatomical differences in the vocal production apparatus (larynx, pharynx, and oral cavity) between chimpanzees and modern humans suggest about the vocal behavior of each species?

Answers

The anatomical differences suggest that humans have evolved specialized vocal structures for complex speech, while chimpanzees have anatomical features suited for simpler vocalizations.

The anatomical differences between chimpanzees and modern humans in their vocal production apparatus provide insights into the vocal behavior of each species. Humans have undergone significant anatomical changes during evolution that have facilitated the development of speech.

One crucial difference lies in the positioning of the larynx, or voice box. In humans, the larynx is positioned lower in the throat, allowing for a longer vocal tract. This elongation of the vocal tract enables the production of a wide range of sounds and phonemes, contributing to the complexity of human speech.

In contrast, chimpanzees have a higher larynx position, resulting in a shorter vocal tract. This anatomical configuration restricts the variety of sounds they can produce and limits the complexity of their vocalizations. While chimpanzees possess the ability to communicate through vocal signals, their vocal repertoire primarily consists of simple calls, such as hoots, grunts, and screams, which serve more immediate and basic communicative functions.

The differences in the pharynx and oral cavity further highlight the distinctions in vocal behavior between the two species. Humans have a descended hyoid bone, which supports the larynx and allows for intricate tongue movements necessary for articulating a wide range of sounds during speech. Additionally, humans have a highly developed oral cavity, including specialized lips, teeth, and tongue, which contribute to the precise articulation of speech sounds.

On the other hand, chimpanzees lack these specialized adaptations in their pharynx and oral cavity, limiting their ability to produce the diverse range of sounds found in human speech. Their vocalizations rely more on facial expressions, gestures, and body postures to convey meaning.

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Refer to the graph pictured below. Allele frequencies for this population are A₁=0.5, A₂=0.5, and assume the population is in Hardy Weinberg equilibrium. What is p* for this population? 1 0.8 11₁ A₁A₁ A₁A₂ A₂A₂ Relative fitness 0.6 0.4 0.2 Refer to the graph pictured below. Allele frequencies for this population are A₁-0.5, A₂-0.5, and assume the population is in Hardy Weinberg equilibrium. What is the average population fitness for this population (round to the nearest tenth or 1 decimal place)?

Answers

The Hardy-Weinberg equilibrium is a mathematical model that is utilized to calculate allele and genotype frequencies in populations. Hardy-Weinberg equilibrium requires five conditions to be met.

These are random mating, large population size, no migration, no mutation, and no natural selection. Let's consider the first part of the question: Allele frequencies for this population are A₁=0.5, A₂=0.5, and assume the population is in Hardy Weinberg equilibrium. What is p* for this population? The formula to calculate p* is:

p* = √p

where: p = frequency of the dominant allele p* = frequency of the homozygous dominant genotype

Thus, in the given case: p* = √0.5 = 0.707Let's consider the second part of the question: Refer to the graph pictured below. Allele frequencies for this population are A₁=0.5, A₂=0.5, and assume the population is in Hardy Weinberg equilibrium.

We are given the following information:

Genotypes Relative Fitness  A₁A₁ 0.6A₁A₂ 0.4A₂A₂ 0.2

The formula to calculate average population fitness is:

average population fitness = [(frequency of A₁A₁) x (relative fitness of A₁A₁)] + [(frequency of A₁A₂) x (relative fitness of A₁A₂)] + [(frequency of A₂A₂) x (relative fitness of A₂A₂)]

We can use the Hardy-Weinberg formula to calculate the frequency of genotypes:

p² + 2pq + q² = 1where:p² = frequency of A₁A₁q² = frequency of A₂A₂2pq = frequency of A₁A₂

Thus, in the given case:p² = 0.5² = 0.25q² = 0.5² = 0.252pq = 2(0.5)(0.5) = 0.5

Now, we can plug these frequencies into the formula for average population fitness:

average population fitness = [(0.25) x (0.6)] + [(0.5) x (0.4)] + [(0.25) x (0.2)]

average population fitness = 0.15 + 0.2 + 0.05average population fitness = 0.4

The average population fitness for this population is 0.4.

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If vision is lost, sensory information relayed through the hands
typically becomes more detailed and nuanced. How might this change
be represented in the primary sensory cortex?

Answers

The brain is able to adapt to the changes in sensory input and allocate more resources to other senses to compensate for the lost sense.

If vision is lost, the sensory information relayed through the hands typically becomes more detailed and nuanced.

This change can be represented in the primary sensory cortex by increasing the size of the hand area within the primary sensory cortex.

The primary sensory cortex is the region of the brain responsible for processing the sensory information relayed to it from the peripheral nervous system.

It receives signals that are generated by the senses and sends them to different parts of the brain for further processing.

When an individual loses vision, they become more attuned to their sense of touch.

This change in the sensory experience can be represented in the primary sensory cortex by increasing the size of the hand area.

This is because the region of the cortex that is responsible for processing tactile information from the hands becomes more active and larger in size.

This phenomenon is known as cortical reorganization, and it is a common occurrence in individuals who have lost one of their senses.

The brain is able to adapt to the changes in sensory input and allocate more resources to other senses to compensate for the lost sense.

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Draw a tRNA with the anticodon 3’ACGUA5’ Given wobble, what two
different codons could it bind to? Draw each codon on an mRNA,
labeling all 5' and 3' ends, the tRNA, and the amino acid it
carries.

Answers

The tRNA with the anticodon 3'ACGUA5' can bind to two different codons: 5'UGCAA3' and 5'UGCAC3'. These codons are complementary to the anticodon sequence of the tRNA. When bound to the codons, the tRNA carries a specific amino acid.

The anticodon sequence of the tRNA is 3'ACGUA5'. In the process of translation, the anticodon of the tRNA pairs with the complementary codon sequence on the mRNA molecule. The anticodon and codon interaction follows the rules of base pairing: adenine (A) pairs with uracil (U) and guanine (G) pairs with cytosine (C).

The tRNA with the anticodon 3'ACGUA5' can bind to two different codons due to the phenomenon known as wobble. Wobble allows for flexibility in the base pairing between the third position of the codon and the corresponding position of the anticodon. In this case, the anticodon has a guanine (G) at the third position, which can pair with either cytosine (C) or adenine (A).

Thus, the tRNA with the anticodon 3'ACGUA5' can bind to the codons 5'UGCAA3' and 5'UGCAC3'. The tRNA molecule carries a specific amino acid attached to its 3' end, and this amino acid is delivered to the ribosome during translation when the tRNA binds to the complementary codon on the mRNA.

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15. Different terminology is used to characterize thermoregulation in animals: Warm-blooded, cold- blooded, homeotherm, poikilotherm, endotherm, ectotherm, etc. Why are these terms usually insufficien

Answers

The terms used to characterize thermoregulation in animals such as warm-blooded, cold-blooded, homeotherm, poikilotherm, endotherm, ectotherm, etc. are usually insufficient because they are either too general or imprecise.

They do not provide a comprehensive or accurate understanding of thermoregulation in animals. Additionally, they have been replaced by more precise terms and concepts in modern biology. The term warm-blooded.

Is imprecise and is commonly used to describe endothermic animals, which generate their body heat internally. However, there are some cold-blooded animals that are capable of maintaining a relatively constant body temperature by using external sources of heat, such as basking in the sun.

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6. Which is not correct regarding the hypothalamo-hypophyseal portal system? a. The system includes two capillary plexuses b. The system carries venous blood c. The system is the circulatory connectio

Answers

The hypothalamo-hypophyseal portal system is the circulatory connection between the hypothalamus and the anterior pituitary gland. This portal system carries venous blood between the two capillary plexuses.The correct answer is option C.

The hypothalamo-hypophyseal portal system is the circulatory connection between the hypothalamus and the anterior pituitary gland. It includes two capillary plexuses and carries venous blood from the hypothalamus to the anterior pituitary gland. In the first capillary plexus, the hypothalamus secretes regulatory hormones into the blood, which then travel through the portal veins to the second capillary plexus, where they stimulate or inhibit the secretion of anterior pituitary hormones. This allows for precise control of hormone secretion by the anterior pituitary gland.The hypothalamus secretes several hormones that regulate the secretion of anterior pituitary hormones. These hormones are referred to as releasing hormones or inhibiting hormones.

For example, the hypothalamus secretes thyrotropin-releasing hormone (TRH), which stimulates the anterior pituitary gland to secrete thyroid-stimulating hormone (TSH). The hypothalamus also secretes prolactin-inhibiting hormone (PIH), which inhibits the anterior pituitary gland from secreting prolactin. The hypothalamus and anterior pituitary gland work together to regulate a wide range of physiological processes, including growth, metabolism, and reproduction.In summary, the hypothalamo-hypophyseal portal system is a specialized circulatory connection that allows for precise control of hormone secretion by the anterior pituitary gland. The system includes two capillary plexuses and carries venous blood from the hypothalamus to the anterior pituitary gland. The hypothalamus secretes regulatory hormones into the blood, which then travel to the second capillary plexus, where they stimulate or inhibit the secretion of anterior pituitary hormones.

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A mutation occurs in the trpC gene. This mutation creates a rho-independent terminator within the 3 prime end of the trpC open reading frame but does not alter the activity of TrpC protein. However, the strain is Trp-. What kind of mutation was this and why is the strain Trp-?

Answers

The mutation is a nonsense mutation. The strain is Trp- because the rho-independent terminator prematurely terminates the synthesis of the trpC mRNA, preventing the production of functional TrpC protein.

The mutation is a nonsense mutation, specifically a premature stop codon, which leads to the termination of translation before the complete TrpC protein is synthesized. The strain is Trp- because the mutation disrupts the normal production of the TrpC protein, which is essential for the biosynthesis of tryptophan.

A premature stop codon is a type of nonsense mutation that introduces a stop signal in the DNA sequence, leading to the premature termination of translation during protein synthesis. In this case, the mutation creates a rho-independent terminator within the trpC gene, causing the synthesis of the TrpC protein to be prematurely halted. Since the TrpC protein is involved in the biosynthesis of tryptophan, a crucial amino acid, the strain carrying this mutation is unable to produce tryptophan, resulting in the Trp- phenotype. The strain will require an exogenous supply of tryptophan to survive and grow.

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Under normal conditions in the kidneys, which substance does not enter the filtrate from the glomerulus?
a. amino acids b. water-soluble vitamins c. minerals d. glucose e. blood proteins

Answers

Under normal conditions in the kidneys, blood proteins do not enter the filtrate from the glomerulus. So, option E is accurate.

The glomerulus is a network of capillaries in the kidney responsible for the initial filtration of blood to form urine. It acts as a selective filter, allowing small molecules and waste products to pass through while retaining larger molecules like blood proteins. Blood proteins, such as albumin and globulins, are too large to pass through the filtration barrier of the glomerulus, which consists of fenestrated capillaries and a filtration membrane. This filtration barrier prevents the entry of blood proteins into the filtrate. On the other hand, substances like amino acids, water-soluble vitamins, minerals, and glucose are small enough to pass through the filtration barrier and enter the filtrate. Therefore, under normal conditions, blood proteins do not enter the filtrate from the glomerulus.

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What must be true for DNA polymerase to work Select one or more: a. There must be a free 3¹ OH for it to attach nucleotides to. b. New nucleotides must be tri-phosphates c. hydrolysis of the bond between the first and second phosphate drives the polymerization reaction d. Continuous replication doesn't need an RNA primer Okazaki fragments only happen on one of the DNA X strands in a replication bubble (that's a fork going in both directions)

Answers

DNA polymerase is a type of enzyme that is responsible for the formation of a new strand of DNA. In order for DNA polymerase to function, there must be a free 3'OH to which nucleotides can be added. It can only attach nucleotides to a strand of DNA that is complementary to the template strand, as per the Watson-Crick base-pairing rules.

The new nucleotides must be tri-phosphates, which means that they have three phosphates attached to them. When a nucleotide is added to the growing DNA strand, the bond between the first and second phosphate groups is hydrolyzed. This reaction provides the energy needed to drive the polymerization reaction. Continuous replication doesn't need an RNA primer. On one of the DNA strands in a replication bubble, Okazaki fragments only occur.

These fragments are synthesized in the opposite direction of the replication fork. The RNA primers, on the other hand, are needed for the synthesis of Okazaki fragments. DNA polymerase is the enzyme that creates new DNA molecules. It adds nucleotides in the 5' to 3' direction to the complementary strand of DNA.

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Explain in you own words why arteriosclerosis and
atherosclerosis can lead to the development of heart diseases
(*list what happens with EACH disease?)

Answers

Arteriosclerosis and atherosclerosis are two related conditions that involve the hardening and narrowing of arteries, which can lead to the development of heart diseases. Here's an explanation of each disease and their respective consequences

Arteriosclerosis: Arteriosclerosis refers to the general thickening and hardening of the arterial walls. This condition occurs due to the buildup of fatty deposits, calcium, and other substances in the arteries over time. As a result, the arteries lose their elasticity and become stiff. This stiffness restricts the normal expansion and contraction of the arteries, making it more difficult for blood to flow through them. The consequences of arteriosclerosis include:

Increased resistance to blood flow: The narrowed and stiffened arteries create resistance to the flow of blood, making it harder for the heart to pump blood effectively. This can lead to increased workload on the heart and elevated blood pressure.

Reduced oxygen and nutrient supply: The narrowed arteries restrict the flow of oxygen-rich blood and essential nutrients to the heart muscle and other organs. This can result in inadequate oxygen supply to the heart, leading to chest pain or angina.

Atherosclerosis: Atherosclerosis is a specific type of arteriosclerosis characterized by the formation of plaques within the arterial walls. These plaques consist of cholesterol, fatty substances, cellular debris, and calcium deposits. Over time, the plaques can become larger and more rigid, further narrowing the arteries. The consequences of atherosclerosis include:

Reduced blood flow: As the plaques grow in size, they progressively obstruct the arteries, restricting the flow of blood. In severe cases, the blood flow may become completely blocked, leading to ischemia (lack of blood supply) in the affected area.

Formation of blood clots: Atherosclerotic plaques can become unstable and prone to rupture. When a plaque ruptures, it exposes its inner contents to the bloodstream, triggering the formation of blood clots. These blood clots can partially or completely block the arteries, causing a sudden interruption of blood flow. If a blood clot completely occludes a coronary artery supplying the heart muscle, it can lead to a heart attack.

Risk of cardiovascular complications: The reduced blood flow and increased formation of blood clots associated with atherosclerosis increase the risk of various cardiovascular complications, including heart attacks, strokes, and peripheral artery disease.

In summary, arteriosclerosis and atherosclerosis contribute to the development of heart diseases by narrowing and hardening the arteries, reducing blood flow, impairing oxygen and nutrient supply to the heart, and increasing the risk of blood clots and cardiovascular complications. These conditions underline the importance of maintaining a healthy lifestyle and managing risk factors such as high blood pressure, high cholesterol, smoking, and diabetes to prevent the progression of arterial diseases and reduce the risk of heart-related complications.

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Which of the following is not an application of PCR?
O I. Amplifying DNA molecules
O II. Amplifying RNA molecules
O III. Synthesis of protein
O IV. Genome sequencing

Answers

Synthesis of protein is not an application of PCR. So, option III is appropriate.

PCR, or Polymerase Chain Reaction, is a widely used technique in molecular biology that allows for the amplification of specific DNA sequences. It is a powerful tool that enables researchers to generate millions or even billions of copies of a target DNA segment, making it easier to analyze and study.

PCR (Polymerase Chain Reaction) is a powerful molecular biology technique used for amplifying DNA molecules. It is not used for directly amplifying RNA molecules or synthesizing proteins. PCR allows for the selective amplification of specific DNA sequences, making it valuable in applications such as DNA cloning, genetic testing, forensic analysis, and diagnostic assays. Genome sequencing, on the other hand, involves determining the complete DNA sequence of an organism's genome and is a separate process from PCR.

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what type of goal is based on measurable and
qualifiable data
66. What type of goal is based on measurable and quantifiable data? A. Motivational goal B. Sersonal goal C. Subjective goal D. Objective goal

Answers

The type of goal based on measurable and quantifiable data is Objective goal.

Goals are the things that a person aims to achieve. They are targets that a person wants to reach. People often set goals to provide themselves with a clear path to follow while working on a specific task. Objectives are one of the most important types of goals. These are goals that are based on measurable and quantifiable data.

Objective goals are specific, measurable, attainable, relevant, and time-bound. They are goals that are based on quantifiable data. Quantifiable data is the data that can be measured using a specific tool or unit of measurement. Objective goals are essential for tracking progress because they allow you to know when you have met your target. If you want to make progress towards your goal, you must track it. By tracking your progress, you can tell whether you are making progress towards your objective goals or not.

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Which of the following is not a part of the positive feedback that drives the rising phase of the action potential? Select one: ONa+ channel gating Ob voltage-gated channels depolarization Od Na+ channel inactivation

Answers

Na+ channel inactivation is not a part of the positive feedback that drives the rising phase of the action potential.

Option (c) is correct.

During the rising phase of an action potential, positive feedback mechanisms drive the depolarization of the cell membrane. These mechanisms involve the opening of voltage-gated Na+ channels (option Na+ channel gating) and the influx of Na+ ions, leading to further depolarization of the membrane.

Option c, Na+ channel inactivation, is not a part of the positive feedback process during the rising phase of the action potential. After the Na+ channels open and allow the influx of Na+ ions, they undergo a process called inactivation. This occurs shortly after the channels open, and it involves a mechanism that blocks further Na+ influx and contributes to the repolarization of the membrane. Inactivation is an essential step that helps regulate the duration and magnitude of the action potential.

In summary, while options Na+ channel gating and depolarization are involved in the positive feedback mechanism during the rising phase of the action potential, Na+ channel inactivation is not part of the positive feedback process but rather plays a role in the subsequent repolarization phase.

Therefore, the correct option is (c).

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Which of the following is not a part of the positive feedback that drives the rising phase of the action potential? Select one:

a) Na+ channel gating

b) voltage-gated channels depolarization

c) Na+ channel inactivation

e) None of the above

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