The animals that belong to the vertebrate category are birds, turtle, lizard, shark, frog and chimpanzee.
Animals belonging to the invertebrate category are scorpion, planaria, lamprey, jellyfish, ascarids, sponge, coral, flatworms, leeches, spearworm, snail, clam, octopus, lobster, starfish, and tunicate.
Vertebrates belong to the animal kingdom and are characterized by having a vertebral column and an articulated skeleton formed by bones, while invertebrates lack these elements.
Within the category of vertebrates are mammals, birds, reptiles, amphibians, and fish. In the invertebrate category are annelids, arthropods, mollusks, cnidarians, and echinoderms.
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As it relates to DNA composition: i. Explain Chargaff's rules ii. What chemical information is found in the major groove? iii. What chemical information is found in the minor groove? iv. What is the significance of access to chemical information in the major and minor grooves?
Chargaff's Rules state that the percentage of guanine is equal to the percentage of cytosine, and the percentage of adenine is equal to the percentage of thymine.
Chemical Information in Major Groove: arrangement of base pairs, allowing recognition of base sequences for transcription and translation. Chemical Information in Minor Groove: hydrogen-bonding pattern of the base pairs, which is important for DNA-protein interactions. Access to chemical information in the major and minor grooves of DNA helps in protein-DNA interaction.
In the year 1950, Erwin Chargaff presented his rules for the bases present in DNA. The rule states that the amount of adenine and thymine should be the same, whereas the amount of guanine and cytosine should also be the same. It means the percentage of adenine equals the percentage of thymine, and the percentage of cytosine equals the percentage of guanine in DNA. The relationship is A=T and C=G.
The major groove in DNA is the larger groove in the double helix structure. The DNA structure contains two grooves; a major groove and a minor groove. The major groove is the site for the binding of most transcription factors that help in regulating gene expression. Major grooves contain the most structural information for identifying specific protein-DNA interactions.
The minor groove is the smaller groove in the DNA structure. It's a place where fewer atoms are visible. The minor groove of DNA is where the DNA-binding proteins can form hydrogen bonds with the DNA sequence. Proteins that bind to the minor groove of DNA typically bind to sequences that lack adenine and thymine residues.
The major and minor grooves in DNA play a vital role in protein-DNA interaction. They provide the necessary space to interact with the DNA sequence. The major groove in DNA is a place where the majority of the proteins that regulate gene expression binds. The minor groove, on the other hand, provides binding sites for transcription factors, enzymes, and DNA binding proteins. Thus, access to chemical information in the major and minor grooves of DNA helps in protein-DNA interaction.
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You are stranded on a boat in the middle of the ocean with no fresh water on board. However, you are not alone, and some of those with you want to drink ocean water. You tell them they should not do that but they don’t believe you. Use your understanding of osmosis to convince them WHY drinking salt water is very bad and a potentially deadly idea.
Answer: Your welcome!
Explanation:
Osmosis is the process by which water moves from an area of low salt concentration to an area of high salt concentration. When you drink salt water, the water in your body has a lower salt concentration than the salt water you are consuming. The salt water will then try to move into your cells, attempting to balance out the salt concentrations. This process causes your cells to swell as they become filled with water, and too much of this can cause cells to rupture. Additionally, salt water causes dehydration because the body can’t process the salt, and the body will expel water to try to dilute the salt. So drinking salt water can be extremely dangerous because it can lead to dehydration and can cause cells to rupture.
1. Read the article 'Deep Mutational Scanning of SARS-CoV-2 Receptor Binding Domain Reveals Constraints on Folding and ACE2 Binding'? Provide a citation for the article.
2. What are the specific objectives of this study and how did the authors address them? (100 words)
1. The citation for the article 'Deep Mutational Scanning of SARS-CoV-2 Receptor Binding Domain Reveals Constraints on Folding and ACE2 Binding' is as follows:
Starr, T. N., Greaney, A. J., Hilton, S. K., Ellis, D., Crawford, K. H. D., Dingens, A. S., ... & Bloom, J. D. (2020). Deep Mutational Scanning of SARS-CoV-2 Receptor Binding Domain Reveals Constraints on Folding and ACE2 Binding. Cell, 182(5), 1295-1310. 2. The specific objectives of this study were to understand the constraints on the folding and binding of the SARS-CoV-2 receptor binding domain (RBD) to the ACE2 receptor, and to identify mutations that could potentially affect the virus's ability to bind to and infect human cells. The authors addressed these objectives by using deep mutational scanning to analyze the effects of all possible amino acid mutations on the folding and binding of the RBD. This allowed them to identify mutations that could potentially affect the virus's ability to bind to and infect human cells, as well as to understand the constraints on the folding and binding of the RBD. The authors also used structural modeling to further understand the effects of these mutations on the RBD's structure and function.
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If a bacterial cell needs to optimize nutrient uptake in its environemt, it probably has
- an appendaged cell
- a filamentous cell
- a helical or spiral shaped cell
If a bacterial cell needs to optimize nutrient uptake in its environment, it probably has a filamentous cell. Thus, Option B holds true.
Filamentous cells have an elongated shape that allows them to maximize their surface area for nutrient absorption. This is especially useful in environments where nutrients are scarce or in competition with other organisms. An appendaged cell may also be beneficial for nutrient uptake, as appendages like pili can aid in the attachment to surfaces and the acquisition of nutrients.
However, a helical or spiral shaped cell would not necessarily optimize nutrient uptake, as this shape is more associated with motility and movement.
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2. T or F
The purpose of test validation is to demonstrate that a procedure is ready to be implemented as a clinical test. 3. T or F
The analytical measurement range is part of the test validation process. AMR is established by showing accurate test results on a range of values or test results.
2. True as The purpose of test validation is to demonstrate that a procedure is ready to be implemented as a clinical test.
3. True as The analytical measurement range (AMR) is part of the test validation process.
This involves showing that the test is accurate, reliable, and consistent, and that it is suitable for the intended use.
AMR is established by showing accurate test results on a range of values or test results. This helps to ensure that the test is accurate and reliable across the entire range of values that it is intended to measure.
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what types of models can show the cycling of matter and energy? create a summary
Answer:
particle of matter
Explanation:
I know this from science
Answer: Food webs
Explanation: are models that demonstrate how matter and energy is transferred between producers (generally plants and other organisms that engage in photosynthesis), consumers, and decomposers as the three groups interact—primarily for food—within an ecosystem.
Describe how you would be able to tell which side of a leaf is
the upper surface. Does the available evidence differ in monocots
and eudicots, and if so, how?
The upper surface of a leaf can be identified by looking at the characteristics of the leaf, including the presence of a waxy cuticle, the number of stomata, and the color of the leaf. The available evidence does differ in monocots and eudicots, with differences in vein structure and stomata distribution helping to identify the upper surface of a leaf.
To tell which side of a leaf is the upper surface, you can look at the characteristics of the leaf. The upper surface of a leaf typically has a waxy cuticle to protect it from drying out, and it also typically has more stomata (small openings that allow for gas exchange) than the lower surface. Additionally, the upper surface of a leaf is usually darker in color than the lower surface, due to the presence of more chlorophyll for photosynthesis.
The available evidence does differ in monocots and eudicots. Monocots typically have leaves with parallel veins, while eudicots have leaves with branching veins. This difference in vein structure can also help to identify the upper surface of a leaf, as the veins are typically more prominent on the upper surface. Additionally, monocots typically have stomata on both the upper and lower surfaces of their leaves, while eudicots typically have more stomata on the lower surface. This difference in stomata distribution can also help to identify the upper surface of a leaf.
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The aim of this assignment is to help you to develop your anatomy, physiology, pathology. and the sort of critical thinking skills that will help you to analyse data and to produce your own well thought conclusions. The topic Below is a description of a number of symptoms that a patient presents to a doctor, together with a brief overview of the patients history. It is your task to diagnose the disease that the patient is suffering from and any possible treatments and/or (if the patient was diagnosed earlier) lifestyle changes that the patient could make to improve her condition and to alleviate her symptoms. Case study 1 Patient History Name: Jane Doe Age: 26 Gender: Female Chief Complaint: Jane presents with a 3-month history of abdominal pain and diarrhoea, which have been increasing in frequency and severity. She also reports feeling fatigued and having lost weight unintentionally. Medical History: Jane has a history of asthma, but she is otherwise healthy. She takes albuterol inhaler as needed for asthma symptoms. Family History: Her father has a history of lung cancer, and her mother has a history of hypertension. Social History: Jane works as a infant school teacher and lives with her husband and two children. She denies smoking, alcohol, or drug use. She reports a healthy diet and regular exercise. Physical Examination: On physical examination, Jane appears pale and fatigued. There is diffuse abdominal tenderness, but no palpable masses or organomegaly. There are no other significant findings on examination. Diagnostic Tests; Blood tests show microcytic anemia, elevated inflammatory markers, and mildly elevated liver enzymes. Stool tests are positive for occult blood and leukocytes. Colonoscopy reveals multiple ulcerations and areas of inflammation throughout the solse Structure of essay The essay should be no more than 1000-1500 words in total and should include the following sections: - A brief introduction - List of symptoms - Diagnosis (including how the diagnosis was made) - Cause (how this condition may have aristy)) - Prognosis (the likely long-term outcome of the patient's condition if she had not had a stroke) - Treatments (how this condition might have been treated if diagnosed earlier) - Conclusion - References (use the Harvard referencing style and inline citations) Use figures and diagrams fo support your work where appropriate.
Diagnosing the disease a patient is suffering from requires a thorough understanding of anatomy, physiology, pathology, and critical thinking skills. Jane Doe likely has IBD, which is an inflammatory condition of the digestive tract. It is important to diagnose and treat this condition promptly in order to reduce inflammation and maintain remission.
Jane Doe, a 26 year-old female, presents with a 3-month history of abdominal pain and diarrhoea, fatigue, and unintentional weight loss. Her medical history includes asthma, and her family history shows her father having a history of lung cancer and her mother having hypertension. Jane is a school teacher and does not have any known history of substance abuse. Physical examination shows diffuse abdominal tenderness but no palpable masses.
Blood tests show microcytic anemia, elevated inflammatory markers, and mildly elevated liver enzymes. Stool tests are positive for occult blood and leukocytes. Colonoscopy reveals multiple ulcerations and areas of inflammation. Based on the symptoms and medical tests, Jane is likely suffering from Inflammatory Bowel Disease (IBD), specifically ulcerative colitis.
The cause of IBD is unknown, but it is believed to be related to a combination of environmental and genetic factors. IBD can cause significant abdominal pain, diarrhoea, fatigue, and weight loss if left untreated. The long-term prognosis is generally good with prompt diagnosis and treatment.
The goal of treatment is to reduce inflammation and maintain remission. This may involve lifestyle changes such as diet modifications, stress management, smoking cessation, and regular exercise. Additionally, medications such as corticosteroids, immunosuppressants, biologic agents, and antibiotics may be prescribed to reduce inflammation and maintain remission.
In conclusion, Lifestyle changes and medication can be effective in managing this condition.
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Factors Affecting the Rate of Enzyme Activity Student Handout Introduction: In this tab, you will study an enzyme that is found in the cells of many living tissues. The name of the enzyme is catalase, it speeds up a reaction which breaks down hydrogen peroxide (H2O2), a toxie chemical, into two harmless substances water and oxygen The reaction is as follows: 2H2O2 → 2 H2O + O2 This reaction is important to cells because H:02 is produced as a by product of many normal cellular reactions. If the cells did not break down the H2O2, they would be poisoned and die. This reaction can be Hetected in the lab by observing the Oz bubbles generated. Purpose: The purpose of this lab is to plan and carry out experiments to test the effects of temperature and pH on activity of the enzyme catalase. 1. What is the effect of different temperatures on the functioning of the enzyme catalase? 2. What is the effect of varying pH on the functioning of the enzyme catalase?
The enzyme catalase is responsible for breaking down hydrogen peroxide into water and oxygen. Like many enzymes, the activity of catalase is sensitive to changes in temperature and pH leading to variations in enzyme activity.
The effect of temperature on catalaseThe optimal temperature for catalase activity in humans is around 37°C, which is the average body temperature. At this temperature, the enzyme works efficiently and can break down hydrogen peroxide quickly. If the temperature is too low, the reaction rate will be slower, and if it's too high, the enzyme can be damaged, resulting in a decrease in activity.
The effect of pH on catalaseThe optimal pH for catalase activity in humans is between pH 7 and 8. At low pH levels (acidic conditions), the enzyme can become denatured, which results in a decrease in activity. At high pH levels (alkaline conditions), the enzyme can become inactive, which also results in a decrease in activity.
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What is an example of a hypothesis relating to human evolution
or what makes us human that can be tested and supported by
empirical evidence?
A hypothesis relating to human evolution and what makes us human that can be tested and supported by empirical evidence is that human behavior is influenced by evolutionary processes, such as natural selection. This hypothesis can be tested through looking at data on human physical and behavioral characteristics in relation to their environment, as well as examining fossils and archaeological evidence.
This hypothesis can be tested by first gathering data on the physical and behavioral characteristics of humans in different environments, and then looking for correlations between those characteristics and the environment. This data can then be used to formulate a hypothesis about how natural selection has acted on humans, which can be tested further by examining fossil and archaeological evidence.
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Explain why cardiac and smooth muscles are involuntary. Why is
this an important characteristic?
Cardiac and smooth muscles are involuntary because they are not under conscious control; they are controlled by the autonomic nervous system. This is an important characteristic because it means that the muscles are able to respond quickly and reflexively to physiological changes.
Cardiac and smooth muscles are not under conscious control. This means that they contract and relax automatically in response to signals from the nervous system or hormones without requiring any conscious thought or effort.Involuntary muscles play an important role in the body by helping to maintain important bodily functions such as breathing and digestion.
Cardiac muscles are responsible for the contractions that drive blood flow through the circulatory system, while smooth muscles help to regulate organ and vessel size and control the movement of materials through the digestive and urinary tracts.
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epigenetic changes may be ___ but they aren't necessarily ___
Epigenetic changes may be inherited but they aren't necessarily reshaped.
What do you mean by epigenetic changes?Epigenetics is the study of stable phenotypic changes that do not involve alterations in the DNA sequence. The epigenetics implies features that are "on top of" or "in addition to" the traditional genetic basis for inheritance.
Several lifestyle factors have been identified that might modify epigenetic patterns, such as diet, obesity, physical activity, tobacco smoking, alcohol consumption, environmental pollutants.
Epigenetic modifications are heritable chemical or physical changes in chromatin. There are two types of epigenetic modifications – DNA methylation and histone modifications.
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1.how is motility accomplished by some bacteria?
2.what type of bacteria is the SIM test specific for?
3.what bacteria structural component permits locomotion?
4.what protein is it made of?
5.what are the different types of arrangement of this structural component?
6.what other types of bacterial structural components permit locomotion?
Motility in bacteria is achieved by the presence of flagella, which are long, whip-like appendages that rotate like propellers to propel the bacteria forward. (find the rest below)
How are bacteria motility described?The SIM (Sulfide-Indole-Motility) test is specific for detecting motility in bacteria, and does not target any specific type of bacteria. It is a test that is commonly used in microbiology to differentiate between motile and non-motile bacteria.
The structural component that permits locomotion in bacteria is the flagellum. It is a long, helical appendage that extends from the surface of the cell and rotates to generate movement.
The flagellum is made up of several different proteins, including the flagellin protein, which forms the filament of the flagellum.
There are different types of arrangements of flagella in bacteria, including:
Monotrichous: a single flagellum at one end of the cell
Amphitrichous: a single flagellum at both ends of the cell
Lophotrichous: a cluster of flagella at one or both ends of the cell
Peritrichous: flagella distributed over the entire surface of the cell
Other structural components that permit locomotion in bacteria include pili (short, hair-like structures that are involved in twitching motility) and slime (which can be used for gliding motility).
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Which best describes the role of enzymes in a chemical reaction ?
Explanation:
Enzymes are catalysts that speed up reactions by helping to lower the activation energy needed to start a reaction.
Q3. Provide a description of the rules of the model including the genetics effects on phenotype. Q4. Briefly discuss how you expect additivity to impact heritability and the response to selection. Q5. Briefly discuss how your result differ from your expectations and why these differences may have occured
Q3. The rules of the model include the genetic effects on phenotype, which describe how the combination of a person's genotype contributes to their physical appearance and other observable characteristics.
Q4. Additivity is expected to have a large impact on heritability and the response to selection.
Q5. The total effect of genetic factors was not quite as large as expected, meaning the heritability and the response to selection were also lower than expected.
In this model, the phenotype is determined by the action of additive genetic variance, meaning that the total effect of genetic factors is the sum of all the effects of each gene.
Heritability is determined by the amount of additive genetic variance, so more additive genetic variance will lead to a higher heritability. Since the effects of selection are proportional to the heritability, more additive genetic variance will also lead to a greater response to selection.
The results were not quite what was expected. This could be due to epistatic effects, where the effect of one gene may be influenced by the presence of another gene, or other environmental factors that affect the phenotype.
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Would detection of E. coli in meat be indicative of
contamination or spoilage of the product? (i.e. which of these two
is it indicative of and why)
Yes, the detection of E. coli in meat would be indicative of contamination. E. coli is a type of bacteria that can cause foodborne illnesses, so its presence is an indication that the meat has been contaminated by pathogens.
What Is E. Coli?E. coli is a type of bacteria that is commonly found in the intestines of animals and humans. It can be transferred to meat during the slaughtering process if the meat comes into contact with fecal matter. This type of contamination is a serious health concern because E. coli can cause severe illness or even death in some cases. Therefore, if E. coli is detected in meat, it is an indication that the meat has been contaminated and is not safe to eat. Spoilage, on the other hand, is the result of the natural decay of food over time, and is typically indicated by changes in color, texture, or odor. While spoilage can make food unappetizing, it is not necessarily a health concern like contamination is.
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Rima shone light into a light meter
Then she put a leaf in front of the meter
How did the leaf affect the reading?
The leaf reduced the light reaching the meter, causing the reading to decrease due to light absorption or scattering properties of the leaf.
The leaf probably decreased the amount of light that reached the light metre, lowering the value.
A lesser light intensity reaches the metre when leaves are present as opposed to when the metre is exposed directly to the light source because leaves are frequently opaque to variable degrees and can absorb or scatter light.
Thus, the reading on the light metre would indicate this change in light intensity.
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Fluid is flowing through a pipe to a bioreactor at a volumetric flow rate (Q) of 5000 cm3/sec. The flow is fully developed (i.e. Poiseuille flow). The fluid then flows into a pipe with a radius (R) that is 2X larger than that of the previous pipe. If all other parameters are equal, what is the new flow rate?
When the fluid flows into a pipe with a radius R that is 2X larger than the previous pipe, the new flow rate is 1250 cm³/sec.
The new flow rate Q is related to the previous flow rate Q₀ as follows:
Q/Q₀ = (R₀/R)²
Where R₀ is the radius of the previous pipe, and R is the radius of the new pipe.
The volumetric flow rate (Q) through the first pipe is 5000 cm³/sec. Since the flow is fully developed (i.e., Poiseuille flow), the volumetric flow rate (Q) is constant throughout the pipe. Therefore, Q₀ = Q = 5000 cm³/sec.
Since the new pipe's radius is 2 times larger than that of the previous pipe, we have
R = 2R₀
Plugging this value of R into the equation above gives:
Q/Q₀ = (R₀/R)² = (R₀/2R₀)² = 1/4 = 0.25
Hence, the new flow rate Q is:
Q = Q₀ x(R₀/2R₀)² = 5000 cm³/sec x 0.25 = 1250 cm³/sec
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difference in rates of survival and reproduction
The difference in rates of survival and reproduction among individuals in a population is due to natural selection.
What factors account for the difference in rates of survival and reproduction?Natural selection accounts for the difference in rates of survival and reproduction.
Natural selection is the process by which traits that provide a survival or reproductive advantage become more common in a population over time, while traits that do not provide an advantage become less common.
Natural selection occurs because individuals with advantageous traits are more likely to survive and reproduce than individuals with less advantageous traits. As a result, the advantageous traits become more common in the population over time, while the less advantageous traits become less common.
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The Illinois corn oil content study shows that you can select for increased or decreased oil content. If one keeps selecting only the highest oil content plants to breed each generation, what will happen to oil content in the long term?
In the long term, the oil content in the corn plants will continue to increase. This is because by selecting only the highest oil content plants to breed each generation, the genes that are responsible for high oil content are being passed on to the next generation.
Over time, this will lead to an increase in the overall oil content in the corn plants. This process is known as artificial selection, where humans select for certain traits in a population and only allow those with the desired traits to reproduce. By continuing to select for high oil content in each generation, the overall oil content in the corn plants will increase over time.
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so what region of an mRNA molecule does an enzyme that removes methylated guanosines during regulation bind to?
An enzyme that removes methylated guanosines during regulation binds to the 5' region of an mRNA molecule. This region is known as the 5' cap and is important for protecting the mRNA from degradation and for promoting its translation into protein. The removal of methylated guanosines by the enzyme is a key step in the regulation of gene expression, as it can influence the stability and translation efficiency of the mRNA.
The enzyme that removes methylated guanosines during regulation binds to the 5' untranslated region (UTR) of the mRNA molecule. Specifically, it binds to the 7-methylguanosine cap, which is a modification added to the 5' end of the mRNA during RNA processing. This cap is important for stabilizing the mRNA molecule and facilitating translation initiation. The removal of the cap by demethylases can affect the stability and translation efficiency of the mRNA, thus impacting gene expression.
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Transporter proteins aid the cells in getting energy substrates across the membrane. Different glucose transport protein have different sites of expression. Mention 5 glucose transport proteins with their sites of expression. There are many organs involved and many more enzymes required for the digestion and the absorption from the oral cavity to the colon. Mention at least five enzymes that are being utilized within oral cavity, stomach, and small intestine. Mention the function of each enzyme.
Many transport proteins help move energy sources across the membrane. Some of these proteins include different glucose transport proteins.
Several enzymes are involved in the digestion and absorption of nutrients from the oral cavity to the colon. These include:
Salivary amylase: This enzyme is found in the oral cavity and is responsible for the breakdown of starch into maltose.Pepsin: This enzyme is found in the stomach and is responsible for the breakdown of proteins into peptides.Lipase: This enzyme is found in the stomach and is responsible for the breakdown of fats into fatty acids and glycerol.Pancreatic amylase: This enzyme is found in the small intestine and is responsible for the breakdown of starch into maltose. Trypsin: This enzyme is found in the small intestine and is responsible for the breakdown of proteins into peptides.Learn more about transport proteins at https://brainly.com/question/24253003.
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Which of the following projects posteriorly of the two laminae of a typical vertebra?
The structure that projects posteriorly of the two laminae of a typical vertebra is the spinous process.
The spinous process is a bony projection that extends posteriorly from the point where the two laminae of the vertebra meet.
It serves as an attachment site for muscles and ligaments that help stabilize the spine and facilitate movement. The spinous process is one of several bony projections found on a typical vertebra, along with the transverse processes, which extend laterally, and the superior and inferior articular processes, which are involved in articulating with adjacent vertebrae. Each of these structures plays an important role in the overall structure and function of the spine.
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Review Questions 3.1 What is a genotype? 3.2 What did Mendel's experiments on the garden pea show us about the nature of genetic transmission? 3.3 What is a neutral mutation? 3.4 What is heritability?
3.1 A genotype is the genetic makeup of an individual or organism.
3.2 Mendel's experiments on the garden pea showed us that genetic transmission.
3.3 A neutral mutation is a change in an organism's DNA that does not affect its phenotype.
3.4 Heritability is the proportion of variation in a trait that is due to genetic factors.
3.1 A genotype is the genetic makeup of an individual or organism, which determines its physical and behavioral traits. It is made up of an individual's DNA, which contains the instructions for how an organism will develop and function.
3.2 Mendel's experiments on the garden pea showed us that genetic transmission follows certain patterns and that traits are inherited from one generation to the next through the passing down of genes. His experiments also helped us understand the concept of dominant and recessive traits, and how they are inherited.
3.3 A neutral mutation is a change in an organism's DNA that does not affect its phenotype, or physical and behavioral traits. These mutations are usually silent, meaning that they do not result in any noticeable changes to the organism.
3.4 Heritability is the proportion of variation in a trait that is due to genetic factors. It is a measure of how much of the variation in a trait within a population is due to differences in genes, as opposed to environmental factors.
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Modern microbiology began in the 17th century with invention of the telescope. a) true b) false
Answer:
false
Explanation:
Microbiology would be studying smaller organisms that are either hard or impossible to see with the bare eye. A telescope was invented to look at the stars. So false, microbiology did NOT begin with the invention of the telescope.
Cells and ToolsName the hypothesized first self-replicating molecule. What characteristics of this molecule led to this hypothesis?
All living cells contain ribonucleic acid (abbreviated RNA), a nucleic acid with properties comparable to those of DNA. Nevertheless, RNA is often single-stranded, unlike DNA. Instead of the deoxyribose present in DNA, the backbone of an RNA molecule is made up of alternating phosphate groups and the sugar ribose.
The hypothesized first self-replicating molecule is RNA. The characteristics of this molecule that led to this hypothesis are:
RNA can store genetic information, similar to DNA.RNA can catalyze chemical reactions, similar to enzymes.RNA can self-replicate, meaning it can create copies of itself.These characteristics suggest that RNA may have been the first self-replicating molecule and played a crucial role in the origin of life on Earth.
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In situ hybridization shows that Lim Homeobox protein
2(Lh×2)
is expessed in the optic cup (OC), telencephalon (T), midbrain (M), hindbrain (H), and limb buds (LB) in
11dpc
mouse embryos. A. All somatic cells of a chicken have the same genome, yet
fgf25
is only expressed by some cells. How can a gene be expressed in one cell but not another? ( 5 pts) A qune can be exprssed in one cen but not another by cul diffirentiation. Even theugh an organim cerries the some set of DNh. defferent genes tan be ceprited in difeernt cetus why, how? B. To identify the cis-regulatory regions that control Lhxz expression, you make a series of GFP reporter constructs. In the diagram below, deleted regulatory regions are shown as thin black lines, while the sequences that are still present are shown as gray boxes. Images on the right show GFP expression. Based on the results below, identify the regulatory regions that control Lhxz expression. Label the regulatory regions fenhancer, repressor or promoteri on the top construct of the diagram below. For each enhancer, you must include the Bonus point: Were the locations of all cis-regulatory regions identified in this experiment? Why or why not? Yes, thy wer identiaed, because in the diagrem if is shown that the protans present ware taigcted
Different genes can be expressed in different cells due to the process of epigenetic modification. Epigenetic modification is the process in which genes can be switched "on" or "off" without changing the underlying genetic code. It involves changing how genes are packaged and regulated in the cell, which is known as chromatin remodeling.
Epigenetic modification allows cells to express different genes depending on their current state or environment. By analyzing the diagram, the regulatory regions that control Lhxz expression can be identified as enhancer, repressor, and promoter.
The enhancer is the thin black line in the top construct, the repressor is the thin black line in the middle construct, and the promoter is the gray box in the bottom construct.
Of all cis-regulatory regions identified, because in the diagram it is shown that the promoters present were targeted. The diagram is a graphical representation of the results from the experiment, showing which regulatory regions were found to control Lhxz expression.
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1. Nurse cells in the insect ovarioles are: A. haploid cells B. polyploid C. eggs D. Leydig 2. Gonadotropin-releasing hormone (GRH) is produced by: A. cortex of the brain B. stem of the brain C. hypothalamus D. anterior pituitary gland E. sella turcica 3. Equivalent of hypothalamus found in amphibian, mammals and other higher animals, in insects is: A. fat body B. follicle cell C. corpus allatum D. ecdysone E. vitellogenin 4. In the process of oocyte maturation in amphibians, progesterone is secreted by: A. oocyte B. follicle cells C. zona granulosa D. jelly layer E. yolk 5. What is the key protein required during the process of maturation of Xenopus oocyte? A. FSH B. MPF C. MPK (mitogen dependent protein kinase) D. mos E. LH throughout oogenesis
1.Nurse cells in the insect ovarioles are B. polyploid.
These cells are responsible for providing nutrients to the developing oocyte.
2. Gonadotropin-releasing hormone (GRH) is produced by C. hypothalamus. This hormone stimulates the release of follicle-stimulating hormone (FSH) and luteinizing hormone (LH) from the anterior pituitary gland.
3. The equivalent of hypothalamus found in amphibian, mammals, and other higher animals, in insects is C. corpus allatum. This gland is responsible for producing juvenile hormone, which regulates growth and development in insects.
4. In the process of oocyte maturation in amphibians, progesterone is secreted by B. follicle cells. This hormone helps to prepare the oocyte for fertilization and subsequent development.
5. The key protein required during the process of maturation of Xenopus oocyte is B. MPF (maturation promoting factor). This protein is responsible for initiating the process of oocyte maturation and is essential for the successful completion of oogenesis.
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Wich of these describes a difference between viruses and cells
Answer:
Viruses
Needs a host to reproduce
Has capsid
No growth
Doesn't get or use energy
Cells
Can reproduce by itself
Has cell membrane
Growth
Get and use energy
Respond to environment
Classification Activity Worksheet Instructions: Read each myth (untruth). Reword it to make a factual statement. Then, give two to three reasons why the myth is untrue. Use complete sentences and support your answer with evidence, using your own words. Myth: A dead organism is the same as a nonliving thing in science. Fact: Evidence: Myth: The Linnaeus system of classification will always stay the same. Fact: Evidence: Myth: Tigers and goldfish are not related. Fact: Evidence: Myth: An organism's kingdom only describes physical characteristics. Fact: Evidence: Myth: Mammals and plants don't belong in the same domain. Fact: Evidence: Your Turn Come up with another myth about the classification of organisms. Then, give two to three reasons why the myth is untrue. Use complete sentences and support your answer with evidence, using your own words. Your myth: Fact: Evidence: I JUST NEED HELP WITH #6
Nonliving things can vary greatly from living things. An organism's realm describes more than just its physical properties.
Distinguish between myth and fact
The capacity to assemble scientific evidence to support a claim is the foundation of a fact. Myths are based on historical ideas and beliefs, whereas facts provide support.
Myth: In terms of science, a living organism is equivalent to a nonliving entity.
Fact: Nonliving things can vary greatly from living things.
Evidence: Once upon a time, living things were composed of distinct materials than non-living ones. Despite not being alive, rocks don't have the same structure as creatures that are. Given that it includes components like cells, proteins, etc., if we were to examine a once-living object that is now dead, we would probably infer that it was a living thing. Dirt or any other similar substance wouldn't have done it.
Myth: The Linnaeus system of classification will always stay the same.
Fact: The reason for the Linnean system of classification's widespread recognition is its ability to transmit intricate connections to scientists around the globe.
Evidence: Despite the fact that many scientists attempted to classify living organisms, Carl Linnaeus was the first to develop a sequential categorization of living organisms under various taxons, including kingdom, phylum, class, order, family, genus, and species.
Myth: Tigers and goldfish are not related.
Fact: As vertebrate creatures, they share some traits and are members of the same Phylum even though they are not closely related.
Tigers and goldfish pertain to different Classes, as explained. The class Mammalia includes tigers (Panthera tigris), while the class Actinopterygii includes goldfish (Carassius auratus). But they are both members of the Phylum Chordata.
Fact: An organism's realm describes more than just its physical properties.
Evidence: Kingdom Plantae: Photosynthesis is how they get their sustenance. mate in an asexual way.
They stalk or eat other living things to get their food, which belongs to the kingdom Animalia. physically reproduce.
Myth: Mammals and plants don't belong in the same domain.
Fact: Plants and mammals share the same domain.
Evidence: Domains are categorized according to their cell intricacy and number of cells.
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