A study was conducted on a population of snowshoe hares in Algonquin Provincial Park, Ontario, Canada. 110-month-old hares were snatched from their nests, micro-transmitters were inserted under their skin, and they were quickly returned to their homes. Scientists monitored their activities on monthly basis, and during the breeding season checked nests and counted the babies produced. Below is a summary of the data for only the female hares that were tagged (60 individuals). For simplicity, the monthly data has been summed into yearly totals.
Survival: 60 were alive at the start the experiment (year = 0), 20 at the start on the next year (year 1), 8 at year 2, 1 at year 3, and 0 at year 4.
Fecundity: during year 0, no babies were produced; year 1- females produced an
average of 4 babies each; year 2 - 8 babies each; and year 3 - 8 babies each.
Q1. Create a life table from this data. Include nx, dx, lx, mx, lxmx, and xlxmx
Q2. What is the net reproductive rate, R0? What is the generation time T?
Q3. Given that Snowshoe Hare populations increase geometrically, if there are 80 female Snowshoe Hares alive at time t, how many Snowshoe Hares will be alive at time t+1?
Q4. If there are 80 female Snowshoe Hares alive at time t, what will the Snowshoe Hares population be at time t + 5?

Answers

Answer 1

Q1. The life table from the data is as follows:

Year (x) | nx | dx | lx | mx | lxmx | xlxmx
-------- | -- | -- | -- | -- | ---- | ------
0 | 60 | 40 | 1.00 | 0 | 0.00 | 0.00
1 | 20 | 12 | 0.33 | 4 | 1.33 | 1.33
2 | 8 | 7 | 0.13 | 8 | 1.07 | 2.13
3 | 1 | 1 | 0.02 | 8 | 0.13 | 0.40
4 | 0 | 0 | 0.00 | 0 | 0.00 | 0.00

Q2. The net reproductive rate is 1.54.

Q3. So if there are 80 female Snowshoe Hares alive at time t, there will be 202.4 Snowshoe Hares alive at time t+1.

Q4. If there are 80 female Snowshoe Hares alive at time t, the population at time t + 5 will be 80 * 2.53^5 = 1654.6 Snowshoe Hares.

Q1 . This is a life table, which shows information about the mortality experience of a hypothetical population.

The table includes the number of individuals alive at the beginning of each age interval (nx), the number of deaths during the interval (dx), and the resulting number of individuals alive at the end of the interval (lx).

The table also includes the probability of dying during each interval (mx), the number of person-years lived during each interval (lxmx), and the expected number of years left to live at the beginning of each interval (xlxmx).

Q2 . The net reproductive rate, R0, is the sum of the lxmx column, so R0 = 0.00 + 1.33 + 1.07 + 0.13 + 0.00 = 2.53.  The generation time T is the sum of the xlxmx column divided by R0, so T = (0.00 + 1.33 + 2.13 + 0.40 + 0.00)/2.53 = 1.54.

Q3 . Given that Snowshoe Hare populations increase geometrically, the number of Snowshoe Hares alive at time t+1 will be the number alive at time t multiplied by the net reproductive rate, R0. So if there are 80 female Snowshoe Hares alive at time t, there will be 80 * 2.53 = 202.4 Snowshoe Hares alive at time t+1.

Q4 . The calculations in this answer are based on the assumption that the population is closed (i.e., there is no immigration or emigration) and that the survival and fecundity rates remain constant over time.

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Related Questions

Can you please make 1 sentence
out of the two words. without just simply adding the
defintions.
1. crossing
over
2. synteny
testing

Answers

The synteny testing is a powerful tool for genetic analysis that allows scientists to gain insights into the complex relationships between different species and the genetic factors that shape their evolution.

Synteny testing is a genetic analysis method that compares gene arrangements and chromosomal organizations to determine evolutionary relationships between species.In the field of genetics, synteny testing is a method of analyzing genetic structures and evolutionary relationships between species. This approach focuses on comparing gene arrangements and chromosomal organizations between different species to determine how they evolved and how closely related they are. By examining patterns of synteny, researchers can identify which genes and regions of the genome are likely to be conserved across different species, and can use this information to better understand how these genes function and how they have evolved over time.Synteny testing has become increasingly important in recent years as genetic research has advanced, and many scientists are now using this approach to study a wide range of biological phenomena. For example, synteny testing is often used to study the evolution of different animal species, as well as to identify key genetic differences between species that may be responsible for differences in traits or behaviors. Overall, synteny testing is a powerful tool for genetic analysis that allows scientists to gain insights into the complex relationships between different species and the genetic factors that shape their evolution.

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a food chain follows the connection between one

Answers

A food chain follows the connection between one producer and a single chain of consumers within an ecosystem.

What is an Ecosystem?

This is a term which consists of all the organisms and the physical environment with which they interact. The biotic and abiotic components which are present are linked together through nutrient cycles and energy flows.

Food chain on the other hand is referred to as the sequence of transfers of matter and energy in the form of food from organism to organism in which the producer makes the food which is the passed to other organisms through various processes and mechanisms.

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How do the properties of water benefit freshwater fish in Ontario during the winter?

Answers

The properties of water, including its high heat capacity, expansion when freezing, and cohesive property, all benefit freshwater fish in Ontario during the winter by helping to maintain a stable and safe environment for them to live in.

The properties of water benefit freshwater fish in Ontario during the winter in several ways. First, the high heat capacity of water helps to keep the water temperature stable, even during extreme temperature fluctuations. This allows fish to maintain their body temperature and metabolism without experiencing stress or harm.

Second, the fact that water expands when it freezes is also beneficial for freshwater fish in Ontario during the winter. This expansion creates a layer of ice on the surface of the water, which acts as an insulator and helps to prevent the water from freezing solid. As a result, fish are able to continue living in the water, even when the air temperature drops below freezing.

Lastly, the cohesive property of water, which allows it to stick together, also benefits freshwater fish in Ontario during the winter. This property helps to keep the water from evaporating, which helps to maintain a stable water level and prevent the fish from becoming stranded or exposed to predators.

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A student has a cell suspension of 7.3x105 cells/mL.
They need to plate 5 mL of volume at a cell density of
4.7x103 cells/mL in a T25 cell culture flask. How are
they going to plate the cells?

Answers

The student will need to add 0.032 mL of the cell suspension to 75 mL of medium and mix it thoroughly before plating 5 mL of the diluted cell suspension into the T25 cell culture flask.

To plate the cells at the desired cell density of 4.7x10³ cells/mL, the student will need to dilute the cell suspension. This can be done using the following formula:

                             C₁V₁ = C₂V₂

Where C₁ is the initial cell concentration, V₁ is the initial volume, C₂ is the final cell concentration, and V₁ is the final volume. Plugging in the values from the question, we get:
(7.3x105 cells/mL)(V₁) = (4.7x103 cells/mL)(5 mL)

Solving for V₁ gives us the volume of the original cell suspension that we need to use:
V₁ = (4.7x103 cells/mL)(5 mL) / (7.3x105 cells/mL)
V₁ = 0.032 mL

So, the student will need to take 0.032 mL of the original cell suspension and add it to 4.968 mL of media to get a final volume of 5 mL at the desired cell density of 4.7x10³ cells/mL. This can be done using a micropipette and a sterile culture flask.

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When a solution outside the cell is hypertonic compared to the inside of the cell, what can we expect to see in the cell?What is the term given to what the cell is experiencing?

Answers

When a solution outside the cell is hypertonic compared to the inside of the cell, the cell will shrink and lose water due to osmosis. This is known as crenation.

When a solution outside the cell is hypertonic compared to the inside of the cell, we can expect to see water moving out of the cell. This causes the cell to shrink and become dehydrated. The term given to what the cell is experiencing is called "crenation" in animal cells and "plasmolysis" in plant cells.
This occurs because the hypertonic solution has a higher concentration of solutes than the inside of the cell, causing the water to move out of the cell in an attempt to balance the concentrations. This process is known as osmosis.
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What is immunity? a condition of being able to resist a particular disease a condition in which an individual becomes infected by a disease causing agent a condition of being susceptible to a disease all of the above

Answers

Immunity is a condition of being able to resist a particular disease. This means that an individual's immune system is able to recognize and effectively fight off a disease-causing agent, preventing them from becoming infected or experiencing severe symptoms.

Immunity can be acquired through natural exposure to a disease, through vaccination, or through the transfer of antibodies from one individual to another. It is an important aspect of overall health and helps to prevent the spread of disease within a population.

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"Place the components of photosynthesis in order
Light-dependent stage - Photosystem I
Calvin Cycle - reduction
Light-dependent stage - Photosystem II
Calvin Cycle - regeneration
Light-dependent stage"

Answers

The correct order of the components of photosynthesis is as follows:

Light-dependent stage - Photosystem IILight-dependent stage - Photosystem ICalvin Cycle - reductionCalvin Cycle - regeneration

Photosynthesis is the process by which organisms with chlorophyll, such as plants, algae and some bacteria, convert light energy from the sun into chemical energy.

In photosynthesis, during the light-dependent stage, energy from sunlight is used to produce ATP and NADPH, which are then used in the Calvin Cycle to produce glucose. Photosystem II occurs first, followed by Photosystem I. The Calvin Cycle then takes place in two stages, the reduction stage and the regeneration stage.

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To get past the endodermis and enter the plant's xylem tissues solutes must? a) Move only through the apoplast b)Enter the symplast and stay there until it reaches the leaves c) Cross a membrane to enter the symplast to get past the casparian strip, and then cross a membrane to enter the xylem d) Move with the gradient caused by electrochemical potential

Answers

Cross a membrane to enter the symplast to get past the casparian strip, and then cross a membrane to enter the xylem. The correct answer is c.

The endodermis is a layer of cells that surrounds the vascular tissue in plant roots. It serves as a barrier to prevent the free flow of water and solutes into the xylem.

The Casparian strip is a band of cell wall material that blocks the apoplast pathway and forces solutes to move through the symplast pathway. In order to get past the endodermis and enter the xylem, solutes must cross a membrane to enter the symplast, move through the symplast until they reach the Casparian strip, and then cross another membrane to enter the xylem. This ensures that only the necessary solutes enter the xylem and are transported to the rest of the plant.

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what are some behavioral response to exercise

Answers

The behavioral response to exercise are smoking like activities affect.

What is health data?

Data support for an organization's business goals is referred to as the data health of that organization. When individuals who need to use it can find, understand, and value the data quickly and consistently throughout its existence, the data is said to be healthy.

What is health care ?

Health care of the highest caliber enhances life quality and aids in disease prevention.

Therefore, behavioral response to exercise are smoking like activities affect.

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Question
12(7
pointed) This table shows the
F2
data from a typical cross of pea plants. The
P
generatic cross was between a plant with tall stems and purple leaves and a plant with sh stems and white leaves. The F1 all had tall stems and purple leaves. Your job is to do a Chi-squared Goodness of Fit analysis- this is a typical ditybri cross with unlinked genes like Mendel would have performed. Do the calculations on a scratch paper and your calculator, then type your calcul numbers in the blanks in this question. Remember, you HAD to show your blank scratch paper to the screen before you started the exam. If you didn't, do it NOV What is the Expected \# of tall steams with purple flowers (the first row)? What is the expected \# for the second row, tall stems white leaves? What is the expected \# for the third row, short stems with purple leaves? What is the expected \# for the fourth row, the short stems with white leaves? Complete the calculation for the chi-square value and enter your
x 2
value it here, round to the nearest hundredth:.

Answers

The chi-square value from table that shows the F2 data from a typical cross of pea plants , rounded to the nearest hundredth, is 28.67.

In order to answer this question, we must first calculate the expected number of tall stems with purple flowers, tall stems with white leaves, short stems with purple leaves, and short stems with white leaves. To do this, you must use the equation:
Expected # = (row total × column total) ÷ grand total

Using this equation, the expected # of tall stems with purple flowers is:
Expected # = (54 × 22) ÷ 130 = 24.92

The expected # of tall stems with white leaves is:
Expected # = (54 × 8) ÷ 130 = 12.31

The expected # of short stems with purple leaves is:
Expected # = (76 × 22) ÷ 130 = 34.08

The expected # of short stems with white leaves is:
Expected # = (76 × 8) ÷ 130 = 14.62

To calculate the chi-square value, use the equation:
X2 = ∑ (O-E)2 / E

Where O is the observed value, and E is the expected value.

Therefore, the chi-square value is:
X2 = (24-24.92)2/24.92 + (14-12.31)2/12.31 + (18-34.08)2/34.08 + (22-14.62)2/14.62 = 28.67

Therefore, the chi-square value, rounded to the nearest hundredth, is 28.67.

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which protein turns oxygen into water?

Answers

The protein that turns oxygen into water is called cytochrome c oxidase, also known as Complex IV.

Proteins explained

Protein is a large biomolecule that is essential for life. It is made up of chains of smaller building blocks called amino acids, which are linked together by peptide bonds.

Proteins have many important functions in the body, including serving as enzymes that catalyze chemical reactions, as structural components of cells and tissues, and as signaling molecules that coordinate biological processes.

Proteins are found in many foods, including meat, fish, eggs, dairy products, legumes, and nuts. When we eat protein, our digestive system breaks it down into its individual amino acids, which are then used by our cells to build new proteins and carry out a wide range of biological processes.

The protein that turns oxygen into water is called cytochrome c oxidase, also known as Complex IV. It is a crucial component of the electron transport chain in aerobic respiration, which takes place in the mitochondria of cells.

Therefore, Cytochrome c oxidase catalyzes the reduction of molecular oxygen (O2) to water (H2O) using electrons from cytochrome c and protons from the mitochondrial matrix.

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The reactive lymphocytosis is due to blastogenic _____ transformation resulting in cytotoxic potential that limits the proliferation of the infected B cells.

Answers

The reactive lymphocytosis is due to blastogenic T-cell transformation resulting in cytotoxic potential that limits the proliferation of the infected B cells.


Lymphocytosis is an increase in the number of lymphocytes in the blood. Reactive lymphocytosis occurs when there is an increase in the number of reactive lymphocytes, which are a type of white blood cell that helps to fight infection. Blastogenic transformation is the process by which lymphocytes are activated and begin to proliferate in response to an infection or other stimulus. T-cells are a type of lymphocyte that play a crucial role in the immune response, including the ability to kill infected cells and limit the proliferation of infected B cells. Thus, reactive lymphocytosis is due to blastogenic T-cell transformation, which results in an increase in the number of cytotoxic T-cells that can help to limit the proliferation of infected B cells.

An increase in the quantity or percentage of lymphocytes in the blood is known as lymphocytosis. Relative lymphocytosis refers to the condition where the proportion of lymphocytes relative to white blood cell count is over the normal range, whereas absolute lymphocytosis refers to an increase in the lymphocyte count above the normal range. Absolute lymphocytosis is defined as the presence of more than 5000 lymphocytes per microliter (5.0 x 109/L) in adults, 7000 or more in older children, and 9000 or more in newborns. 20% to 40% of the white blood cells that are in circulation typically are lymphocytes. Relative lymphocytosis is defined as the presence of more than 40% lymphocytes.

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The antibiotic rifampicin targets the bacterial RNA polymerase and inhibits RNA synthesis by physically blocking the elongation step. Resistance to rifampicin occurs by target modification (mutations occur in the gene that encodes one of the protein subunits of the RNA polymerase). Would you expect rifampicin resistance to exhibit a trade-off with bacterial growth rate? Explain why or why not.

Answers

Yes, rifampicin resistance is likely to exhibit a trade-off with bacterial growth rate.

Rifampicin is a potent antibiotic that works by targeting the RNA polymerase enzyme, which is responsible for gene transcription. Rifampicin binds to the β-subunit of RNA polymerase and prevents it from elongating the growing RNA chain by interfering with the enzyme's ability to bind nucleoside triphosphates.

In the presence of rifampicin, cells with an altered RNA polymerase subunit gene (rpoB) that results in reduced rifampicin binding are expected to have a selective advantage over cells with the wild-type gene. The cells with the mutated gene will be able to transcribe genes at a faster rate than cells with the wild-type gene, allowing them to grow faster.

When cells are exposed to rifampicin, resistant cells will have a growth advantage over susceptible cells. However, when rifampicin is removed, cells with mutated rpoB genes may have reduced RNA polymerase efficiency, resulting in a decreased growth rate. The reduced RNA polymerase efficiency may result in a trade-off between rifampicin resistance and bacterial growth rate.

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A global positioning system (GPS) is a navigation tool that can provide a user’s exact location any time of day in any weather condition. The system sends and receives radio signals from Earth to satellites in space. Explain why Einstein’s general relativity theory is important to the makers of GPS systems.

Answers

A Global Positioning System (GPS) explains the gravitational force of massive bodies, like the Earth, affects the transition of time.

What is a Global Positioning System (GPS)?

It is a system that uses space satellites to provide positioning, navigation, and timing information. It is a navigation tool, that provides the user’s correct location at any time of day in any weather condition.

As suggested by Einstein's theory, clocks experience the force of gravity running at a slower rate than clocks seen from a distant region undergoing weaker gravity.

Therefore, it suggests that clocks on Earth found from orbiting satellites run at a slower rate.

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which bacteria group has a genomes similar to that of mitochondrial
DNA?
a. escherichia coli
b. rickettsia spp.
c. mycobacterium spp.

Answers

The bacteria group that has a genome similar to that of mitochondrial DNA is rickettsia spp. (Option B).

Mitochondria are organelles found in eukaryotic cells that are responsible for producing energy in the form of ATP. They contain their own DNA, which is circular and similar to that of bacteria. This has led scientists to believe that mitochondria were once free-living bacteria that were engulfed by a host cell and became endosymbionts.

Rickettsia spp. are a group of bacteria that are known to be intracellular parasites, meaning they live and reproduce inside host cells. Their genome is similar to that of mitochondrial DNA, which supports the endosymbiotic theory of mitochondrial evolution.

In contrast, Escherichia coli (Option A) and Mycobacterium spp. (Option C) are both free-living bacteria with genomes that are not similar to mitochondrial DNA.

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Examine this image of bacteria cells.

3D image of probiotic bacterial cells

Which of the following statements best describes a mutualistic relationship between human health and certain types of bacteria?

Intestinal bacteria obtain nutrients from the gut and aid in human digestion.
Bacteria in improperly prepared food are consumed by humans, causing food poisoning.
Humans treat infections with antibacterial medication, which bacteria become resistant to.
Invasive bacteria in an area of injury produce toxins that damage healthy tissues of the human body.

Answers

The following statement is  the mutualistic relationship that exists between certain kinds of bacteria and human health: Human digestion is aided by intestinal bacteria that obtain nutrients from the gut.

This statement explains how beneficial bacteria in the human gut play a crucial role in the breakdown of food, the extraction of nutrients, and the assistance in the synthesis of certain vitamins.

What are bacteria in the intestine?

Microorganisms known as intestinal bacteria can be found throughout the human digestive system, particularly the large intestine. The group of these bacteria is referred to as the gut flora or microbiota. There are hundreds of different species of bacteria, viruses, fungi, and other microorganisms in the complex ecosystem known as the gut microbiota.

What function do intestinal bacteria play?

By assisting in the digestion and absorption of nutrients, producing certain vitamins, and assisting in the maintenance of a healthy immune system, intestinal bacteria contribute significantly to human health.

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Answer:

A. Intestinal bacteria obtain nutrients from the gut and aid in human digestion.

Explanation:

i got it right on my quiz

Identify and explain the differences in the fate of NADH molecules produced in glycolysis AND their respective energy yields during the following conditions:
A. aerobic catabolism in a skeletal muscle fiber (cell)
B. catabolism in an erythrocyte (red blood cell, which lack mitochondria)
C. aerobic catabolism in a hepatocyte (liver cell)

Answers

The fate of NADH molecules produced in glycolysis AND their respective energy yields differ during the following conditions: Aerobic catabolism, Catabolism in an erythrocyte and hepatocyte.

A. Aerobic catabolism in a skeletal muscle fiber (cell): In this condition, the NADH molecules produced in glycolysis are transported into the mitochondria, where they are used in the electron transport chain to produce ATP. The energy yield in this condition is high, as each NADH molecule can produce up to 3 ATP molecules.

B. Catabolism in an erythrocyte (red blood cell, which lack mitochondria): In this condition, the NADH molecules produced in glycolysis are used to reduce pyruvate to lactate in a process called fermentation. The energy yield in this condition is low, as no ATP is produced from the NADH molecules.

C. Aerobic catabolism in a hepatocyte (liver cell): In this condition, the NADH molecules produced in glycolysis are transported into the mitochondria, where they are used in the electron transport chain to produce ATP. The energy yield in this condition is also high, as each NADH molecule can produce up to 3 ATP molecules.

In summary, the fate of NADH molecules and their respective energy yields differ depending on the presence or absence of mitochondria and the type of catabolism (aerobic or anaerobic) occurring in the cell.

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In glycolysis, NADH molecules are produced, and their fates and energy yields depend on the type of cell and the presence or absence of mitochondria.

The fate of NADH molecules produced in glycolysis and their respective energy yields are different under the following conditions:Aerobic catabolism in a skeletal muscle fiber (cell): In this condition, the NADH molecules produced in glycolysis are shuttled into the mitochondria, where they are used in the electron transport chain to produce ATP. The energy yield from this process is approximately 2.5 ATP molecules per NADH molecule.Catabolism in an erythrocyte (red blood cell, which lacks mitochondria): Since erythrocytes lack mitochondria, the NADH molecules produced in glycolysis cannot be used in the electron transport chain. Instead, they are used to reduce pyruvate to lactate, which is then transported out of the cell. The energy yield from this process is 0 ATP molecules per NADH molecule.Aerobic catabolism in a hepatocyte (liver cell): Like skeletal muscle fibers, hepatocytes have mitochondria and can use the NADH molecules produced in glycolysis in the electron transport chain to produce ATP. However, the energy yield in hepatocytes is slightly lower, at approximately 2.3 ATP molecules per NADH molecule, due to the presence of uncoupling proteins in the mitochondria that allow protons to leak back into the matrix without producing ATP.

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What process is food to amino acid

Answers

The process of food to amino acid is referred to as protein digestion.

What is Digestion?

This is referred to as the breakdown of large insoluble food molecules into small water-soluble food molecules so that they can be absorbed into the watery blood plasma.

Protein digestion begins when you first start chewing but once a protein source reaches your stomach, hydrochloric acid and enzymes called proteases break it down into smaller chains of amino acids which is used in the growth and replacement of worn out tissues in the body which is therefore the reason why it was chosen as the correct choice.

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Environmentalists are monitoring an area of tropical forest that is being
deforested because of human activities. The graph here shows the scientists'
predictions based on the data they have collected.

How is the ecosystem likely to change as a result? Select the two correct
answers.
MULTIPLE ANSWERS
A. Decrease in biodiversity
B. Gain of species
C. Gain in oxygen
D. Loss of habitat

Answers

Hi my guess is a and D

PLEASE HELP ME THIS IS DUE IN LESS THAN A HOUR!!

Answers

3. Frequency of the dominant allele= 0.41

Frequency of the recessive allele= 0.59

17% of the homozygous dominant

35% of the homozygous recessive.

48% heterozygous

What is the meaning of homozygous?

When two paired chromosomes harbour the same or identical alleles for a given characteristic at nearby loci, this condition is referred to as homozygosity (i.e. homologous chromosomes). An entity with two sets of chromosomes is said to be diploid. Both sets are inherited; one set is from the mother and the other from the father. Based on their locations, each maternal chromosome can be matched with a corresponding paternal chromosome. Homozygous occurs when the same alleles are present at the loci in the corresponding chromosomes. It indicates that the same trait is coded for by both alleles.

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What is DNA replication?

Answers

Answer:

the process by which the genome's DNA is copied in cells.

Explanation:

Answer:

DNA replication is the process by which DNA makes a copy of itself during cell division

Explanation:

Hope this helps UwU

List four factors that drive water scarcity for human societies in
different regions of the globe, and for each write a sentence
explaining the factor

Answers

Water scarcity is a major issue for human societies in different regions of the globe. There are several factors that drive water scarcity, including: Population growth, Climate change, Pollution, Overuse.

Population growth: As the global population continues to increase, the demand for water also increases. This puts a strain on available water resources and can lead to scarcity in regions with high population growth.Climate change: Changes in climate can affect water availability and distribution. For example, regions that experience droughts may have less available water for human societies to use.Pollution: The contamination of water sources by human activities such as industrial waste and agricultural runoff can reduce the amount of clean, usable water for human societies.Overuse: The overuse of water resources by human societies can lead to depletion of water sources and contribute to water scarcity in regions where water is already limited.

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The stage of mitosis depicted in the image is...
A anaphaseanaphase
B telophasetelophase
C prophaseprophase
D interphase

Answers

The stage of mitosis depicted in the image is anaphase.

What is mitosis?

Mitosis is a type of cell division that is essential for the growth and development of organisms. It involves the duplication of the genetic material in a cell’s nucleus, and the subsequent division of the nucleus into two new nuclei, each of which contains the same genetic material as the parent cell. During mitosis, the chromosomes (structures that contain the genetic material) are duplicated and then divided equally between the two new nuclei. The two new nuclei then separate from each other, resulting in two new cells, each with the same number of chromosomes as the parent cell.

This can be determined by the presence of the two sets of chromosomes migrating to opposite ends of the cell as part of the process of separating the sister chromatids. Anaphase is the fourth and final stage of mitosis, following prophase, metaphase, and interphase.

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Answer:The answer is prophase, I just did it.

Explanation:

PLS ANSWER MY QUESTION (I WILL MARK THE BRAINLIEST IF ANSWERED CORRECTLY)

Answers

Answer:

Explanation:

There is not a trend in this graph. It is scattered.

So you know that the cloudier the broth is, the more microbes are growing in the broth. How can you use this information to determine if a newly discovered species is a psychrophile, mesophile, or thermophile. Describe a very simple experiment to use the information in the first sentence to determine their temperature preference. (you'll have to look these up from a source you trust)..

Answers

To determine if a newly discovered species is a psychrophile, mesophile, or thermophile based on broth cloudiness, one could incubate the species in broth at various temperatures and observe the level of cloudiness to determine their temperature preference.

To conduct this experiment, a sample of the newly discovered species would be inoculated into separate flasks of nutrient broth and incubated at different temperatures - such as 4°C, 37°C, and 60°C - for a set period of time. The level of cloudiness would then be observed and compared between the different temperature conditions.

A psychrophile would show increased cloudiness at the lower temperature, a mesophile would show increased cloudiness at the middle temperature, and a thermophile would show increased cloudiness at the higher temperature.

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Let’s return to pigeons for our last set of questions. However, this time, we will look at 2 genes at once.Recall the slipper gene from question 2 that affects feathering on the feet. It displays incomplete dominance in that the heterozygote has an intermediate level of foot feathering. There are two main alleles for this trait. S1 = slipper = foot feathering S2 = no slipper = no feathers on feet Pigeons can also have a crest of upturned feathers on their head or not. The main gene controlling this has 2 alleles and complete dominance:
N = No crest = dominant n = Crest = recessive
What is the phenotype of a pigeon with genotype S1S2Nn?​​​​​​​
What is the genotype of a pigeon with Full foot feathering and a crest?
Do a Punnett square for the following cross and indicate what fraction of offspring would be expected to have Partial foot feathering AND no crest. S1S2Nn X S2S2Nn

Answers

The phenotype of a pigeon with genotype S1S2Nn is partial foot feathering and no crest. This is because the slipper gene displays incomplete dominance, so the heterozygote (S1S2) has an intermediate level of foot feathering.

The no crest gene (N) is dominant, so even though the pigeon has one copy of the recessive crest gene (n), it will not have a crest.

The genotype of a pigeon with full foot feathering and a crest is S1S1nn. This is because full foot feathering is only seen in pigeons that are homozygous for the slipper gene (S1S1), and a crest is only seen in pigeons that are homozygous for the recessive crest gene (nn).

Punnett square for the cross S1S2Nn X S2S2Nn:

|   | S1N | S1n | S2N | S2n |
|---|-----|-----|-----|-----|
| S2N | S1S2NN | S1S2Nn | S2S2NN | S2S2Nn |
| S2n | S1S2Nn | S1S2nn | S2S2Nn | S2S2nn |

The fraction of offspring expected to have partial foot feathering and no crest is 2/8 or 1/4. This is because there are two offspring with the genotype S1S2Nn (which results in the desired phenotype) out of a total of eight offspring.

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I have to put together a presentation of a specific scientific paper. What is the difference between a "revised diagnosis" and a "diagnosis" when both are included in the paper? Is the revised diagnosis this currently accepted diagnosis?

Answers

The difference between a "revised diagnosis" and a "diagnosis" in a scientific paper is that a revised diagnosis is an updated or corrected version of the original diagnosis.

What's diagnosis

A diagnosis is an initial identification of a condition or disease based on a patient's symptoms and medical history. However, as new information or evidence becomes available, the diagnosis may be revised or updated to reflect a more accurate understanding of the patient's condition.

In a scientific paper, a revised diagnosis is typically included to reflect changes or updates to the original diagnosis based on new information or evidence.

This may include new symptoms, test results, or other data that were not available at the time of the original diagnosis. The revised diagnosis is typically the currently accepted diagnosis, as it reflects the most up-to-date understanding of the patient's condition. In summary, a revised diagnosis is an updated or corrected version of the original diagnosis, and is typically the currently accepted diagnosis in a scientific paper.

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The first documented registry helped in caring and controlling which disease:
a) Black Plague. b) Malaria. c) Leprosy.

Answers

The first documented registry helped in caring and controlling the disease of Leprosy. The correct answer is option c) Leprosy.

Leprosy is a chronic infectious disease caused by the bacteria Mycobacterium leprae. It primarily affects the skin, peripheral nerves, mucosal surfaces of the upper respiratory tract, and the eyes. Leprosy can be cured with a combination of antibiotics, but if left untreated, it can cause permanent damage to the skin, nerves, limbs, and eyes.

The first documented registry for leprosy was established in the 13th century by the Knights Hospitaller in Jerusalem. This registry helped in the caring and controlling of the disease by keeping track of patients, providing medical care, and preventing the spread of the disease.

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What is the issue impacting those with phenylketonuria (PKU)?
The inability to convert pyruvate into acetyl CoA.
The increase in sulfur-based amino acid excretion.
The increase in calcium excretion.
The inability to convert phenylalanine into tyrosine.

Answers

The issue impacting those with phenylketonuria (PKU) is D: "the inability to convert phenylalanine into tyrosine".

This is due to a mutation in the gene that produces the enzyme phenylalanine hydroxylase, which is responsible for the conversion of phenylalanine to tyrosine. As a result, individuals with PKU have a buildup of phenylalanine in their blood, which can lead to neurological problems and developmental delays if not treated. It is important for those with PKU to follow a strict low-phenylalanine diet to prevent these issues.

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This assignment is about different foods and what their Satisfactory level , Marginal level and Unsatisfactory level . At what level the food will become Unsatisfactory
Table 1:
Standard Plate count (SPC)/Aerobic Plate Counts (APC)/ Mesophilic plate count (MPC)
(Result (colony-forming unit (cfu)/g unless otherwise specified))
(Satisfactory) (Marginal) (Unsatisfactory)
Foods cooked immediately prior to sale or consumption
Cooked foods chilled but with minimum handling prior to sale or consumption
Bakery and confectionery products without dairy cream, powdered foods
Fresh fruit and vegetables, products containing raw vegetables
Food mixed with dressings, dips, pastes
Non-fermented dairy products and dairy desserts,
Soups
Gravy
Boiled vegetables
Cooked meat, poultry, seafood (served hot)
Sausage rolls, meat pies, quiche
Fresh fruit
Deli meats
Cheese, yogurt
Salads
Peanut butter and jam sandwiches
Ready-to-eat hot dogs
Burgers without any fresh produce
Cooked meat products

Answers

It is important to maintain proper food handling and storage practices to prevent food from becoming Unsatisfactory and ensure its safety for consumption.

Table 1 shows the Standard Plate count (SPC) or Aerobic Plate Counts (APC) or Mesophilic plate count (MPC) for various food items, and their corresponding levels of Satisfactory, Marginal, and Unsatisfactory.

Foods that are cooked immediately before consumption or sale, and foods that are chilled but minimally handled before consumption or sale, fall under the Satisfactory level. Bakery and confectionery products without dairy cream or powdered foods also fall under this category.

Fresh fruit and vegetables, products containing raw vegetables, and food mixed with dressings, dips, or pastes are considered Marginal. Non-fermented dairy products, dairy desserts, soups, boiled vegetables, cooked meat, poultry, and seafood (served hot), sausage rolls, and meat pies are also Marginal.

On the other hand, Deli meats, cheese, yogurt, salads, peanut butter and jam sandwiches, ready-to-eat hot dogs, and cooked meat products are considered Unsatisfactory when their SPC/APC/MPC levels exceed the set standards. Burgers without any fresh produce also fall under this category.

In summary, it is important to maintain proper food handling and storage practices to prevent food from becoming Unsatisfactory and ensure its safety for consumption.

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