Many new species of desert have arisen from a population which was .pdf
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Many new species of desert have arisen from a population which was introduced into the American southwest. This example best describes: allopatric speciation. sympatric speciation. microevolution. adaptive radiation. the effect of climatic change. What is the only source of all new ? genetic drift gene flew genetic polymorphism mutation natural selection Extensive data from hospitals have shown that born with intermediate weights are the ones most likely to survive. This is a classic example of which type of selection? stabilizing selection directional selection no selection disruptive selection natural selection What are considered to be the two most important factors that have been involved in past mass ? glaciers and earthquakes disease and parasites plate tectonics and climatic factors fire and weather events environmental and biological factors Which was a possible source of high energy no early. Earth that likely contributed to the origin of Solution 11. Adaptive radiation: Adaptive radiation is when a species diversify to many different species from one ancestral species when there is a change in the environment. Change in environment would be when they are exposed to a different environment like in this case, to a new land. Change in the environment presents new challenges and advantages in term of resources, climate, predators etc. Such changes force a species to evolve to adapt. 12. Mutation: Mutation is change in the genetic code which can occur due to many reasons which change the basic DNA code as compared to what the parent cell has. Mutation can give the cell new characteristics or even hamper basic functions of a cell or an organism. Hence, mutation is the only source of change in alleles Genetic drift can happen only with mutation. Gene flow is basically the inheritance of genes from the parents to offspring. Genetic polymorphism does not guarantee change in phenotype, hence not change in alleles. Natural selection is basically survival of the fittest. 13. Stabilizing selection : It is when for a particular trait, the genetic diversity decreases and the population mean stabilizes. It is assumed to be the most common type of natural selection. When it comes to phenotypes it supports the intermediate variations. Human babies who are underweight are susceptible to diseases and lose heat more quickly and have high mortality rate and similarly for the babies with more than normal weight, which are difficult to deliver..
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Name __________________________________________Date ___________.docx
Name __________________________________________ Date _______________
Chapter 5 Reading Questions
1. Explain the biological significance of the Dung of the Devil.
Extracts from the plant have strong antiviral properties that are very helpful for fighting disease.
2. Give 2 reasons conservation of biodiversity may be beneficial to our health.
There may be some organisms that can provide health benefits, such as the Dung of the Devil. Also, greater biodiversity allows for greater stabilization of the climate, which is necessary for our well-being.
3. Give 3 examples of how natural substances have been used to treat disease.
The rosy periwinkle is used to treat childhood leukemia and Hodgkin’s disease. The mayapple is the source of two anti-cancer drugs. Other medicines come from invertebrates such as sponges, corals, and sea squirts.
4. Which has more biodiversity a carefully tended lawn or a small plot of untended land? Explain.
A small plot of untended land has greater biodiversity, since there is a greater variety of species there than on the well-kempt and uniform lawn.
5. Define species.
A species is a group of organisms that is distinct from other groups in terms of characteristics, and that can breed with each other.
6. Approximately how many species have scientists named? How many species are estimated to live on Earth?
About 2 million have been named, and about 10 million are thought to live on Earth.
7. Compare and contrast species richness and species evenness. Use an example to explain.
Species richness – the number of species in a given area, such as a pond or a tree
Species evenness – the proportion of the number of species in an ecosystem
Two communities may have an equal species richness, but one community may be dominated by one species, while the other may have an equal number of each species.
8. How are species phylogenies determined?
They are determined by the similarity of different species (morphology and behavior)
9. List the classification system from the most broad to the most specific.
Domain, Kingdom, Phylum, Class, Order, Family, Genus, Species
10. Speciation is an example of microevolution or macroevolution? Explain.
Speciation is an example of macroevolution. Both result in the creation of new species. Microevolution is small-scale changes in a species that may eventually result in speciation.
11. Explain 2 ways mutations may occur. Are they good or bad? Explain.
Mutations may occur by UV radiation. It can also occur through mistakes made while DNA is copied.
12. Explain how recombination occurs.
It is the switching of one small part of a chromosome with another during meiosis.
13. Give 2 examples of how humans are purposefully or inadvertently directing evolution of organisms.
The breeding of plants and animals for human use is done purposefully. The evolution of weeds after being sprayed with herbicides is inadvertent.
14. List the 5 key ideas of Darwin’s Theory of Evolution by Natural Selection.
1. Individuals.
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Ecosystem And Natural Selection - Biology
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Extracts from the plant have strong antiviral properties that are very helpful for fighting disease.
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4. Which has more biodiversity a carefully tended lawn or a small plot of untended land? Explain.
A small plot of untended land has greater biodiversity, since there is a greater variety of species there than on the well-kempt and uniform lawn.
5. Define species.
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6. Approximately how many species have scientists named? How many species are estimated to live on Earth?
About 2 million have been named, and about 10 million are thought to live on Earth.
7. Compare and contrast species richness and species evenness. Use an example to explain.
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Mutations may occur by UV radiation. It can also occur through mistakes made while DNA is copied.
12. Explain how recombination occurs.
It is the switching of one small part of a chromosome with another during meiosis.
13. Give 2 examples of how humans are purposefully or inadvertently directing evolution of organisms.
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14. List the 5 key ideas of Darwin’s Theory of Evolution by Natural Selection.
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Genotype-By-Environment Interaction (VG X E) wth Examples
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Phenotypic variation can be caused by the combination of genotypes and environments in a population. Genotypes are all equally sensitive to their environments, meaning that a change of environment would impact the phenotype of all genotypes to the same extent. In fact, genotypes very often have different degrees of sensitivity to environmental conditions. This cause of phenotypic variance is called genotype by- environment interaction and is symbolized by VG x E. This adds another term to the expression for the independent causes of total phenotypic variation in a population
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All fungi have tiny tube-like structures call hyphae. One is labeled within the collection. What
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Solution
Answer:
Hyphae is a long, branching, filamentous structure of a fungus, it contain the cytoplasm or cell
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denying access to those who lack authorization. What are the top five policies that you would
implement? How would you enforce compliance with these policies? How would you educate
your users to convince them to adhere to your policies?
Solution
The policies that system administrator use to provide the correct level of accessibility to
authorized users while denying access to unauthorized users are-
1. Using antivirus software and running them regularly to protect systems from attack by
malicious softwares.
2. Using firewalls to maintain access control based on sender\'s and reciever\'s ip addresses.
3. By providing Login system policy to ensure authorized access to system by individuals (using
login ID and passwords).
4. Using high level encryption techniques to secure access of information on transit between two
users.
5. By monitoring network communication.
These policies can be enforced by system administrator by using distributed operating system.
Because DO/S works on the property of global control of all assets.
Users can be provided with individual login and password as unauthorized access may lead to
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Answer:
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What would the predicted result be if Griffith injected mice with a mixture of heat-killed R
bacteria and living S bacteria?
a. All mice would live, but only if DNAse was added to the mixture prior to injection
b. All mice would die.
c. All mice would live.
d. All mice would die, but only if DNAse was added to the mixture prior to injection.
Solution
Griffith’s Experiment-A Quick Review: Fredrick Griffith, an English physician, first reported
transformation of one strain of Streptococcus pneumoniae into other strain in 1928. The bacteria
cause pneumonia. A strain of this bacterium produces capsule and gives a smooth, shiny colony
on culture media, thus called S-strain (smooth strain). The S-strain exhibits pathogenicity in
laboratory mice. Another strain without capsule produces rough and coarse colony on culture
medium, thus called R-strain (rough strain). The R-strain is avirulent (non-pathogenic).
Mice injected with
Outcome
1
Live S-strain (virulent)
Pneumonia mice dead
2
Live R-strain (avirulent)
No pneumonia mice alive
3
Heat killed S-strain + live R-strain
Pneumonia mice dead
When injected with live S-strain bacteria, mice developed pneumonia and died. When injected
with live R-strain bacteria, mice did not develop pneumonia and survived. When heat killed S-
strain injected to mice, they remain unaffected. However, when heat killed S-strain bacteria
mixed with live R-strain injected to mice, they developed pneumonia and died. He was also able
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no way been revived, he concluded that some molecules from heat killed S-strain had
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Now,
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killing them at last. It is noteworthy that transformation of R-strain into S-strain occurs only
when there is alive R-strain and fragments of DNA from heat-killed S-strain.
Option A: Incorrect. DNAse is a large protein incapable of entering into bacterial cytoplasm
through bacterial membrane of living S-strain. So, this treatment further cleaves DNA fragments
of heat killed R-strain BUT leaves the S-strain unaffected. So, injecting the mixture to mice kills
them as live S-strain causes pneumonia.
Option B: Correct. Presence of living S-strain causes pneumonia development in mice killing
them at last.
Option C: Incorrect. All mice die because S-strain cause pneumonia to them.
Option D: Incorrect. All mice would die, BUT NOT because of addition of DNAse to the
mixture. As mentioned in option A explanation, DNAse does not penetrate into live S-strain,
thus leave them unaffected. Therefore, mice die irrespective of DNAse treatment.
Mice injected with
Outcome
1
Live S-strain (virulent)
Pneumonia mice dead
2
Live R-strain (avirulent)
No pneumonia mice alive
3
Heat killed S-strain + live R-strain
Pneumonia mice dead.
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Write a function called myfind that will search for a key in a vector and return the indices of all
occurrences of the key, like the built-in find function. It will receive two arguments— the vector
and the key—and will return a vector of indices or the empty vector [ ] if the key is not found.
The function will be invoked as follows:
idx = myfind(vec, key);
Following is an example of correct function behavior:
>> vec = randi([5, 10], 1, 10)
vec = 5 10 10 7 9 5 7 10 9 10
>> key = 10;
>> idx = myfind(vec, key)
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>> key = 2;
>> idx = myfind(vec, key)
idx = Empty matrix: 1-by-0
Do not use the in-built find function.
Solution
% matlab code find all index of key in vector
function idx = myfind(vec, key)
index = 1;
for i=1:length(vec)
if vec(i) == key
idx(index) = i;
index = index + 1;
end
end
end
vec = randi([5, 10], 1, 10);
disp(\"Vector: \");
disp(vec);
key = 10;
fprintf(\"\ Key: %d\ \",key);
idx = myfind(vec, key);
disp(\"\ Index: \");
disp(idx);
%{
sample output:
Vector:
10 10 8 10 5 10 5 8 9 5
Key: 10
Index:
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%}.
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4. To understand link between personal and public one must use critical thinking skills such as
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Solution
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Superheated vapor
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Saturated liquid
Saturated mixturA)
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syndrome is anX-inked, recessive condition caused by mutation of the HPRT1 muscle
movements, The syndrome is characterized byjuvenile-onset gout, in and behavioral disorders. A
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syndrome, but neither one of her parents was diagnosed with condition. Select all the
descriptions that could explain how their daughter inherited Lesch-Nyhan syndrome. The
daughter could not have Nyhan because her maternal grandparents did not have disease. The girl
must have acquired the disease through a random mutation. The daughter inherited two
xchromosomes with the of the mothers X mutation. Nondisjunction chromosome chromosomes
occurred during meiosis, and the daughter inherited from her father. The father did not contribute
a sex chromosome to his daughter due to nondisjunction of the sex chromosomes. The daughter
is xo and her only X chromosome came from her mother, who was a O The d from her father,
who was a carrier. camies the Lesch-Nyhan mutation and was Her inherited There is a hint
available! bottom divider bar. Click on the divider bar again to hide the hint. Previous ® Give
Up & View solution Check Answer 0 Next HExi Hint
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1) the hint here is that the mother\'s maternal was diagnosed with Lesch-Nyhan syndrome. Since
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The starship Epidemiology Team has cone many missions exploring the universe. The captain
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Question 3)
Perform a regression analysis using compa as the dependent variable and the same independent
variables as used in question 2. (Question #2 found below the table). Show the result, and
interpret your findings by answering the same questions.
Note: be sure to include the appropriate hypothesis statements.
A ) Regression hypotheses
Ho:
Ha:
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Ho:
Ha:
B )Interpretation:
For the Regression as a whole:
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What is the p-value associated with this value:
Is the p-value < 0.05?
Do you reject or not reject the null hypothesis:
What does this decision mean for our equal pay question (Do M and F get paid equaly?):
C )For each of the coefficients: Intercept, Midpoint, Age, Perf. rating, Service, Gender, Degree
What is the coefficient\'s p-value for each of the variables:
Is the p-value < 0.05?
Do you reject or not reject each null hypothesis:
What are the coefficients for the significant variables?
Using only the significant variables, what is the equation? Compa =
Is gender a significant factor in compa:
If so, who gets paid more with all other things being equal?
How do we know?
ID
Salary
Compa
Midpoint
Age
Performance Rating
Service
Gender
Raise
Degree
Gender1
Gr
1
66.1
1.159
57
34
85
8
0
5.7
0
M
E
2
25.9
0.834
31
52
80
7
0
3.9
0
M
B
3
35.2
1.135
31
30
75
5
1
3.6
1
F
B
4
55.3
0.971
57
42
100
16
0
5.5
1
M
E
5
49.6
1.033
48
36
90
16
0
5.7
1
M
D
6
78.3
1.168
67
36
70
12
0
4.5
1
M
F
7
42.3
1.058
40
32
100
8
1
5.7
1
F
C
8
22.8
0.990
23
32
90
9
1
5.8
1
F
A
9
78
1.164
67
49
100
10
0
4
1
M
F
10
23.3
1.014
23
30
80
7
1
4.7
1
F
A
11
23.6
1.025
23
41
100
19
1
4.8
1
F
A
12
60.8
1.067
57
52
95
22
0
4.5
0
M
E
13
40.6
1.014
40
30
100
2
1
4.7
0
F
C
14
21.7
0.943
23
32
90
12
1
6
1
F
A
15
21.8
0.949
23
32
80
8
1
4.9
1
F
A
16
37.4
0.934
40
44
90
4
0
5.7
0
M
C
17
57
1.000
57
27
55
3
1
3
1
F
E
18
33.5
1.081
31
31
80
11
1
5.6
0
F
B
19
23
1.000
23
32
85
1
0
4.6
1
M
A
20
36
1.162
31
44
70
16
1
4.8
0
F
B
21
76
1.135
67
43
95
13
0
6.3
1
M
F
22
43.7
0.911
48
48
65
6
1
3.8
1
F
D
23
25.3
1.098
23
36
65
6
1
3.3
0
F
A
24
48.9
1.019
48
30
75
9
1
3.8
0
F
D
25
25.8
1.122
23
41
70
4
0
4
0
M
A
26
23.3
1.013
23
22
95
2
1
6.2
0
F
A
27
42.3
1.057
40
35
80
7
0
3.9
1
M
C
28
75.2
1.122
67
44
95
9
1
4.4
0
F
F
29
80.9
1.208
67
52
95
5
0
5.4
0
M
F
30
49
1.020
48
45
90
18
0
4.3
0
M
D
31
24.2
1.054
23
29
60
4
1
3.9
1
F
A
32
27.5
0.886
31
25
95
4
0
5.6
0
M
B
33
63.6
1.115
57
35
90
9
0
5.5
1
M
E
34
28.6
0.922
31
26
80
2
0
4.9
1
M
B
35
22.4
0.976
23
23
90
4
1
5.3
0
F
A
36
23.6
1.026
23
27
75
3
1
4.3
0
F
A
37
24.3
1.057
23
22
95
2
1
6.2
0
F
A
38
63
1.105
57
45
95
11
0
4.5
0
M
E
39
34.8
1.123
31
27
90
6
1
5.5
0
F
B
40
24.3
1.057
23
24
90
2
0
6.3
0
M
A
41
42.8
1.071
40
25
80
5
0
4.3
0
M
C
42
23
0.998
23
32
100
8
1
5.7
1
F
A
43
75.4
1.125
67
42
95
20
1
5.5
0
F
F
44
60.7
1.065
57
45
90
16
0
5.2
1
M
E
45
57.9
1.206
48
36
95
8
1
5.2
1
F
D
46
62.2
1.091
57
39
75
20
0
3.9
1
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E
47
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1.091
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37
.
Problems in PYTHON3.5 1. Define a function sumOfDigits that has one .pdf
Problems in PYTHON3.5 1. Define a function sumOfDigits that has one parameter of type
string. The function should return the sum of the digits in the string. Do not treat multiple digit
string as one number – “2014” should be treated as 4 different digits of 2, 0, 1, 4. The function
should return 17 for the string “Today’s date is 09/01/2014”. Then write a main program that
asks a user to enter a sentence and then displays the sum of digits that are in the sentence by
using the function, sumOfDigits. The function sumOfDigits should only do what’s asked to do. It
should not do any additional tasks such as print or input!!!
Solution
Before implementing the function, we need to check whether the integer comes into the string or
not.
To do that there is an in-built function in Python, isdigit() which checks whether the input is the
integer
or some other character.
def sumOfDigits(input_str):
summation = 0
for number in list(input_str):
if number.isdigit():
summation += int(number)
return summation
Hence, this will loop through the entire string and check for each of them whether it is integer or
some other character.For all the integer it is summing up the values and returning it at the end of
the function definition..
Only a very small percentage of organisms actually fossilize. Based o.pdf
Only a very small percentage of organisms actually fossilize. Based on the information in the
table below, rank the likely abundance of fossils of organisms A-E. Rank the fossils of
organisms A-E from most common to least common. If two fossils have the same commonness,
overlap them.
Solution
Answer:
Ranking based on the abundance of fossils of organisms, A-E:
1. Organism E is most likely to fossilize
This organism has a hard shell and lives with sediments near the shore and has a large population
with long existence.
2. Organism A
It has the larger population and longer existence time. This can lead to A having a better chance
at fossilizing than B.
3. Organism B
Organism B has a small population with a short existence time.
4. Organism C
It lives with no sediment, has a large population with a long existence thus leaving it more
susceptible to becoming fossilize than D.
5. Organism D (being the least common fossil)
It has a soft body, lives with no sediment deposits, has a short life and a small population..
package lab7; import java.util.Scanner; Sammy Student, Program.pdf
package lab7;
import java.util.Scanner;
//Sammy Student, Programmer
public class Maze
{
static Scanner sc = new Scanner(System.in);
// maze movements
static char myMove = \'\\0\';
// cell position
static int currentCell = 0;
static int score = 0;
static boolean advance = true;
static boolean checkThis = false;
public static void main(String args[])
{
// the local variables declared and initialized
char answer = \'Y\';
displayMenu();
while(answer == \'Y\' || answer == \'y\')
{
displayMovement();
makeYourMove();
checkThis = checkYourMove();
mazeStatus();
System.out.println(\"move again(Y or N)?\");
answer = sc.next().charAt(0);
}
System.out.println(\"***************************\");
}// end main() method
static void displayMenu()
{
System.out.println(\"\");
System.out.println(\"***************************\");
System.out.println(\"----The Maze Strategy---\");
System.out.println(\"\");
}// end method
static void displayMovement()
{
if(currentCell == 0)
{
System.out.println(\"You have entered the maze!!\");
System.out.println(\"There is no turning back!!\");
currentCell = 1;
mazeStatus();
advance = true;
}
System.out.println(\"make your move (W, A, S, D)\");
System.out.println(\"W = up, A = left, S = down, D = right)\");
}// end method
static void makeYourMove()
{
myMove = sc.next().charAt(0);
switch(myMove)
{
case \'W\': { MoveUp(); break; }
case \'A\': { MoveLeft(); break; }
case \'S\': { MoveDown(); break; }
case \'D\': { MoveRight(); break; }
}
// end program menu
}// end method
static boolean checkYourMove()
{
if(currentCell == 1 && advance == true)
{
if (myMove == \'W\')
{
advance = false;
System.out.println(\"try again\");
return advance;
}
if (myMove == \'A\')
{
advance = false;
System.out.println(\"SORRY, there is no return\");
return advance;
}
if (myMove == \'D\')
{
currentCell = 2;
advance = true;
System.out.println(\"continue through the maze\");
return advance;
}
if (myMove == \'S\')
{
advance = false;
System.out.println(\"continue through the maze\");
return advance;
}
}
if(currentCell == 2 && advance == true)
{
if (myMove == \'W\')
{
advance = false;
System.out.println(\"try again\");
return advance;
}
if (myMove == \'A\')
{
advance = false;
System.out.println(\"SORRY, there is no return\");
return advance;
}
if (myMove == \'D\')
{
advance = false;
System.out.println(\"continue through the maze\");
return advance;
}
if (myMove == \'S\')
{
currentCell = 5;
advance = true;
System.out.println(\"continue through the maze\");
return advance;
}
}
return advance;
// end program menu
}// end method
static void MoveLeft()
{
System.out.println(\"you moved to the left\");
}//end method
static void MoveRight()
{
System.out.println(\"you moved to the right\");
}//end method
static void MoveUp()
{
System.out.println(\"you moved up (forward)\");
}//end method
static void MoveDown()
{
System.out.println(\"you moved down (downward)\");
}//end method
static void mazeStatus()
{
System.out.println(\"current position: cell \" + currentCell);
}//end meth.
Odin and Zeus travel to the outermost reaches of our galaxy using wa.pdf
Odin and Zeus travel to the outermost reaches of our galaxy using warp drive spaceship
searching for other forms of life. They make contact with highly advanced civilization on planet
X329. However, while trying to land at a guided landing site, they are exposed to high energy
gamma rays which produces defects in their classical complement and some cytokine genes, and
obliterates monocyte mactrophages, and granulocytes. Their mission mandates staying with their
friends for three days and then journey back to earth. The trip back to Earth will take between 6-
7 days. Will Odin and Zeus survive meeting their new friends on planet X329? If so what
immune mechanisms will help their survival?
Solution
In the present case, Odin and Zeus are exposed to high energy gamma radiations. Gamma rays
are highly penetrating, and can cause genetic damage leading to detrimental effects. Its ionizing
energy breaks up bonds of genetci material. In the case presented here, gamma rays have caused
defects in classical complement system and also the cytokine genes. Classical complement is the
term given group of more than 30 serum proteins that enhance the ability to produce antibodies
and phagocytic cells, that are critically important in our defense against infections. Similarly,
cytokines which include chemokines, interferons, interleukins, lymphokines and TNFs etc are
small proteins involved in autocrine signalling, paracrine signalling and endocrine signalling,
acting as immunomodulating agents.Thus defects produced by gammma radiations in phagocytic
cells, classical complement, cytokines, cytokine receptors or in molecules that mediate effector
responses might lead to various immunodeficiency diseases. Also, the individuals have suffered
from obliteration of monocyte, macrophages, and granulocytes due to gamma rays exposure.
Thus they have practically no defence against (no cells to eat up) cellular debris, foreign
substances, microbes, cancer cells.
In such cases of severe exposure and damage, immediate medical aid is required. Delay of 6-7
days (as per the given mission mandate) could be fatal for the individuals exposed, especially in
case of acute exposure to ioninzing radiations. Odin and Zeus may not survive for meeting their
new friends on planet X329 without emergency medical help, change of clothes and
radioprotecive drugs..
List and analyze some of the most salient social factors of group un.pdf
List and analyze some of the most salient social factors of group unrest AND deviance in cities.
Solution
Some of the important social factors form group unrest are listed below.
1. Social inequality
2. Corruption
3. Inflation
4. Government apathy
If the people of the group feels that all members of the group are not being treated equal, sense of
inferiority arises and the member of the group starts feeling harassed. If the same amount of
resources are not available to all for exploitation, there will always be group unrest. When people
are not treated equal among different sections of the society unrest happens. They can be
harassed on the basis of colour, caste, ethnicity and creed.
Corrpution is also an important factor for group unrest. When there is corruption in the society
people of the group feels being cheated. There is impression that they are paying more than what
they should have been. The exploited lots will cause social unrest and raise awareness.
Inflation affects whole section of the society and all members of the group in a society. The rise
in the expenditure on the same amount of commodities bought in a month will provoke the
buyers to take action, sometimes result in social unrest. Usually when the authority is unable to
control inflation and when there is apathy of the authorities towards the members of the group,
the social unrest happens..
J. A. Moore investigated the inheritance of spotting patterns in leop.pdf
J. A. Moore investigated the inheritance of spotting patterns in leopard frogs (J. A. Moore. 1943.
Journal of Heredity 34:3-7). The pipiens phenotype had the normal spots that give leopard frogs
their name. In contrast, the burnsi phenotype lacked spots on its back. Moore earned out the
following crosses, producing the progeny indicated. Parent phenotypes Progeny phenotypes
burnsi X burnsi 42 burnsi, 8 pipiens burnsi X pipiens 25 burnsi, 34 pipiens burnsi X pipiens 194
bumsi, 208 pipiens On the basis of these results, what is the most likely mode of inheritance of
the burnsi phenotype? Give the most likely genotypes of the parent in each cross. Use a chi-
square test to evaluate the fit of the observed numbers of progeny to the number expected on the
basis of your proposed genotypes for each of the three matings shown.
Solution
a. On the basis of the above crosses, the mode of inheritance of burnsi phenotype is
DOMINANT PHENOTYPE.
b. The Genotypes of the parents
Burnsi X Burnsi = Bb X Bb
Burnsi X pipiens = Bb X bb
Burnsi X pipiens = Bb X bb
C.
1.
Null hypothesis: The observed values are not deviating from 3:1 ratio.
Chisquare test:
Category
Burnsi
pipiens
Total
Observed values
42
8
50
Exptected Ratio
3
1
4
Exprected Values
37.5
12.5
Deviation
4.5
-4.5
D^2
20.25
20.25
D^2/E
0.54
1.62
2.16
X^2
2.16
Degrees of freedom
-
1
Inference: The calculated chisquare values is less than the table i.e. 3.84 at 1 DF and 0.05
probability, hence null hypothesis accepted.
2.
Null hypothesis: The observed values are not deviating from 1:1 ratio.
Chisquare test:
Category
Burnsi
pipiens
Total
Observed values
28
34
62
Exptected Ratio
1
1
2
Exprected Values
31
31
Deviation
-3
3
D^2
9
9
D^2/E
0.290323
0.290323
0.580645
X^2
0.580645
Degrees of freedom
-
1
Inference: The calculated chisquare values is less than the table i.e. 3.84 at 1 DF and 0.05
probability, hence null hypothesis accepted.
3.
Null hypothesis: The observed values are not deviating from 1:1 ratio.
Chisquare test:
Category
Burnsi
pipiens
Total
Observed values
194
208
402
Exptected Ratio
1
1
2
Exprected Values
201
201
Deviation
-7
7
D^2
49
49
D^2/E
0.243781
0.243781
0.487562
X^2
0.487562
Degrees of freedom
-
1
Inference: The calculated chisquare values is less than the table i.e. 3.84 at 1 DF and 0.05
probability, hence null hypothesis accepted.
Category
Burnsi
pipiens
Total
Observed values
42
8
50
Exptected Ratio
3
1
4
Exprected Values
37.5
12.5
Deviation
4.5
-4.5
D^2
20.25
20.25
D^2/E
0.54
1.62
2.16
X^2
2.16
Degrees of freedom
-
1.
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Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...
Many new species of desert have arisen from a population which was .pdf
- 1. Many new species of desert have arisen from a population which was introduced into the American southwest. This example best describes: allopatric speciation. sympatric speciation. microevolution. adaptive radiation. the effect of climatic change. What is the only source of all new ? genetic drift gene flew genetic polymorphism mutation natural selection Extensive data from hospitals have shown that born with intermediate weights are the ones most likely to survive. This is a classic example of which type of selection? stabilizing selection directional selection no selection disruptive selection natural selection What are considered to be the two most important factors that have been involved in past mass ? glaciers and earthquakes disease and parasites plate tectonics and climatic factors fire and weather events environmental and biological factors Which was a possible source of high energy no early. Earth that likely contributed to the origin of Solution 11. Adaptive radiation: Adaptive radiation is when a species diversify to many different species from one ancestral species when there is a change in the environment. Change in environment would be when they are exposed to a different environment like in this case, to a new land. Change in the environment presents new challenges and advantages in term of resources, climate, predators etc. Such changes force a species to evolve to adapt. 12. Mutation: Mutation is change in the genetic code which can occur due to many reasons which change the basic DNA code as compared to what the parent cell has. Mutation can give the cell new characteristics or even hamper basic functions of a cell or an organism. Hence, mutation is the only source of change in alleles Genetic drift can happen only with mutation. Gene flow is basically the inheritance of genes from the parents to offspring. Genetic polymorphism does not guarantee change in phenotype, hence not change in alleles. Natural selection is basically survival of the fittest. 13. Stabilizing selection : It is when for a particular trait, the genetic diversity decreases and the population mean stabilizes. It is assumed to be the most common type of natural selection. When it comes to phenotypes it supports the intermediate variations. Human babies who are underweight are susceptible to diseases and lose heat more quickly and have high mortality rate and similarly for the babies with more than normal weight, which are difficult to deliver.