Problems in PYTHON3.5 1. Define a function sumOfDigits that has one .pdf
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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..
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76
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28
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29
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63.6
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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
M
E
47
62.2
1.091
57
37
.
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..
Many new species of desert have arisen from a population which was .pdf
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..
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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