If 0.150 mL of 10^-7.5 dilution yielded an average of 256 coloniesp.pdf
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If 0.150 mL of 10^-7.5 dilution yielded an average of 256 colonies/plate, calculate the number of bacteria per mL of the original culture. Show all calculations. Solution The number of colony-forming units (CFU) = [no of cells]/ [volume of dilution x volume of diluted sample added to the plate] Now CFU per ml of dilution is 256 colonies divided by 0.150 ml plated = 1706.7 colony- forming units/ml of dilution. Dividing this number by the 10-7.5 dilution = 1706.7 / 10-7.5 = 1706.7 x 107.5 colony-forming units/ml of original culture sample.
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![If 0.150 mL of 10^-7.5 dilution yielded an average of 256 colonies/plate, calculate the number of
bacteria per mL of the original culture. Show all calculations.
Solution
The number of colony-forming units (CFU) = [no of cells]/ [volume of dilution x volume of
diluted sample added to the plate]
Now CFU per ml of dilution is 256 colonies divided by 0.150 ml plated = 1706.7 colony-
forming units/ml of dilution.
Dividing this number by the 10-7.5 dilution = 1706.7 / 10-7.5 = 1706.7 x 107.5 colony-forming
units/ml of original culture sample](https://image.slidesharecdn.com/if0-230702063449-b55d12e1/85/If-0-150-mL-of-10-7-5-dilution-yielded-an-average-of-256-coloniesp-pdf-1-320.jpg)
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In the equation below f is a frequency, and L is a length. What are .pdf
In the equation below f is a frequency, and L is a length. What are the dimensions of a? In the
equation below f is a frequency, and L is a length. What are the dimensions of a 1/f = 2.pi.root
L/a
Solution
On the left hand side:
Dimension of frequency is T-1. So, dimension of 1/f will be T.
Now on the right hand side,
2pi is dimension less constant and L is length
To obtain the dimension of T on the right side, we have to divide the length dimension by
another length dimension and multiply by T2.
Hence, dimension of 1/a = T2/L
So, dimensions of a = L/T2 = LT-2.
how are the axial and appendicular skeletons attachedSolutionA.pdf
how are the axial and appendicular skeletons attached
Solution
Answer : Axial and appendicular skeleton are the two principal kinds of bones found in the adult
humans in which both divisions join to form a complete skeleton. Axial skeleton consists of the
bones that lie around the longitudinal axis of the human body. Examples of axial bons are skull
bones, ear bones, ribs, sterum and bones of vertebral column. Appendicular skeleton consists of
the bones of the upper and the lower limbs. Appendicular bones also consists of bones forming
the girdles that connects the limbs to the axial skeleton.
The pectoral girdle which is also called as shoulder girdle connects the upper limbs to the body
where as the pelvic girdle/Hip girdle of the bony pelvis connects the bones of the lower limbs to
the axial skeleton.
How many base pairs of DNA represent the haploid human genome How m.pdf
How many base pairs of DNA represent the haploid human genome? How many base pairs of
DNA represent the haploid chimpanzee genome? Assume the similarity between human and
chimpanzee genomes is 97.5%. Based on 97.5% similarity how many base pairs do the
chimpanzee and human genomes differ by?
Solution
3 billion DNA base pairs represent haploid genome
there are 24 base pairs of DNA in the haploid chimpanzee genome.
humans and chimpanzee differ by 1 base pair .The humans have 23 base pairs and chimpanzee
have 24.
From a group of cards that has 13 hearts, 6 spades, 2 diamonds and 4.pdf
From a group of cards that has 13 hearts, 6 spades, 2 diamonds and 4 clubs, the number of ways
that all four dealt cards could be from different suits is
Solution
there will be total 25 cards and it will be a problem without replacement type
therefore getting 4 card of different suit will be
the total ways in wich this can ve done is
13*6*2*4 ( as there can be 1 card from13 cards at 1st and similarly 6,2,4 cards at respective
times)
= 624 ways.
Find and classify all singularities of the gamma and riemann zeta fu.pdf
Find and classify all singularities of the gamma and riemann zeta function
Determine the residues at any isolated singularities. Hence find the value of
please answer this with step by step information or in clear image
Solution
sin(z)(z)(z) = i 2 e iz Z C w z1 ew 1 dw. On the other hand, we can calculate using the residue
theorem the integral along the path drawn in the following figure. There the outside square
passes thru points (2m + 1)i and (2m + 1)i. -1 2i -2i 2mi 2(m+1)i -2mi -2(m+1)i NOTES ON
RIEMANN’S ZETA FUNCTION 9 We have Z w z1 ew 1 dw = 2iXm n=1 (2n) z1 e i (z1) 2 +
(2n) z1 e 3i (z1) 2 = 2z z iei(z1) e i (z1) 2 + e i (z1) 2 Xm n=1 n z1 = 2 z+1 z ieiz cos (z 1) 2 Xm
n=1 1 n1z = 2 z+1 z ieiz sin z 2 Xm n=1 1 n1z . Now we want to estimate the integral along the
sides of the square for Re z < 0. For a fixed z, we see as before that there exists M > 0, such that
on the right vertical side of the square the integrand satisfies the estimate w z1 ew 1 M |w| Re
z1 |ew 1| . Moreover, we have |e w 1| ||e w| 1| = |e Re w 1| for any w, hence we have w z1 ew
1 M |w| Re z1 |eRe w 1| M | Re w| Re z1 |eRe w 1| and the right side of the square. This
expression clearly tends to zero faster than 1 Re w as Re w +. Hence, the integral over the right
side tends to zero as the square grows. On the left side of the square we have the same estimate
w z1 ew 1 M | Re w| Re z1 |eRe w 1| . Since Re w is negative in this case, for large | Re w| we
have w z1 ew 1 2M| Re w| Re z1 . Since Re z < 0, this bound goes to zero faster that 1 | Re w| .
On the other hand, the length of the side of the square is 2| Re w|. Hence, the integral along the
left side also tends to zero as the square grows. It remains to treat the top and bottom side. As
before, we see that on these sides we have w z1 ew 1 M |w| Re z1 |ew 1| M |Im w| Re z1 |ew
1| . On the other hand, the function w 7 |e w 1| on the horizontal lines tends to as Re w + and to
1 as Re w . Hence it is bounded from below. Moreover, since it is periodic with period 2i,
shifting the line up and down by 2i doesn’t change that bound. Hence, in our estimates we can
assume that 1 |ew 1| C 10 D. MILICI C´ on top and bottom sides of all squares. This implies
that w z1 ew 1 MC|Im w| Re z1 on top and bottom sides of all squares. Since Re z < 0 and the
length of these sides is 2|Im w|, we see that the integrals over these sides tend to zero as the
squares grow. Hence, for Re z < 0, as the squares grow, i.e., as m tends to infinity, the integral
over the sides of the square tends to 0 and the integral along converges to the integral along the
path C from 2.2.2. In addition, the sum on the right side of the above expression converges to the
series for (1 z), i.e., we have 2ieiz sin(z)(z)(z) = 2 z+1 z ieiz sin z 2 (z 1). Hence, we have
sin(z)(z)(z) = 2z z sin z 2 (1 z). This yields 2 sin z 2 cos z 2 (z)(z) = 2z z sin z 2 (1 z). and cos z
2 (z)(z) = 2z1 z (1 z). for all z C. By substituting 1 z for z we get 2 z 1z (z) = cos (1 z) .
Discuss the two group classifications that exist in Windows Server 2.pdf
Discuss the two group classifications that exist in Windows Server 2012 R2 describing both the
group type and group scope. Is there a reason you would use one scope over another?
Solution
A group is a collection of users and accouts that can be managed as a single unit. The 2 group
classification in Windows 2012 are
1. Distribution group
2. Security group
Distribution groups are only created to use email applications (like Exchange Server) to send
emails to a group of users. It is a feature introduced by Office 365 and allows sending emails to
large number of people both inside and outside the organization. It eliminates sending individual
emails to users. Bulk emails can be sent to a large group of users. The only purpose of
distribution group is to share information
Security groups are sets of users who have the same access rights, permission to resources and
performs same kind of tasks. There can be many security groups and a user can belong to more
than 1 security group. It facilitates administrators to assign same set of permissions to users
rather than individually assigning permissions to each one of them.
There are 3 group scopes as follows:
1. Universal scope: It can have users, accounts, groups or other universal groups from any
domain. Users can add members from any domain to an universal group and can assign
permission to access any resource located in any domain.
2. Global scope: These are containers for users and accounts which can be assigned permission
to objects in any domain. However, it can only have members from the domain in which it is
created.
3. Domain local group: These can have users and accounts from any domain as member.
However, only domain local groups can assign permissions to local resources or to resources that
reside in the domain in which the domain local group was created.
It is preferable to choose domain local group scope as it provides more security than the other 2
scopes. Also, the applied restrictions mean limited resources are utilized by the group..
Describe the differences between allelic heterogeneity and locus het.pdf
Describe the differences between allelic heterogeneity and locus heterogeneity
Solution
Ans.) The differences between allelic heterogeneity and locus heterogeneity
In general terms, allelic heterogeneity defines the different mutation in the same gene locus that
displays different phenotype. The variation occur due to natural selection as a consequence of
genetic drift, mutation etc.
The genetic or locus heterogeneity is the single trait that caused due to mutation in different gene
loci. Only a single mutant locus is needed to display the phenotype.
Allelic heterogeneity is the mutation in same loci that showed different phenotypic character
whereas locus heterogeneity is the only one trait occurs due to mutation in different loci..
Decision Tree Table for choosing present and future value equations. .pdf
Decision Tree Table for choosing present and future value equations. ltem to Number of Present
or Annual or Perpetual or evaluate occurrences Future Equation to use Periodi Terminal Series
na na 0 value payment:s Future Present na na Periodic Annual Periodic Vn Vo(1+i)\"
Vo=Vn/(1+i)\" Vn=pl(1+i)n-1)/[(1+1)t-1)] Single Revenue Future Terminal Perpetual Vo=
p/[(1+i)\'-1] Series of Equal payments Present Terminal Vo (1-1(1+)-1](1+i)) Cost PerpetualVo
Annual Terminal Vo=p[(1+i)n-1)/[i(1+i)\") interest rate per year (in decimals) Vo = initial value
(in time 0) V- value in time n n total number of years of compounding or discounting considered
p = amount of payment occurring every t years t number of years in one multiyear group or
interval of years (length of the interval) Payments occur at the end of the period or multiyear
interval Please note: these equations can be used for months or weeks or any other length of time
as long as the interest rate is defined at that period, n is defined as total number of periods, and t
is the number of periods in one multiperiod group or interval. In this class, we will stick
primarily to using the annual rates. Source: Figure 7.3 from Davis, et al. 2001 Forest
Management, p. 341
Solution
a. As seen from the table,
Present value of perpetual periodic CF is = P/[(1+i)^t - 1]= 50000/[1.07^45 - 1] = 50000/(21.002
- 1)
= $2,499.69
b. Present value of a perpetual $100 tax = P/i = 100/0.07 = $1,428.57
Present value of the cash flow = 2,499.69 - 1,428.57 = $1,071.12
c. AS the asking price of the land is $1000 and is less than the NPV of $1,071.12 you should buy
it..
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In the equation below f is a frequency, and L is a length. What are .pdf
In the equation below f is a frequency, and L is a length. What are the dimensions of a? In the
equation below f is a frequency, and L is a length. What are the dimensions of a 1/f = 2.pi.root
L/a
Solution
On the left hand side:
Dimension of frequency is T-1. So, dimension of 1/f will be T.
Now on the right hand side,
2pi is dimension less constant and L is length
To obtain the dimension of T on the right side, we have to divide the length dimension by
another length dimension and multiply by T2.
Hence, dimension of 1/a = T2/L
So, dimensions of a = L/T2 = LT-2.
how are the axial and appendicular skeletons attachedSolutionA.pdf
how are the axial and appendicular skeletons attached
Solution
Answer : Axial and appendicular skeleton are the two principal kinds of bones found in the adult
humans in which both divisions join to form a complete skeleton. Axial skeleton consists of the
bones that lie around the longitudinal axis of the human body. Examples of axial bons are skull
bones, ear bones, ribs, sterum and bones of vertebral column. Appendicular skeleton consists of
the bones of the upper and the lower limbs. Appendicular bones also consists of bones forming
the girdles that connects the limbs to the axial skeleton.
The pectoral girdle which is also called as shoulder girdle connects the upper limbs to the body
where as the pelvic girdle/Hip girdle of the bony pelvis connects the bones of the lower limbs to
the axial skeleton.
How many base pairs of DNA represent the haploid human genome How m.pdf
How many base pairs of DNA represent the haploid human genome? How many base pairs of
DNA represent the haploid chimpanzee genome? Assume the similarity between human and
chimpanzee genomes is 97.5%. Based on 97.5% similarity how many base pairs do the
chimpanzee and human genomes differ by?
Solution
3 billion DNA base pairs represent haploid genome
there are 24 base pairs of DNA in the haploid chimpanzee genome.
humans and chimpanzee differ by 1 base pair .The humans have 23 base pairs and chimpanzee
have 24.
From a group of cards that has 13 hearts, 6 spades, 2 diamonds and 4.pdf
From a group of cards that has 13 hearts, 6 spades, 2 diamonds and 4 clubs, the number of ways
that all four dealt cards could be from different suits is
Solution
there will be total 25 cards and it will be a problem without replacement type
therefore getting 4 card of different suit will be
the total ways in wich this can ve done is
13*6*2*4 ( as there can be 1 card from13 cards at 1st and similarly 6,2,4 cards at respective
times)
= 624 ways.
Find and classify all singularities of the gamma and riemann zeta fu.pdf
Find and classify all singularities of the gamma and riemann zeta function
Determine the residues at any isolated singularities. Hence find the value of
please answer this with step by step information or in clear image
Solution
sin(z)(z)(z) = i 2 e iz Z C w z1 ew 1 dw. On the other hand, we can calculate using the residue
theorem the integral along the path drawn in the following figure. There the outside square
passes thru points (2m + 1)i and (2m + 1)i. -1 2i -2i 2mi 2(m+1)i -2mi -2(m+1)i NOTES ON
RIEMANN’S ZETA FUNCTION 9 We have Z w z1 ew 1 dw = 2iXm n=1 (2n) z1 e i (z1) 2 +
(2n) z1 e 3i (z1) 2 = 2z z iei(z1) e i (z1) 2 + e i (z1) 2 Xm n=1 n z1 = 2 z+1 z ieiz cos (z 1) 2 Xm
n=1 1 n1z = 2 z+1 z ieiz sin z 2 Xm n=1 1 n1z . Now we want to estimate the integral along the
sides of the square for Re z < 0. For a fixed z, we see as before that there exists M > 0, such that
on the right vertical side of the square the integrand satisfies the estimate w z1 ew 1 M |w| Re
z1 |ew 1| . Moreover, we have |e w 1| ||e w| 1| = |e Re w 1| for any w, hence we have w z1 ew
1 M |w| Re z1 |eRe w 1| M | Re w| Re z1 |eRe w 1| and the right side of the square. This
expression clearly tends to zero faster than 1 Re w as Re w +. Hence, the integral over the right
side tends to zero as the square grows. On the left side of the square we have the same estimate
w z1 ew 1 M | Re w| Re z1 |eRe w 1| . Since Re w is negative in this case, for large | Re w| we
have w z1 ew 1 2M| Re w| Re z1 . Since Re z < 0, this bound goes to zero faster that 1 | Re w| .
On the other hand, the length of the side of the square is 2| Re w|. Hence, the integral along the
left side also tends to zero as the square grows. It remains to treat the top and bottom side. As
before, we see that on these sides we have w z1 ew 1 M |w| Re z1 |ew 1| M |Im w| Re z1 |ew
1| . On the other hand, the function w 7 |e w 1| on the horizontal lines tends to as Re w + and to
1 as Re w . Hence it is bounded from below. Moreover, since it is periodic with period 2i,
shifting the line up and down by 2i doesn’t change that bound. Hence, in our estimates we can
assume that 1 |ew 1| C 10 D. MILICI C´ on top and bottom sides of all squares. This implies
that w z1 ew 1 MC|Im w| Re z1 on top and bottom sides of all squares. Since Re z < 0 and the
length of these sides is 2|Im w|, we see that the integrals over these sides tend to zero as the
squares grow. Hence, for Re z < 0, as the squares grow, i.e., as m tends to infinity, the integral
over the sides of the square tends to 0 and the integral along converges to the integral along the
path C from 2.2.2. In addition, the sum on the right side of the above expression converges to the
series for (1 z), i.e., we have 2ieiz sin(z)(z)(z) = 2 z+1 z ieiz sin z 2 (z 1). Hence, we have
sin(z)(z)(z) = 2z z sin z 2 (1 z). This yields 2 sin z 2 cos z 2 (z)(z) = 2z z sin z 2 (1 z). and cos z
2 (z)(z) = 2z1 z (1 z). for all z C. By substituting 1 z for z we get 2 z 1z (z) = cos (1 z) .
Discuss the two group classifications that exist in Windows Server 2.pdf
Discuss the two group classifications that exist in Windows Server 2012 R2 describing both the
group type and group scope. Is there a reason you would use one scope over another?
Solution
A group is a collection of users and accouts that can be managed as a single unit. The 2 group
classification in Windows 2012 are
1. Distribution group
2. Security group
Distribution groups are only created to use email applications (like Exchange Server) to send
emails to a group of users. It is a feature introduced by Office 365 and allows sending emails to
large number of people both inside and outside the organization. It eliminates sending individual
emails to users. Bulk emails can be sent to a large group of users. The only purpose of
distribution group is to share information
Security groups are sets of users who have the same access rights, permission to resources and
performs same kind of tasks. There can be many security groups and a user can belong to more
than 1 security group. It facilitates administrators to assign same set of permissions to users
rather than individually assigning permissions to each one of them.
There are 3 group scopes as follows:
1. Universal scope: It can have users, accounts, groups or other universal groups from any
domain. Users can add members from any domain to an universal group and can assign
permission to access any resource located in any domain.
2. Global scope: These are containers for users and accounts which can be assigned permission
to objects in any domain. However, it can only have members from the domain in which it is
created.
3. Domain local group: These can have users and accounts from any domain as member.
However, only domain local groups can assign permissions to local resources or to resources that
reside in the domain in which the domain local group was created.
It is preferable to choose domain local group scope as it provides more security than the other 2
scopes. Also, the applied restrictions mean limited resources are utilized by the group..
Describe the differences between allelic heterogeneity and locus het.pdf
Describe the differences between allelic heterogeneity and locus heterogeneity
Solution
Ans.) The differences between allelic heterogeneity and locus heterogeneity
In general terms, allelic heterogeneity defines the different mutation in the same gene locus that
displays different phenotype. The variation occur due to natural selection as a consequence of
genetic drift, mutation etc.
The genetic or locus heterogeneity is the single trait that caused due to mutation in different gene
loci. Only a single mutant locus is needed to display the phenotype.
Allelic heterogeneity is the mutation in same loci that showed different phenotypic character
whereas locus heterogeneity is the only one trait occurs due to mutation in different loci..
Decision Tree Table for choosing present and future value equations. .pdf
Decision Tree Table for choosing present and future value equations. ltem to Number of Present
or Annual or Perpetual or evaluate occurrences Future Equation to use Periodi Terminal Series
na na 0 value payment:s Future Present na na Periodic Annual Periodic Vn Vo(1+i)\"
Vo=Vn/(1+i)\" Vn=pl(1+i)n-1)/[(1+1)t-1)] Single Revenue Future Terminal Perpetual Vo=
p/[(1+i)\'-1] Series of Equal payments Present Terminal Vo (1-1(1+)-1](1+i)) Cost PerpetualVo
Annual Terminal Vo=p[(1+i)n-1)/[i(1+i)\") interest rate per year (in decimals) Vo = initial value
(in time 0) V- value in time n n total number of years of compounding or discounting considered
p = amount of payment occurring every t years t number of years in one multiyear group or
interval of years (length of the interval) Payments occur at the end of the period or multiyear
interval Please note: these equations can be used for months or weeks or any other length of time
as long as the interest rate is defined at that period, n is defined as total number of periods, and t
is the number of periods in one multiperiod group or interval. In this class, we will stick
primarily to using the annual rates. Source: Figure 7.3 from Davis, et al. 2001 Forest
Management, p. 341
Solution
a. As seen from the table,
Present value of perpetual periodic CF is = P/[(1+i)^t - 1]= 50000/[1.07^45 - 1] = 50000/(21.002
- 1)
= $2,499.69
b. Present value of a perpetual $100 tax = P/i = 100/0.07 = $1,428.57
Present value of the cash flow = 2,499.69 - 1,428.57 = $1,071.12
c. AS the asking price of the land is $1000 and is less than the NPV of $1,071.12 you should buy
it..
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congenital malformations. why do you think they occurSolution.pdf
congenital malformations. why do you think they occur?
Solution
Congenital malformations
It is a morphological defect of an organ, part of an organ or a larger region of the body resulting
from an intrinsically abnormal developmental process. Structural, functional, metabolic or
behavioral disorders those are present at birth
The causes of congenital malformations divided into five broad groups.
Considering the cultures used to inoculate each medium in this labora.pdf
Considering the cultures used to inoculate each medium in this laboratory, how many different
microbial types should you expect to see in/on each medium? Which medium was most difficult
for you to transfer from? which medium was most difficult for you to inoculate? Explain your
difficulties. What is the purpose of flaming in the aseptic techniques? Provide 3 (three) reasons
why the use of aseptic techniques is essential when handling microbial cultures. Methoos of
isolation What is a microbial colony? Can a pure culture by prepared from a mixd broth or
mixed agar slant? Explain.
Solution
Answers are :
When the cultures are inoculated onto the surface of the medium, different colonies are
obtained, based on the mother culture we have chosen colonies are obtained. Until unless it is not
contaminated so.
Preparation of agar slants is difficult to transfer, and inoculate, because it needs to place
correct angle while solidifying, less surface area. Though among the medium we can say some
difficult but not so rather difficult. Nutrient broth is also difficult to transfer because more prone
of contamination and determination of the bacterial growth is also difficult.
Purpose of flaming during aseptic transfer :- Flaming causes direct burning of the cultures
samples which stick onto the inoculating loop to avoid from contamination, which technique is
called Incineration i.e., Flame sterilization.
Aseptic techniques are safe transfer of culture onto the fresh medium this is very essential
because :-
To avoid contamination to grow unwanted cultures and spoiling the medium components.
To know the perfect study of the unknown characteristics, contamination shouldn’t be seen .
To avoid infecting the person who handling the sample, some may cause severe infections and
death of the investigator.
To obtain the pure culture techniques to study the characteristics of the cultures.
Microbial colony means, group of the similar cells which shows same characteristic features ,
which obtained on the surface of the solid agar medium. These are varied in shapes and surface
but shows same morphological and physiological characteristics.
Yes, we can prepare pure culture from mixed sample either broth or solid agar, in case of
broth prepare serial dilution of the sample to obtain the pure cultures. In case of the solid slant
prepare quadrant streaking on the fresh agar plate..
Are the transfermations linear Are they an isomorphism WhySol.pdf
Are the transfermations linear? Are they an isomorphism? Why?
Solution
1. T(f)=f\'
T(f+g)=(f+g)\'=f\'+g\'
T(cf)=(cf)\'=cf\'
So, T is linear
P2 is space of polynomials of degree atmost 2 so T(P2) can have polynomials of degree atmost 1
So it cannot be an isomorphism because P2 has dimension 2 and T(P2) has dimension atmost 1
2.
T(f+g)=(f+g)\'+(f+g)\'\'=f\'+f\'\'+g\'+g\'\'=T(f)+T(g)
T(cf)=(cf)\'+(cf)\'\'=cf\'+cf\'\'=cT(f)
HEnce, T is linear
But T is not an isomorphism
As T maps a polynomial to a polynomial of degree stricly smaller than itself.
HEnce P3 and T(P3) have different dimension so T cannot be an isomorphism..
bessavillabessavillabessavillabessavillaSoluti.pdf
bessavilla
bessavilla
bessavilla
bessavilla
Solution
Besavilla Review Center, formerly called Besavilla Engineering Review Center was the oldest
review center (32 years) for civil engineering up to now. It was founded by Engr. Venancio I.
Besavilla, Jr. at Cebu City on 1980 and expanded its operations in Manila and other key cities in
the country (Baguio, Cagayan de Oro, General Santos City and Davao)
Besavilla Review Center offer comprehensive review courses to Civil and Geodetic Engineering
Students using innovative and updated teaching technique. It is an intensive lecture/discussion
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ciny Univer Laboratory es and crossing over assortment of independent.pdf
ciny Univer Laboratory es and crossing over assortment of independent the processes of Explain
Meiosis. 3 Compare and contrast meiosis with mitosis. Number of times Number of cell
divisions Number of daughter cells Genetic relationship of daughter cells to What structures
separate during anaphase (Anaphase I in Meiosis)? Name the function(s) each process performs
4. Describe the events ofspermatogenesis and oogenesis. s. Define the followin Study suggestion
1. Explain
Solution
Answer for Question 3.
A population of rabbits quadruples every 2 years. If the initial num.pdf
A population of rabbits quadruples every 2 years. If the initial number of rabbits was 25, how
long will it take for the population to reach 700??
I believe I understand how to solve the logarithm, I just don\'t understand how to extract the log
it from the question.
Solution
The population of rabbits quadruples every 2 years. Let the number of years taken to reach 700
be T. The initial population is 25. This gives an equation:
700 = 25*(4)^(T/2)
=> 700/25 = 4^(T/2)
=> 28 = 4^(T/2)
take the log of both the sides
=> log 28 = (T/2)*log 4
=> T/2 = log 28/log 4
=> T = 2*(log 28/log 4)
=> T = 4.807 years.
The population of rabbits grows to 700 from 25 in 4.807 years..
7. describe and correct the error. Describe and correct the error. .pdf
7. describe and correct the error. Describe and correct the error. (-2)^2/43 middot 2^5 = (-
2)^2/(2^2)^3 middot 2^5 (-2)^2/2^6 middot 2^5 =(-2)^2/2^11 = 1/2^2 middot 2^11 1/2^13
Solution
the error is at step 1/2^2 .2^11
since (-2)^2 is not equal to 1/2^2
so correcting (-2)^2/2^11 = 2^2/2^11 = 1/2^9
ANSWER: 1/2^9.
3. Where would you expect to find fenestrated capillaries (More tha.pdf
3. Where would you expect to find fenestrated capillaries? (More than one answer may be
correct)
1. Thyroid
2. Skeletal muscle
3. Cardiac muscle
4. Spleen
Solution
Answer : The correct answer is 1 ( Thyroid glands)
Explanation : The other location where fenestrated capillaries found are kidneys, villi of small
instetine, choroid plexus of the ventricles of the brain and in some endocrine glands..
Assembly LanguageReverse an array.Use a loop with indirect or i.pdf
Assembly Language:
Reverse an array.
Use a loop with indirect or indexed addressing to reverse the elements of an integer array in
place. Do not copy the elements to any other array. Do not use stack operations. Use the
SIZEOF, TYPE, LENGTHOF operators.
Fibonacci Numbers.
Write a code that fills the array of 15 elements with consecutive Fibonacci numbers: Fib(1) =
Fib(2) = 1; Fib (n) = Fib(n-1) + Fib(n-2) for each N > 2.
corrent code:
Solution
data barray BYTE 6 DUP (?) dwarray DWORD 3 DUP (?) TOTAL_LENGTH = $-barray
bdmessage BYTE \"Today as stored in a double word\", 0dh, 0ah, 0 dwmessage BYTE 0dh,
0ah, \"Seeing bytes in a single word\", 0dh, 0ah, 0 lengthmessage BYTE 0dh, 0ah, \"total bytes
used: \", 0 finalmessage BYTE 0dh, 0ah, \"Press any key to continue ...\", 0 .code main PROC
CALL Crlf MOV edx, OFFSET bdmessage CALL WriteString MOV eax, DWORD
PTR [barray + 2] CALL WriteDec MOV edx, OFFSET dwmessage CALL WriteString
MOVZX eax, WORD PTR [dwarray + 6] CALL WriteDec MOV edx, OFFSET
lengthmessage CALL WriteString MOV eax, TOTAL_LENGTH CALL WriteDec
MOV edx, OFFSET finalmessage CALL WriteString CALL ReadChar INVOKE
ExitProcess, 0 main ENDP END main.
wo alternative investment proposals are under consideration for a va.pdf
wo alternative investment proposals are under consideration for a vacant owner by Urban
Development Corporation. Plan A would require an immediate investment of $120,000 and first-
year expenditure for property taxes, maintenance, and insurance of $4,000, with this amount
expected to increase at a rate of $1,000 per year. Plan B would have a first cost of $170,000 and
total first-year expenses of $9,000, with an increase of $1,000 per year. The economic life of
each project is forecast to be 10 years; and at the end of this time, only the facilities from Plan B
with a value of $50,000 are expected to salvage. During the life of the project, the facility in plan
A is expected to produce $34,000 annually, whereas Plan B is expected to produce $42,000 . a)
Determine the rate of return of each plan. b) Determine the rate of return of the Additional
investment required in Plan B compared with Plan A. c) Which plan should Urban Development
select if the company uses a MARR of 12 percent?
Solution
Working notes:
(1) For each plan,
Net Annual Benefit = Annual Benefit - Annual Expense
(2) For plan B, Net Annual Benefit in terminal year (year 10) is higher by $50,000 (Salvage
Value).
(3) Rate of Return is the Internal Rate of Return (IRR) computed using Excel function
=IRR(Value 1, Value 2,.....Value n) where \"Value\" is the Net Annual Benefit, undiscounted.
(4) Based on MARR of 12%, the plan choice will be made based on NPV of the plans. NPV is
the sum of all cash outflows and inflows discounted at 12%. The plan with higher positive NPV
should be chosen.
Calculations as follow:
Plan A
Year
Initial Cost ($)
Annual Benefit ($)
Annual Cost ($)
Net Annual Benefit ($)
Net Cash Flow ($)
Discount Factor @12%
Discounted Net Annual Benefit ($)
(A)
(B)
(C)
(D) = (B) - (C)
(E) = (D) + (A)
(F)
(G) = (E) x (F)
0
-1,20,000
-1,20,000
1.0000
-1,20,000
1
34,000
4,000
30,000
30,000
0.8929
26,786
2
34,000
5,000
29,000
29,000
0.7972
23,119
3
34,000
6,000
28,000
28,000
0.7118
19,930
4
34,000
7,000
27,000
27,000
0.6355
17,159
5
34,000
8,000
26,000
26,000
0.5674
14,753
6
34,000
9,000
25,000
25,000
0.5066
12,666
7
34,000
10,000
24,000
24,000
0.4523
10,856
8
34,000
11,000
23,000
23,000
0.4039
9,289
9
34,000
12,000
22,000
22,000
0.3606
7,933
10
34,000
13,000
21,000
21,000
0.3220
6,761
IRR (%) =
18.12%
NPV ($) =
29,253
Plan B
Year
Initial Cost ($)
Annual Benefit ($)
Annual Cost ($)
Net Annual Benefit ($)
Net Cash Flow ($)
Discount Factor @12%
Discounted Net Annual Benefit ($)
(A)
(B)
(C)
(D) = (B) - (C)
(E) = (D) + (A)
(F)
(G) = (E) x (F)
0
-1,70,000
-1,70,000
1.0000
-1,70,000
1
42,000
9,000
33,000
33,000
0.8929
29,464
2
42,000
10,000
32,000
32,000
0.7972
25,510
3
42,000
11,000
31,000
31,000
0.7118
22,065
4
42,000
12,000
30,000
30,000
0.6355
19,066
5
42,000
13,000
29,000
29,000
0.5674
16,455
6
42,000
14,000
28,000
28,000
0.5066
14,186
7
42,000
15,000
27,000
27,000
0.4523
12,213
8
42,000
16,000
26,000
26,000
0.4039
10,501
9
42,000
17,000
25,000
25,000
0.3606
9,015
10
92,000
18,.
You are searching for new species that are highly tolerant to acidic.pdf
You are searching for new species that are highly tolerant to acidic conditions and find a new
single celled organism that you call WERID. The organism is ~5 um, does not have a cell wall,
but your portable field microscope is broken, and you are unable to look at the internal
structures. You do have portable equipment that will allow you to analyze the chemical content
of the WEIRD cells along with control prokaryotic and eukaryotic cells. Out of protein, DNA,
and lipid, which analysis would be most useful if you were interested in figuring out if your cells
are prokaryotic or eukaryotic? Explain your answer.
Solution
The organism that we have discovered is single celled, and devoid of a cell wall. It is ~5
micrometers in length. Since the organism has no cell wall, we cannot reach a conclusion based
or the presence/absence of peptidoglycan. Also, morphological analysis cannot be carried out
since the microscope is broken.
However, we do have portable equipment to analyze the chemical content of the cells: protein,
DNA and lipid, and compare it with control prokaryotic and eukaryotic cells. The protein and
lipid type and content may vary between prokaryotes and eukaryotes.
However, the DNA has certain differences that are absolutely unique to prokaryotes and
eukaryotes. For example, the ribosomes-part of the protein synthesis machinery differ between
the prokaryotes and eukaryotes in that the prokaryotes have a 16S rRNA while the eukaryotes
have an 18S rRNA. This difference can be exploited using our portable equipment to check for
presence of genes in the WEIRD DNA that code for either of these rRNAs, and comparing them
with the controls. Also, eukaryotes usually have DNA in the form of multiple chromosomes,
while the prokaryotes have a single circular DNA molecule.
The above differences are irrefutable, and unique to prokaryotes and eukaryotes. Therefore, our
choice for identification of the WEIRD cells would be DNA analysis.
P.S. It is also important to take into account that eukaryotes are usually multicellular, while
prokaryotes are unicellular. Also, eukaryotic cells are generally larger in size than that of the
prokaryotes..
Why must some nutrients be absorbed by active transport rather than .pdf
Why must some nutrients be absorbed by active transport rather than passive diffusion?
Solution
Passive transport is the absorption of molecules down their concentration gradients, which does
not require energy. Active transport is the absorption of nutrients, which require energy for the
process to occur. Only small molecules and water can pass through the cell membrane by passive
transport. The transport of many nutrients require energy (active transport) because of their
larger molecular size and hydrophilic nature..
Who first discovered the properties of equalitySolutionEqual.pdf
Who first discovered the properties of equality??
Solution
Equality is a relationship between two mathematical expressions. The equality between the
mathematical expressions A and B is written as A = B. The word “equality” is derived from the
Latin word “aequalis” which means equal. The symbol = was used for the first time in 1557 by
Robert Recorde. It was Leibiniz who first characterized its notion..
What happens when a bacterium is placed in a HYPOTONIC solution Wha.pdf
What happens when a bacterium is placed in a HYPOTONIC solution? What about
HYPERTONIC?
Solution
Hypertonic solution is a solution which has a higher concentrate of solute and a solution with a
low concentration of dissolved solute.When a bacterial cell placed in a hypertonic solution will
shrink and die due to the water content in bacteria goes to outside having lower concentration of
water.This process is called plasmolysis.At the other case,process by which a bacterial cell when
placed in a hypotonic solution will swell and burst by absorbing high water concentration outside
into bacterial cell is called plasmoptysis..
Why does the stack use a FILO (first in last out) schemeSolutio.pdf
Why does the stack use a FILO (first in last out) scheme?
Solution
When a program starts executing, a certain contiguous section of memory is set aside for the
program called the stack.
A stack is a limited access data structure - elements can be added and removed from the stack
only at the top. push adds an item to the top of the stack, pop removes the item from the top. A
helpful analogy is to think of a stack of books; you can remove only the top book, also you can
add a new book on the top.
Hence when the elements are pushed onto the stack using push operations, the first element
pushes will be at the bottom of stack and the last element pushed in will be at top of stack.
Therefore when pop operation is executed , top element ( last element pushed in )of stack is
popped out first and hence first element pushed in popped out last.
Thereby Stack operates in LIFO (last in first out ) or FILO ( First In Last Out) Scheme.
Trace the path followed by a sperm and by and egg from the area of t.pdf
Trace the path followed by a sperm and by and egg from the area of their formation to the
outside of the body.
Solution
The path followed by sperm self from formation to ejaculation:
The sperm cells are produced in the testicles. They travel through the seminiferous tubules into
epididymis. Epididymis stores them until the get matured. 10 day travel through West difference
where the pickup the fluid from seminal vesicles prostrate gland and cowper\'s gland. This fluid
and sperm cells is called semen. It is released from the body during ejaculation.
SEVEN UP!!
S- seminiferous tubules
E- epididymis
V- vas deferens
E- ejaculatory duct
N-nothing
U-urethra
P-penis
path of ovum:
Female is born with all the eggs that she will have in her lifetime one egg matures every month
and the egg is either fertilized are discarded.
The egg starts in the ovaries and moves into the fallopian tubes, from the fallopian tubes the egg
travels to the uterus The egg is fertilized in the uterus or comes out the cervix and through the
vagine as a period..
The potential distribution of organism can be influenced by all of th.pdf
The potential distribution of an organism can be influenced by salinity, environmental temperature, and availability of environmental oxygen, but not competition with other species of organisms. According to the document, an organism's potential distribution depends on its own physiological capacity, which is affected by temperature, salinity, and oxygen availability in the environment.
The function f(x,y)= -x^4-256x+y^3-75y+9 has two critical points. En.pdf
The function f(x,y)= -x^4-256x+y^3-75y+9 has two critical points. Enter them in order of
increasing y-coordinate and classify each as a local minimum, local maximum, or saddle point.
First point?
Second point?
Solution
df/dx = 4x^3 - 256 = 0 => x = -4 df/dy = 3y^2 - 75 => y = +5,-5 Thus (-4,-5) is
local max (-4,5) is saddle point..
1. What are telomeres What distinct mechanisms protect the ends of .pdf
1. What are telomeres? What distinct mechanisms protect the ends of telomeres from being
“repaired” by complexes that recognize double-strand breaks in nuclear DNA?
Solution
Telomeres are the very end part of the chromosomes or the capped region of chromosomes with
the repetitive nucleotide sequences TTAGGG. Sheltering protein complex helps telomere from
being repaired. The 3\' single strand telomere end forms a T-loop and it enclose in the double
stranded DNA..
DRUGS AND ITS classification slide share
Any substance (other than food) that is used to prevent, diagnose, treat, or relieve symptoms of a
disease or abnormal condition
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Aberdeen
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Consider this experiment I have 20-sided die where each face has a .pdf
The document describes an experiment where a 20-sided die is rolled once and the number on the top face is recorded. The sample space, or all possible outcomes, of this experiment is the set of numbers {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20} that could appear on the die.
congenital malformations. why do you think they occurSolution.pdf
congenital malformations. why do you think they occur?
Solution
Congenital malformations
It is a morphological defect of an organ, part of an organ or a larger region of the body resulting
from an intrinsically abnormal developmental process. Structural, functional, metabolic or
behavioral disorders those are present at birth
The causes of congenital malformations divided into five broad groups.
Considering the cultures used to inoculate each medium in this labora.pdf
Considering the cultures used to inoculate each medium in this laboratory, how many different
microbial types should you expect to see in/on each medium? Which medium was most difficult
for you to transfer from? which medium was most difficult for you to inoculate? Explain your
difficulties. What is the purpose of flaming in the aseptic techniques? Provide 3 (three) reasons
why the use of aseptic techniques is essential when handling microbial cultures. Methoos of
isolation What is a microbial colony? Can a pure culture by prepared from a mixd broth or
mixed agar slant? Explain.
Solution
Answers are :
When the cultures are inoculated onto the surface of the medium, different colonies are
obtained, based on the mother culture we have chosen colonies are obtained. Until unless it is not
contaminated so.
Preparation of agar slants is difficult to transfer, and inoculate, because it needs to place
correct angle while solidifying, less surface area. Though among the medium we can say some
difficult but not so rather difficult. Nutrient broth is also difficult to transfer because more prone
of contamination and determination of the bacterial growth is also difficult.
Purpose of flaming during aseptic transfer :- Flaming causes direct burning of the cultures
samples which stick onto the inoculating loop to avoid from contamination, which technique is
called Incineration i.e., Flame sterilization.
Aseptic techniques are safe transfer of culture onto the fresh medium this is very essential
because :-
To avoid contamination to grow unwanted cultures and spoiling the medium components.
To know the perfect study of the unknown characteristics, contamination shouldn’t be seen .
To avoid infecting the person who handling the sample, some may cause severe infections and
death of the investigator.
To obtain the pure culture techniques to study the characteristics of the cultures.
Microbial colony means, group of the similar cells which shows same characteristic features ,
which obtained on the surface of the solid agar medium. These are varied in shapes and surface
but shows same morphological and physiological characteristics.
Yes, we can prepare pure culture from mixed sample either broth or solid agar, in case of
broth prepare serial dilution of the sample to obtain the pure cultures. In case of the solid slant
prepare quadrant streaking on the fresh agar plate..
Are the transfermations linear Are they an isomorphism WhySol.pdf
Are the transfermations linear? Are they an isomorphism? Why?
Solution
1. T(f)=f\'
T(f+g)=(f+g)\'=f\'+g\'
T(cf)=(cf)\'=cf\'
So, T is linear
P2 is space of polynomials of degree atmost 2 so T(P2) can have polynomials of degree atmost 1
So it cannot be an isomorphism because P2 has dimension 2 and T(P2) has dimension atmost 1
2.
T(f+g)=(f+g)\'+(f+g)\'\'=f\'+f\'\'+g\'+g\'\'=T(f)+T(g)
T(cf)=(cf)\'+(cf)\'\'=cf\'+cf\'\'=cT(f)
HEnce, T is linear
But T is not an isomorphism
As T maps a polynomial to a polynomial of degree stricly smaller than itself.
HEnce P3 and T(P3) have different dimension so T cannot be an isomorphism..
bessavillabessavillabessavillabessavillaSoluti.pdf
bessavilla
bessavilla
bessavilla
bessavilla
Solution
Besavilla Review Center, formerly called Besavilla Engineering Review Center was the oldest
review center (32 years) for civil engineering up to now. It was founded by Engr. Venancio I.
Besavilla, Jr. at Cebu City on 1980 and expanded its operations in Manila and other key cities in
the country (Baguio, Cagayan de Oro, General Santos City and Davao)
Besavilla Review Center offer comprehensive review courses to Civil and Geodetic Engineering
Students using innovative and updated teaching technique. It is an intensive lecture/discussion
type covering basic theories, principles, problems and applications..
ciny Univer Laboratory es and crossing over assortment of independent.pdf
ciny Univer Laboratory es and crossing over assortment of independent the processes of Explain
Meiosis. 3 Compare and contrast meiosis with mitosis. Number of times Number of cell
divisions Number of daughter cells Genetic relationship of daughter cells to What structures
separate during anaphase (Anaphase I in Meiosis)? Name the function(s) each process performs
4. Describe the events ofspermatogenesis and oogenesis. s. Define the followin Study suggestion
1. Explain
Solution
Answer for Question 3.
A population of rabbits quadruples every 2 years. If the initial num.pdf
A population of rabbits quadruples every 2 years. If the initial number of rabbits was 25, how
long will it take for the population to reach 700??
I believe I understand how to solve the logarithm, I just don\'t understand how to extract the log
it from the question.
Solution
The population of rabbits quadruples every 2 years. Let the number of years taken to reach 700
be T. The initial population is 25. This gives an equation:
700 = 25*(4)^(T/2)
=> 700/25 = 4^(T/2)
=> 28 = 4^(T/2)
take the log of both the sides
=> log 28 = (T/2)*log 4
=> T/2 = log 28/log 4
=> T = 2*(log 28/log 4)
=> T = 4.807 years.
The population of rabbits grows to 700 from 25 in 4.807 years..
7. describe and correct the error. Describe and correct the error. .pdf
7. describe and correct the error. Describe and correct the error. (-2)^2/43 middot 2^5 = (-
2)^2/(2^2)^3 middot 2^5 (-2)^2/2^6 middot 2^5 =(-2)^2/2^11 = 1/2^2 middot 2^11 1/2^13
Solution
the error is at step 1/2^2 .2^11
since (-2)^2 is not equal to 1/2^2
so correcting (-2)^2/2^11 = 2^2/2^11 = 1/2^9
ANSWER: 1/2^9.
3. Where would you expect to find fenestrated capillaries (More tha.pdf
3. Where would you expect to find fenestrated capillaries? (More than one answer may be
correct)
1. Thyroid
2. Skeletal muscle
3. Cardiac muscle
4. Spleen
Solution
Answer : The correct answer is 1 ( Thyroid glands)
Explanation : The other location where fenestrated capillaries found are kidneys, villi of small
instetine, choroid plexus of the ventricles of the brain and in some endocrine glands..
Assembly LanguageReverse an array.Use a loop with indirect or i.pdf
Assembly Language:
Reverse an array.
Use a loop with indirect or indexed addressing to reverse the elements of an integer array in
place. Do not copy the elements to any other array. Do not use stack operations. Use the
SIZEOF, TYPE, LENGTHOF operators.
Fibonacci Numbers.
Write a code that fills the array of 15 elements with consecutive Fibonacci numbers: Fib(1) =
Fib(2) = 1; Fib (n) = Fib(n-1) + Fib(n-2) for each N > 2.
corrent code:
Solution
data barray BYTE 6 DUP (?) dwarray DWORD 3 DUP (?) TOTAL_LENGTH = $-barray
bdmessage BYTE \"Today as stored in a double word\", 0dh, 0ah, 0 dwmessage BYTE 0dh,
0ah, \"Seeing bytes in a single word\", 0dh, 0ah, 0 lengthmessage BYTE 0dh, 0ah, \"total bytes
used: \", 0 finalmessage BYTE 0dh, 0ah, \"Press any key to continue ...\", 0 .code main PROC
CALL Crlf MOV edx, OFFSET bdmessage CALL WriteString MOV eax, DWORD
PTR [barray + 2] CALL WriteDec MOV edx, OFFSET dwmessage CALL WriteString
MOVZX eax, WORD PTR [dwarray + 6] CALL WriteDec MOV edx, OFFSET
lengthmessage CALL WriteString MOV eax, TOTAL_LENGTH CALL WriteDec
MOV edx, OFFSET finalmessage CALL WriteString CALL ReadChar INVOKE
ExitProcess, 0 main ENDP END main.
wo alternative investment proposals are under consideration for a va.pdf
wo alternative investment proposals are under consideration for a vacant owner by Urban
Development Corporation. Plan A would require an immediate investment of $120,000 and first-
year expenditure for property taxes, maintenance, and insurance of $4,000, with this amount
expected to increase at a rate of $1,000 per year. Plan B would have a first cost of $170,000 and
total first-year expenses of $9,000, with an increase of $1,000 per year. The economic life of
each project is forecast to be 10 years; and at the end of this time, only the facilities from Plan B
with a value of $50,000 are expected to salvage. During the life of the project, the facility in plan
A is expected to produce $34,000 annually, whereas Plan B is expected to produce $42,000 . a)
Determine the rate of return of each plan. b) Determine the rate of return of the Additional
investment required in Plan B compared with Plan A. c) Which plan should Urban Development
select if the company uses a MARR of 12 percent?
Solution
Working notes:
(1) For each plan,
Net Annual Benefit = Annual Benefit - Annual Expense
(2) For plan B, Net Annual Benefit in terminal year (year 10) is higher by $50,000 (Salvage
Value).
(3) Rate of Return is the Internal Rate of Return (IRR) computed using Excel function
=IRR(Value 1, Value 2,.....Value n) where \"Value\" is the Net Annual Benefit, undiscounted.
(4) Based on MARR of 12%, the plan choice will be made based on NPV of the plans. NPV is
the sum of all cash outflows and inflows discounted at 12%. The plan with higher positive NPV
should be chosen.
Calculations as follow:
Plan A
Year
Initial Cost ($)
Annual Benefit ($)
Annual Cost ($)
Net Annual Benefit ($)
Net Cash Flow ($)
Discount Factor @12%
Discounted Net Annual Benefit ($)
(A)
(B)
(C)
(D) = (B) - (C)
(E) = (D) + (A)
(F)
(G) = (E) x (F)
0
-1,20,000
-1,20,000
1.0000
-1,20,000
1
34,000
4,000
30,000
30,000
0.8929
26,786
2
34,000
5,000
29,000
29,000
0.7972
23,119
3
34,000
6,000
28,000
28,000
0.7118
19,930
4
34,000
7,000
27,000
27,000
0.6355
17,159
5
34,000
8,000
26,000
26,000
0.5674
14,753
6
34,000
9,000
25,000
25,000
0.5066
12,666
7
34,000
10,000
24,000
24,000
0.4523
10,856
8
34,000
11,000
23,000
23,000
0.4039
9,289
9
34,000
12,000
22,000
22,000
0.3606
7,933
10
34,000
13,000
21,000
21,000
0.3220
6,761
IRR (%) =
18.12%
NPV ($) =
29,253
Plan B
Year
Initial Cost ($)
Annual Benefit ($)
Annual Cost ($)
Net Annual Benefit ($)
Net Cash Flow ($)
Discount Factor @12%
Discounted Net Annual Benefit ($)
(A)
(B)
(C)
(D) = (B) - (C)
(E) = (D) + (A)
(F)
(G) = (E) x (F)
0
-1,70,000
-1,70,000
1.0000
-1,70,000
1
42,000
9,000
33,000
33,000
0.8929
29,464
2
42,000
10,000
32,000
32,000
0.7972
25,510
3
42,000
11,000
31,000
31,000
0.7118
22,065
4
42,000
12,000
30,000
30,000
0.6355
19,066
5
42,000
13,000
29,000
29,000
0.5674
16,455
6
42,000
14,000
28,000
28,000
0.5066
14,186
7
42,000
15,000
27,000
27,000
0.4523
12,213
8
42,000
16,000
26,000
26,000
0.4039
10,501
9
42,000
17,000
25,000
25,000
0.3606
9,015
10
92,000
18,.
You are searching for new species that are highly tolerant to acidic.pdf
You are searching for new species that are highly tolerant to acidic conditions and find a new
single celled organism that you call WERID. The organism is ~5 um, does not have a cell wall,
but your portable field microscope is broken, and you are unable to look at the internal
structures. You do have portable equipment that will allow you to analyze the chemical content
of the WEIRD cells along with control prokaryotic and eukaryotic cells. Out of protein, DNA,
and lipid, which analysis would be most useful if you were interested in figuring out if your cells
are prokaryotic or eukaryotic? Explain your answer.
Solution
The organism that we have discovered is single celled, and devoid of a cell wall. It is ~5
micrometers in length. Since the organism has no cell wall, we cannot reach a conclusion based
or the presence/absence of peptidoglycan. Also, morphological analysis cannot be carried out
since the microscope is broken.
However, we do have portable equipment to analyze the chemical content of the cells: protein,
DNA and lipid, and compare it with control prokaryotic and eukaryotic cells. The protein and
lipid type and content may vary between prokaryotes and eukaryotes.
However, the DNA has certain differences that are absolutely unique to prokaryotes and
eukaryotes. For example, the ribosomes-part of the protein synthesis machinery differ between
the prokaryotes and eukaryotes in that the prokaryotes have a 16S rRNA while the eukaryotes
have an 18S rRNA. This difference can be exploited using our portable equipment to check for
presence of genes in the WEIRD DNA that code for either of these rRNAs, and comparing them
with the controls. Also, eukaryotes usually have DNA in the form of multiple chromosomes,
while the prokaryotes have a single circular DNA molecule.
The above differences are irrefutable, and unique to prokaryotes and eukaryotes. Therefore, our
choice for identification of the WEIRD cells would be DNA analysis.
P.S. It is also important to take into account that eukaryotes are usually multicellular, while
prokaryotes are unicellular. Also, eukaryotic cells are generally larger in size than that of the
prokaryotes..
Why must some nutrients be absorbed by active transport rather than .pdf
Why must some nutrients be absorbed by active transport rather than passive diffusion?
Solution
Passive transport is the absorption of molecules down their concentration gradients, which does
not require energy. Active transport is the absorption of nutrients, which require energy for the
process to occur. Only small molecules and water can pass through the cell membrane by passive
transport. The transport of many nutrients require energy (active transport) because of their
larger molecular size and hydrophilic nature..
Who first discovered the properties of equalitySolutionEqual.pdf
Who first discovered the properties of equality??
Solution
Equality is a relationship between two mathematical expressions. The equality between the
mathematical expressions A and B is written as A = B. The word “equality” is derived from the
Latin word “aequalis” which means equal. The symbol = was used for the first time in 1557 by
Robert Recorde. It was Leibiniz who first characterized its notion..
What happens when a bacterium is placed in a HYPOTONIC solution Wha.pdf
What happens when a bacterium is placed in a HYPOTONIC solution? What about
HYPERTONIC?
Solution
Hypertonic solution is a solution which has a higher concentrate of solute and a solution with a
low concentration of dissolved solute.When a bacterial cell placed in a hypertonic solution will
shrink and die due to the water content in bacteria goes to outside having lower concentration of
water.This process is called plasmolysis.At the other case,process by which a bacterial cell when
placed in a hypotonic solution will swell and burst by absorbing high water concentration outside
into bacterial cell is called plasmoptysis..
Why does the stack use a FILO (first in last out) schemeSolutio.pdf
Why does the stack use a FILO (first in last out) scheme?
Solution
When a program starts executing, a certain contiguous section of memory is set aside for the
program called the stack.
A stack is a limited access data structure - elements can be added and removed from the stack
only at the top. push adds an item to the top of the stack, pop removes the item from the top. A
helpful analogy is to think of a stack of books; you can remove only the top book, also you can
add a new book on the top.
Hence when the elements are pushed onto the stack using push operations, the first element
pushes will be at the bottom of stack and the last element pushed in will be at top of stack.
Therefore when pop operation is executed , top element ( last element pushed in )of stack is
popped out first and hence first element pushed in popped out last.
Thereby Stack operates in LIFO (last in first out ) or FILO ( First In Last Out) Scheme.
Trace the path followed by a sperm and by and egg from the area of t.pdf
Trace the path followed by a sperm and by and egg from the area of their formation to the
outside of the body.
Solution
The path followed by sperm self from formation to ejaculation:
The sperm cells are produced in the testicles. They travel through the seminiferous tubules into
epididymis. Epididymis stores them until the get matured. 10 day travel through West difference
where the pickup the fluid from seminal vesicles prostrate gland and cowper\'s gland. This fluid
and sperm cells is called semen. It is released from the body during ejaculation.
SEVEN UP!!
S- seminiferous tubules
E- epididymis
V- vas deferens
E- ejaculatory duct
N-nothing
U-urethra
P-penis
path of ovum:
Female is born with all the eggs that she will have in her lifetime one egg matures every month
and the egg is either fertilized are discarded.
The egg starts in the ovaries and moves into the fallopian tubes, from the fallopian tubes the egg
travels to the uterus The egg is fertilized in the uterus or comes out the cervix and through the
vagine as a period..
The potential distribution of organism can be influenced by all of th.pdf
The potential distribution of an organism can be influenced by salinity, environmental temperature, and availability of environmental oxygen, but not competition with other species of organisms. According to the document, an organism's potential distribution depends on its own physiological capacity, which is affected by temperature, salinity, and oxygen availability in the environment.
The function f(x,y)= -x^4-256x+y^3-75y+9 has two critical points. En.pdf
The function f(x,y)= -x^4-256x+y^3-75y+9 has two critical points. Enter them in order of
increasing y-coordinate and classify each as a local minimum, local maximum, or saddle point.
First point?
Second point?
Solution
df/dx = 4x^3 - 256 = 0 => x = -4 df/dy = 3y^2 - 75 => y = +5,-5 Thus (-4,-5) is
local max (-4,5) is saddle point..
1. What are telomeres What distinct mechanisms protect the ends of .pdf
1. What are telomeres? What distinct mechanisms protect the ends of telomeres from being
“repaired” by complexes that recognize double-strand breaks in nuclear DNA?
Solution
Telomeres are the very end part of the chromosomes or the capped region of chromosomes with
the repetitive nucleotide sequences TTAGGG. Sheltering protein complex helps telomere from
being repaired. The 3\' single strand telomere end forms a T-loop and it enclose in the double
stranded DNA..
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