how are the axial and appendicular skeletons attachedSolutionA.pdf

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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.

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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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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.

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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.

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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.

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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) .

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