2a) The inheritance pattern is Autosomal recessive.Parents are not.pdf
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B) Based on the allele frequency for this trait in the population, w.pdf
B) Based on the allele frequency for this trait in the population, what assumptions would you
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recessive trait? Explain your reasoning.
C) You collect phenotypic data from the living family member and determine that the following
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Affected member of this family are: II2, II4, II7, II8, III2, III4, III5, III7, III8, IV1, IV2
Which, if any, modes of inheritance may be ruled out for this family? Briefly explain your
reasoning for each mode you rule out (use table below).
D) Are the two sets of twins in this family identical or fraternal twins? What is the basis of your
answer?
E) If III-2 and III-7 married and had a child, what is the probability that their child would have
tooth agenesis? State the mode of inheritance that you think is most likely for this tree. Show
your calculations based on that mode of inheritance.
Mode of Inheritance:
Probability:Rule out? (Yes or No)Rationaleautosomal dominantautosomal recessiveX-linked
dominantX-linked recessive
Solution
A) The outsiders of the family tree are
II-1, II-5, II-6, II-9, III-1
B) The genotype of outsiders are as under
II-1, a carrier, because one of the offspring is affected (III-2).
II-5, affected, because both of its offsprings are affected (III-4, III-5).
II-6, normal, also not a carrier, because its offspring is normal (III-6).
II-9, affected, because both of its offsprings are affected (III-7, III-8)
III-1, affected, because both of its offsprings are affected (IV-1, IV-2).
C)
The mode of inheritance is not autosomal dominant, because the offspring of the affected parent
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offspring of the affected parent (II-7) is normal.
The inheitance is not X-linked dominant nor X-linked recessive, because both male and female
are affected.
D) The two sets of twins in this family- one set is identical and other is fraternal twins
Twins (II-1, II-2) are fraternal twins, because one is affeted (II-2) and other is unaffected (II-1).
Twins (II-7, II-8) are identical twins, because both the twins are affected.
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NaCN(aq) ? Na+(aq) + CN-(aq) Assuming that NaCN is an ionic compound
(composed of a monatomic cation and a polyatomic anion) it will completely dissociate in water,
therefore [NaCN] = [CN-] Then, since HCN (Prussic Acid) is a weak acid, write the equilibrium
equation between the acid and its conjugate base: ..... H2O(l) + CN-(aq) ? HCN(aq) + OH-(aq)
Pure solids and liquids are ignored in equilibrium reactions, so water is ignored. I ........ ..........
0.1 . . | . . . . 0 . . . . . 0 C ...... ............ -x . . | . . . .+x . . . . +x E ...... .......... 0.1 . . | . . . . x . . . . . x
Since Ka was given, change it to Kb by using: Ka * Kb = Kw 4.9*10^(-10) * Kb = 10^(-14) Kb
= 2.0*10^(-5) Then set up the base dissociation constant equation (since it was written with a
CN- (base) on the left and HCN (acid) on the right). It is assumed that x is significantly smaller
than [CN-], just to make the calculations easier (avoiding quadratics). Kb = [HCN] [OH-] / [CN-
] 2.0*10^(-5) = x² / 0.1 x² = 2.0*10^(-6) x = [OH-] = 1.4*10^(-3) M pOH = -log[OH-] = -
log(1.4*10^(-3)) = 2.8 pH = 14 - pOH = 14 - 2.8 = 11.2
Solution
NaCN(aq) ? Na+(aq) + CN-(aq) Assuming that NaCN is an ionic compound
(composed of a monatomic cation and a polyatomic anion) it will completely dissociate in water,
therefore [NaCN] = [CN-] Then, since HCN (Prussic Acid) is a weak acid, write the equilibrium
equation between the acid and its conjugate base: ..... H2O(l) + CN-(aq) ? HCN(aq) + OH-(aq)
Pure solids and liquids are ignored in equilibrium reactions, so water is ignored. I ........ ..........
0.1 . . | . . . . 0 . . . . . 0 C ...... ............ -x . . | . . . .+x . . . . +x E ...... .......... 0.1 . . | . . . . x . . . . . x
Since Ka was given, change it to Kb by using: Ka * Kb = Kw 4.9*10^(-10) * Kb = 10^(-14) Kb
= 2.0*10^(-5) Then set up the base dissociation constant equation (since it was written with a
CN- (base) on the left and HCN (acid) on the right). It is assumed that x is significantly smaller
than [CN-], just to make the calculations easier (avoiding quadratics). Kb = [HCN] [OH-] / [CN-
] 2.0*10^(-5) = x² / 0.1 x² = 2.0*10^(-6) x = [OH-] = 1.4*10^(-3) M pOH = -log[OH-] = -
log(1.4*10^(-3)) = 2.8 pH = 14 - pOH = 14 - 2.8 = 11.2.
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1. A pedigree is presented where a woman has a 100% chance of passing on a defective gene to her daughter based on her genotype. Calculations are shown for determining the probability that her grandson will have the defective gene.
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B) Based on the allele frequency for this trait in the population, w.pdf
B) Based on the allele frequency for this trait in the population, what assumptions would you
make about the genotype of each outsider in this family tree if you were expecting it to be a
recessive trait? Explain your reasoning.
C) You collect phenotypic data from the living family member and determine that the following
family members have tooth agenesis
Affected member of this family are: II2, II4, II7, II8, III2, III4, III5, III7, III8, IV1, IV2
Which, if any, modes of inheritance may be ruled out for this family? Briefly explain your
reasoning for each mode you rule out (use table below).
D) Are the two sets of twins in this family identical or fraternal twins? What is the basis of your
answer?
E) If III-2 and III-7 married and had a child, what is the probability that their child would have
tooth agenesis? State the mode of inheritance that you think is most likely for this tree. Show
your calculations based on that mode of inheritance.
Mode of Inheritance:
Probability:Rule out? (Yes or No)Rationaleautosomal dominantautosomal recessiveX-linked
dominantX-linked recessive
Solution
A) The outsiders of the family tree are
II-1, II-5, II-6, II-9, III-1
B) The genotype of outsiders are as under
II-1, a carrier, because one of the offspring is affected (III-2).
II-5, affected, because both of its offsprings are affected (III-4, III-5).
II-6, normal, also not a carrier, because its offspring is normal (III-6).
II-9, affected, because both of its offsprings are affected (III-7, III-8)
III-1, affected, because both of its offsprings are affected (IV-1, IV-2).
C)
The mode of inheritance is not autosomal dominant, because the offspring of the affected parent
(II-7) is normal.
The mode of inheritance is outosomal recessive. Because the parents I-1, I-2 are carriers and
offspring of the affected parent (II-7) is normal.
The inheitance is not X-linked dominant nor X-linked recessive, because both male and female
are affected.
D) The two sets of twins in this family- one set is identical and other is fraternal twins
Twins (II-1, II-2) are fraternal twins, because one is affeted (II-2) and other is unaffected (II-1).
Twins (II-7, II-8) are identical twins, because both the twins are affected.
E) If III-2 and III-7 married, their 100% probability that the child would have tooth agenesis.
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a.) Topoisomerase-Topoisomerases are enzymes used in the overwindin.pdf
NaCN(aq) ? Na+(aq) + CN-(aq) Assuming that NaCN is an ionic compound
(composed of a monatomic cation and a polyatomic anion) it will completely dissociate in water,
therefore [NaCN] = [CN-] Then, since HCN (Prussic Acid) is a weak acid, write the equilibrium
equation between the acid and its conjugate base: ..... H2O(l) + CN-(aq) ? HCN(aq) + OH-(aq)
Pure solids and liquids are ignored in equilibrium reactions, so water is ignored. I ........ ..........
0.1 . . | . . . . 0 . . . . . 0 C ...... ............ -x . . | . . . .+x . . . . +x E ...... .......... 0.1 . . | . . . . x . . . . . x
Since Ka was given, change it to Kb by using: Ka * Kb = Kw 4.9*10^(-10) * Kb = 10^(-14) Kb
= 2.0*10^(-5) Then set up the base dissociation constant equation (since it was written with a
CN- (base) on the left and HCN (acid) on the right). It is assumed that x is significantly smaller
than [CN-], just to make the calculations easier (avoiding quadratics). Kb = [HCN] [OH-] / [CN-
] 2.0*10^(-5) = x² / 0.1 x² = 2.0*10^(-6) x = [OH-] = 1.4*10^(-3) M pOH = -log[OH-] = -
log(1.4*10^(-3)) = 2.8 pH = 14 - pOH = 14 - 2.8 = 11.2
Solution
NaCN(aq) ? Na+(aq) + CN-(aq) Assuming that NaCN is an ionic compound
(composed of a monatomic cation and a polyatomic anion) it will completely dissociate in water,
therefore [NaCN] = [CN-] Then, since HCN (Prussic Acid) is a weak acid, write the equilibrium
equation between the acid and its conjugate base: ..... H2O(l) + CN-(aq) ? HCN(aq) + OH-(aq)
Pure solids and liquids are ignored in equilibrium reactions, so water is ignored. I ........ ..........
0.1 . . | . . . . 0 . . . . . 0 C ...... ............ -x . . | . . . .+x . . . . +x E ...... .......... 0.1 . . | . . . . x . . . . . x
Since Ka was given, change it to Kb by using: Ka * Kb = Kw 4.9*10^(-10) * Kb = 10^(-14) Kb
= 2.0*10^(-5) Then set up the base dissociation constant equation (since it was written with a
CN- (base) on the left and HCN (acid) on the right). It is assumed that x is significantly smaller
than [CN-], just to make the calculations easier (avoiding quadratics). Kb = [HCN] [OH-] / [CN-
] 2.0*10^(-5) = x² / 0.1 x² = 2.0*10^(-6) x = [OH-] = 1.4*10^(-3) M pOH = -log[OH-] = -
log(1.4*10^(-3)) = 2.8 pH = 14 - pOH = 14 - 2.8 = 11.2.
A) There are several factors taken into consideration before giving .pdf
Tanzania is seeking a credit rating to raise capital for infrastructure projects through bond issues. It has shown strong economic growth of 7% in 2014, reducing poverty and increasing incomes, allowing it to advance from low- to middle-income status. Tanzania recently issued a $500 million bond with a 5-year maturity to be repaid in semi-annual installments, though the price faced speculation. Further bond issues may have higher coupons given Tanzania's continued growth, with proceeds still intended for infrastructure development.
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here heat capacity of calorimeter = (50 x 4.186 x ( 37.92 - 29.11 ) - 50 x 4.186 x (
29.11 - 20.91 ))/dt = 127.673/dt where dt is the change in temp .. it is not given so i have
mentioned it as dt
Solution
here heat capacity of calorimeter = (50 x 4.186 x ( 37.92 - 29.11 ) - 50 x 4.186 x (
29.11 - 20.91 ))/dt = 127.673/dt where dt is the change in temp .. it is not given so i have
mentioned it as dt.
3.) 2 methyl 2 penteneSolution3.) 2 methyl 2 pentene.pdf
H2CrO4 ? 2 H+ + CrO42- Ka = 5.83 x 10-8 Initial .0375 2(.150) 0 Change - x +x
+x Eq. .0375 -x .3 + x x Ka = [H+][A-]/[HA] 5.83E-8 = (.3 + x)(x)/(.0375 - x) Let\'s assume
that the value of [H+] is within +- 5%, so we can simplify the above eq. to make calculations
simpler. 5.83E-8 = (.3x + x^2)/(.0375) x = [CrO4^2-] = 7.3E-9 mol/L
Solution
H2CrO4 ? 2 H+ + CrO42- Ka = 5.83 x 10-8 Initial .0375 2(.150) 0 Change - x +x
+x Eq. .0375 -x .3 + x x Ka = [H+][A-]/[HA] 5.83E-8 = (.3 + x)(x)/(.0375 - x) Let\'s assume
that the value of [H+] is within +- 5%, so we can simplify the above eq. to make calculations
simpler. 5.83E-8 = (.3x + x^2)/(.0375) x = [CrO4^2-] = 7.3E-9 mol/L.
16. Innate immunity is the bodys first lie of defence against foreig.pdf
Fe2O3(s)+3CO(g)=======>2Fe(s)+3CO2(g) 50.0g Fe2O3*(1 mol Fe2O3/
159.69 g Fe2O3)*(2 mol Fe/1 mol Fe2O3)*(55.85 g Fe/1 mol Fe) 34.97 g Fe [ theoretical yield]
% of yield = practical/theoretical * 100 = 32.8/34.97 *100 = 93.80 %
Solution
Fe2O3(s)+3CO(g)=======>2Fe(s)+3CO2(g) 50.0g Fe2O3*(1 mol Fe2O3/
159.69 g Fe2O3)*(2 mol Fe/1 mol Fe2O3)*(55.85 g Fe/1 mol Fe) 34.97 g Fe [ theoretical yield]
% of yield = practical/theoretical * 100 = 32.8/34.97 *100 = 93.80 %.
2) Answer is The cloud storage provider ensures that you have a bac.pdf
FOR ACIDIC MEDIUM: First separate into half-reactions. Second balance all
elements other than \"O\" and \"H\" Third balance \"O\" with \"H2O\" Fourth balance \"H\" with
\"H+\" ions Fifth balance charge with electrons then repeat for the second half reaction. make
sure the electrons are on opposite sides of you half reactions. (one reaction should have them in
the reactants, the other the products) make the electrons \"cancel\" each other by multiplying the
other half reaction by the number of the electrons. then add the two half reactions together by
combining like terms.
Solution
FOR ACIDIC MEDIUM: First separate into half-reactions. Second balance all
elements other than \"O\" and \"H\" Third balance \"O\" with \"H2O\" Fourth balance \"H\" with
\"H+\" ions Fifth balance charge with electrons then repeat for the second half reaction. make
sure the electrons are on opposite sides of you half reactions. (one reaction should have them in
the reactants, the other the products) make the electrons \"cancel\" each other by multiplying the
other half reaction by the number of the electrons. then add the two half reactions together by
combining like terms..
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a. Sn2/E2 mixture (secondary; strong nucleophile) b. Mostly Sn2 with some E2
(primary; strong nucleophile) c. E2 (secondary and very hindered, blocking Sn2) d. Sn1/E1
mixture (tertiary; water indicates it\'s going Sn1/E1) a. cyanide b. alcohol c. ester
Solution
a. Sn2/E2 mixture (secondary; strong nucleophile) b. Mostly Sn2 with some E2
(primary; strong nucleophile) c. E2 (secondary and very hindered, blocking Sn2) d. Sn1/E1
mixture (tertiary; water indicates it\'s going Sn1/E1) a. cyanide b. alcohol c. ester.
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- 1. 2a) The inheritance pattern is Autosomal recessive. Parents are not affected (generation 1) In Generation 2, son is affected In generation 3, no one is affected. In Autosomal recessive, both recessive genes should be present to cause a disease. If the parents in generation 1 carry one recessive gene each, son of generation 2 might get two recessive gene (each from one parent) to get affected. The son carries two recessive genes from which one gene will be passed on to third generation which is not sufficient to get affected. So autosomal recessive inheritance pattern skip generation that is not seen in every generation of an affected family. Solution 2a) The inheritance pattern is Autosomal recessive. Parents are not affected (generation 1) In Generation 2, son is affected In generation 3, no one is affected. In Autosomal recessive, both recessive genes should be present to cause a disease. If the parents in generation 1 carry one recessive gene each, son of generation 2 might get two recessive gene (each from one parent) to get affected. The son carries two recessive genes from which one gene will be passed on to third generation which is not sufficient to get affected. So autosomal recessive inheritance pattern skip generation that is not seen in every generation of an affected family.