A. ANSAutosomal dominant (a trait could passed down from previous.pdf

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A. ANS: Autosomal dominant (a trait could passed down from previous generation), X – linked recessive (in this disease the dominant gene is not expressed in female, it only expressed in male) and Autosomal recessive in female (Females are only carriers) effects would be included according to pedigree. B. ANS: Leber hereditary optic atrophy which causes blindness in adulthood predominantly occurs in male, in this the inheritance a point mutation was arisen in mitochondrial DNA and this affect causing gene is directly transmitted from mother to offspring. It is mitochondrial inherited disease in which a retinal ganglion cell was degenerated, and it leads to lose of central vision. C. ANS: The individuals II 3 (Male) and II 4 (Female) both are normal, so the offspring always should be normal (According to the pedigree in offspring 4 (Female) is normal). D. ANS: The individuals II 2 (Male) and II 1 (Female) both are not in normal condition. According to the pedigree male is normal and the female is not normal so the entire offspring is affected in this disease. (According to the pedigree in offspring 1 (Female) is carrier and 2 & 3 (Males) are affected). Solution A. ANS: Autosomal dominant (a trait could passed down from previous generation), X – linked recessive (in this disease the dominant gene is not expressed in female, it only expressed in male) and Autosomal recessive in female (Females are only carriers) effects would be included according to pedigree. B. ANS: Leber hereditary optic atrophy which causes blindness in adulthood predominantly occurs in male, in this the inheritance a point mutation was arisen in mitochondrial DNA and this affect causing gene is directly transmitted from mother to offspring. It is mitochondrial inherited disease in which a retinal ganglion cell was degenerated, and it leads to lose of central vision. C. ANS: The individuals II 3 (Male) and II 4 (Female) both are normal, so the offspring always should be normal (According to the pedigree in offspring 4 (Female) is normal). D. ANS: The individuals II 2 (Male) and II 1 (Female) both are not in normal condition. According to the pedigree male is normal and the female is not normal so the entire offspring is affected in this disease. (According to the pedigree in offspring 1 (Female) is carrier and 2 & 3 (Males) are affected)..

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The individuals II 2 (Male) and II 1 (Female) both are not in normal condition. According to the
pedigree male is normal and the female is not normal so the entire offspring is affected in this
disease. (According to the pedigree in offspring 1 (Female) is carrier and 2 & 3 (Males) are
affected).

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Cu(H2O)42+(aq) + 4CN-(aq) <=> Cu(CN)42-(aq) +4 H2O(l) Cu+2 is a Lewis acid as it has vacantd-orbitals that can accomodate an electron pair(Electron pairacceptor) CN- is a Lewis base as it has lone pair ofelectron that is donated to Cu+2 in the reaction(Electron pair donor) H2O is a Lewis base as the Oxygen atom haslone pair of electrons and therefre, it acts as an electronpair donor. Solution Cu(H2O)42+(aq) + 4CN-(aq) <=> Cu(CN)42-(aq) +4 H2O(l) Cu+2 is a Lewis acid as it has vacantd-orbitals that can accomodate an electron pair(Electron pairacceptor) CN- is a Lewis base as it has lone pair ofelectron that is donated to Cu+2 in the reaction(Electron pair donor) H2O is a Lewis base as the Oxygen atom haslone pair of electrons and therefre, it acts as an electronpair donor..

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Assets=Liabilities+Stockholder\'s Equity Cash given $1000 on the same day and $1000 is Accounts Payable Cash given $1000 on the same day accounting equation -$1000=Liabilities-$1000 Accounts payable for remaining half of the expenses accounting equation Assets= +$1000-$1000 Solution Assets=Liabilities+Stockholder\'s Equity Cash given $1000 on the same day and $1000 is Accounts Payable Cash given $1000 on the same day accounting equation -$1000=Liabilities-$1000 Accounts payable for remaining half of the expenses accounting equation Assets= +$1000-$1000.

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1)Interest per year 810/6= 135 Simple interest = Amount * Rate*Time 135 = 3750 *r * 1 135 = 3750*r r = 135/3750 = .036 or 3.6% 2) Quarterly rate = 4.9/4 = 1.225% [4 quarters in a year] Term (number of quarters) = IN[Future value/present value]/IN(1+r) = IN[8600/2000]/IN[1+.01225] = IN 4.3/ IN 1.01225 = 1.45862/ .012176 = 119.79468 Number of years =119.79468 / 4 = 29.95 years Term : 29.95 Solution 1)Interest per year 810/6= 135 Simple interest = Amount * Rate*Time 135 = 3750 *r * 1 135 = 3750*r r = 135/3750 = .036 or 3.6% 2) Quarterly rate = 4.9/4 = 1.225% [4 quarters in a year] Term (number of quarters) = IN[Future value/present value]/IN(1+r) = IN[8600/2000]/IN[1+.01225] = IN 4.3/ IN 1.01225 = 1.45862/ .012176 = 119.79468 Number of years =119.79468 / 4 = 29.95 years Term : 29.95 .

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Answer: Answer: 1. The body of a fungus is in the form of solid sheets: True The body of fungus is called mycelium, it is not differentiated into roots, stem or leaf. 2. Yeasts are a kind of fungus: True Yeast are unicellular fungus 3. False Spore are haploid, the need to fuse with another to produce diploid, multicellular individual. 4. False Phloem is alive at maturity because movement of material within the phloem requires energy. 5. True 6. True 7. True Ferns have motile sperms and non motile egg. 8. False Most of the plant species are angiosperms. 9. False Angiosperms are divided into monocots and dicots 10. Charophyte---e. not a plant 11. fern---d. seedless vascular plant 12. pine tree—b. gymnosperm 13. moss—c. non-vascular plant 14. palm tree—a. angiosperm Solution Answer: Answer: 1. The body of a fungus is in the form of solid sheets: True The body of fungus is called mycelium, it is not differentiated into roots, stem or leaf. 2. Yeasts are a kind of fungus: True Yeast are unicellular fungus 3. False Spore are haploid, the need to fuse with another to produce diploid, multicellular individual. 4. False Phloem is alive at maturity because movement of material within the phloem requires energy. 5. True 6. True 7. True Ferns have motile sperms and non motile egg. 8. False Most of the plant species are angiosperms. 9. False Angiosperms are divided into monocots and dicots 10. Charophyte---e. not a plant 11. fern---d. seedless vascular plant 12. pine tree—b. gymnosperm 13. moss—c. non-vascular plant 14. palm tree—a. angiosperm.

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The key activity areas for securities firms are i. Investing Se.pdf

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The finger like projections of intestine known as villi increases surface area of small intestines. The surface of small intestine is lined up with epithelial cells containing brush borders or microvilli that help to absorb nutrients into the blood or lymph, which later moves to the body cells. Due to the presence of specialized proteins within the microvilli known as porters, the nutrients move into the cells. Human gut flora contains several types of bacterial species in the small intestines that are involved in the production of vitamin K, B as well as metabolism of bile acid and xenobiotics. So, correct option is d.transferring nutrients across intestinal membranes Solution The finger like projections of intestine known as villi increases surface area of small intestines. The surface of small intestine is lined up with epithelial cells containing brush borders or microvilli that help to absorb nutrients into the blood or lymph, which later moves to the body cells. Due to the presence of specialized proteins within the microvilli known as porters, the nutrients move into the cells. Human gut flora contains several types of bacterial species in the small intestines that are involved in the production of vitamin K, B as well as metabolism of bile acid and xenobiotics. So, correct option is d.transferring nutrients across intestinal membranes.

The finger like projections of intestine known as villi increases su.pdf

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The answer isC. 12In establishing paternity by dna fingerprintin.pdf

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