LAMBIE 스토어

LAMBIE

개인인증

팔로워0 팔로우

소개

등록된 소개글이 없습니다.

전문분야 등록된 전문분야가 없습니다.

판매자 정보

학교정보

입력된 정보가 없습니다.

직장정보

입력된 정보가 없습니다.

자격증

입력된 정보가 없습니다.

판매지수

판매중 자료수

12개
전체 판매량

2개
최근 3개월 판매량

0개
자료후기 점수

-
자료문의 응답률

-

전체자료 12개

황산 콘드로이틴, 황산 콘드로이친, Chondroitin sulfate

Chondroitin SulphateCarbohydrate C x (H 2 O) y Most of them cause sweetness characteristics Used as a source of energy and as a building block of body. 2021-05-23CarbohydrateProteoglycan Protein cores and negatively charged disaccharide continuum (Glycosaminoglycan, GAG) are combined. Once known as “muco-polysaccharide” A macromolecule 2021-05-23Proteoglycan Secretory Transmembrane 2021-05-23Aggrecan aggregate 2021-05-23Protein cores + GAG 2021-05-23GAG(Glycosaminoglycan) Linear heterosaccharides with large molecular weight Formed in Golgi apparatus during O-linked glycosylation by linking with protein cores 2021-05-23Chondroitin sulfate Abbreviated to ‘CS’ GAG( Glycosaminglycan ) Proteoglycan N- acetylgalactosamine + glucuronic acid + sulfate A(4), B(dermatan sulfate), C(6), D(2,6), E(4,6) types 2021-05-232021-05-23 Ichijo H, Sugiura N, Kimata K. Application of Chondroitin Sulfate Derivatives for Understanding Axonal Guidance in the Nervous System during Development. Polymers. 2013; 5(1):254-268. https://doi.org/10.3390/polym5010254Chondroitin sulfate Along with protein collagen, it is a major component of extracellular matrix Important in maintaining the structural integrity of the tissue It is widely distributed in skin, umbilical cord, and other connective tissue. Cartilage contains 20–40% of the chontroitin sulfate 2021-05-23Guess! 2021-05-23Beneficial Effects improvement in the symptoms of osteoarthritis with chondroitin sulfate. CS-E helps treat CNS damage Chondroitin sulfate proteoglycan (CSPG); chondroitin sulphate E (CS-E). Science-Business eXchange 5, 314 (2012). https://doi.org/10.1038/scibx.2012.314 2021-05-23{nameOfApplication=Show}

자연과학| 2021.05.23| 13페이지| 2,500원| 조회(88)

탄수화물, Glycosaminoglycan, 세포외기질, GAG, 황산 콘드로이..

미리보기

닫기
해양산성화, Ocean acidification, 산성화, Acidification

• Importance of phytoplankton underestimated• Petroleum is made from phytoplankton• Diatom ooze is used for reflective paints, polishing material and insulation• Oxygen in earth‘s atmosphere was made originally by cyanobacteria

자연과학| 2021.05.23| 16페이지| 2,500원| 조회(97)

해양학, 해양산성화, 산성화, Oceanology, ocean acidifica..

미리보기

닫기
판매자 표지

유전학, 소집단의 유전학, genetics in small population

Genetics and Breeding Science Chapter 7 Problems in Small PopulationQuantification of Genetic Variability Average Heterozygosity Average Number of AllelesQuantification of Genetic Variability Average Heterozygosity Average Number of AllelesIt was discovered that the genetic variation is exist in protein and DNA level( Isozyme , Microsatellite) . ( Markert and Moller 1959) The observed amount of genetic variation is larger than the expected Value estimated from the balance of selection and mutation. ( Lewontin and Hubby 1966) It is necesarry to quantify the genetic diversity Average Heterozygosity （ H e) ： A probability the locus in an individual is heterozygote h ( Heterozygosity ) = 1-Σ p 2 H e = 1/n ･ Σ h （ Average of examined all loci ） p = Allele frequency in a locus n = Number of examined loci Average Heterozygosity ( He )Calculation of heterozygosity in Japanese Scallop Abashiri Akkeshi Oshamanbe Hiranai Wakinosawa Sohma Mean Aat A 0.006 0 0 0.005 0.005 0 B 0.827 0.867 0.864 0.75of the Bottleneck of N=2 Large number of individual N=100 He = 0.5 Na = 200 N=2 Decrease Large number of individual N = 100 He = 0.375 Na = 4 ExpandDecrease of heterozygosity in finite population In the infinite population, heterozygosity does not change h t = h 0 But, in the finite population, heterozygosity decrease according to the number of individuals in the population Δh = -1 / 2N → - Δh is equal to the increase of Coefficient of inbreeding The heterozygosity after t- th generation is h t = ( 1 - 1/2N) t h 0Differences of the decrease rate of heterozygosity in the different population size Number of individual = 10 h 0 = 0.5 The heterozygosity after 2 generations is h t = Number of individual = 100, h 0 = 0.5 The heterozygosity after 2 generations is h t =Differences of the decrease rate of heterozygosity in the different population size Number of individual = 10 h 0 = 0.5 The heterozygosity after 2 generations is h t = (1 – 1 / 2 ・ 10) 2 ・ h 0 = 0.451 Number of individual = 100,hi Oshamanbe Hiranai Wakinosawa Sohma Mean Aat A 0.006 0 0 0.005 0.005 0 B 0.827 0.867 0.864 0.752 0.688 0.715 C 0.161 0.128 0.136 0.242 0.307 0.285 D 0.006 0.005 0 0 0 0 Allele 4 3 2 3 3 2Average number of allele in Japanese flounder population Hokkaido Fukushima Fukuoka Miyazaki Average Number of Allele 1.739 1.739 1.652 1.609 Examined 79 145 72 100 individuals Examined number of individual affects the average number of alleleEffective number of allele ( n e ) n e = 1 / ∑ x i 2 It is proposed by Kimura and Crow (1964 ) as the reciprocal Of homozygosity Hokkaido Fukushima Fukuoka Miyazaki Average Number of allele 1.739 1.739 1.652 1.609 n e 1.049 1.063 1.061 1.036The number of allele in the next generation The examined number of individual and allele frequency are influence to the change of the number of allele. In general, the number of allele in next generation is expected by following formula A ’ = A - ∑ ( 1 – p j ) 2NDifferences of the change of the number of allele in the differe’ =The effect of population size on the decrease rate of number of allele Number of individual = 2, p = 0.5, q = 0.5 A’ = 2 – [(1 – 0.5) 2 ・ 2 + (1 – 0.5) 2 ・ 2 ] = 1.875 Number of individual = 10, p = 0.5, q = 0.5 A’ = 2 – [( 1 – 0.5) 2 ・ 10 + (1 – 0.5) 2 ・ 10 ] ≒ 2.0 Number of individual = 2, p = 0.8, q = 0.2 A’ = 2 – [( 1 – 0.8) 2 ・ 2 + (1 – 0.2) 2 ・ 2 ] ≒ 1.589Which is the better marker to reflect the passed history of population, heterozygosity or average number of allele? In the case of heterozygosity, about 25% of variations will decrease after the bottleneck of N = 2. Allele frequency = 0.5 and two alleles were assumed, A ’ = 2 – (1-0.5) 4 – (1-0.5) 4 = 1.88 Allele frequency = 0.9 and 0.1, two alleles were assumed, A ’ = 2 – (1-0.9) 4 – (1-0.1) 4 = 1.34Number of allele It is sensitive marker to reflect the passed population size and its fluctuation. But, the existing number of individual and allele frequency are influence to the number of allele. It is difficult to use for the opulation size after a bottleneck can estimate used logistic growth equiation N ( t ) = K / (1 + be - rt ) K : the equilibrium size of the population r : growth rate b : ( K - N 0 ) / N 0 Population Size after a BottleneckAssume that the equilibrium population size = 500 In the case of growth rate = 1.0 The population size will increase 2.72 times per generation In the case of growth rate = 0.5 The population size will increase 1.65 times per generation In the case of growth rate = 0.2 のとき The population size will increase 1.22 times per generation Growth Rate and Population SizeThe change of heterozygosity after a bottleneck of N=2 in each growth rateGenetic differences between original population and the combination of subpopulations Number of Combined Subpopulation Number of Combination of Subpopulation Number of combinations that i ndicated significant differences f rom original populationMinimum Number of individual to keep the genetic diversity The minimum number of individual how}

자연과학| 2021.05.23| 36페이지| 5,000원| 조회(83)

유전학, 병목현상, genetics, population, 집단유전학, 유전학개론, 소집단유전

미리보기

닫기
판매자 표지

유전학, 자연 선택, natural selection

Genetics and Breeding Science Chapter 6 Natural SelectionAllele and genotype frequencies does not change in the case of following phenomenon does not occur. Genetic drift, Natural selection, Inbreeding, Mutation, Migration The relationship between allele frequency and genotype frequency can express as follows. ( p + q ) 2 = p 2 + 2 pq + q 2 Even if the equilibrium was disturbed, it will be recovered after one generation. Hardy-Weinberg’s lawThe Concept of Natural Selection Natural Selection is simply defined as “The differential reproduction of alternative genetic variants”. It would occur not only by competition, but also by “the physical condition of life”.Computation of the fitness of three genotypes when the number of progeny produced by each genotype is known Genotype A 1 A 1 A 1 A 2 A 2 A 2 Total Number of zygotes in one generation 40 50 10 100 Number of zygotes produced by each genotype in next generation 80 90 10 180 Average number of progeny per individual in next generation 8 0.9 1.0 × 0.5 = 0.5Genotype AA Aa aa Total Fitness （ W ）１１ 1-s Initial Frequency p 2 2 pq q 2 1.0 Selection Against Recessive HomozygotesGenotype AA Aa aa Total Fitness （ W ）１１ 1-s Initial Frequency p 2 2 pq q 2 1.0 Next Generation p 2 2 pq q 2 （ 1-s ） 1-s q 2 Normalized p 2 / （ 1-s q 2 ） 2 pq / （ 1-s q 2 ） q 2 （ 1-s ） / （ 1-s q 2 ） Frequency Selection Against Recessive HomozygotesGenotype AA Aa aa Total Fitness （ W ）１１ 1-s Initial p 2 2 pq q 2 1.0 Frequency Next Generation p 2 2 pq q 2 （ 1-s ） 1-s q 2 Normalized p 2 / （ 1-s q 2 ） 2 pq / （ 1-s q 2 ） q 2 （ 1-s ） / （ 1-s q 2 ） Frequency q 1 = （ 1/2 ）･（ 2 pq / （ 1-s q 2 ）） + q 2 （ 1-s ） / （ 1-s q 2 ） = （ q -s q 2 ） / （ 1-s q 2 ） Δq = q 1 – q = -s pq 2 / （ 1-s q 2 ） Selection Against Recessive Homozygotes⊿q = - spq 2 /(1- sq 2 ) If the a is recessive lethal, s = 1 = - pq 2 /(1- q 2 ) = -(1- q ) q 2 /(1+ q )(1- q ) = - q 2 /(1+ q ) The allele frequency of a in next generation is calculated as follows. q 1 = ( q 0 - sq 0 2 )/(1- sq 0 itial Frequency of 0.5 Generation FrequencyQ = q (1 - q ) F / ( q 2 + q (1- q ) F ) Rate of Homozygote increased by inbreedingF = 0 のとき q = 0.0098 Probability of homozygotes is occur = 0.000096 F = 0.125(Half-sib) Q = 0.927 Probability of homozygotes is occur= 0.00131 ( 13.6 times) F = 0.250(Full-sib) Q = 0.962 Probability of homozygotes is occur= 0.00252 ( 26.3 times) F = 0.5( Selfing ) Q = 0.981 Probability of homozygotes is occur= 0.00495 ( 51.6 times) The probability of the recessive lethal allele become homozygote by inbreeding, after the 100 generations of natural selectionGenotype AA Aa aa Total Fitness （ W ）１ -s １ 1-t Initial p 2 2 pq q 2 1.0 Frequency Next generation p 2 （ 1-s ） 2 pq q 2 （ 1-t ） 1-s p 2 -t q 2 Normalized p 2 （ 1-s ） / （ 1-s p 2 - tq 2 ） q 2 （ 1- ｔ） / （ 1-s p 2 - tq 2 ） Frequency 2 pq / （ 1-s p 2 - tq 2 ） Change in allele frequency Δq = （ pq （ s p - t q ） / （ 1-s p 2 - tq 2 ） p = t / ( s+t ) q = s / ( s+t ) Heterozygote Advantage - Over Dominance -0.54 Initialhere a certain form of malaria, caused by the parasite Plasmdium falciparum, is common.1 2 3 4 5 6 7 8 9 10 ATG AGT AAA CCA GCT GAA GCC AAT GCT GCT Met Ser Lys Pro Ala Glu Ala Asn Ala Ala 1 2 3 4 5 6 7 8 9 10 ATG AGT AAA CCA GCT G T A GCC AAT GCT GCT Met Ser Lys Pro Ala Val Ala Asn Ala Ala Nucleotide Amino Acid Nucleotide Amino Acid The change of amino acid sequence in Hemoglobin-β chain caused by the change of nucleotide sequence The characteristics of protein was changed which was caused by the change of amino acid from Glutamic Acid which has polarity to Valine which has not polaroty . Normal hemoglobin Sickle-cell hemoglobinGenotype distribution of Hemoglobin Types in The population from Nigeria Normal Hb A Hb A = 9365 人 Heterozygote Hb A Hb S = 2993 Sick-Cell Anemia Hb S Hb S = 29 Total 12387 Calculation of Fitness for the Normal and Anemia Hemoglobin ( Ayala 1982 )observed Expected Ob/Ex Relative Fitness Hb A Hb A Hb A Hb S Hb S Hb S 9365 2993 29 9523.8 2675.3 187.9 0.983 1.119 0 equilibrium frequency of Hb A estimated from selection coefficient qHb A = 0.863 / (0.863 + 0.122) = 0.876 The equilibrium frequency of Hb S estimated from selection coefficient qHb S = 0.122 / (0.863 + 0.122) = 0.124The goodness of fit to the Hardy-Weinberg’s law is checked by Chi-squire test Observed Expected （ Ob-Ex) 2 /Ex Χ 2 test Hb A Hb A Hb A Hb S Hb S Hb S 9365 2993 29 9523.8 2675.3 187.9 q Hb A = （（ 9365×2 ） + 2993 ） / （ 12387×2 ） = 0.877 q Hb S = （（ 29×2 ） + 2993 ） / （ 12387×2 ） = 0.123 Expected of Hb A Hb A = Expected of Hb A Hb S = Expected of Hb S Hb S =The goodness of fit to the Hardy-Weinberg’s law is checked by Chi-squire test Observed Expected （ Ob-Ex) 2 /Ex Χ 2 test Hb A Hb A Hb A Hb S Hb S Hb S 9365 2993 29 9523.8 2675.3 187.9 q Hb A = （（ 9365×2 ） + 2993 ） / （ 12387×2 ） = 0.877 q Hb S = （（ 29×2 ） + 2993 ） / （ 12387×2 ） = 0.123 Expected of Hb A Hb A = 0.877×0.877×12387 = 9523.8 Expected of Hb A Hb S = 0.877×0.123×2×12387 = 2675.3 Expected of Hb S Hb S = 0.123×0.123×1w}

자연과학| 2021.05.23| 24페이지| 5,000원| 조회(135)

유전학, 다윈, 자연선택설, 진화, 자연선택, 찰스 다윈, 자연 선택, Natural..

미리보기

닫기
판매자 표지

유전학, 유전변이와 근교, genetic drift and inbreeding

Genetics and Breeding Science Chapter 5 Genetic drift and inbreedingGenetic drift is random change in allele frequencies from generation to generation because of sampling error. Genetic drift is an example of a stochastic process. Standard Error of allele frequency SE=√ （ p(1-p)/ ２ N ） For Example. ： When N=50 and p=0.5, SE=0.05 The fluctuation level is depend on the number of individuals Genetic DriftFluctuation of Allele Frequency in the Case of N=10Fluctuation of Allele Frequency in the Case of N=500Effective population size ( N e) is define as the size of the ideal population ( N ) that will result in the same amount of genetic drift as in the actual population being considered. Another way to say, Ne is the population size affect to increase coefficient of inbreeding in the population. Concept of Effective Population size （ N e ）Differences of number of female and male which contributing to next generation Differences of number of individual in each generations Differences of numbNf / ( Nm + Nf ) is ledN e = ( 4 N m ･ N f ) / ( N m + N f ) N m : Number of male N f : Number of female The case of N m=50, N f =50 N e = ( 4 ･ 50 ･ 50 ) / ( 50+50 ) = 100 The case of N m=30, N f =70 N e = (4 ･ 30 ･ 70 ) / ( 30+70 ) = 84 Unequal Sex ratioNe when the number of the male and female are extremely different Number of Female Number of Male N e 1000 1000 1000 1000 1000 1 2 3 4 5 4.0 8.0 12.0 15.9 19.9Sex ratio and N e1/ N e = ( 1/n ) ･ ( 1/ N 1 + 1/ N 2 + ･･･ +1/ N n ) n ： Number of generation N 1 、 N 2 、 N n ： Number of individuals in each generation The case of N 1 = N 2 = N 3 = N 4 =100 1/ N e=(1/4) ･ (1/100+1/100+1/100+1/100)=1/100 N e=100 The case of N 1 = N 2 = N 4 =100 、 N 3 =30 1/ N e=(1/4) ･ (1/100+1/100+1/30+1/100)=1/63.2 N e=63.2 Fluctuating Population SizeThe fluctuation of Population Size of Banana fly in a year March April May June July August September October November N e 100 500 1000 3000 5000 3000 1000 500 100 1/ N e = 1/9(1/100 + 1/500 + 1/1000 + 1/3000 + 1/2+1/150+1/3+1/100+1/10+1/130+1/15 +1/80+1/2+1/70+1/20+1/70+1/5+1/80+1/2+1/80) =1/17×2.352 = 1/7.23 N e ≒ 7.2×1000 individuals ○ ○ ○ ○ ○ ○ ○ ○ ○ ● ● ● ● ● ● ● ● Year Rabbit LynxV K = ∑ ( k i – 2) 2 N The variance in the population in which the mean number of gametes to translate to next generation is 2 is represented as follows On the other hand, the probability of the two gametes in an individual are originated from same parental individual is represented as follows ∑ k i ( k i – 1) 2 N (2 N – 1) = 2 + V K 4 N – 2 Differences of number of gamete which contributing to next generation in each pair （１）The N e is detected as the reciprocal of the probability of two gametes originated from same parent. So, it is represented as follows N e = 2 + V K 4 N – 2 Differences of number of gamete which contributing to next generation in each pair （２）Distribution of number of offspring under the Poison distribution of average is 2Estimation of Ne in three experimental populations （１） A B C No. of237 Pop. B 0 0 19.00 0.026 Pop. C 20 2 9.50 0.053The crossing between closely related individuals. There is common ancestor of mother and father. InbreedingAn image of Autozygous , Allozygous , Homozygous and Heterozygous♂ ？ ♀ ♂ ♀ Ⅰ Ⅲ Ⅳ Ⅱ Ⅴ AB CD The probability in which A goes to Ⅲ from Ⅰ ： 1/2 A goes to Ⅴ from Ⅲ ： 1/2 A goes to Ⅳ from Ⅰ ： 1/2 A goes to Ⅴ from Ⅳ ： 1/2 ↓ （ 1/2 ） 4 The probability in which A , B , C and D become homozygote, respectively. ↓ （ 1/2 ） 4 × ４ =0.25 Calculation of Coefficient of Inbreeding In the case of full-sib matingCalculation of Coefficient of Inbreeding from Genealogy F = ( 1/2 ) N ( 1 + F CA ) F = ∑ [( 1/2 ) N ( 1 + F CA )] If there are several roots, add all F . If there is the ancestor which has common ancestor, the coefficient of inbreeding of the ancestor has to be added.Complicated pedigree of the population of great tits in the NetherlandsJ ● N ● M ● K ● H ● F ● Common Ancestor in the population of great tits in the NetherlandsJ ● Path ： B-C-D-I- 8 F ：（ 1/2 ） 8 =0.0039M ● Common Ancestor “ M ” Path ： B-C-F-G-L- M -L-G-F-A Number ： 10 F ：（ 1/2 ） 10 = 0.0010K ● Common Ancestor “ K ” Path ： B-C-D-I- K -H-E-A Number ： 8 F ：（ 1/2 ） 8 =0.0039H ● Common Ancestor “ H ” Path ： B- H -E-A Number ： 4 F ：（ 1/2 ） 4 =0.0625F ● Common Ancestor “ F ” Path ： B-C- F -A Number ： 4 F ：（ 1/2 ） 4 =0.0625Coefficient of Inbreeding in the individual “ ？ ”It was discovered that the genetic variation is exist in protein and DNA level( Isozyme , Microsatellite) . ( Markert and Moller 1959) The observed amount of genetic variation is larger than the expected value estimated from the balance of selection and mutation. ( Lewontin and Hubby 1966) It is necesarry to quantify the genetic diversity Average Heterozygosity （ H e) ： A probability the locus in an individual is heterozygote h ( Heterozygosity ) = 1-Σ p 2 H e = 1/n ･ Σ h （ Average of examined all loci ） p = Allele frequency in a locus n = Number of examined loci Average Heterozygosity ( He )Genothow}

자연과학| 2021.05.23| 37페이지| 5,000원| 조회(87)

유전학, statistics, 근교, 유전변이, genetic drift, Inb..

미리보기

닫기