B\u00e0i 5 (trang 75 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0Th\u1ee7y ph\u00e2n ho\u00e0n to\u00e0n m\u1ed9t pentapeptit A thu \u0111\u01b0\u1ee3c 3 mol glyxin, 1 mol alanin v\u00e0 1 mol valin. Khi th\u1ee7y ph\u00e2n kh\u00f4ng ho\u00e0n to\u00e0n A th\u00ec trong h\u1ed7n h\u1ee3p s\u1ea3n ph\u1ea9m th\u1ea5y c\u00f3 c\u00e1c \u0111ipeptit Ala-Gly, Gly-Ala v\u00e0 tripeptit Gly-Gly-Val<\/strong><\/p>\na) H\u00e3y x\u00e1c \u0111\u1ecbnh tr\u00ecnh t\u1ef1 c\u00e1c \u03b1-amino axit trong pentapeptit A.<\/p>\n
b) H\u00e3y ch\u1ec9 ra \u0111\u00e2u l\u00e0 aminoaxit \u0111\u1ea7u N, d\u00e2u l\u00e0 aminoaxit \u0111\u1ea7u C \u1edf pentapeptit A.<\/p>\n
L\u1eddi gi\u1ea3i:<\/b><\/p>\n
a.X\u00e1c \u0111\u1ecbnh tr\u00ecnh t\u1ef1 c\u00e1c \u03b1-amino axit trong pentapeptit A<\/p>\n
– Pentapeptit A g\u1ed3m Gly,Ala, Val<\/p>\n
– Th\u1ee7y ph\u00e2n kh\u00f4ng ho\u00e0n to\u00e0n A thu \u0111\u01b0\u1ee3c 2 \u0111ipeptit Ala-Gly v\u00e0 Gly-Ala v\u00e0 m\u1ed9t tripeptit Gly-Gly-Val<\/p>\n
T\u1eeb c\u00e1c d\u1eef ki\u1ec7n tr\u00ean suy ra v\u1ecb tr\u00ed c\u1ee7a Ala l\u00e0 th\u1ee9 2 sau Gly v\u00e0 Val \u0111\u1ee9ng \u1edf cu\u1ed1i : Gly-Ala-Gly-Gly-Val<\/p>\n
b. Amino axit \u0111\u1ea7u N l\u00e0 Gly; Amino axit \u0111\u1ea7u C l\u00e0 Val<\/p>\n
B\u00e0i 6 (trang 75 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0Thu\u1ed1c th\u1eed n\u00e0o sau \u0111\u00e2y c\u00f3 th\u1ec3 d\u00f9ng \u0111\u1ec3 ph\u00e2n bi\u1ec7t \u0111\u01b0\u1ee3c c\u00e1c dung d\u1ecbch glucozo, glixerol, etanol v\u00e0 l\u00f2ng tr\u1eafng tr\u1ee9ng?<\/strong><\/p>\nA. dd NaOH<\/p>\n
B. dd AgNO3<\/sub><\/p>\nC. CuOH)2<\/sub><\/p>\nD. dd HNO3<\/sub><\/p>\nL\u1eddi gi\u1ea3i:<\/b><\/p>\n
\u0110\u00e1p \u00e1n C<\/p>\n
\n
\n
B\u00e0i 7 (trang 75 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0Ph\u00e2n bi\u1ec7t c\u00e1c kh\u00e1i ni\u1ec7m<\/strong><\/p>\na. Peptit v\u00e0 protein<\/strong><\/p>\nb. Protein \u0111\u01a1n gi\u1ea3n v\u00e0 protein ph\u1ee9c t\u1ea1p<\/strong><\/p>\nL\u1eddi gi\u1ea3i:<\/strong><\/p>\na. Peptit l\u00e0 nh\u1eefng h\u1ee3p ch\u1ea5t ch\u1ee9a c\u00e1c g\u1ed1c \u03b1-amin axit li\u00ean k\u1ebft v\u1edbi nhau b\u1eb1ng nh\u1eefng li\u00ean k\u1ebft peptit \u2013CO-NH-<\/p>\n
Protein l\u00e0 nh\u1eefng polipeptit cao ph\u00e2n t\u1eed<\/p>\n
b. Protein \u0111\u01a1n gi\u1ea3n : \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh ch\u1ec9 t\u1eeb c\u00e1c \u03b1-amino axit<\/p>\n
Protein ph\u1ee9c t\u1ea1p : t\u1ea1o th\u00e0nh t\u1eeb c\u00e1c protein \u0111\u01a1n gi\u1ea3n c\u1ed9ng v\u1edbi th\u00e0nh ph\u1ea7n \u201cphi protein\u201d<\/p>\n
B\u00e0i 8 (trang 75 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0H\u00e3y ph\u00e2n bi\u1ec7t c\u00e1c dung d\u1ecbch keo sau \u0111\u00e2y b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p h\u00f3a h\u1ecdc : n\u01b0\u1edbc x\u00e0 ph\u00f2ng, h\u1ed3 tinh b\u1ed9t, l\u00f2ng tr\u1eafng tr\u1ee9ng<\/strong><\/p>\nL\u1eddi gi\u1ea3i:<\/b><\/p>\n
Ph\u00e2n bi\u1ec7t n\u01b0\u1edbc x\u00e0 ph\u00f2ng, h\u1ed3 tinh b\u1ed9t, l\u00f2ng tr\u1eafng tr\u1ee9ng.<\/p>\n
– D\u00f9ng I2<\/sub>\u00a0nh\u1eadn ra tinh b\u1ed9t<\/p>\n– D\u00f9ng Cu(OH)2<\/sub>\u00a0do c\u00f3 ph\u1ea3n \u1ee9ng m\u00e0u v\u1edbi protein \u0111\u1ec3 nh\u1eadn ra l\u00f2ng tr\u1eafng tr\u1ee9ng<\/p>\nB\u00e0i 9 (trang 75 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0X\u00e1c \u0111\u1ecbnh ph\u00e2n t\u1eed kh\u1ed1i g\u1ea7n \u0111\u00fang c\u1ee7a m\u1ed9t hemoglobin (h\u1ed3ng c\u1ea7u c\u1ee7a m\u00e1u) ch\u1ee9a 0,4% Fe(m\u1ed7i ph\u00e2n t\u1eed hemoglobin ch\u1ec9 ch\u1ee9a m\u1ed9t nguy\u00ean t\u1eed Fe)<\/strong><\/p>\nL\u1eddi gi\u1ea3i:<\/b><\/p>\n
Do m\u1ed7i ph\u00e2n t\u1eed hemoglobin ch\u1ec9 ch\u1ee9a 1 Fe<\/p>\n
=> Mhemoglobin<\/sub>\u00a0= (56 x 100) : 0,4 = 140000 (\u0111vC)<\/p>\nB\u00e0i 10 (trang 75 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0Khi th\u1ee7y ph\u00e2n 500 gam protein A thu \u0111\u01b0\u1ee3c 170 g alanin. N\u1ebfu ph\u00e2n t\u1eed kh\u1ed1i c\u1ee7a A l\u00e0 50.000 th\u00ec s\u1ed1 m\u1eaft x\u00edch alanin trong ph\u00e2n t\u1eed A l\u00e0 bao nhi\u00eau?<\/strong><\/p>\nL\u1eddi gi\u1ea3i:<\/b><\/p>\n
S\u1ed1 mol alanin = 170 : 89 = 1,91 (mol)<\/p>\n
V\u1edbi MA<\/sub>\u00a0= 50.000 => nA = 500 : 50000 = 0,01 (mol)<\/p>\nTrong 0,01 mol ph\u00e2n t\u1eed A c\u00f3 1,91 mol alanin<\/p>\n
=> Trong 1 mol ph\u00e2n t\u1eed A c\u00f3 191 mol alanin<\/p>\n
S\u1ed1 m\u1eaft x\u00edch alanin c\u00f3 trong ph\u00e2n t\u1eed A l\u00e0 191<\/p>\n
<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"
B\u00e0i 1 (trang 75 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0T\u1eeb ba \u03b1-aminoaxit X, Y, Z c\u00f3 th\u1ec3 t\u1ea1o th\u00e0nh m\u1ea5y tripeptit trong \u0111\u00f3 X, Y, Z? A. 2 B. 3 C. 4 D. 6 L\u1eddi gi\u1ea3i: \u0110\u00e1p \u00e1n D B\u00e0i 2 (trang 75 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0Ph\u00e1t bi\u1ec3u n\u00e0o sau \u0111\u00e2y \u0111\u00fang? A. Ph\u00e2n t\u1eed […]<\/p>\n","protected":false},"author":3,"featured_media":25290,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1306],"tags":[1436,1435],"yoast_head":"\n
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A. 2 B. 3 C. 4 D. 6 L\u1eddi gi\u1ea3i: \u0110\u00e1p \u00e1n D B\u00e0i 2 (trang 75 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0Ph\u00e1t bi\u1ec3u n\u00e0o sau \u0111\u00e2y \u0111\u00fang? A. 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