护理入学考试生物
方法
- 说出这道题的主题:细胞结构、物质运输、细胞分裂、DNA 与蛋白质合成、遗传、生物分子、微生物或实验设计。
- 回忆关键规则:A–T 和 G–C,扩散由高浓度向低浓度,水向溶质更多的一侧移动,有丝分裂保持而减数分裂减半染色体数目。
- 对于杂交,先写出亲本的基因型,列出配子,并填好庞尼特方格再计数。
- 对于实验,标出自变量、因变量和控制变量,并问一问证据是支持还是仅仅提示了结论。
- 把每个选项与所问的问题对照,而不是只看它是不是一句真话。
公式表、每个专题的关键方法步骤和 8 道练习题,完整解答附在最后。
腕管综合征中受压的是哪条神经?
垂体肿瘤向上生长,导致双眼外侧(颞侧)半视野缺失。视觉通路在哪里受压?
Frank–Starling 机制描述的是:
一名惊恐发作时过度通气的患者手指麻刺。动脉血气最可能显示什么?
What is a random sample?
Why do cells get far more ATP from glucose when oxygen is present than when it is absent?
0.5 mol of KCl is dissolved to make 250 mL of solution. What is its molarity?
Which pair of substances could form a buffer solution?
答案: D · 正中神经
正中神经在屈肌支持带下方穿过腕管;受压会引起拇指、示指和中指麻刺感。
答案: B · 视交叉
双颞侧偏盲是由于视交叉处交叉的鼻侧纤维受压所致,视交叉正好位于垂体上方。
答案: A · 心室充盈增加时,每搏输出量增加
舒张末期容积越大,心肌被拉伸得越多,收缩力和每搏输出量就越大,从而使输出与静脉回流相匹配。
答案: A · pH 升高,PaCO₂ 降低
过度通气排出 CO₂,引起急性呼吸性碱中毒。碱中毒使钙与白蛋白的结合增加,离子钙降低,从而引起麻刺感。
答案: A · A group where everyone in the population has an equal chance of being chosen
In a random sample, chance decides who is chosen, so the sample is more likely to represent the whole population. Convenience and volunteer samples can be biased.
答案: A · Without oxygen only glycolysis runs, giving a net of 2 ATP per glucose
Glycolysis gives a net of 2 ATP. With oxygen, pyruvate is fully oxidised in the Krebs cycle and electron transport chain, which make many more ATP. Without oxygen these stages stop, and fermentation only recycles NAD⁺ so that glycolysis can continue.
答案: C · 2 M
250 mL = 0.250 L. M = mol ÷ L = 0.5 ÷ 0.250 = 2 M. Dividing by 250 without converting would give a wrong answer.
答案: D · Carbonic acid (H₂CO₃) and bicarbonate (HCO₃⁻)
A buffer contains a weak acid and its conjugate base, which can absorb added base or acid. H₂CO₃/HCO₃⁻ is the main buffer in blood. HCl is a strong acid, so Cl⁻ cannot pick up H⁺ well enough to buffer.
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