113 年 國立臺灣大學農藝系暨研究所作物乙組《遺傳學(D)》
第 1 題15 分
For each of the terms in the upper section (a to o), choose the best matching phrase in the lower section (1 to 15). (15 points, 1 point each)
Upper section:
a. Alleles
b. Law of segregation
c. Epistasis
d. Chiasma
e. Okazaki fragments
f. Hydrogen bond
g. Locus
h. Nondisjunction
i. Interphase
j. Incomplete dominance
k. Law of independent assortment
l. Synapsis
m. Anaphase
n. Pleiotropy
o. Hemizygous
Lower section:
- Failure of homolog separation in meiosis I or chromatid separation in meiosis II
- Describes genes present in only one copy such as those on the X in males
- Alternate forms of a gene
- Alleles of one gene separate into gametes randomly with respect to alleles of other genes
- The alleles of one gene mask the effects of alleles of another gene
- The heterozygote resembles neither homozygote
- One gene affects more than one trait
- The part of the cell cycle during which the chromosomes are not visible
- The time during mitosis when sister chromatids separate
- Pairing of homologous chromosomes
- Structure formed at the spot where crossing-over occurs between homologs
- The relative chromosomal location of a gene
- Short DNA fragments formed by discontinuous replication of one of the strands
- Noncovalent bonds that hold the two strands of the double helix together
- The separation of the two alleles of a gene into different gametes
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本題為基礎遺傳學名詞解釋配對題,旨在測試考生對遺傳學基本術語及其定義的掌握程度。
配對結果如下:
a. Alleles - 3. Alternate forms of a gene (基因的等位基因是同一基因的不同形式)
b. Law of segregation - 15. The separation of the two alleles of a gene into different gametes (孟德爾分離定律指出,控制同一性狀的等位基因在配子形成時會彼此分離)
c. Epistasis - 5. The alleles of one gene mask the effects of alleles of another gene (上位性是指一個基因的等位基因會掩蓋另一個基因的表現)
d. Chiasma - 11. Structure formed at the spot where crossing-over occurs between homologs (交叉點是同源染色體發生互換時形成的結構)
e. Okazaki fragments - 13. Short DNA fragments formed by discontinuous replication of one of the strands (岡崎片段是 DNA ગ discontinuous replication 過程中形成的短 DNA 片段)
f. Hydrogen bond - 14. Noncovalent bonds that hold the two strands of the double helix together (氫鍵是維持 DNA 雙股螺旋結構的非共價鍵)
g. Locus - 12. The relative chromosomal location of a gene (基因座是基因在染色體上的相對位置)
第 2 題6 分
In a cross between a tall pea plant with yellow and round seeds (TT Yy Rr) and a tall pea plant with yellow and wrinkled seeds (Tt Yy rr), what proportion of the following offspring could be expected: (6 points, 2 points each)
(a) Tall, yellow, round.
(b) Tall, green wrinkled
(c) Dwarf, green, round
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本題為連鎖與自由組合的孟德爾遺傳問題,涉及三個基因的遺傳。題目給定親代基因型,要求計算特定表現型後代的比例。
親代基因型:
父本:TT Yy Rr (高莖, 黃色,圓粒)
母本:Tt Yy rr (高莖, 黃色, 皺粒)
我們將分別分析三個基因 (T/t, Y/y, R/r) 的遺傳。
-
莖高 (T/t):
親代:TT x Tt
後代基因型比例:1 TT : 1 Tt
表現型:所有後代皆為高莖 (Tall)。 -
種子顏色 (Y/y):
親代:Yy x Yy
後代基因型比例:1 YY : 2 Yy : 1 yy
表現型:3/4 黃色 (Yellow), 1/4 綠色 (green)。 -
種子形狀 (R/r):
親代:Rr x rr
後代基因型比例:1 Rr : 1 rr
表現型:1/2 圓粒 (round), 1/2 皺粒 (wrinkled)。
由於這三個基因位於不同的染色體上,因此遵循獨立分配定律。
(a) Tall, yellow, round:
要得到 "Tall, yellow, round" 的後代,需要同時滿足以下條件:
- 莖高為 Tall (T_)
- 種子顏色為 Yellow (Y_)
- 種子形狀為 round (Rr)
從上面的分析我們知道:
- Tall 的機率是 1 (因為 TT x Tt 後代都是 Tall)
- Yellow 的機率是 3/4 (Yy x Yy 產生 3/4 Y_)
- round 的機率是 1/2 (Rr x rr 產生 1/2 Rr)
第 3 題3 分
Considering two pairs of alleles, A/a and B/b, each pair located on a different pair of chromosomes in the somatic cells of an organism. With regard to these alleles, what "allele combination" would you find in cells that are in the Metaphase stage of the cell cycle? (3 points)
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本題考查細胞週期的中期(Metaphase)染色體結構與基因組成。題目指出 A/a 和 B/b 兩對基因位於不同對的染色體上,代表它們遵循獨立分配定律。
細胞週期的中期(Metaphase)是細胞分裂(有絲分裂 Mitosis 或減數分裂 Meiosis II)中,染色體排列在細胞中央(赤道板)的階段。此時,每個染色體都已經複製完成,並由兩個緊密連結的姊妹染色分體(sister chromatids)組成。
假設該生物的體細胞基因型為 AaBb。
在細胞進行有絲分裂或減數分裂 II 的中期時:
- 每個染色體都包含兩個 DNA 分子,這兩個 DNA 分子是完全相同的姊妹染色分體。
- 因此,如果一個染色體帶有 A 等位基因,那麼其姊妹染色分體也帶有 A 等位基因。同樣地,如果帶有 a,則姊妹染色分體也帶有 a。
- 由於 A/a 和 B/b 基因位於不同對的染色體上,這兩對基因會獨立排列。
第 4 題2 分
Which of the following is true regarding the genetic code? (2 points)
(a) The genetic code is overlapping
(b) The genetic code is degenerate
(c) One genetic code has 4 unique nucleotides
(d) The genetic code has three start codons and one stop codon
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本題考查遺傳密碼(genetic code)的基本性質。遺傳密碼是將 DNA 或 RNA 中的核苷酸序列翻譯成蛋白質中胺基酸序列的規則。
讓我們逐一分析選項:
(a) The genetic code is overlapping (遺傳密碼是重疊的):這是錯誤的。遺傳密碼是連續的(non-overlapping),也就是說,三個核苷酸組成一個密碼子,密碼子之間不重疊,例如 AUG CGA UUU...。重疊密碼子的意思是,一個核苷酸可能同時屬於兩個或多個密碼子,這在生物遺傳密碼中不存在。
(b) The genetic code is degenerate (遺傳密碼是簡併的):這是正確的。簡併性(degeneracy)指的是,有超過一種密碼子可以對應到同一個胺基酸。例如,亮胺酸(Leucine)就有六個不同的密碼子可以編碼。這主要是因為有 64 種可能的密碼子(),但只有 20 種常見的胺基酸和 3 個終止密碼子。
第 5 題2 分
The classic Hershey and Chase (1952) experiment that offered evidence in support of DNA being the genetic material in bacteriophages made use of which of the following radioactive labelling component(s)? (2 points)
(a) Nitrogen and oxygen
(b) Hydrogen and sulfate
(c) Phosphorus and sulfur
(d) Oxygen and phosphorus
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本題考查 Hershey 和 Chase 的經典實驗,該實驗利用放射性標記證明 DNA 是遺傳物質。
Hershey 和 Chase 的實驗使用了噬菌體(bacteriophage),這是一種感染細菌的病毒。噬菌體主要由 DNA 和蛋白質組成。實驗的關鍵在於利用放射性同位素標記 DNA 和蛋白質中的特定元素,然後觀察這些標記物在感染過程中如何進入細菌。
- DNA 的組成:DNA 包含磷(Phosphorus)和脫氧核糖(一種碳水化合物,含有碳、氫、氧)。
- 蛋白質的組成:蛋白質由胺基酸組成,胺基酸含有碳、氫、氧、氮。有些胺基酸的側鏈含有硫(Sulfur)。
實驗設計如下:
- 標記 DNA:利用放射性同位素 標記噬菌體的 DNA。磷是 DNA 的骨架成分(磷酸二酯鍵),而蛋白質中不含磷,因此 只能標記 DNA。
第 6 題22 分
Please find the proper term from the left list (A to N) to describe [1] to [11] in the figure. Each term can only be used once in the figure, same term is not allowed to appear in multiple places (22 points, 2 points each)
List of terms:
A. Longer fragments
B. Phosphodiester bond
C. 5' end
D. Anode (+)
E. Guanine
F. Thymine
G. Hydrogen bond
H. Cathode (-)
I. Shorter fragments
J. Cytosine
K. 3' end
L. Adenine
M. Uracil
N. ATP
🖼️【此處有附圖,請對照原卷】
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核心觀念
本題考查三部分:
- DNA 單股的方向性:由 端至 端。
- DNA 結構與鹼基互補配對:A–T 有 2 個氫鍵,G–C 有 3 個氫鍵。
- 凝膠電泳:DNA 帶負電,向正極移動;短片段移動較快、距離較遠。
DNA 骨架中的核苷酸以磷酸二酯鍵連接,DNA 鹼基配對為:
解題方法
依圖中各標示位置判斷:
- 左側 DNA 鏈:判斷 端、磷酸二酯鍵與 端。
- 中間雙股 DNA:依鹼基大小、互補配對及氫鍵數判斷 A–T 與 G–C。
- 下方電泳圖:依 DNA 負電性判斷電極,再依片段移動距離判斷長短。
各標示答案
| 標示 | 答案 | 判斷理由 |
|---|---|---|
| [1] | C. end | DNA 鏈的磷酸端為 端。 |
| [2] | B. Phosphodiester bond | 相鄰核苷酸之間由磷酸二酯鍵連接。 |
| [3] | K. end | 具有游離 -OH 的末端為 端。 |
| [4] | L. Adenine | 與 thymine 配對,形成 2 個氫鍵。 |
| [5] | F. Thymine | DNA 中與 adenine 互補配對的鹼基。 |
| [6] | E. Guanine | 與 cytosine 配對,形成 3 個氫鍵。 |
| [7] | J. Cytosine | DNA 中與 guanine 互補配對的鹼基。 |
| [8] | H. Cathode (−) | DNA 帶負電,靠近加樣孔的一側為負極。 |
| [9] | D. Anode (+) | DNA 向正極移動,因此下方為正極。 |
| [10] | A. Longer fragments | 長片段移動速度慢,停留在較上方。 |
| [11] | I. Shorter fragments | 短片段移動較快,跑到較下方。 |
選項分析
第 7 題20 分
224 F1 cowpea plants were derived from a cross of a plant with inflated green pod and a plant with restricted green pod. These plants can be classified as four distinct types, based on pod phenotypes: 42 yellow inflated; 14 yellow restricted; 70 green inflated; 98 green restricted. We assign , , , and genotypes represent phenotypes for yellow pod, green pod, inflated pod, and restricted pod, respectively. Also, and are two loci controlling pod color and pod shape, respectively.
(a) Write down the genotypes of two parents and each of four classes of F1 plants. (10 points)
(b) Please use the chi-square test of independence to check whether locus and locus are in independent assortment. (10 points)
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核心觀念
本題考查由子代表型比例推定親代基因型,以及用列聯表卡方獨立性檢定判斷兩基因座是否獨立分配。
題目指定 表現黃色豆莢、 表現綠色豆莢; 表現膨脹豆莢、 表現受限豆莢。依子代比例可知,雜合型 表現綠色,雜合型 表現膨脹。因此,黃色與受限分別是 與 的表型;綠色包含 、,膨脹則包含 。
(a) 親代與四種子代表型的基因型
先分別看兩個性狀的子代比例:
- 豆莢顏色:黃色 株,綠色 株,比例為 。這符合 ,其中 為黃色, 與 為綠色。
- 豆莢形狀:膨脹 株,受限 株,比例為 。這符合 ,其中 為膨脹, 為受限。
因此,兩親本的基因型為:
四種子代表型的基因型如下:
| 子代表型 | 基因型 |
|---|---|
| 黃色、膨脹 | |
| 黃色、受限 | |
| 綠色、膨脹 | 或 |
| 綠色、受限 | 或 |
綠色子代無法只從表型區分 與 ,所以應以兩種基因型表示該表型類別。
(b) 卡方獨立性檢定
假設
- 虛無假設 :基因座 與基因座 獨立分配,豆莢顏色與形狀相互獨立。
- 對立假設 :兩性狀不獨立。
第 8 題30 分
You are working on a marker-assisted backcross breeding project in rice, and the goal is to pyramid three genes (, , and ) from three rice near-isogenic lines (NILs). These lines have the same genetic background, but different favorable alleles were introduced:
- The (red pericarp) NIL: a dominant allele on chromosome 7 (the phenotype for the recessive allele is colorless pericarp).
- The (strong seed dormancy) NIL: a dominant allele on chromosome 7.
- The (bacterial blight resistance) NIL: a dominant allele on chromosome 11.
All three NILs were validated as pure lines. The recombination frequency between and is 0.16. The working plan is as follows.
The first step is to make a cross between the NIL and the NIL, and another cross between the NIL and the NIL simultaneously.
The second step is to make a cross between the two F1 hybrids produced from the first step.
The third step is to select individuals with heterozygous genotypes at the loci and , and with homozygous genotypes at the locus . Then, produce self-pollinated seeds.
The fourth step is to select individuals with homozygous genotypes at all three loci , , and .
(a) Please draw the crossing scheme, and add annotation of correct genotype(s) that were selected in each generation. Each generation corresponds to each step described above. (10 points)
(b) Please estimate how many progenies are required with 95% chance to obtain at least one plant in each step (generation). (20 points)
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