2020 遗传学（双语课）(福建农林大学)1458664452 最新满分章节测试答案

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本课程起止时间为:2020-04-27到2020-06-30
本篇答案更新状态:已完结

第一章 绪论 绪论 单元测验

1、 问题: In 1900, whose law was rediscovered, which marked the birth of genetics.
选项：
A:C·R·Darwin
B:G·J·Mendel
C:A·Weismann
D:J·Lamarck
答案: 【G·J·Mendel】

2、 问题:Who was recognized as the founder of genetics?
选项：
A:J·Lamarck
B:T·H·Morgan
C:G·J·Mendel
D:C·R·Darwin
答案: 【G·J·Mendel】

3、 问题:Who was recognized as the founder of cytogenetics（细胞遗传学）?
选项：
A:J·Lamarck
B:T·H·Morgan
C:G·J·Mendel
D:C·R·Darwin
答案: 【T·H·Morgan】

4、 问题:Acquired traits （后天获得的性状）can be inherited.
选项：
A:正确
B:错误
答案: 【错误】

5、 问题:The two genetic law proposed by Mendel are the law of segregation and the law of linkage.
选项：
A:正确
B:错误
答案: 【错误】

6、 问题:The third genetic law proposed by Morgan is the law of linkage.
选项：
A:正确
B:错误
答案: 【正确】

【作业】第二章 孟德尔式遗传分析 孟德尔遗传分析 课后习题

1、 问题: Consider genes P, R and A are inherited independently and dominance is complete. In the population contructed by cross ppRRAA×PPrraa, we want to selected 10 lines with PPRRAA genotype in F3 population, how many plants we should plant in F2 poupulation? In this population how many plants show P_R_A_ phenotype?
评分规则: 【 10 ÷（1/41/41/4）=640株
10÷（1/31/31/3）=270株
】

2、 问题:In the following family, there is a disease controlled by a recessive gene A on autosome. Please answer the following questions:(1) Write the genotype of individual I-1, I-2, II-4, III-2, III-5, IV-1 and V-1. (2) What is the probability that the brother of V-1 is heterozygote?(3) What is the probability that the two sisters of V-1 are both heterozygote?(4) If V-1 married V-5, what is the probability that their first child is suffered by disease? If we know that the first child is affected（患病）, what is the probability that their second child is also affected?
评分规则: 【 I-1: Aa, I-2: Aa, II-4: aa, III-2: Aa, III-5: Aa, IV-1: Aa, V-1: aa
V-1弟弟是杂合体的概率2/3
V-1两个妹妹都是杂合体的概率：4/9
第一个孩子患病的概率1/2；第二个孩子患病的概率也是1/2.
】

3、 问题:In barley, husk grain (带壳，N）is dominant to naked grain(裸粒，n) and loose spike (散穗,L) is dominant to dense spike(密穗,l). When a husk grain and loose spike homozygous plant cross with a naked grain and dense spike homozygous plant, in F2 population, there are 265 plants with husk grain and loose spike, 90 plants with husk grian dense spike, 86 plants with naked grain and loose spike, 59 plants with naked grian and dense spike. Question: Are these two triats in accordance with 9:3:3:1 ratio? Please check it in chi-square test.
评分规则: 【 Observed number: N_L_ 265 nnL_ 100 N_ll 90 nnll 45 Expected number: N_L_ 281.25 nnL_ 93.75 N_ll 93.75 nnll 31.25 Degree freedom: N=4-1=3
χ2=(265-281.25）2/281.5+(100-93.75)2/93.75+(90-93.75)2/93.75+(45-31.25)2/31.25 =0.938+0.417+0.15+6.05=7.555
From the chi-square table, we can know that P≧0.05, no significant difference , so these two traits are accroding with the ratio 9:3:3:1.
】

4、 问题:In tomato, red fruit (R) is dominant to yellow fruit (r). In the following cross, what are the phenotype and their proportion in the progeny (后代)？(1) RR×rr(2) Rr×rr(3) Rr×Rr(4) Rr×RR(5) rr×rr
评分规则: 【 Rr (red color)
Rr:rr=1:1 (red : yellow=1:1)
RR:Rr:rr=1:2:1 (red:yellow=3:1)
RR:Rr=1:1(red color)
rr (yellow color)
】

5、 问题:In tomato, there are two pairs of genes, A(purple leave)/a(green leave) and B(red fruit) /b (yellow fruit). When the homozygous plant with purple leave and yellow fruit corss with a homozygous plant with green leave and red fruit, in the F2 population, we found the segreation ratio was 9:3:3:1. If the F1 heterzyous plant (1) back cross with the purple leave and yellow fruit parent; (2) back cross with the green leave and red fruit parent; (3) test cross with the double recessive homozygous plants.Question: What are the phenotype and their proportion of the progeny of the above cross? Please write the genotype of the cross.
评分规则: 【 AaBb × AAbb → 1A_ B_ : 1A_bb
AaBb × aaBB → 1A_B_ : 1 aaB_
AaBb × aabb → 1A_B_ :1 A_bb: 1aaB_:1aabb
】

【作业】第二章 孟德尔式遗传分析 孟德尔遗传分析 课后习题-1

1、 问题: In tomato, red fruit (R) is dominant to yellow fruit (r). In the following cross, what are the phenotype and their proportion in the progeny (后代)？(1) RR×rr(2) Rr×rr(3) Rr×Rr(4) Rr×RR(5) rr×rr
评分规则: 【 Rr (red color)
Rr:rr=1:1 (red : yellow=1:1)
RR:Rr:rr=1:2:1 (red:yellow=3:1)
RR:Rr=1:1(red color)
rr (yellow color)
】

2、 问题:In tomato, there are two pairs of genes, A(purple leave)/a(green leave) and B(red fruit) /b (yellow fruit). When the homozygous plant with purple leave and yellow fruit corss with a homozygous plant with green leave and red fruit, in the F2 population, we found the segreation ratio was 9:3:3:1. If the F1 heterzyous plant (1) back cross with the purple leave and yellow fruit parent; (2) back cross with the green leave and red fruit parent; (3) test cross with the double recessive homozygous plants.Question: What are the phenotype and their proportion of the progeny of the above cross? Please write the genotype of the cross.
评分规则: 【 AaBb × AAbb → 1A_ B_ : 1A_bb
AaBb × aaBB → 1A_B_ : 1 aaB_
AaBb × aabb → 1A_B_ :1 A_bb: 1aaB_:1aabb
】

第二章 孟德尔式遗传分析 单元测验2

1、 问题:1. It is known that albinism is controlled by one recessive gene located on the euchromosome. When a couple who both carry the albinism genes, what is the probability that their child suffer from the albinism?
选项：
A:1
B:0.75
C:0.5
D:0.25
答案: 【0.25】

2、 问题:2. In a cross Aabb and AaBb the frequency of genotype A_B_ is
选项：
A:3/8
B:5/8
C:1/16
D:9/16
答案: 【3/8】

3、 问题:3. Consider four independently assorting gene pairs, A/a, B/b, C/c and D/d, where each demonstrates typical dominance. What is the probability of obtaining an offspring that is AABbCcDD from parents that are AaBbCcDd and AABbCcDD?
选项：
A:1/64
B:1/8
C:1/16
D:1/32
答案: 【1/16】

4、 问题:In human, the albinism is controlled by a recessive gene on the autosome. The parents of the sufferer should be
选项：
A:Both of the parents are sufferer.
B:At least one of the parents is sufferer.
C:Both of the parents are normal but carriers.
D:At least one of the parents is normal.
答案: 【Both of the parents are normal but carriers.】

5、 问题:4. A man was affected by the disease of hemophilia, which is controlled a recessive gene on X chromosome. He can inherit this disease to his son.
选项：
A:正确
B:错误
答案: 【错误】

6、 问题:5. A man was affected by the disease of hemophilia, which is controlled a recessive gene on X chromosome. He can inherit this disease to his son.
选项：
A:正确
B:错误
答案: 【错误】

7、 问题:Consider the cross AaBb ×aabb. If the alleles for both genes exhibit complete dominance, the phenotypic ratio is 1:1:1:1.
选项：
A:正确
B:错误
答案: 【正确】

8、 问题:In human, brown eyes (B) is dominant to blue eyes (b). The gene is located on autosome. A couple both with brown eyes has a son with blue eyes. We can know that the genotype of the couple are heterozygotes.
选项：
A:正确
B:错误
答案: 【正确】

第二章 孟德尔式遗传分析 单元测验1

1、 问题:The law of segregation proves that when F1 hybrids form gametes, the paired genes
选项：
A:Segregate, enter the same gamete
B:Segregate, enter the different gamete
C:Not segregate, enter the same gamete
D:Not segregate, enter the different gamete
答案: 【Segregate, enter the different gamete】

2、 问题:An individual heterozygous with four different genes can produce how many genetically different gametes?
选项：
A:4
B:8
C:16
D:32
答案: 【16】

3、 问题:After meiosis, what gametes will be formed by AaBb?
选项：
A:Aa Ab aB Bb
B:A a B b
C:AB Ab aB ab
D:Aa aa BB bb Bb AA
答案: 【AB Ab aB ab】

4、 问题:How many kinds of gametes can AaBbCc produce？
选项：
A:2
B:4
C:6
D:8