What are the answers and detailed solutions for the 1st-semester question papers for 12th-grade Biology in Vietnam? What does the equipment used for practical teaching in 12th-grade Biology include?

What are the answers and detailed solutions for the 1st-semester question papers for 12th-grade Biology in Vietnam?

Students can refer to the following 1st-semester question papers for 12th-grade Biology in Vietnam for preparation for the upcoming exam:

Answers and Solutions

Question 1: Answer c

Explanation: Cyanobacteria (cyanobacteria are prokaryotic organisms, their genes are not fragmented and do not contain introns (non-coding segments))

Question 2: Answer d

Explanation: Forms taxonomic groups above the species level: genus, family, order, class, phylum

Question 3: Answer a

Explanation: TXX (in transcription, nucleotides pair according to the complementary principle: A – U, G – C, so on mRNA aggregated by the gene, the triplet TXX cannot appear)

Question 4: Answer c. UAG, UAA, UGA

Question 5: Answer b. thymine.

Question 6: Answer b

Explanation: basic filament – chromatin – supercoiled filament – chromatid.

Question 7: Answer b

Explanation: 11/12 (genotype Aaaa produces gamete aa at a ratio of 1/2; genotype AAaa produces gamete aa at a ratio of 1/6. Thus, the ratio of offspring differing in phenotype from the parents (with genotype aaaa – defining the recessive phenotype) is 1/2.1/6 = 1/12. Hence the ratio of offspring with the same phenotype as the parent is: 1-1/12=11/12)

Question 8: Answer d

Explanation: 6 (the fruit fly has 2n = 8, so a double monosomic (2n – 1 – 1) has 8 – 1 – 1 = 6 chromosomes in each somatic cell)Question 9: Answer b

Explanation: 2500 (the number of G-type nucleotides accounts for 60% of the nucleotides in one strand, which corresponds to 1/2 of 60% = 30% of the total nucleotides of the gene. Since %G + %A = 50% of the total nucleotides of the gene, %A = 20% of the total nucleotides of the gene.

We also have 2A + 3G = 3250, so if N is the total number of nucleotides of the gene, then 2.20%N + 3.30%N = 3250. Thus, N = 2500.)

Question 10: Answer a

Explanation: 10% (in silkworms, gene crossover only occurs in males. Individuals with genotype Ab/ab are formed from the combination of crossover gamete (Ab) of the male and fully linked gamete (ab) of the female, and the proportion of this genotype is: 50%.40%(Ab).50%(ab) = 10%)

Question 11: Answer d

Explanation:

25% (crossing a yellow-flowered plant (A-) and a purple-flowered plant (aa) results in offspring with both yellow and purple flowers, indicating that the yellow-flowered plant in P must have genotype Aa.

P: Aa x aa

Gametes: 1/2A, 1/2a a

F1: 1/2Aa : 1/2aa

Thus, F1 has a genotype segregation ratio of 1/2Aa: 1/2aa

When F1 is self-pollinated, the proportion of non-pure yellow-flowered plants (Aa) is: 1/2 (initial Aa).(1/2)^1 = 1/4 or 25%)

Question 12: Answer d

Explanation: 0.3AA : 0.2Aa : 0.5aa (allele frequency A = frequency of genotype AA + 1/2.frequency of genotype Aa, thus allele frequency AA = 0.4-1/2.0.2 = 0.3, frequency of genotype aa = 1 – 0.2 – 0.3 = 0.5)

Question 13: Answer d

Explanation: Chromosomal deletion and chromosomal translocation (non-reciprocal translocation)

Question 14: Answer a

Explanation: 2 (in the case of heterozygous genotype, from one spermatogenic cell undergoing normal meiosis, genes segregate independently producing only 4 sperm with two different types of gametes with different genotypes)

Question 15: Answer b

Explanation: Loss of 3 nucleotide pairs (point mutation only involves 1 nucleotide pair)

Question 16: Answer b. somatic cell.

Question 17: Answer c. 27

Explanation:

The number of genotypes that can be formed from the third gene is: (2.(2+1))/2 = 3

The number of genotypes in the heterogametic sex (XY) that can be formed from the first and second genes is: 3.3 = 9

Thus, the number of genotypes in the heterogametic sex regarding the genes being considered is: 3.9 = 27.

Question 18: Answer b

Explanation: 1/16 ( =1/2(aa).1/2(BB,bb).1/2(CC).1/2(DD,dd))

Question 19: Answer b

Explanation: 5/16 (AaBbCc x AabbCc. Each individual carries 6 alleles. Notice that in the Bb x bb gene pair, the offspring always contains a recessive allele b (received from bb pair). Thus, the ratio of genotype with 3 recessive alleles must always include this allele b. So for the remaining 5 alleles, the probability for the offspring to carry 2 recessive alleles out of the 5 alleles is: Combinatorial choice of 2 out of 5.(1/2)^5 = 5/16)

Question 20: Answer a. 50%

Explanation:

A normal couple has siblings with albinism (aa), and normal parents (A-) indicating this couple will have genotypes AA or Aa with the probability 1/3AA : 2/3Aa (producing gametes with the ratio: 2/3A : 1/3a).

Thus, their normal son will have genotype AA or Aa with the probability 4/9:8/9AA : 4/9:8/9Aa or 1/2AA : 1/2Aa (producing gametes with the ratio: 3/4A : 1/4a).

The normal daughter with a father having albinism (aa) will have genotype Aa (producing gametes with the ratio 1/2A : 1/2a).

Therefore, the probability of this couple's child being normal but carrying the disease gene (Aa) is: 3/4(A).1/2(a) + 1/4(a).1/2(A) = 1/2 =50%

Question 21: Answer b

Explanation: 9 : 3 : 3 : 1 (9A-B- : 3 A-bb : 3aaB- : 1aabb)

Question 22: Answer a

Explanation:

The ratio of individuals with pure genotypes in the offspring is 25%.

P is purebred contrasting, so F1 has genotype AB/ab or Ab/aB.

The short-stemmed, red-flowered plant in F2 has a genotype form aB/a-. Because the distance between the two genes is 30cM, if recombination occurs (corresponding to 30%) in both parental bodies, the genotype ratio ab/ab must be 35%.35% (with F1 genotype as AB/ab) or 15%.15% (with F1 genotype as Ab/aB) and correspondingly the ratio of aB/a- genotype = 50%-35%.35% or 50% - 15%.15% (all results are different from 7.5%(=50%.15%) indicating gene recombination occurs only in male or female gamete-forming cells).

Moreover, in F2, a plant homozygous for the two genes considered appears (ab/ab), indicating that the fully linked gamete-forming cell must produce allele ab. This also means the genotype in F1 is AB/ab and the recombination frequency corresponds to the distance between the two genes, equaling 30%.

We have the cross: AB/ab x AB/ab

Gametes: 35%AB, 35%ab, 15%Ab, 15%aB 50%AB, 50%ab

Therefore, the proportion of genotype purebred for both genes considered will be: 35%.50%(AB/AB) + 35%.50%(ab/ab) = 35%

Thus, among the given options, the incorrect one is: a. The proportion of individuals with purebred genotype for both gene pairs in offspring is 25%.

Question 23: Answer a

Explanation: Ab/ab x aB/ab (1Ab/aB : 1Ab/ab : 1aB/ab : 1ab/ab)

Question 24: Answer d

Explanation: All other options are correct (all are mainly controlled by genotype)

Question 25: Answer c. Modification

Question 26: Answer b

Explanation: Replace the A – T pair with the G – C pair

A point mutation is a mutation involving only 1 pair of nucleotides without changing the number of nucleotides, indicating the mutation was a substitution. Moreover, the number of H bonds increases indicating the replacement of an A – T pair (with 2 H bonds) by a G – C pair (with 3 H bonds).

Question 27: Answer c

Explanation: Because at this stage chromosomes are maximally coiled and condensed

Question 28: Answer b. Lethal or reduces vitality

Question 29: Answer b

Explanation: Chromosomal inversion (The segment of chromosome carrying genes: gene for stem height – gene for seed type – gene for leaf type – gene for fruit sweetness has been cut out, reversed 180 degrees and reattached to its original position)

Question 30: Answer c

Explanation: Random factors (genetic drift occurs randomly before the impact of natural disasters and diseases and can randomly eliminate beneficial alleles or retain harmful alleles for the population)

Question 31: Answer d

Explanation: 0.09AA : 0.42Aa : 0.49aa (because the population has reached a genetic equilibrium state)

Question 32: Answer d

Explanation: 180 (allelic frequency A is 0.8(AA) + 0.5.0.2 (Aa) = 0.9; allelic frequency a is: 1 – 0.9 = 0.1. When the population is panmictic, the genetic structure of the population at equilibrium state with genotype components is: 0.9^2AA : 2.0,9.0,1Aa : 0.1^2aa. Thus theoretically, the number of individuals with Aa genotype in the offspring is: 2.0,9.0,1.1000 = 0.18.1000 = 180 individuals)

Question 33: Answer a. 37%

Explanation:

Allele frequencies A, a are 0.6 and 0.4, respectively. After undergoing panmixia (whether one or many generations), the population reaches genetic equilibrium following Hardy-Weinberg law with genotype makeup: 0.6^2AA : 2.0,6.0,4Aa : 0.4^2aa = 0.36AA : 0.48Aa : 0.16aa

After 3 generations of self-pollination, allele frequency Aa will be: 0,48.(1/2)^3=0,06. The frequency of genotype aa will be: 0,16 + (0,48-0,06)/2 = 0,37 or 37%

Question 34: Answer b

Explanation: Gene migration (because it can introduce new alleles and genotypes into the population)

Question 35: Answer c

Explanation: Turner Syndrome (XO sex chromosome form)

Question 36: Answer b. Perform the same function

Explanation:

Question 37: Answer b

Explanation: Artificial selection (humans control the other factors or filter, direct them in favor of economic activities and production)

Question 38: Answer b. Creating purebred lines

Question 39: Answer d. All other options are correct

Question 40: Answer b. Insert the nucleus from a mammary cell into an ovum from which the nucleus was removed

Note: The information is for reference only!

What are the answers and detailed solutions for the 1st-semester question papers for 12th-grade Biology in Vietnam? (Image from Internet)

What does the equipment used for practical teaching in 12th-grade Biology in Vietnam include?

Under sub-section 3 of Section 8 of the General Education Program for Biology issued together with Circular 32/2018/TT-BGDDT, equipment used for practical teaching in 12th-grade Biology in Vietnam includes:

- Microscopic slides set: cellular.

- Practical tools set for: cells; cell division; microorganisms and viruses; exchange and energy transformation in plants; frog heart dissection; dressing and hemorrhage control; growth and development in plants; observing chromosomal mutations.- Specimen box: organism classification, adaptation forms, ...

- Measuring equipment set: measuring respiratory capacity and diaphragm activity in animals, blood pressure, heart rate, pH levels, ...

What is the maximum number of students in a 12th-grade class in Vietnam?

According to Clause 3, Article 16 of the lower secondary school, upper secondary school and multi-level school charter issued together with Circular 32/2020/TT-BGDDT which prescribes the maximum number of students in a class at the upper secondary level:

Classes

...

3. Chairpersons of provincial People’s Committees shall stipulate number of students in each class with the aim of reducing number of students in one class; ensure that each class at the lower secondary or upper secondary level has a maximum of 45 students.

4. Number of students in each class of a special school is provided for in organizational and operational regulations for special schools.

Thus, according to the regulations, the maximum number of students in a 12th-grade class is not to exceed 45 students.