Lesson plan for developing students' capacity in Biology 12-14

We have: p + q + r = 1 ( * ) Blood O accounts for 0.090 => r(i) = 0.30 Ratio of blood A: I A I A + I A I O = 0.2125 => p2 + 2 pr = 0.2125

* p 2 + 2 pr + r 2 = ( p + r ) 2 = 0.2125 + 0.090 = 0, 3025 = ( 0.55 ) 2

( p + r ) 2 = ( 0.55 ) 2 => p + r = 0.55 => p = 0.55 – 0.30 = 0.25 From: ( * ) => q = 1 – ( p + r ) = 1 - ( 0.25 + 0.30 ) = 0.45

So: The relative frequency of each allele is: p(I A ) = 0.25; q(I B ) = 0.45; r(i) = 0.30 

select A

Lesson 9: Given the genetic structure of a human population regarding blood groups A, B, AB, O:

0.25 I A I A + 0.20 I A I O + 0.09 I B I B + 0.12 I B I O + 0.30 I A I B + 0.04I O I O = 1

The relative frequency of each allele I A , I B , I O is:

A) 0.3 : 0.5 : 0.2 B) 0.5 : 0.2 : 0.3 C) 0.5 : 0.3 : 0.2 D) 0.2 : 0.5 : 0 ,3

Solution: Relative frequency of allele I A : 0.25 + ( 0.2 : 2 ) + ( 0.3 : 2 ) = 0.5 Relative frequency of allele I B : 0.05 + ( 0.12 : 2 ) + ( 0.3 : 2 ) = 0.5

Relative frequency of allele I O : 1 - ( 0.5 + 0.3 ) = 0.2  choose C

A)0.128. B)0.287. C)0.504. D)0.209.

Solution: Blood type O accounts for 0.483 => r(i) = 0.695

Blood ratio A: I A I A + I A I O = 0.194 => p 2 + 2 pr = 0.194

* p 2 + 2 pr + r 2 = ( p + r ) 2 = 0.194 + 0.483 = 0, 677 = ( 0.823 ) 2

( p + r ) 2 = ( 0.823 ) 2 => p + r = 0.823 => p = 0.823 – 0.695 = 0.128 

select A

Lesson 11: About blood groups A, O, B of a human population in genetic equilibrium. Allele frequencies I A = 0.1, I B = 0.7, I o = 0.2. The frequencies of blood groups A, B, AB, O are respectively:

A. 0.3; 0.4; 0.26; 0.04 B. 0.05; 0.7 ; 0.21; 0.04

C. 0.05; 0.77; 0, 14; 0.04 D. 0.05; 0.81; 0.10; 0.04

Solution: Frequency of blood group O: r 2 = ( 0.2) 2 = 0.04

Frequency of blood group A: p 2 + 2pr = ( 0,1) 2 + 2(( 0,1 ) x ( 0,2 )) = 0,05 Frequency of blood group B: q 2 + 2qr = ( 0,7 ) 2 + 2(( 0,7 ) x ( 0,2 )) = 0,77 Frequency of blood group AB: 2pq = 2(( 0,1 ) x ( 0,7 )) = 0,14  choose C

Lesson 12: A population has 4 genes I, II, III. IV; the number of alleles of each gene is respectively: 2, 3, 4, 5. The number of genotypes obtained in the above random mating population is:

A. 2700 B. 370 C. 120

D. 81

Solution: Gene I has: ((2(2+1) : 2 ) 1 = 3 genotypes; Gene II has: ((3(3+1) : 2 ) 1 = 6 genotypes; Gene III has: ((4(4+1) : 2 ) 1 = 10 genotypes; Gene IV has: ((5(5+1) : 2 ) 1 = 15 genotypes

The total number of genotypes obtained in a random mating population is: 3 x 6 x 10 x 15 = 2700 

Select A

Lesson 13: A population has the following structure P: 17.34%AA: 59.32%Aa: 23.34%aa. In the above population, after 3 generations of random mating, which of the following results does not appear in F 3 ?

D. The frequency of allele A decreases and the frequency of allele a increases compared to P.

Solution: We have: P: 17.34%AA: 59.32%Aa: 23.34%aa.

Frequency of allele A: ( pA) = 0.1734 + ( 0.5932 : 2 ) = 0.47 Frequency of allele a ( qa ) = 0.2334 + ( 0.5932 : 2 ) = 0.53

Through 1 generation of random mating: ( 0.47) 2 AA : 2 x ( 0.47) x ( 0.53 ) : ( 0.53 ) 2 aa

 Genotype ratio 22.09%AA : 49.82%Aa : 28.09%aa. After 3 generations of random mating (F 3 ) the genotype ratio is still 22.09%AA : 49.82%Aa :

28.09%aa.

Thus: answers A, B, C are all correct  the frequency of allele A decreases and the frequency of allele a increases compared to P

not appear in F 3 . Select D

Lesson 14: In humans, the gene that determines eye color has 2 alleles (A, a), the gene that determines hair type has 2 alleles (B,

b) The gene that determines blood type has 3 alleles (I A . I B , I O). The genes are located on different chromosomes. The number of different genotypes that can be created from the above 3 genes in the human population is: A.54 B.24 C.10 D.64

Solution: The gene that determines eye color has: (2 (2+1) : 2) 1 = 3 genotypes; the gene that determines hair type has: (2 (2+1) : 2) 1 = 3 genotypes; the gene that determines blood type has: (3 (3+1) : 2) 1 = 6 genotypes.

The number of different genotypes in a human population is: 3 x 3 x 6 = 54  Choose A Exercise 15: In an animal population, consider a gene with 3 alleles located on the autosome and a gene with 2 alleles located on the sex chromosome with no corresponding allele on the Y. This population has the maximum number of genotypes for the above 2 genes: A.30 B.60 C. 18

D.32

Solution: 1 gene has 3 alleles located on an autosome: (3(3+1) : 2 ) 1 = 6 types of genotypes.

1 gene has 2 alleles located on the sex chromosome with no corresponding allele on the Y: there are 5 types of genotypes

- The number of genotypes on Y is 2: X A Y, X a Y

- The number of genotypes on X is 3: X A X A , X a X a , X A X a

So: This population has the maximum number of genotypes for the above 2 genes: 6 x 5 = 30  Choose A

Lesson 16: In humans, gene A determines normal color vision, allele a determines red-green color blindness; gene B determines normal blood clotting, allele b determines hemophilia. These genes are located on the X sex chromosome and have no corresponding allele on the Y. Gene D determines right-handedness, allele d determines left-handedness and is located on the autosome. The maximum number of genotypes for the above 3 loci in the human population is:

A.42 B.36 C.39 D.27

Solution: The genes (AaBb) located on the X sex chromosome have no corresponding allele on the Y: there are 14 genotypes.

- The number of genotypes on Y is 4: X A Y, X a Y, X A Y, X a Y

B bb B

- The number of genotypes on X is 10: X A X A , X a X a , X A X a

X A X A , X a

X a , X A

BBBB

X

a

b,

BB, B b B bb

X A X A , X a

X a , X A X a

X A X a

bbbb B b, b B

Genes located on autosomes (D and d) have: (2(2+1) : 2 ) 1 = 3 genotypes

So: The human population has the maximum number of genotypes for the above 3 loci: 14 x 3 = 42  Choose A

Lesson 17: An initial population has the genetic structure: 0.7AA + 0.3Aa. After one generation of random mating, 4000 individuals are obtained in the offspring. Theoretically, the number of individuals with heterozygous genotypes is

The offspring are: A.90 B.2890 C.1020 D.7680

Solution: P. 0.7AA + 0.3Aa => pA = 0.7 + (0.3 / 2 ) = 0.85 ; qa = 0 + (0.3 / 2 ) = 0.15 F 1 .( 0.85 ) 2 AA + ( 2 x 0.85 x 0.15 ) Aa + ( 0.15 ) 2 aa = 1

 F 1 . 0.7225 AA + 0.255 Aa + 0.0225 aa = 1.

So: The number of individuals with heterozygous genotype in the offspring (F1 ) is: 0.255 x 4000 = 1020  Choose C Exercise 18: Suppose a population in genetic equilibrium has 10,000 individuals, of which 100 individuals have homozygous recessive genotype (aa), then the number of individuals with heterozygous genotype (Aa) in the population will be:

A. 9900 B. 900 C. 8100 D. 1800

Solution: We have: q 2 aa = 100 / 10000 = 0.01 => qa = 0.1

The population is in genetic equilibrium => pA = 1 - 0.1 = 0.9 ; 2pqAa = 2 x 0.1 x 0.9 = 0.18

So: The number of individuals with heterozygous genotype (Aa) is: 0.18 x 10000 = 1800  Choose D

Lesson 19: In chickens, A determines black feathers incompletely dominant over a determines white feathers, genotype Aa determines spotted feathers. A population of jungle fowl in genetic equilibrium has 10,000 individuals, of which 4,800 are spotted feathers, the number of black feathers and white feathers in the population are respectively

A.3600, 1600. B.400, 4800. C.900, 4300.

D.4900, 300.

Solution: Genotype ratio of spotted chicken (Aa) = 4800 / 10000 = 0.48

Let p: frequency of allele A (black fur), q: frequency of allele a (white fur)

The jungle fowl population is in genetic equilibrium, according to the Hardy-Weinberg law: ( p + q ) = 1 and 2pq = 0.48  p + q = 1 (1) and pq = 0.24 (2)

According to Viet's law (1), (2) we have the equation: X 2 - X + 0.24 = 0. Solving we get: x 1 = 0.6; x 2 = 0.4 ( x 1 is p; x 2 is q ).

Therefore: Genotype frequency AA (black fur): (0.6) 2 = 0.36 Genotype frequency aa (white fur): (0.4) 2 = 0.16

So: Number of black feathered chickens: 0.36 x 10000 = 3600

Number of white feathered chickens: 0.16 x 10000 = 1600  Choose A

Lesson 20: A mating population is in genetic equilibrium. Considering a gene with 2 alleles (A and a), we see that the number of homozygous dominant individuals is 9 times greater than the number of homozygous recessive individuals. The percentage of heterozygous individuals in this population is:

A.37.5 % B.18.75 % C.3.75 % D.56.25 %

Solution : Let p 2 be the frequency of homozygous dominant genotype, q 2 be the frequency of homozygous recessive genotype.

We have: p 2 = 9 q 2 or p = 3q

Population in genetic equilibrium: p + q = 1

So: 3q + q = 1 => q = 1 / 4 = 0.25 and p = 3 x 0.25 = 0.75

So: The percentage of heterozygous individuals in this population is: 2pq = 2 x 0.25 x 0.75 = 0.375 = 37.5 %  Choose A

Lesson 21: In a balanced population, consider 2 pairs of alleles AaBb on 2 different pairs of homologous chromosomes. Allele A has a relative frequency of 0.4 and Allele B has a relative frequency of 0.6. The frequency of each type of gamete in this population is:

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• Identify Rating Levels and Rating Scales zt2i3t4l5ee zt2a3gstourism,quan lan,quang ninh,ecology,ecotourism,minh chau,van don,geography,geographical basis,tourism development,science zt2a3ge zc2o3n4t5e6n7ts of the islanders. Therefore, this indicator will be divided into two sub-indicators: a1. Natural tourism attractiveness a2. Cultural tourism attractiveness b. Tourist capacity The two island communes in Quan Lan have different capacities to receive tourists. Minh Chau Commune is home to many standard hotels and resorts, attracting high-income domestic and international tourists. Meanwhile, Quan Lan Commune has many motels mainly built and operated by local people, so the scale and quality are not high, and will be suitable for ordinary tourists such as students. c. Time of exploitation of Quan Lan Island Commune: Quan Lan tourism is seasonal due to weather and climate conditions and festivals only take place on certain days of the year, specifically in spring. In Quan Lan commune, the period from April to June and from September to November is considered the best time to visit Quan Lan because the cultural tourism activities are mainly associated with festivals taking place during this time. Minh Chau island commune: Tourism exploitation time is all year round, because this is a place with a number of tourist attractions with diverse ecosystems such as Bai Tu Long National Park Research Center, Tram forest, Turtle Laying Beach, so besides coming to the beach for tourism and vacation in the summer, Minh Chau will attract research groups to come for tourism combined with research at other times of the year. d. Sustainability The sustainability of ecotourism sites in Quan Lan and Minh Chau communes depends on the sensitivity of the ecosystems to climate changes. landscape. In general, these tourist destinations have a fairly high level of sustainability, because they are natural ecosystems, planned and protected. However, if a large number of tourists gather at certain times, it can exceed the carrying capacity and affect the sustainability of the environment (polluted beaches, damaged trees, animals moving away from their habitats, etc.), then the sustainability of the above ecosystems (natural ecosystems, human ecosystems) will also be affected and become less sustainable. e. Location and accessibility Both island communes have ports to take tourists to visit from Van Don wharf: - Quan Lan – Van Don traffic route: Phuc Thinh – Viet Anh high-speed boat and Quang Minh high-speed boat, depart at 8am and 2pm from Van Don to Quan Lan, and at 7am and 1pm from Quan Lan to Van Don. There are also wooden boats departing at 7am and 1pm. - Van Don - Minh Chau traffic route: Chung Huong high-speed train, Minh Chau train, morning 7:30 and afternoon 13:30 from Van Don to Minh Chau, morning 6:30 and afternoon 13:00 from Minh Chau to Van Don. f. Infrastructure Despite receiving investment attention, the issue of infrastructure and technical facilities for tourism on Quan Lan Island is still an issue that needs to be resolved because it has a direct impact on the implementation of ecotourism activities. The minimum conditions for serving tourists such as accommodation, electricity, water, communication, especially medical services, and security work need to be given top priority. Ecotourism spots in Minh Chau commune are assessed to have better infrastructure and technical facilities for tourism because there are quite complete and synchronous conditions for serving tourists, meeting many needs of domestic and foreign tourists. 3.2.1.4. Determine assessment levels and assessment scales Corresponding to the levels of each criterion, the index is the score of those levels in the order of 4, 3, 2, 1 decreasing according to the standard of each level: very attractive (4), attractive (3), average (2), less attractive (1). 3.2.1.5. Determining the coefficients of the criteria For the assessment of DLST in the two communes of Quan Lan and Minh Chau islands, the students added evaluation coefficients to show the importance of the criteria and indicators as follows: Coefficient 3 with criteria: Attractiveness, Exploitation time. These are the 2 most important criteria for attracting tourists to tourism in general and eco-tourism in particular, so they have the highest coefficient. Coefficient 2 with criteria: Capacity, Infrastructure, Location and accessibility . Because the assessment area is an island commune of Van Don district, the above criteria are selected by the author with appropriate coefficients at the average level. Coefficient 1 with criteria: Sustainability. Quan Lan has natural and human-made ecotourism sites, with high biodiversity and little impact from local human factors. Most of the ecotourism sites are still wild, so they are highly sustainable. 3.2.1.6. Results of DLST assessment on Quan Lan island a. Assessment of the potential for natural tourism development  For Minh Chau commune: + Natural tourism attractiveness is determined to be very attractive (4 points) and the most important coefficient (coefficient 3), so the score of the Attractiveness criterion is 4 x 3 = 12. + Capacity is determined as average (2 points) and the coefficient is quite important (coefficient 2), then the score of Capacity criterion is 2 x 2 = 4. + Exploitation time is long (4 points), the most important coefficient (coefficient 3) so the score of the Exploitation time criterion is 4 x 3 = 12. + Sustainability is determined as sustainable (4 points), the important coefficient is the average coefficient (coefficient 1), so the score of the Sustainability criterion is 4 x 1 = 4 points + Location and accessibility are determined to be quite favorable (2 points), the coefficient is quite important (coefficient 2), the criterion score is 2 x 2 = 4 points. + Infrastructure is assessed as good (3 points), the coefficient is quite important (coefficient 2), then the score of the Infrastructure criterion is 3 x 2 = 6 points. The total score for evaluating DLST in Minh Chau commune according to 6 evaluation criteria is determined as: 12 + 4 + 12 + 4 + 4 + 6 = 42 points Similar assessment for Quan Lan commune, we have the following table: Table 3.3: Assessment of the potential for natural ecotourism development in Quan Lan and Minh Chau communes Attractiveness of self-tourismof course Capacity Mining time Sustainability Location and accessibility Infrastructure Result Point DarkMulti Point DarkMulti Point DarkMulti Point DarkMulti Point DarkMulti Point DarkMulti CommuneMinh Chau 12 12 4 8 12 12 4 4 4 8 6 8 42/52 Quan CommuneLan 6 12 6 8 9 12 4 4 4 8 4 8 33/52 b. Assessment of the potential for humanistic tourism development  For Quan Lan commune: + The attractiveness of human tourism is determined to be very attractive (4 points) and the most important coefficient (coefficient 3), so the score of the Attractiveness criterion is 4 x 3 = 12. + Capacity is determined to be large (3 points) and the coefficient is quite important (coefficient 2), then the score of the Capacity criterion is 3 x 2 = 6. + Mining time is average (3 points), the most important coefficient (coefficient 3) so the score of the Mining time criterion is 3 x 3 = 9. + Sustainability is determined as sustainable (4 points), the important coefficient is the average coefficient (coefficient 1), so the score of the Sustainability criterion is 4 x 1 = 4 points. + Location and accessibility are determined to be quite favorable (2 points), the coefficient is quite important (coefficient 2), the criterion score is 2 x 2 = 4 points. + Infrastructure is rated as average (2 points), the coefficient is quite important (coefficient 2), then the score of the Infrastructure criterion is 2 x 2 = 4 points. The total score for evaluating DLST in Quan Lan commune according to 6 evaluation criteria is determined as: 12 + 6 + 6 + 4 + 4 + 4 = 36 points. Similar assessment with Minh Chau commune we have the following table: Table 3.4: Assessment of the potential for developing humanistic eco-tourism in Quan Lan and Minh Chau communes Attractiveness of human tourismliterature Capacity Mining time Sustainability Location and accessibility Infrastructure Result Point DarkMulti Point DarkMulti Point DarkMulti Point DarkMulti Point DarkMulti Point DarkMulti Quan CommuneLan 12 12 6 8 9 12 4 4 4 8 4 8 39/52 Minh CommuneChau 6 12 4 8 12 12 4 4 4 8 6 8 36/52 Basically, both Minh Chau and Quan Lan localities have quite favorable conditions for developing ecotourism. However, Quan Lan commune has more advantages to develop ecotourism in a humanistic direction, because this is an area with many famous historical relics such as Quan Lan Communal House, Quan Lan Pagoda, Temple worshiping the hero Tran Khanh Du, ... along with local festivals held annually such as the wind praying ceremony (March 15), Quan Lan festival (June 10-19); due to its location near the port and long exploitation time, the beaches in Quan Lan commune (especially Quan Lan beach) are no longer hygienic and clean to ensure the needs of tourists coming to relax and swim; this is also an area with many beautiful landscapes such as Got Beo wind pass, Ong Phong head, Voi Voi cave, but the ability to access these places is still very limited (dirt hill road, lots of gravel and rocks), especially during rainy and windy times; In addition, other natural resources such as mangrove forests and sea worms have not been really exploited for tourism purposes and ecotourism development. On the contrary, Minh Chau commune has more advantages in developing ecotourism in the direction of natural tourism, this is an area with diverse ecosystems such as at Rua De Beach, Bai Tu Long National Park Conservation Center...; Minh Chau beach is highly appreciated for its natural beauty and cleanliness, ranked in the top ten most beautiful beaches in Vietnam; Minh Chau commune is also home to Tram forest with a large area and a purity of up to 90%, suitable for building bridges through the forest (a very effective type of natural ecotourism currently applied by many countries) for tourists to sightsee, as well as for the purpose of studying and researching. Figure 3.1: Thenmala Forest Bridge (India) Source: https://www.thenmalaecotourism.com/(August 21, 2019) 3.2.2. Using SWOT matrix to evaluate Quan Lan island tourism General assessment of current tourism activities of Quan Lan island is shown through the following SWOT matrix: Table 3.5: SWOT matrix evaluating tourism activities on Quan Lan island Internal agent Strengths- There is a lot of potential for tourism development, especially natural ecotourism and humanistic ecotourism.- The unskilled labor force is relatively abundant.- resource environmentunpolluted, still Weaknesses- Poorly developed infrastructure, especially traffic routes to tourist destinations on the island.- The team of professional staff is still weak.- Tourism products in general quite wild, originalintact general and DLST in particularalone is monotonous. External agents Opportunity- Tourism is a key industry in the socio-economic development strategy of the province and Van Don economic zone.- Quan Lan was selected as a pilot area for eco-tourism development within the framework of the green growth project between Quang Ninh province and the Japanese organization JICA.- The flow of tourists and especially ecotourism in the world tends toincreasing Challenge- Weather and climate change abnormally.- Competition in tourism products is increasingly fierce, especially with other localities in the province such as Ha Long, Mong Cai...- Awareness of tourists, especially domestic tourists, about ecotourism and nature conservation is not high. Through summary analysis using SWOT matrix we see that:  To exploit strengths and take advantage of opportunities, it is necessary to: - Diversify products and service types (build more tourism routes aimed at specific needs of tourists: experiential tourism immersed in nature, spiritual cultural tourism...) - Effective exploitation of resources and differentiated products (natural resources and human resources) div.maincontent .p { color: black; font-family:"Times New Roman", serif; font-style: normal; font-weight: normal; text-decoration: none; font-size: 14pt; margin:0pt; } div.maincontent p { color: black; font-family:"Times New Roman", serif; font-style: normal; font-weight: normal; text-decoration: none; font-size: 14pt; margin:0pt; } div.maincontent .s1 { color: black; font-family:"Times New Roman", serif; font-style: normal; font-weight: normal; text-decoration: none; font-size: 13pt; } div.maincontent .s2 { color: black; font-family:"Times New Roman", serif; font-style: normal; font-weight: normal; text-decoration: none; font-size: 13pt; } div.maincontent .s3 { color: #0D0D0D; font-family:"Times New Roman", serif; font-style: normal; font-weight: bold; text-decoration: none; font-size: 14pt; } div.maincontent .s4 { color: black; font-family:"Times New Roman", serif; font-style: italic; font-weight: normal; text-decoration: none; font-size: 14pt; } div.maincontent .s5 { color: black; font-family:"Times New Roman", serif; font-style: italic; font-weight: bold; text-decoration: none; font-size: 14pt; } div.maincontent .s6 { color: black; font-family:"Times New Roman", serif; font-style: italic; font-weight: normal; text-decoration: none; font-size: 14pt; vertical-align: -3pt; } div.maincontent .s7 { color: black; font-family:"Times New Roman", serif; font-style: italic; font-weight: normal; text-decoration: none; font-size: 14pt; vertical-align: -2pt; } div.maincontent .s8 { color: black; font-family:"Times New Roman", serif; font-style: italic; font-weight: normal; text-decoration: none; font-size: 14pt; vertical-align: -1pt; } div.maincontent .s9 { color: black; font-family:"Times New Roman", serif; font-style: normal; font-weight: normal; text-decoration: none; font-size: 14pt; } div.maincontent .s10 { color: black; font-family:"Times New Roman", serif; font-style: normal; font-weight: bold; text-decoration: none; font-size: 14pt; } div.maincontent .s11 { color: black; font-family:"Times New Roman", serif; font-style: normal; font-weight: normal; text-decoration: none; font-size: 14pt; } div.maincontent .s12 { color: black; font-family:Symbol, serif; font-style: normal; font-weight: normal; text-decoration: none; font-size: 14pt; } div.maincontent .s13 { color: black; font-family:Wingdings; font-style: normal; font-weight: normal; text-decoration: none; font-size: 14pt; } div.maincontent .s14 { color: black; font-family:"Times New Roman", serif; font-style: normal; font-weight: normal; text-decoration: none; font-size: 9pt; vertical-align: 5pt; } div.maincontent .s15 { color: black; font-family:"Times New Roman", serif; font-style: normal; font-weight: normal; text-decoration: none; font-size: 9pt; vertical-align: 5pt; } div.maincontent .s16 { color: black; font-family:Cambria, serif; font-style: italic; font-weight: normal; text-decoration: none; font-size: 14pt; } div.maincontent .s17 { color: #080808; font-family:"Times New Roman", serif; font-style: normal; font-weight: bold; text-decoration: none; font-size: 14pt; } div.maincontent .s18 { color: #080808; font-family:"Times New Roman", serif; font-style: normal; font-weight: normal; text-decoration: none; font-size: 14pt; } div.maincontent .s19 { color: black; font-family:"Times New Roman", serif; font-style: normal; font-weight: normal; text-decoration: none; font-size: 11pt; } div.maincontent .s20 { color: black; font-family:"Times New Roman", serif; font-style: normal; font-weight: normal; text-decoration: none; font-size: 10pt; } div.maincontent .s21 { color: black; font-family:"Times New Roman", serif; font-style: normal; font-weight: bold; text-decoration: none; font-size: 11pt; } div.maincontent .s22 { color: black; font-family:"Times New Roman", serif; font-style: normal; font-weight: normal; text-decoration: none; font-size: 11pt; } div.maincontent .s23 { color: black; font-family:"Times New Roman", serif; font-style: italic; font-weight: normal; text-decoration: none; font-size: 14pt; } div.maincontent .s24 { color: #212121; font-family:"Times New Roman", serif; font-style: normal; font-weight: normal; tex
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C. AB = 0.48

Solution : The population is in genetic equilibrium: p + q = 1

-Allele A: pA = 0.4 => qa = 0.6.

-Allele B: pB = 0.6 => qb = 0.4

So: The frequency of each type of gamete in this population is:

AB = pA x pB = 0.4 x 0.6 = 0.24; Ab = pA x qb = 0.4 x 0.4 = 0.16

aB = qa x pB = 0.6 x 0.6 = 0.36; ab = qa x qb = 0.6 x 0.4 = 0.24  Choose B

Lesson 22: In cats, coat color is determined by the gene located on the X sex chromosome. Red coat color is determined by allele d, black coat: D, heterozygous female cats: Dd have tricolor coat color. When examining 691 cats, the frequency of allele D was determined to be: 89.3%; allele d: 10.7%; the number of tricolor cats counted was 64. Knowing that: determining allele frequency follows the Hardy-Weinberg law. The number of male and female cats with different coat colors in order is:

A.335, 356 B.356, 335 C. 271, 356 D.356,

271

Solution : We have: ( 0.893 ) 2 DD + 2 ( 0.893 x 0.107 ) Dd + ( 0.107 ) 2 dd = 1

2 ( 0.893 x 0.107 ) Dd = 64 => Dd = 64 / 0.191102 = 335 children

Therefore: Number of male cats: 691 – 335 = 356 cats,

Number of female cats with different fur colors: 335 – 64 = 271 cats  Choose D

Lesson 23: A population at the time of statistics has the ratio of genotypes as 0.7AA: 0.3aa. Let the population mate randomly for 4 generations, then let it self-mating continuously for 3 generations. What is the ratio of heterozygous individuals in the population? Knowing that there is no mutation, no gene migration, the individuals have the same vitality and fertility:

A. 0.0525 B.0.60 C.0.06 D.0.40

Solution: pA = 0.7; qa = 0.3. Genetic diversity of the population after 4 generations of random mating: 0.49AA; 0.42Aa: 0.09aa

Self-breeding through 3 generations: Aa = (1/2 ) 3 x 0.42 = 0.0525  Choose A

Lesson 24: In humans, A determines black eyes, a: blue eyes, B: curly hair, b: straight hair; related to ABO blood type, there are 4 phenotypes:

- Blood type AB corresponds to genotype I A I B .

- Blood type O corresponds to genotype ii.

Knowing that I A and I B are completely dominant over i, the gene pairs that determine the above traits are located on different pairs of autosomes. The number of different possible genotypes (for the above traits) is: A. 32 B. 54 C. 16

D. 24

Prize :

- The gene that determines blood type has 3 alleles: I A , I B , I 0 => Number of genotypes: (3(3+1) : 2) 1 = 6 genotypes

- The gene that determines eye color has 2 alleles: A, a => Number of genotypes: (2(2+1) : 2 ) 1 = 3 genotypes

- The gene that determines hair type has 2 alleles: B, b => Number of genotypes: (2(2+1) : 2 ) 1 = 3 genotypes So: Number of different possible genotypes (for the above traits): 6 x 3 x 3 = 54

 Choose B

Chapter IV- GENETIC APPLICATIONS

Lesson 21 - Lesson 18: SELECTING LIVESTOCK AND PLANT BREEDS BASED ON SOURCES OF COMBINATIONAL VARIATIONS

I. Objective:

After completing this lesson, students will be able to:

1. Knowledge

- Explain the mechanism of emergence and the role of combinatorial variation in the process of creating pure lines.

- State the concept of hybrid vigor and present methods of creating hybrids for hybrid vigor.

- Explain why hybrid advantage is usually highest in F1 and gradually decreases in the next generation.

2. Skills

- Develop analytical skills on visual channels, comparison, analysis, generalization and synthesis skills

- Ability to work independently with textbooks

- Improve the skills of analyzing phenomena to understand the nature of the event through selecting new varieties from the source of combined variations.

3. Attitude

- Forming faith in science and human intelligence through achievements in breeding by hybridization methods

4. Capacity development

a/ Knowledge capacity:

- Practice and develop analytical and generalizing thinking skills.

- Students can ask many questions about the learning topic/ Life skills:

episode

- Ability to demonstrate confidence when presenting ideas in front of groups, teams, and classes.

- Ability to present thoughts/ideas; collaborate; manage time and assume responsibility, in group activities.

- Ability to search and process information.

learning process such as friends, learning tools, teachers...

- Correctly identify the rights and obligations of learning the subject...