Presented by Amit Kumar Gaur L-2014-A-93-M PAU, Ludhiana (Punjab) Analysis of Basic generation means and variances and its utility in plant breeding Presented by Amit Kumar Gaur L-2014-A-93-M PAU, Ludhiana (Punjab)

What is a quantitative trait? Some traits do not fall in discrete or qualitative categories but have a continuous spectrum of various possible values. Height is a great example: people are not just short or tall, but can measure anywhere from 74.6 cm, the smallest man (He Ping Ping) to 272 cm, the tallest man (Robert Wadlow). If we look at the distribution of height of men we would find that it makes a bell shape curve. There are a number of traits in plants showing such kind of behavior. For example: Kernel colour in wheat, Flower length in tobacco etc. Such characters are called as Quantitative characters or continuous traits.

Need of statistical methods in quantitative genetics The methods used for analysing a qualitative characteristics (examining the phenotypic ratios of progeny from a genetic cross) will not work with quantitative characteristics due to following reasons: Quantitative traits are controlled by many genes or polygenes i.e. genes with small indistinguishable effects It is not possible to follow the segregation of an individual gene Quantitative characteristics are highly affected by environmental factors Due to these reasons the inheritance of quantitative character is analyzed using statistical models. These models helps in Finding the degree of similarity or difference among families. These models also help in estimating that how much of the variation in a character results from genetic factors and how much results from environmental factors.

Generation mean and variance analysis is a statistical model which generally helps us to find the components of genetic variation Genetic Variance is the genotypic differences among individuals within a population Genetic Variance is divided into different Components by different scientists like: Fisher(1918) 1. Additive variance 2. Dominance variance 3. Epistatic variance: it is of three types additive x additive additive x dominance dominance x dominance Mather (1949) 1. Heritable – fixable Additive variance and A x A component of epistatis 2. Heritable non – fixable Dominance variance and A x D and D x D types of epistatis

Additive genetic variance It is due to additive gene action. It is heritable and fixable. In this gene action each allele adds a small and specific value to the phenotype. It is denoted by symbol d. Suppose we have a gene A having two alleles A + and A − A + contributes 2 value A − reduce value by 1 or A − = 1 value If their effects adds additively than A + A + = 2+2 = 4 A + A − = 2+1 = 3 A − A − = 1+1 = 2 So heterozygous ( A + A − ) is equal to mid parent value. Additive gene effects are also used to define the breeding value of an individual Breeding value of an individuals can be described as the sum of the average effect of the individuals alleles. For example, if an A + allele is worth +2 and an A − allele is -1, then an A + A − (heterozygote) has a breeding value of +1.

Dominance genetic variance It is due to dominance gene action. It is Heritable but Non-Fixable. It is denoted by symbol h It is an intra-allelic interaction means Interaction between allele of same gene If dominance gene action is present than in a heterozygote one allele will express itself (dominant) while the other do not express (recessive) A + A + (4) x A − A − (2) A + A − (4) In case of complete dominance hybrid is equal to the dominant parent

Epistasis genetic variance It is a form of non-allelic interaction. Heritable but Non-Fixable. It is an Interaction between allele of two or more different genes It is of three types 1. Additive x additive Denoted by symbol (i) Interaction between two or more loci each exhibit a lack of dominance, only additive effects are present It is fixable 2. Additive x dominance Denoted by (j) Interaction between two or more loci one exhibit lack of dominance while other shows dominance individually Non fixable 3. Dominance x Dominance Denoted by (l) Interaction between two or more loci each exhibits dominance

The concept of Generation Mean analysis was given by Hayman(1958), jinks and jones(1958). Procedure For Generation Mean analysis Development of Basic Generations Recording the observations Development and Testing of a model

Development of Basic Generations There are six basic generations used in this analysis: P 1 , P 2 , F 1 , F 2 , BC 1.1 , BC 1.2 P 1 and P 2 are parents they may be an inbred, homozygous or true breeding lines but they should be diverse. F 1 is the hybrid produced by crossing P 1 with P 2 F 2 is the segregating generation produced by selfing of F 1 BC 1.1 is first backcross generation ( F 1 x P 1 ) BC 1.2 is second backcross generation ( F 1 x P 2 )

Flow chart showing relation between various basic generation P 1 x P 2 Backcross Backcross 𝑩𝑪 𝟏.𝟏 F 1 𝑩𝑪 𝟏.𝟐 Selfing F 2 Observations are recorded from these generations for the traits under study

Additive-dominance model for single gene We take two homozygous lines P 1 and P 2 as parents which are identical except for one gene say A, having two allele A + and A − . We consider P 1 as high scoring parent and P 2 as low scoring parent So we can calculate the genetic value of P 1 and P 2 genotypes as follows P 1 ( A + A + ) = m + d P 2 ( A − A − ) = m - d F 1 ( A + A − ) = m + h d h P 1 P 2 F 1 m A − A − A + A + A + A −

Mid- Parent values It is defined as the average or the mean of two homozygotes (parents).It is donated by m P 1 m P 2 m=1/2( P 1 + P 2 ) e.g. if P 1 is 15 and P 2 is 5 than m is 10

Mathematical equation for additive gene effects Mathematically, additive gene effects is the difference between the two homozygotes. It is donated by symbol d d= 1 2 ( P 1 − P 2 ) 2d d d P 1 m P 2 A + A + A + A − A − A − 15 10 5 d = 5

Mathematical formula for dominance It is the deviation of heterozygote from the mean of the homozygotes (parents).It is denoted by the symbol h h= F 1 - ( P 1 + P 2 /2) h P 1 F 1 m P 2 A + A + A + A − A − A − 15 13 5 h= 3

Direction of dominance h P 1 P 2 F 1 m A + A + A − A − A + A − Case1. If A + is completely or partially dominant to A − than h is positive Case2.If A − is completely or partially dominant to A + h is negative Case3. when no allele is dominant h=0 Other conclusion from this figure h less than d incomplete or partial dominance h equal to d complete dominance h greater than d superior overdominance

Degree of dominance or dominance ratio It is the ratio of dominance effects to the additive effects. Degree of dominance = ℎ(𝑑𝑜𝑚𝑖𝑛𝑎𝑛𝑡 𝑒𝑓𝑓𝑒𝑐𝑡) 𝑑(𝑎𝑑𝑑𝑖𝑡𝑖𝑣𝑒 𝑒𝑓𝑓𝑒𝑐𝑡) If ℎ 𝑑 = +1, allele A + is completely dominant to A − If ℎ 𝑑 = -1, allele A − is completely dominant to A + If 0 < ℎ 𝑑 < +1, allele A + is partially dominant to A − If -1 < ℎ 𝑑 < 0, allele A − is partially dominant to A + If ℎ 𝑑 = 0, there is no dominance If ℎ 𝑑 > +1, , allele A + is over-dominant to A − If ℎ 𝑑 < -1, , allele A − is over-dominant to A +

Mean of F 2 generation In F 2 population we have following genotypes A + A + , A + A − , A − A − Frequency 1/4 1/2 1/4 MEAN 1/4 A + A + + 1/2 A + A − + 1/4 A − A − Or 1/4(m+d) + 1/2(m+h) + 1/4(m-d) F 2 = m + 1/2h

Mean of BC 1.1 and BC 1.2 In BC 1.1 (Parents) A + A − x A + A + Genotypes(Progeny) A + A + A + A − Frequency 1/2 1/2 Mean 1/2 A + A + + 1/2 A + A − Or 1/2 (m + d) + 1/2 (m + h) BC 1.1 = m + 1/2d + 1/2h In BC 1.2 (Parents) A + A − x A − A − Genotypes(Progeny) A − A − A + A − Frequency 1/2 1/2 Mean 1/2 A − A − + 1/2 A + A − Or 1/2(m - d) + 1/2(m + h) BC 1.2 = m - 1/2d + 1/2h

Additive-dominance model for two genes In this we take two homozygous lines P 1 and P 2 as parents which are identical except for two gene say A and B, having two allele A + and A − and B, having two allele B + and B − We consider P 1 as high scoring parent and P 2 as low scoring parent. Here two case can be possible Case1. Association mean a parent contain same nature allele means all positive or all negative A + A + B + B + or A − A − B − B − Case2. Dispersion means a parent have allele of both nature A + A + B − B − or A − A − B + B +

In case of association P 1 = m + da+ db P 2 = m - da - db So we can calculate the genetic value of P 1 and P 2 genotypes as follows: P 1 = m + da+ db P 2 = m - da - db F 1 = m + ha + hb F 2 =m + 1/2ha + 1/2hb BC 1.1 =m + 1/2da + 1/2db + 1/2ha + 1/2hb BC 1.2 =m - 1/2da - 1/2db + 1/2ha + 1/2hb P 1 F 1 P 2 m A + A + B + B + A + A − B + B − A − A − B − B −

In case of dispersion So we can calculate the genetic value of P 1 and P 2 genotypes as follows: P 1 = m + da - db P 2 = m – da + db F 1 = m + ha + hb F 2 =m + 1/2ha + 1/2hb BC 1.1 =m + 1/2da - 1/2db + 1/2ha + 1/2hb BC 1.2 =m + 1/2da - 1/2db + 1/2ha + 1/2hb P 1 F 1 P 2 m A + A + B − B − A + A − B + B − A − A − B + B +

Additive-dominance for multiple gene If in P 1 we have a large number of genes say K controlling a character and K’ is the number of gene dispersed. Than increaser allele will be equal to K-K’ we can write P 1 mean as 𝑖 𝑘−𝐾′ 𝑑 𝑖 𝑘′ 𝑑 𝑖 𝐾 𝑑 𝑖 𝐾′ 𝑑 - - P 1 = m + or + P 1 = m 2 m 𝑖 𝐾 𝑑 - 𝑖 𝐾′ 𝑑 + 2 𝑖 𝐾 𝑑 P 1 = x 𝑖 𝐾 𝑑 Called as 𝑟 𝑑 𝑖 𝐾 𝑑 + 𝑟 𝑑 P 1 = m Written as [ d ]

𝑟 𝑑 Is called as coefficient of gene association/ dispersion = m + [d] 𝑟 𝑑 Is called as coefficient of gene association/ dispersion If it is 1 mean complete association if 0 means maximum dispersion Similary, P 2 = m −[d] 𝑖 𝐾 ℎ F 1 = m + [h] [h] = 𝐅 𝟐 = m + 1/2 [h] 𝐁𝐂 𝟏.𝟏 = m + 1/2 [d] + 1/2[h] 𝐁𝐂 𝟏.𝟐 =m − 1/2[d] + 1/2[h]

Conclusion drawn from this model The expected mean of any generation holds a very simple relationship with the parental and F 1 mean. For example, The mean of backcross generation first is: BC 1.1 = m + 1/2[d] + 1/2[h] or m + 1/2 P 1 + 1/2 F 1 It means average of P 1 and F 1 And BC 1.2 =m - 1/2[d] + 1/2[h] or m - 1/2 P 2 + 1/2 F 1 It means average of P 2 and F 1 Similarly, we can easily find m= 1/2 P 1 + 1/2 P 2 d=1/2 P 1 - 1/2 P 2 h= F 1 - 1/2 P 1 - 1/2 P 2

Effect of epistasis components on means The earlier calculated relationships holds good only if the generation mean depends only on additive and dominance effects of genes This model is not adequate if epistasis is present So it is very important for us to check the adequacy of this model To check the adequacy of the model we can use three scaling tests given by Mather (1949) Three scaling tests are A, B and C

Scaling tests- a way to check presence or absence of epistasis A= 2 BC 1.1 - P 1 - F 1 B=2 BC 1.2 - P 2 - F 1 C=4 F 2 - 2 F 1 - P 1 - P 2 In 1955, hayman had added a fourth test called as D scaling test D=4 F 3 - 2 F 2 - P 1 - P 2 ** D=2 F 2 - B 1 - B 2 If the values of A,B,C,D comes zero(0) it means the additive - dominance model is adequate and epistasis is not present. But , if value is significantly different from zero, it means epistasis is present and additive-dominance model is not adequate.

Test of significance t- test Suppose we want to test the scaling test A than 𝑡 𝑐𝑎𝑙 = A / S.E. (A) S.E. (A) means standard error of A Steps to find standard error of A Step1. Take Square of the A scaling test A= 2 BC 1.1 - P 1 - F 1 𝑉 𝐴 = 4 𝑉 𝐵1 + 𝑉 𝑃1 + 𝑉 𝑃2 equation 1 𝑉 𝐵1 = variance of BC 1.1 generation 𝑉 𝑃1 = variance of P 1 generation 𝑉 𝑃2 = variance of F 1 generation Step2. Now divide each generation variance by n(number of individual in that population) put that value in equation 1 Step3. S.E (A)= 𝑉𝐴

Now apply student’s t-test 𝑡 𝑐𝑎𝑙 = A / S.E. (A) Compare this 𝑡 𝑐𝑎𝑙 with 𝑡 𝑡𝑎𝑏 at 5% level of significance In each test the df (degree of freedom) is the sum of df of various generation involve. If 𝑡 𝑐𝑎𝑙 > 𝑡 𝑡𝑎𝑏 , it means significant difference. If 𝑡 𝑐𝑎𝑙 < 𝑡 𝑡𝑎𝑏 , it means no significant difference. Similarly we have to perform the t-test for for B, C, D scaling test also

Interpretation of the results on the basis of t-tests IF A or B is significant it indicate the presence of all the three types of non allelic gene interactions Additive x Additive(i), Additive x Dominance(j), Dominance x Dominance(l) If C scaling tests are significant it indicates Dominance x Dominance(l) If D is significant indicates Additive x Additive(i) If both C and D are significant it indicates Additive x Additive(i) Dominance x Dominance(l)

Drawback of the scaling tests The main drawback of the scaling tests are Out of the six population only three or four are included at a time. It do not provide estimates of m [d] [h]. It do not tests goodness of fit of a model.

Solution to these problems In 1952, Cavalli devised a method known as joint scaling test to overcome these limitations It is based on weighted least square method. Here is a numerical example The data is given on tiller number in sorghum Here P 1 is low scoring parent and P 2 is high scoring hence P 1 is 𝐴 − 𝐴 − and P 2 is 𝐴 + 𝐴 + Weight is reciprocal of squared standard error Generation Mean S.E. Weight m d h P 1 1.4 0.2449 16.6666 1 -1 P 2 2.2 0.2 25 F 1 2.0 0.6325 2.5 F 2 1.85 0.2828 12.5 ½ BC 1.1 2.1 0.3742 7.14 -1/2 BC 1.2 0.6708 2.2222 1/2

To find equation of m 66.0288m+5.8745d+13.4311h=127.0078 equation 1 Generation mean Coefficient of m Weight of m Coefficient x weight ( cw) m Coefficient of m x cw d Coefficient of d x cw h Coefficient of h x cw Mean* P 1 1.4 1 16.666 -16.666 23.333 P 2 2.2 25 55 F 1 2 2.5 5 F 2 1.85 12.5 6.25 23.1250 BC 1.1 2.1 7.14 -3.57 3.57 14.9940 BC 1.2 2.222 1.111 5.555 Total 66.028 5.8745 13.431 127.007 66.0288m+5.8745d+13.4311h=127.0078 equation 1 Mean* = mean x cw

To find equation of d 5.8745m+44.0070d-1.2295h=26.9477 equation 2 P 1 Generation mean Coefficient of d Weight of d Coefficient x weight M D h Mean* P 1 1.4 -1 16.666 -16.666 -23.333 P 2 2.2 1 25 55 F 1 2 2.5 F 2 1.85 12.5 BC 1.1 2.1 -1/2 7.14 -3.57 1.785 -1.785 -7.4970 BC 1.2 1/2 2.222 1.111 0.555 0.5555 2.7777 Total 5.8745 44.0071 -1.229 26.947 5.8745m+44.0070d-1.2295h=26.9477 equation 2

To find equation of h 13.4311m-1.2295d+7.9655h=26.8372 equation 3 P 1 Generation mean Coefficient of h Weight of h Coefficient x weight M D h Mean* P 1 1.4 16.666 P 2 2.2 25 F 1 2 1 2.5 5 F 2 1.85 1/2 12.5 6.25 3.125 11.5625 BC 1.1 2.1 7.14 3.57 -1.785 1.785 7.4970 BC 1.2 2.222 1.111 0.555 0.5555 2.7777 Total 13.431 -1.2295 7.9655 26.8372 13.4311m-1.2295d+7.9655h=26.8372 equation 3

Matrix formation Now, the three equations are as follows 66.0288m + 5.8745d + 13.4311h = 127.0078 5.8745m + 44.0070d - 1.2295h = 26.9477 13.4311m - 1.2295d + 7.9655h = 26.8372 From above three equations we can estimates the parameters in the following way 66.0288 5.8745 13.4311 m 127.0078 5.8745 44.0070 -1.2295 d = 26.947 13.4311 -1.2295 7.9655 h 26.8372 A B C

Say A.B=C Hence, B= 𝐴 −1 .C m 0.0237 -0.0042 -0.0406 127.008 d = -0.0042 0.0235 0.0108 26.9477 h -0.0406 0.0108 0.1957 26.8372 Hence, m 1.81 d = 0.39 h 0.39

Again apply student’s t-test The standard error of each of these elements is obtained as the under root of the diagonal elements of the inverse matrix as given below S.E.(m)= 0.0237 =0.15 S.E.(d)= 0.0235 =0.15 S.E.(h)= 0.1957 =0.44 Now the three parameter with their standard deviation are: m=1.81 ±0.15 d=0.39 ± 0.15 h=0.39 ± 0.44 𝑡 𝑐𝑎𝑙 = estimated value/standard error of the parameter Than compare with table value All are significant except h

Test of goodness of fit of the model It includes two steps Steps 1. Calculate the expected mean of families using estimates of m, d, h in a following manner P 1 =m - d = 1.81-0.39 =1.42 P 2 =m + d = 1.81+0.39 =2.20 F 1 =m + h = 1.81+0.39 =2.20 F 2 =m + 1/2h = 1.81+1/2x0.39 =2.005 BC 1.1 =m - 1/2d + 1/2h = 1.81-1/2x0.39+1/2x0.39 =1.81 BC 1.2 =m + 1/2d + 1/2h = 1.81+1/2x0.39+1/2x0.39 =2.20

Analysis of result on basis of chi square test Step 2. Find 𝑥 2 (chi square) we have six families from which we have estimated three parameters m, d, h. Hence, only three (6-3)df are left for testing the adequacy of the model. The calculated chi square at 3 df is 1.1084, less than the table value hence model is adequate and there is nothing beyond additive and dominance effects Generation Observed (o) Expected (E) (o- E) (𝒐−𝑬) 𝟐 x weight P 1 1.4 1.42 -0.02 0.0067 P 2 2.2 F 1 2.0 -0.2 0.0010 F 2 1.85 2.005 -0.155 0.3003 BC 1.1 2.1 1.81 0.29 0.6005 BC 1.2 2.5 0.3 0.1999 𝑥 2 = 1.1084

Effects of Epistasis on Generations mean Digenic epistasis Change in population mean in presence of epistasis Here, 𝑑𝑑 𝑎𝑏 refers to additive x additive interaction or (i) ℎℎ 𝑎𝑏 refers to dominance x dominance interaction or (j) 𝑑ℎ 𝑎𝑏 refers to additive x dominance interaction or (l) Lines Genetic value in absence of epistasis Genetic value in presence of epistasis A + A + B + B + m + 𝑑 𝑎 + 𝑑 𝑏 m + 𝑑 𝑎 + 𝑑 𝑏 + 𝑑𝑑 𝑎𝑏 A − A − B − B − m - 𝑑 𝑎 - 𝑑 𝑏 m - 𝑑 𝑎 - 𝑑 𝑏 + 𝑑𝑑 𝑎𝑏 A + A − B + B − m + ℎ 𝑎 + ℎ 𝑏 m + ℎ 𝑎 + ℎ 𝑏 + ℎℎ 𝑎𝑏 A + A + B − B − m + 𝑑 𝑎 - 𝑑 𝑏 m + 𝑑 𝑎 - 𝑑 𝑏 - 𝑑𝑑 𝑎𝑏 A − A − B + B + m - 𝑑 𝑎 + 𝑑 𝑏 m - 𝑑 𝑎 + 𝑑 𝑏 - 𝑑𝑑 𝑎𝑏 A + A + B + B − m + 𝑑 𝑎 + ℎ 𝑏 m + 𝑑 𝑎 + ℎ 𝑏 + 𝑑ℎ 𝑎𝑏 A − A − B + B − m - 𝑑 𝑎 + ℎ 𝑏 m - 𝑑 𝑎 + ℎ 𝑏 - 𝑑ℎ 𝑎𝑏

Effects of epistasis on multiple genes Families Mean in additive – dominance model Mean in epistasis model P 1 m + [d] m + [d] + [dd] P 2 m − [d] m − [d] + [dd] F 1 m + [h] m + [h] + [hh] F 2 m + ½[h] m + ½ [h] + ¼ [hh] BC 1.1 m + ½[d] + ½[h] m + ½ [d] + ½ [h] +1/4 [dd]+ ¼ [dh] + 1/4 [hh] BC 1.2 m − ½ [d] +[h] m − ½ [d] + ½ [h] + 1/4 [dd] − ¼ [dh] + 1/4 [hh]

Estimation of different parameters on basis of epistasis m= 1/2 P 1 +1/2 P 2 +4 F 2 − 2 BC 1.1 −2 BC 1.2 [d]= 1/2 P 1 −1/2 P 2 [h]= 6 BC 1.1 + 6 BC 1.2 − F 1 −8 F 2 −3/2 P 1 −3/2 P 2 [dd]= 2 BC 1.1 +2 BC 1.2 −4 F 2 [dh]= 2 BC 1.1 − 2 BC 1.2 − P 1 + P 2 [hh]= P 1 + P 2 +2 F 1 +4 F 2 −4 BC 1.1 −4 BC 1.2

Basic generation variances On the basis of variance the six basic generations are divided into two parts 1. Non-segregating generation: it includes those generations that consists of genetically identical individuals. For example P 1 , P 2 and F 1 2. Segregating generation: it includes those generation which contains genetically different individuals. For example F 2, BC 1.1 and BC 1.2

Non-Segregating generation A + A + A + A − A − A − A − A − A + A + A + A + A + A − A − A − A + A − A + A − A − A − A + A + A − A − A + A + A + A − P 𝟐 generation P 1 generation F 𝟏 generation Segregating generation A + A + A + A + A − A − A + A − A + A + A − A − A + A − A − A − A + A − A + A − A + A − BC 1.1 generation F 2 generation BC 1.2 generation

Variation in Non segregating generations Because the individuals within the P 1 , P 2 and F 1 families are genetically identical hence any variation within these families cannot be genetical and such variation is refer to as environmental variation within families Denoted by 𝑉 𝐸 It is non heritable in nature P 1 variance = 𝑉 𝐸 P 2 variance = 𝑉 𝐸 F 1 variance = 𝑉 𝐸

Variation in segregating generation The individuals in segregating generation are subjected to both type of variation means Environmental as well as genetical Derivation of equation for the genetic variation of F 2 generation In F 2 population we have following genotypes A + A + , A + A − , A − A − Frequency 1/4 1/2 1/4 MEAN 1/4 A + A + + 1/2 A + A − + 1/4 A − A − Or ¼ (m+d) +1/2(m+h) +1/4(m-d) Or F 2 mean = 𝑓𝑖𝑔𝑖 (fi refer to frequency of gi genotype) F 2 mean = m + ½ ℎ 𝑎 Hence, F 2 variance = 𝑓𝑖𝑔𝑖 2 − ( 𝑓𝑖𝑔𝑖 ) 2

F 2 variance = 𝑓𝑖𝑔𝑖 2 − ( 𝑓𝑖𝑔𝑖 ) 2 1/4 ( A + A + ) 2 + 1/2 ( A + A − ) 2 + 1/4 ( A − A − ) 2 − (𝑚𝑒𝑎𝑛) 2 1/4 (𝑚+𝑑) 2 + 1/2 (𝑚+ℎ) 2 + 1/4 (𝑚−𝑑) 2 −1/2 (𝑚+ 1 2 ℎ) 2 F 2 variance = 1/2 𝑑 2 + 1/4 ℎ 2 If we call, 𝑉 𝑑 = 1/2 𝑑 2 And , 𝑉 ℎ = 1/4 ℎ 2 So, 𝑉 𝑑 represent genetic component of variance 𝑉 ℎ represent dominance component of variance Hence total F 2 variance = 𝑉 𝑑 + 𝑉 ℎ + 𝑉 𝐸

BC 1.1 generation Variance In BC 1.1 we have A + A − x A + A + Genotypes A + A + A + A − Frequency ½ ½ Mean 1/2 A + A + + 1/2 A + A − Or 1/2 (m + d) +1/2(m + h) BC 1.1 = m+1/2d+1/2h BC 1.1 variance = 𝑓𝑖𝑔𝑖 2 − ( 𝑓𝑖𝑔𝑖 ) 2 1/2 (𝑚+𝑑) 2 +1/2 (𝑚+ℎ) 2 − (𝑚+ 1 2 𝑑+ 1 2 ℎ) 2 BC 1.1 Variance= 1/4 𝑑 2 + 1/4 ℎ 2 − 1/2 dh BC 1.1 Variance= 1/2 𝑉 𝑑 + 𝑉 ℎ − 𝑉 𝑑ℎ Similarly BC 1.2 Variance is BC 1.2 Variance= 1/2 𝑉 𝑑 + 𝑉 ℎ + 𝑉 𝑑ℎ

Equations for estimation of 𝑉 𝑑 , 𝑉 ℎ , 𝑉 𝑑ℎ in absence and presence of epistasis In absence of epistasis 𝑉 𝑑 = 2 𝑉 𝐹2 - 𝑉 𝐵𝐶1.1 - 𝑉 𝐵𝐶1.2 𝑉 ℎ = 𝑉 𝐵𝐶1.1 - 𝑉 𝐵𝐶1.2 - 𝑉 𝐹2 - 𝑉 𝑒 𝑉 𝑑ℎ =1/2 ( 𝑉 𝐵𝐶1.2 - 𝑉 𝐵𝐶1.1 ) In presence of epistasis 𝑉 𝑑 = 𝑉 𝐵𝐶1.1 - 𝑉 𝐵𝐶1.2 𝑉 ℎ = 𝑉 𝐹1 +16 𝑉 𝐹2 +1/4 V P1 + 1/4 V P2 +4 𝑉 𝐵𝐶1.1 + 4 𝑉 𝐵𝐶1.2 𝑉 𝑖 =4 𝑉 𝐵𝐶1.1 + 1/4 𝑉 𝐵𝐶1.2 +16 𝑉 𝐹2 𝑉 𝑗 = 𝑉 𝐵𝐶1.1 + 1/4 V P1 + 𝑉 𝐵𝐶1.2 +1/4 V P2 𝑉 𝑙 = V P1 + V P2 +4 𝑉 𝐹1 16 𝑉 𝐹2 +16 𝑉 𝐵𝐶1.1 + 16 𝑉 𝐵𝐶1.2

Utility of generation mean analysis in Plant breeding 1st Utility It is an important tool which help us to choose a suitable breeding procedure for the development of a variety

Breeding procedures in relation to gene action Type of gene action Breeding procedure to be adapted A. Self –pollinated species 1. Additive Pure line selection Mass selection Hybridization and selection 2. Non - additive Heterosis breeding B. Cross pollinated species Recurrent selection for gca Synthetic breeding Composite breeding 2. Non- additive Recurrent selection for sca 3. Both additive and non- additive Reciprocal recurrent selection

2nd utility It can be used to calculate Heritability of a character

Heritability Total Phenotypic variance = genotypic variance + environmental variance + interaction between genotype and environment 𝑉 𝑃 = 𝑉 𝑔 + 𝑉 𝑒 + 𝑉 𝑔𝑒 The total genetic variance( 𝑉 𝑔 ) is composed of three major components: additive genetic variance ( 𝑉 𝑑 ), dominance variance ( 𝑉 ℎ ), and nonallelic interactions or epistasis variance ( 𝑉 𝑛 ). 𝑉 𝑔 = 𝑉 𝑑 + 𝑉 ℎ + 𝑉 𝑛 Heritability is the proportion of the observed variation (phenotypic variance) in a progeny that is inherited. If the genetic variation is large in relation to the environmental variation, then heritability will be high if genetic variation is small in relation to the environmental variation, then heritability will be low. Selection is more effective when genetic variation in relation to environmental variation is high.

Types of heritability Heritability is of two type: broad sense heritability Narrow sense heritability Broad sense heritability: It is the ratio of total genetic variation to the total phenotypic variation = 𝑉 𝑔 / 𝑉 𝑃 = 𝑉 𝑔 / 𝑉 𝑔 + 𝑉 𝑒 + 𝑉 𝑔𝑒 = 𝑉 𝑔 / 𝑉 𝑃 x100

Heritability and Selection Narrow sense heritability: it measures the proportion of variance which is due to the additive effects of genes = 𝑉 𝑑 / 𝑉 𝑃 x100 Heritability and Selection Quantitatively inherited characters differ in heritability. A character such as yield that is greatly influenced by the environment have a low heritability. Characters not greatly influenced by environment usually have a high heritability. This may influence the choice of selection procedure used by the plant breeder. Selection in the 𝐹 2 population will not be very effective for characters that have low heritability. Selection in the 𝐹 2 is more effective if it is limited to characters that have a high heritability.

3rd utility We can estimate genetic advance under selection

Genetic Advance under selection It is improvement in the mean genotypic value of the selected families over that of the base population. 𝐺 𝑠 = i x 𝑉 𝑃 x ℎ 2 𝐺 𝑠 is the predicted genetic advance i is a constant based on selection intensity V P is the square root of the phenotypic variance h 2 the narrow-sense heritability Interpretation of Results for Genetic Advance If value of GA is high, it shows that the character is control by additive genes and selection would be effective for improvement of such trait If the value of GA is low, it indicates that the character is governed by non- additive genes and heterosis breeding will be rewarding It depends upon 3 factors 1) Phenotypic variability among different plants or families in the base population 2) Heritability of the character under selection 3) Intensity of selection i.e. the proportion of plants or families selected It is calculated by following formula Gs = k ∂p H Where, Gs – Genetic advance under selection k – Selection differential ∂p – Phenotypic standard deviation of base population which is being subjected to selection H – Heritability of the character under selection

4th utility To explain Heterosis

Heterosis The most useful and widely accepted meaning of heterosis is the superiority of hybrid or F 1 over the better parent F 1 > P 1 heterosis is simply m+[h] > m+[d] [h]> [d] ℎ > 𝑟 𝑑 𝑑 It means if dominance is greater than the degree of gene dispersion than there is heterosis

Example….. Let us say, better parent have 70% of the increaser allele than 𝑟 𝑑 will be 0.4 means the average dominance need to be just greater than 0.4 to cause F 1 to outperform the P 1 Means if 𝑟 𝑑 is near to 0.5 than even small amount of dominance can cause heterosis Only if 𝑟 𝑑 =1 than it is necessary to invoke overdominance to explain the heterosis

References Books: i). Biometrical Genetics: The study continuous variation by Kenneth Mather and John L. Jinks ii). The Genetical Analysis of Quantitative Traits by Michael J. Kearsey and Harpal S. Pooni iii). Biometrical techniques in Plant Breeding by Singh and Narayan iv). Quantitative Genetics by Phundan Singh v). Biometrical Methods in Quantitative Genetic analysis by Singh and Chaudhary

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