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Principles of Plant Genetics and Breeding

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<strong>and</strong>omization. The blocks should be laid across the<br />

fertility gradient.<br />

Probability<br />

Statistical probability is a procedure for predicting the<br />

outcome <strong>of</strong> events. Probabilities range from 0 (an event<br />

is certain not to occur) to 1.0 (an event is certain to<br />

occur). There are two basic laws <strong>of</strong> probability – product<br />

<strong>and</strong> sum laws. The probability <strong>of</strong> two or more outcomes<br />

occurring simultaneously is equal to the product<br />

<strong>of</strong> their individual probabilities. Two events are said to<br />

be independent <strong>of</strong> one another if the outcome <strong>of</strong> each<br />

one does not affect the outcome <strong>of</strong> the other. Genetic<br />

ratios may be expressed as probabilities. Consider a<br />

heterozygous plant (Rr). The probability that a gamete<br />

will carry the R allele is one-half. In a cross, Rr × Rr<br />

(selfing), the probability <strong>of</strong> a homozygous recessive<br />

(rr <strong>of</strong>fspring) is 1 / 2 × 1 / 2 = 1 / 4 . The probability that<br />

one or another <strong>of</strong> several mutually exclusive outcomes<br />

will occur is the sum <strong>of</strong> their individual probabilities.<br />

Using the cross Rr × Rr, the F 2 will produce<br />

RR : Rr : rr in the ratio 1 / 4 : 1 / 2 : 1 / 4 . The probability<br />

that a progeny will exhibit a dominant phenotype<br />

(RR, Rr) = 1 / 4 + 1 / 2 = 3 / 4 . Other examples were discussed<br />

in Chapter 3.<br />

In using probabilities for prediction, it is important<br />

to note that a large population size is needed for accurate<br />

prediction. For example, in a dihybrid cross, the<br />

F 2 progeny will have a 9 :3:3:1 phenotypic ratio,<br />

indicating 9/16 will have the dominant phenotype.<br />

However, in a sample <strong>of</strong> exactly 16 plants, it is unlikely<br />

that exactly nine plants will have the dominant phenotype.<br />

A larger sample is needed.<br />

Measures <strong>of</strong> central tendency<br />

The distribution <strong>of</strong> a set <strong>of</strong> phenotypic values tends<br />

to cluster around a central value. The most common<br />

measure <strong>of</strong> this clustering is the arithmetic mean. <strong>Plant</strong><br />

breeders use this statistical procedure very frequently in<br />

their work. The formulae for calculating means are:<br />

Sample mean, X ¯ =∑X/n<br />

Population mean, µ=∑X/N<br />

where X = measured value <strong>of</strong> the item, X ¯ = sample<br />

mean, µ =population mean, n = sample size, <strong>and</strong> N =<br />

population sample.<br />

COMMON STATISTICAL METHODS IN PLANT BREEDING 149<br />

Table 9.1 Ungrouped data for distribution <strong>of</strong> plant<br />

seedling height.<br />

or:<br />

Distribution <strong>of</strong> seedling height (cm)<br />

5 6 7 8 9 10 11 12 13 14 15 16 17 Total<br />

F 1 5 14 16 12 3 50<br />

F 2 4 10 13 17 20 28 25 18 17 13 11 10 7 193<br />

Table 9.2 Grouped data for frequency calculation.<br />

f 1 x i f i f i x i f i x i 2 f i x i f i x i 2<br />

5 4 20 100<br />

6 10 60 360<br />

7 13 91 637<br />

5 8 17 136 1,088 40 320<br />

14 9 20 180 1,620 126 1,134<br />

16 10 28 280 2,800 160 1,600<br />

12 11 25 275 3,029 132 1,452<br />

3 12 18 216 2,592 36 432<br />

13 17 221 2,873 494 4,938<br />

14 13 182 2,548<br />

15 11 165 2,475<br />

16 10 160 2,560<br />

17 7 119 2,023<br />

The sample mean is calculated as:<br />

X¯ = X / n (for ungrouped data; Table 9.1)<br />

X ¯ =∑X i f i /n (for grouped data; Table 9.2)<br />

where X i = value <strong>of</strong> the ith unit included in the sample,<br />

f i = frequency <strong>of</strong> the ith class, <strong>and</strong> n =∑f i .<br />

The sample mean <strong>of</strong> seed size <strong>of</strong> navy beans is:<br />

X ¯ =<br />

n<br />

∑<br />

i=<br />

1<br />

n<br />

∑<br />

i=<br />

1<br />

i<br />

X / n<br />

i<br />

n ∑f i x i ∑f i x i 2<br />

193 2,105 24,705<br />

= (17.2 + 18.1 + ...+ 19.7)/10<br />

= 190.9/10<br />

= 19.01 g per 100 seed

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