basic unit of inheritance

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What is a Gene? A gene is the basic unit of inheritance. Gregor Mendel, 1865. Experiments in Plant Hybridization. Verhandlungen des naturforschenden Vereines in Brünn, 4:3-47

description

What genes are, their purpose and how gene expression and regulation are controlled.

Transcript of basic unit of inheritance

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What is a Gene? A gene is the basic unit of inheritance.

Gregor Mendel, 1865. Experiments in Plant Hybridization. Verhandlungen des naturforschenden Vereines in Brünn, 4:3-47

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Probability # of times outcome is expected/ # of opportunities Given as a fraction, percentage, or proportion of 1

Product Rule: The probability that two independent events both occurring is the product of each event occurring separately.

Using a fair die, the chance of rolling a 1 is 1/6, 16.7%, or .167

Using a pair of fair dice, the chance of rolling two 1s is 1/6 X 1/6 = 1/36 or 0.027

Sum Rule: The probability that either of two mutually exclusive events will occur is the sum of the probability of each event.

Using a fair die, the chance of rolling either a 2 or a 3 1/6 + 1/6 = 1/3 or 0.333

and

or

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1/36

Using a pair of fair dice, the chance of rolling two 1s is 1/6 X 1/6 = 1/36 or 0.027

Product Rule: The probability that two independent events both occurring is the product of each event occurring separately.

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1/36

1/36

Sum Rule: The probability that either of two mutually exclusive events will occur is the sum of the probability of each event.

Using a pair of fair dice, the chance of rolling a 2 and a 3 or A 3 and a 2 are 1/36 + 1/36= 2/36 or 1/18

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You are playing backgammon and your opponent has you on the ropes. However you can knock him off the board with a role of a ten. What is the probability of rolling a ten with two die?

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The seven phenotypic pairs studied by Mendel

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Gamete – specialized haploid cells that function during reproduction

Egg ♀ Sperm ♂

Plants: Egg is located in Ovule Sperms cells are located in Pollen

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Cross-pollination and selfing are two types of crosses

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Figure 2-12 part 2

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Figure 2-12 part 3

Yellow trait is dominant Green is recessive

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Figure 2-12 part 4

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Figure 2-12 part 5

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Figure 2-12 part 6

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Figure 2-12 part 7

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Figure 2-12 part 8

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Figure 2-12 part 9

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Figure 2-12 part 10

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Figure 2-12 part 11

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Figure 2-12 part 12

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Figure 2-12 part 13

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Figure 2-12 part 14

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Figure 2-12 part 15

Punnett Squares

Test Cross

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Table 2-1

These data are pretty darn close to the expected 3:1, but are they close enough to support the hypothesis?

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χ2 (chi-square) Test Goodness of Fit

χ2 = Σ(0bserved-Expected)2/Expected

Class Observed Expected (O-E) (O-E)2 (O-E)2/E

round 5474 5493

(75%) -19 361

361/5493= 0.065

wrinkled 1850 1831 (25%)

19 361 361/1831 = 0.197

Total 7324 7325 χ2 = 0.262

Degrees of Freedom = # of independent variables i.e., # of groups -1

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χ2 Distribution - degrees of freedom

Prob

abili

ty

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χ2 = Σ(0bserved-Expected)2/Expected

Class Observed Expected (O-E) (O-E)2 (O-E)2/E

Yellow 105

Green 95

Total 200 χ2 =

Degrees of Freedom = # of independent variables i.e., # of groups -1

Yellow X Green Test the Hypothesis that it is a Test-Cross

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χ2 (chi-square) Test

Important considerations: * the P value is the probability of a deviation at least as large as what was found. A p of 0.05 means that 1 in 20 times, you may find a deviation this large or larger– not never. * having passed the test means that the results are consistent, not that the results prove the hypothesis. * Since the chi-square value is related to 1/# of expected, the test becomes more accurate with larger numbers.

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1)Recessive trait present in F1 hybrid -heterozygotes. One maternal copy and One Paternal copy.

2) Recessive trait in progeny was same as in grandparent.

3) Traits found in a 3:1 ratio in F2

generation

Mendel’s First Law of Inheritance

Begins to differentiate Genotype (internally coded trait) from Phenotype (outwardly manifested trait).

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Figure 3-3 step4

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Forked-Line Diagram for Gamete Frequencies

RrYy Dihybrid

R

r

Y

y

Y

y

1/2

1/2

1/2

1/2

1/2

1/2

Gamete Frequency

RY ½ X ½ = ¼

Ry ½ X ½ = ¼

ry ½ X ½ = ¼

rY ½ X ½ = ¼

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χ2 (chi-square) Test χ2 = Σ(0bserved-Expected)2/Expected

Phenotype Observed Expected (O-E) (O-E)2 (O-E)2/E

Round, Yellow

315 313 2 4 0.013

Round, Green

108 104 4 16 0.154

Wrinkled, Yellow

101 104 -3 9 0.087

Wrinkled, Green

32 35 -3 9 0.257

Total 55566

556 χ2 = 0.511

Df = 3

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Data are consistent with Independent Segregation

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Rules of Inheritance

1) The units of inheritance (genes) come in pairs. Each gamete will get only one of the two units of inheritance . The progeny then gets one maternal and one paternal copy.

2) The segregation of one gene pair is most often independent of the segregation of another gene pair – independent assortment.