Lecture 7 LINKAGE Part I(1).pdf

LINKAGE CHAPTER 6  2 Introduction Each species must contain thousands to tens of thousands of genes, yet most have at most a few dozen chromosomes Each chromosome is therefore lik...

LINKAGE CHAPTER 6  2 Introduction Each species must contain thousands to tens of thousands of genes, yet most have at most a few dozen chromosomes Each chromosome is therefore likely to carry many hundred or even thousands of different genes The transmission of genes close to one another on the same chromosome violates Mendel’s law of independent assortment INTRODUCTION The term synteny means two or more genes are located on the same chromosome and are physically linked Genetic Linkage is the phenomenon that genes close together on a chromosome tend to be transmitted as a unit, which influences inheritance patterns Chromosomes are called linkage groups They contain a group of genes that are linked together The number of linkage groups is the number of types of chromosomes of the species For example, in humans 22 autosomal linkage groups An X chromosome linkage group A Y chromosome linkage group Genes that are far apart on the same chromosome may independently assort from each other due to crossing over A two-factor cross studies linkage between two genes A three-factor cross studies linkage between three genes BATESON and PUNNETT u Cross involving two traits flower color and pollen shape u This did not yield the expected 9:3:3:1 phenotypic ratio Independent Assortment does not Always Occur Independent Assortment does not Always Occur 2 8 Observed Expected F2 generation Ratio Ratio number number Purple flowers, long pollen 296 15.6 240 9 Purple flowers, round pollen 19 1.0 80 3 Red flowers, long pollen 27 1.4 80 3 Red flowers, round pollen 85 4.5 27 1 u From the results, they concluded that they are not assorted in an independent manner u They failed to realize that the coupling is due to linkage on the same chromosome WHEN IS LINKAGE ALTERED? u During Meiosis as a result of crossing over u Crossing over occurs during Prophase of Meiosis I u Non-sister chromatids of homologous chromosomes exchange DNA segments u Replicated sister chromatids homologues associate as bivalents u Crossing over may produce recombinant genotypes  No Crossing over Linked genes segregate together  CROSSING OVER Haploid cells that contain a combination of alleles NOT found in the original chromosomes are call nonparental or recombinant cells Recombinant offspring are produced by the exchange of DNA between two homologous chromosomes during meiosis in one or both parents, leading to a novel combination of genetic material Recombinant offspring are those that have been produced due to a crossover event during gamete formation in (at least) one of their parents MORGAN’s EXPERIMENTS The first direct evidence of linkage came from studies conducted by Thomas Hunt Morgan Morgan investigated several traits that followed an X-linked pattern of inheritance in Drosophila experiment involving three traits Body color Eye color Wing length The cross results had significant number of non-parental combinations – WHY? Morgan's three-factor Cross © McGraw-Hill Education. 6-14 Morgan's three-factor Cross Morgan observed a much higher proportion of the combinations of traits found in the parental generation Morgan’s explanation: All three genes are located on the X chromosome Therefore, they tend to be transmitted together as a unit © McGraw-Hill Education. 6-15 Morgan’s Three-Factor Cross 2 16 F2 generation Females Males Total Gray body, red eyes, long wings 439 319 758 Gray body, red eyes, miniature wings 208 193 401 Gray body, white eyes, long wings 1 0 1 Gray body, white eyes, miniature wings 5 11 16 Yellow body, red eyes, long wings 7 5 12 Yellow body, red eyes, miniature wings 0 0 0 Yellow body, white eyes, long wings 178 139 317 Yellow body, white eyes, miniature wings 365 335 700 Morgan’s Evidence for X-linked Genes 2 17 u Morgan had to interpret two key observations: u Why did the F2 generation have a significant number of recombinant offspring? u Why was there a quantitative difference between the different types of F2 recombinant offspring? Morgan’s Evidence for X-linked Genes 3 18 u Morgan observed a much higher proportion of the combinations of traits found in the parental generation u Morgan’s explanation: All three genes are located on the X chromosome The three genes tend to be transmitted together as a unit Morgan’s Data 19 u Morgan considered the previous studies of the cytologist F.A. Janssens to explain these data u Janssens had microscopically observed chiasmata and proposed that crossing over involves a physical exchange between homologous chromosomes u Morgan realized that crossing over between homologous X chromosomes was consistent with his data And also assumed crossing over did not occur between the X and Y chromosomes The three genes were not found on the Y chromosome  Morgan’s Hypotheses Morgan made three important hypotheses to explain his results The genes for body color, eye color and wing length are all located on the X- chromosome The alleles will tend to be inherited together Due to crossing over, the homologous X chromosomes (in the female) can exchange pieces of the chromosomes. This created new combinations of alleles The likelihood of crossing over depends on the distance between the two genes Crossing over is more likely to occur between two genes that are far apart from each other. © McGraw-Hill Education. 6-20 Morgan’s Data Simplified into Gene Pairs 21 u Let’s first consider Morgan’s experiments by simplifying the data into gene pairs Gray body, red eyes 1159 Yellow body, white eyes 1017 Gray body, white eyes 17 u Recombinant Yellow body, red eyes 12 offspring Total 2205 Red eyes, long wings 770 White eyes, miniature wings 716 Red eyes, miniature wings 401 u Recombinant offspring White eyes, long wings 318 Total 2205 Nonrecombinant and Recombinant Offspring (a) No crossing over, nonrecombinant offspring © McGraw-Hill Education. 6-22 Nonrecombinant and Recombinant Offspring (b) Crossing over, recombinant offspring © McGraw-Hill Education. 6-23  Morgan’s Drosophila Experiment Explanation for different proportions of recombinant offspring (a) No crossing over in this region, very common © McGraw-Hill Education. 6-24  Morgan’s Drosophila Experiment Explanation for different proportions of recombinant offspring (b) Cross over between eye color and wing length genes, fairly common © McGraw-Hill Education. 6-25 (c) Crossover between body color and eye color genes, uncommon © McGraw-Hill Education. 6-26 (d) Double crossover, very uncommon © McGraw-Hill Education. 6-27 Chi Square Analysis This statistical method is frequently used to determine if the outcome of a dihybrid cross is consistent with linkage or independent assortment Let’s consider the data concerning body color and eye color An example of a chi square approach to determine linkage is shown next © McGraw-Hill Education. 6-28 Chi Square Analysis 2 29 u Step 1: Propose a hypothesis The genes for eye color and body color are assorting independently u Even though the observed data appear inconsistent with this hypothesis, it allows us to calculate expected values u Since we do not know the probability of crossing over, we cannot calculate an expected value for a hypothesis based on linkage u We anticipate that the chi square analysis will allow us to reject this hypothesis in favor of a linkage hypothesis Step 1: Propose a Hypothesis 30 u Access the text alternative for slide images. Step 2: List the Observed Data 31 Include total number of offspring Gray body, red eyes 1,159 Yellow body, white eyes 1,017 Gray body, white eyes 17 Yellow body, red eyes 12 Total 2,205 Data from the cross is shown in the table Total offspring equals 2,205 Therefore, the expected number of each phenotype (combining males and females) is 1/4 × 2,205 = 551  32 Step 3: Based on the hypothesis, calculate the expected values of each of the four phenotypes Each phenotype has an equal probability of occurring. Therefore, the probability of each phenotype is 1 ∕ 4 Total offspring (observed F2 generation) = 2,205 The expected number of each phenotype (combining males and females) = 1 ∕ 4 × 2,205 = 551 Step 4: Apply the Chi Square Formula 33 Use the observed values in Step 2 and expected values calculated in Step 3 ( O1 - E1 ) ( O2 - E2 ) ( O3 - E3 ) ( O4 - E4 ) 2 2 2 2 χ 2 = + + + E1 E2 E3 E4 (1159 - 551) (17 - 551) (12 - 551) (1017 - 551) 2 2 2 2 χ 2 = + + + 551 551 551 551 χ 2 = 670.9 + 517.5 + 527.3 + 394.1 = 2109.8 Step 5: Determine the Degrees of Freedom 34 u The four phenotypes are based on the law of segregation and the law of independent assortment. u The law of independent assortment predicts four categories (n), recombinant and nonrecombinant. So, n = 4 u Therefore, based on a hypothesis of independent assortment alone, u Degrees of freedom (df) equal n − 1 u So, 4 -1 = 3 u Degrees of freedom (df) = 3 Step 6: Determine the P value for the chi square value calculated in step 4 35 u The P value is obtained using: 1. a chi square table (refer to Table 3.1 on next slide) and 2. the degrees of freedom calculated in step 5. u The calculated chi square value of 2109.8 (calculated in Step 4) is enormous which means the deviation between observed and expected values is very large u According to the chi square table, with 3 degrees of freedom, such a large deviation is expected to occur by chance alone less than 1% of the time Chi Square Values and Probability (Table 3.1) 36 Degrees of Null Hypothesis Null Hypothesis P 0.99 0.95 0.80 0.50 0.20 Freedom Rejected 0.05 Rejected 0.01 1 0.000157 0.00393 0.0642 0.455 1.642 3.841 6.635 2 0.020 0.103 0.446 1.386 3.219 5.991 9.210 3 0.115 0.352 1.005 2.366 4.642 7.815 11.345 4 0.297 0.711 1.649 3.357 5.989 9.488 13.277 5 0.554 1.145 2.343 4.351 7.289 11.070 15.086 6 0.872 1.635 3.070 5.348 8.558 12.592 16.812 7 1.239 2.167 3.822 6.346 9.803 14.067 18.475 8 1.646 2.733 4.594 7.344 11.030 15.507 20.090 9 2.088 3.325 5.380 8.343 12.242 16.919 21.666 10 2.558 3.940 6.179 9.342 13.442 18.307 23.209 15 5.229 7.261 10.307 14.339 19.311 24.996 30.578 20 8.260 10.851 14.578 19.337 25.038 31.410 37.566 25 11.524 14.611 18.940 24.337 30.675 37.652 44.314 30 14.953 18.493 23.364 29.336 36.250 43.773 50.892 u Source: Fisher, Ronald A., and Yates, Frank, Statistical Tables for Biological, Agricultural, and Medical Research. London: Oliver and Boyd, 1943. Step 7: Interpret the calculated chi square value 37 According to the calculations, the P value is less than 0.01 (or 1%) Therefore, we reject the hypothesis that the two genes assort independently Alternatively, we accept the hypothesis that the genes are linked Note that rejecting the null hypothesis does not prove that the linkage hypothesis is correct. The hypothesis is not proven; just accepted