I. Allelic, Genic, and Environmental Interactions

I. Allelic, Genic, and Environmental Interactions
II. Sex Determination and Sex Linkage III. Linkage - Overview: ‘Linkage’ is a pattern of correlated inheritance between traits governed by genes on the same chromosome. Because the genes are part of the same physical entity, they are inherited together rather than independently. INDEPENDENT ASSORTMENT (IA) LINKED A a AB ab B b ab A AB A a a B b B b

III. Linkage - Overview: ‘Linkage’ is a pattern of correlated inheritance between traits governed by genes on the same chromosome. Because the genes are part of the same physical entity, they are inherited together rather than independently. Only ‘crossing-over’ can cause them to be inherited in new combinations. Cross-over products

III. Linkage A. ‘Complete’ Linkage - if genes are immediate neighbors, they are almost never separated by crossing over and are ‘always’ inherited together. The pattern mimics that of a single gene. AABB aabb AB ab X ab AB

III. Linkage A. ‘Complete’ Linkage - if genes are immediate neighbors, they are almost never separated by crossing over and are ‘always’ inherited together. The pattern mimics that of a single gene. AABB aabb AB ab X ab AB Gametes AB ab ab Double Heterozygote in F1; no difference in phenotypic ratios compared to IA F1 AB

III. Linkage A. ‘Complete’ Linkage - if genes are immediate neighbors, they are almost never separated by crossing over and are ‘always’ inherited together. The pattern mimics that of a single gene. AB ab ab F1 x F1 X AB Gametes AB ab AB ab

III. Linkage A. ‘Complete’ Linkage - if genes are immediate neighbors, they are almost never separated by crossing over and are ‘always’ inherited together. The pattern mimics that of a single gene. ab ab 1:1 ratio A:a 1:1 ratio B:b 1:1 ratio AB:ab Test cross X AB ab ab Phenotype Gametes AB AaBb AB – 50% aabb Ab – 50% ab Ab Aabb Ab If it was I.A., also: aB aaBb aB

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage - Crossing over in a region is rare

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage - Crossing over in a region is rare - Crossing over events increase as the distance between genes increases A B C a b c LESS LIKELY IN HERE MORE LIKELY IN HERE

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage - Crossing over in a region is rare - Crossing over events increase as the distance between genes increases - So, the frequency of crossing over (‘CO’) gametes can be used as an index of distance between genes! (Thus, genes can be ‘mapped’ through crosses…) A B C a b c FEWER ‘CO’ GAMETES: Ab, aB MORE ‘CO’ GAMETES: bC, Bc

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage - Crossing over in a region is rare - Crossing over events increase as the distance between genes increases - So, the frequency of crossing over (‘CO’) gametes can be used as an index of distance between genes! (Thus, genes can be ‘mapped’ through crosses…) - How can we measure the frequency of recombinant (‘cross-over’) gametes? Is there a type of cross where we can ‘see’ the frequency of different types of gametes produced by the heterozygote as they are expressed as the phenotypes of the offspring?

TEST CROSS !!! III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage B a b b a b A a

TEST CROSS !!! III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage - So, since crossing-over is rare (in a particular region), most of the time it WON’T occur and the homologous chromosomes will be passed to gametes with these genes in their original combination…these gametes are the ‘parental types’ and they should be the most common types of gametes produced. B a b b a b A a a b A B

TEST CROSS !!! III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage - Sometimes, crossing over WILL occur between these loci – creating new combinations of genes… This produces the ‘recombinant types’ B a b b a b A a a b A B a B A b

TEST CROSS !!! III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage As the other parent only contributed recessive alleles (whether crossing over occurs or not)… B a b b a b A a b a a b A B a B A b

B. ‘Incomplete’ Linkage
TEST CROSS !!! III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage As the other parent only contributed recessive alleles (whether crossing over occurs or not)… SO: the phenotype of the offspring is determined by the gamete received from the heterozygote… B a b b a b A a gamete genotype phenotype ab aabb AaBb AB aaBb aB Aabb Ab a b A B a B A b

B. ‘Incomplete’ Linkage B a b
TEST CROSS !!! III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage B a b b a b A a ALTERNATIVES ‘IA’ LINKAGE gamete genotype phenotype ab aabb AaBb AB aaBb aB Aabb Ab a b LOTS of PARENTALS A B FREQUENCIES EQUAL TO PRODUCT OF INDEPENDENT PROBABILITIES a B FEWER CO’S A b

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Example: How do we discriminate between these two alternatives? Conduct a Chi-Square Test of Independence AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Example: How do we discriminate between these two alternatives? Conduct a Chi-Square Test of Independence - Compare the observed results with what you would expect if the genes assorted independently AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Example: How do we discriminate between these two alternatives? Conduct a Chi-Square Test of Independence - Compare the observed results with what you would expect if the genes assorted independently AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37 The frequency of ‘AB’ should = f(A) x f(B) x N = 55/100 x 51/100 x 100 = 28

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Example: How do we discriminate between these two alternatives? Conduct a Chi-Square Test of Independence - Compare the observed results with what you would expect if the genes assorted independently AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37 The frequency of ‘AB’ should = f(A) x f(B) x N = 55/100 x 51/100 x 100 = 28 The frequency of ‘Ab’ should = f(A) x f(B) x N = 55/100 x 49/100 x 100 = 27 The frequency of ‘aB’ should = f(a) x f(B) x N = 45/100 x 51/100 x 100 = 23 The frequency of ‘ab’ should = f(a) x f(b) x N = 45/100 x 49/100 x 100 = 22

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Example: How do we discriminate between these two alternatives? Conduct a Chi-Square Test of Independence This is fairly easy to do by creating a contingency table: AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37 B b Row Total A 43 12 a 8 37 Col. Total

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Example: How do we discriminate between these two alternatives? Conduct a Chi-Square Test of Independence This is fairly easy to do by creating a contingency table: Add across and down… This gives the totals for each trait independently. AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37 B b Row Total A 43 12 55 a 8 37 45 Col. Total 51 49 100

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Example: How do we discriminate between these two alternatives? Conduct a Chi-Square Test of Independence This is fairly easy to do by creating a contingency table: Then, to calculate an expected value based on independent assortment (for ‘AB’, for example), you multiple ‘Row Total’ x ‘Column Total’ and divide by ‘Grand Total’. 55 x 51 / 100 = 28 AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37 B Exp. b Row Total A 43 28 12 55 a 8 37 45 Col. Total 51 49 100

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Example: Repeat to calculate the other expected values… (This is just an easy way to set it up and do the calculations, but you should appreciate it is the same as the product rule: F(A) x f(B) x N AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37 B Exp. b Row Total A 43 28 12 27 55 a 8 23 37 22 45 Col. Total 51 49 100 55 X x = 55x51

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Compare our observed results with what we would expect if the genes assort independently. AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37 B Exp. b Row Total A 43 28 12 27 55 a 8 23 37 22 45 Col. Total 51 49 100

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Compare our observed results with what we would expect if the genes assort independently. . If our results are close to the expectations, then they support the hypothesis of independence. AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37 B Exp. b Row Total A 43 28 12 27 55 a 8 23 37 22 45 Col. Total 51 49 100

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Compare our observed results with what we would expect if the genes assort independently. . If our results are close to the expectations, then they support the hypothesis of independence. If they are far apart from the expected results, then they refute that hypothesis and support the alternative: linkage. AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37 B Exp. b Row Total A 43 28 12 27 55 a 8 23 37 22 45 Col. Total 51 49 100

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Typically, we reject the hypothesis of independent assortment (and accept the hypothesis of linkage) if our observed results are so different from expectations that independently assorting genes would only produce results as unusual as ours less than 5% of the time… AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37 B Exp. b Row Total A 43 28 12 27 55 a 8 23 37 22 45 Col. Total 51 49 100

B Exp. b Row Total A 43 28 12 27 55 a 8 23 37 22 45 Col. Total 51 49 100 III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage We determine this probability with a Chi-Square Test of Independence. Phenotype Obs Exp (o-e) (o-e)2/e AB 43 28 15 8.04 Ab 12 27 -15 8.33 aB 8 23 9.78 37 22 10.23 X2 = 36.38

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Our X2 = 36.38 First, we determine the ‘degrees of freedom’ = (r-1)(c-1) = 1 B b Row Total A 43 12 55 a 8 37 45 Col. Total 51 49 100

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Our X2 = 36.38 First, we determine the ‘degrees of freedom’ = (r-1)(c-1) = 1 Now, we read across the first row in the table, corresponding to df = 1. The column headings are the probability that a number in that column (diff. between observed and expected) would occur by chance. In our case, it is the probability that our hypothesis of independent assortment (expected values are based on that hypothesis) is true. (and the diff is just due to chance).

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Our X2 = 36.38 Note that larger values have a lower probability of occurring by chance… This should make sense, and the value increases as the difference between observed and expected values increases.

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Our X2 = 36.38 So, for instance, a value of 2.71 will occur by chance 10% of the time.

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Our X2 = 36.38 So, for instance, a value of 2.71 will occur by chance 10% of the time. But a value of 6.63 will only occur 1% of the time... (if the hypothesis is true and this deviation between observed and expected values is only due to chance).

B. ‘Incomplete’ Linkage Our X2 = 36.38
III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Our X2 = 36.38 So, for instance, a value of 2.71 will occur by chance 10% of the time. For us, we are interested in the 5% level. The table value is 3.84. Our calculated value is much greater than this; so the chance that independently assorting genes would yield our results is WAY LESS THAN 5%. Our results are REALLY UNUSUAL for independently assorting genes. 36.38

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Our X2 = 36.38 Our results are REALLY UNUSUAL for independently assorting genes. So, either our results are wrong, or the hypothesis of independent assortment is wrong. If you did a good experiment, then you should have confidence in your results; reject the hypothesis of IA and conclude the alternative – the genes are LINKED.

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage Our X2 = 36.38 Our results are REALLY UNUSUAL for independently assorting genes. So, either our results are wrong, or the hypothesis of independent assortment is wrong. If you did a good experiment, then you should have confidence in your results; reject the hypothesis of IA and conclude the alternative – the genes are LINKED. Of course, there is still that 5% probability that chance is responsible for the deviation you observed, and your conclusion is WRONG.

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage OK… so we conclude the genes are linked… NOW WHAT? AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage OK… so we conclude the genes are linked… NOW WHAT? We map the genes using the knowledge that crossing-over is rare, and the frequency of crossing-over correlates with the distance between genes. AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage OK… so we conclude the genes are linked… NOW WHAT? 1. Crossing-over is rare; so the RARE COMBINATIONS must be the products of Crossing-Over. The OTHERS, the MOST COMMON products, represent the PARENTAL TYPES. AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage OK… so we conclude the genes are linked… NOW WHAT? Crossing-over is rare; so the RARE COMBINATIONS must be the products of Crossing-Over. The OTHERS, the MOST COMMON products, represent the PARENTAL TYPES. This tells us the original arrangement of alleles in the heterozygous parent: AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37 A B a b

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage OK… so we conclude the genes are linked… NOW WHAT? Crossing-over is rare; so the RARE COMBINATIONS must be the products of Crossing-Over. The OTHERS, the MOST COMMON products, represent the PARENTAL TYPES. This tells us the original arrangement of alleles in the heterozygous parent: Segregation without crossing over produces lots of AB and ab gametes. AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37 A B a b

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage OK… so we conclude the genes are linked… NOW WHAT? Crossing-over is rare; so the RARE COMBINATIONS must be the products of Crossing-Over. The OTHERS, the MOST COMMON products, represent the PARENTAL TYPES. 2. The frequency of crossing-over is used as an index of the distance between genes: The other progeny are the products of crossing over, and they occurred 20 times in 100 progeny, for a frequency of Multiply that by 100 to free the decimal, and this becomes 20 map units (CentiMorgans). AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37 A B a b 20 map units

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage So, we used the chi-square test to determine whether genes are on the chromosome. Then, we mapped the distance between genes on the chromosome; all by looking at the frequency of phenotypes in the offspring of a test cross. AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37 A B a b 20 map units

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage So, we used the chi-square test to determine whether genes are on the chromosome. Then, we mapped the distance between genes on the chromosome; all by looking at the frequency of phenotypes in the offspring of a test cross. LIMITATION: Genes that are 50 map units apart (or more) will recombine 50% of the time…. Producing parentals and recombinants in equal frequency… just like independent assortment predicts!!! AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37 A B a b 20 map units

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage So, we used the chi-square test to determine whether genes are on the chromosome. Then, we mapped the distance between genes on the chromosome; all by looking at the frequency of phenotypes in the offspring of a test cross. LIMITATION: Genes that are 50 map units apart (or more) will recombine 50% of the time…. Producing parentals and recombinants in equal frequency… just like independent assortment predicts!!! So, genes that are > 50 mu apart assort independently (even though they are on the same chromosome). AaBb x aabb Offspring Number AB 43 Ab 12 aB 8 37 A B a b 20 map units

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping Suppose you conduct two-point mapping and find: A - C are 23 centiMorgans apart C – B are 10 centiMorgans apart: A C B OR A B C

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping Three Point Test Cross AaBbCc x aabbcc Phenotypic Ratio: ABC = 25 ABc = 3 Abc = 42 AbC = 85 aBC = 79 aBc = 39 abc = 27 abC = 5

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping - combine complementary sets Three Point Test Cross AaBbCc x aabbcc Phenotypic Ratio: ABC = 25 ABc = 3 Abc = 42 AbC = 85 aBC = 79 aBc = 39 abc = 27 abC = 5 ABC = 25 abc = 27 52 ABc = 3 abC = 5 = 8 Abc = 42 aBC = 39 81 AbC = 85 aBc = 79 164

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping - combine complementary sets Three Point Test Cross AaBbCc x aabbcc Phenotypic Ratio: ABC = 25 ABc = 3 Abc = 42 AbC = 85 aBC = 79 aBc = 39 abc = 27 abC = 5 ABC = 25 abc = 27 52 ABc = 3 abC = 5 = 8 Abc = 42 aBC = 39 81 AbC = 85 aBc = 79 164 MOST ABUNDANT ARE PARENTAL TYPES

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping - combine complementary sets Three Point Test Cross AaBbCc x aabbcc Phenotypic Ratio: ABC = 25 ABc = 3 Abc = 42 AbC = 85 aBC = 79 aBc = 39 abc = 27 abC = 5 ABC = 25 abc = 27 52 ABc = 3 abC = 5 = 8 LEAST ABUNDANT ARE “DOUBLE CROSSOVERS” D E F Abc = 42 aBC = 39 81 AbC = 85 aBc = 79 164 d e f MOST ABUNDANT ARE PARENTAL TYPES DCO’s = very rare, and gene in middle switches chromosomes

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping - combine complementary sets Three Point Test Cross AaBbCc x aabbcc Compare Parentals and DCO’s to determine which gene is in the middle… Parentals: AbC aBc DCO’s: abC Abc B and C alleles stay together; A switches and so is in middle. REMAKE YOUR DRAWING TO DEPICT ALLELES AND ORDER: ABC = 25 abc = 27 52 ABc = 3 abC = 5 = 8 LEAST ABUNDANT ARE “DOUBLE CROSSOVERS” b A C Abc = 42 aBC = 39 81 AbC = 85 aBc = 79 164 B a c MOST ABUNDANT ARE PARENTAL TYPES

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping - combine complementary sets Three Point Test Cross AaBbCc x aabbcc So, a “single cross-over” (SCO) between the B and A locus will produce the gametes/phenotypes: BAC and bac – find their frequency: 52… plus add DCO’s (8) = 60/305 = x 100 = 19.7 cM ABC = 25 abc = 27 52 ABc = 3 abC = 5 = 8 LEAST ABUNDANT ARE “DOUBLE CROSSOVERS” b A C Abc = 42 aBC = 39 81 AbC = 85 aBc = 79 164 B a c MOST ABUNDANT ARE PARENTAL TYPES 19.7 cM

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping - combine complementary sets Three Point Test Cross AaBbCc x aabbcc “Coefficient of Coincidence” – are SCO’s occurring independently? If so, then DCO’s = product of SCO’s (0.197) x (0.292) x 305 = 17 We only observed 8. C.O.C. = obs/exp = 8/17 = 0.47 Interference = 1 – c.o.c. = 0.53 ABC = 25 abc = 27 52 ABc = 3 abC = 5 = 8 LEAST ABUNDANT ARE “DOUBLE CROSSOVERS” b A C Abc = 42 aBC = 39 81 AbC = 85 aBc = 79 164 B a c MOST ABUNDANT ARE PARENTAL TYPES 19.7 cM 29.2 cM

Bacterial Genetics

Bacterial Genetics Overview - Domains of Life

Bacterial Genetics Overview - Domains of Life II. Prokaryotic Reproduction A. fission

Bacterial Genetics Overview - Domains of Life II. Prokaryotic Reproduction A. fission = 10 billion cells

Bacterial Genetics Overview - Domains of Life II. Prokaryotic Reproduction A. fission The rapid production of new organisms creates genetic diversity by mutation, alone; even though the rates of mutation are low for any given gene. Consider an average gene mutation rate = 1 x 10-5 (meaning a new mutation is produced in every 100,000 copies… or descendants). In 10 billion (1010) descendants, there would be 105 different mutations at this one gene. This is happening independently across 4000 (4 x 103) genes in the E. coli genome. So, in that population of 10 billion cells, there might be as many as 4 x 108 different genomes. About 1/3 will be “silent” (not change the AA), and many will result in a lethal mutation so they won’t occur. But still….. VARIATION.

Bacterial Genetics Overview - Domains of Life II. Prokaryotic Reproduction A. fission B. “sex” – genetic recombination 1. conjugation

Lederberg and Tatum – 1946 - certain strains of bacteria are able to donate genes to other strains – they have a “fertility factor” (F+). Other strains lack this factor (F-).

Bacterial Genetics Overview - Domains of Life II. Prokaryotic Reproduction A. fission B. “sex” – genetic recombination 1. conjugation Davis demonstrated that cell-cell contact was required…

Figure 8-6 And Cavalli Sforza isolated a strain that would cause genetic change at a very high rate: Hfr (High frequency recombination). He recognized that the acquisition of traits was related to the duration of the conjugation event.

He hypothesized that the time between transfer to the recipient cell was related to the distance between genes. As such, if he interrupted mating at specific intervals, he could use time between trait acquisition as an index of distance between genes.

Bacterial Genetics Overview - Domains of Life II. Prokaryotic Reproduction A. fission B. “sex” – genetic recombination 1. conjugation He isolated different strains that transferred genes in different order, suggesting that the transfer process could begin at different places.

He isolated different strains that transferred genes in different order, suggesting that the transfer process could begin at different places. Figure 8-8b

Figure 8-9 part 1

Figure 8-9 part 2

Figure 8-9 part 3

Figure 8-9 part 4

Figure 8-9 part 5

An integrated Hfr plasmid can revert to a free F+ plasmid, and take chromosomal genes along, too.

An integrated Hfr plasmid can revert to a free F+ plasmid, and take chromosomal genes along, too.
It is now an F’ plasmid. Conjugation can now occur.

Figure 8-10 part 4

Figure 8-10 part 5

Bacterial Genetics Overview - Domains of Life II. Prokaryotic Reproduction A. fission B. “sex” – genetic recombination 1. conjugation 2. transformation – absorption of DNA from the environment.

Figure 8-12 part 1

Figure 8-12 part 2

Figure 8-12 part 3

Figure 8-12 part 4

Figure 8-12 part 5

Bacterial Genetics Overview - Domains of Life II. Prokaryotic Reproduction A. fission B. “sex” – genetic recombination 1. conjugation 2. transformation – absorption of DNA from the environment. 3. viral transduction

Figure 8-14

Figure 8-17

Bacterial Genetics Overview - Domains of Life II. Prokaryotic Reproduction The Use of Bacteria in Recombinant DNA Technology

Bacterial Genetics Overview - Domains of Life II. Prokaryotic Reproduction The Use of Bacteria in Recombinant DNA Technology Plasmid ‘vector’ with ampicillin-resistance gene

Bacterial Genetics Overview - Domains of Life II. Prokaryotic Reproduction The Use of Bacteria in Recombinant DNA Technology Plasmid ‘vector’ with ampicillin-resistance gene Transformation: absorption of plasmids

Bacterial Genetics Overview - Domains of Life II. Prokaryotic Reproduction The Use of Bacteria in Recombinant DNA Technology Plasmid ‘vector’ with ampicillin-resistance gene Transformation: absorption of plasmids Grow on selective media with ampicillin; only bacteria that have absorbed plasmids will grow.

Bacterial Genetics Overview - Domains of Life II. Prokaryotic Reproduction The Use of Bacteria in Recombinant DNA Technology Plasmid ‘vector’ with ampicillin-resistance gene Transformation: absorption of plasmids Grow on selective media with ampicillin; only bacteria that have absorbed plasmids will grow. Fission produces millions of cells in a day that have each plasmid – colonies.

Bacterial Genetics Overview - Domains of Life II. Prokaryotic Reproduction The Use of Bacteria in Recombinant DNA Technology Transfer to a piece of filter paper with a radiolabeled probe specific to the gene in question Show the location of all ampicillin-resistant colonies Take an x-ray to identify colonies that have absorbed the plasmid with the gene of interest. Culture the bacteria, cloning the gene for study.

I. Allelic, Genic, and Environmental Interactions

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