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Chapter 5: Genetics Outline A. How many genes per chromosome? 1. Divide number of genes by number of chromosomes a. Human? (25,000/ 23) b. Drosophila? (14,000/ 4) c. Baker’s yeast? (6,000/ 16) B. 5.1 Linked Genes Do Not Assort Independently 1. Principle of Segregation: each individual diploid organism possesses two alleles at a locus that separate in meiosis, with one allele going into each gamete. 2. Principle of Independent Assortment: the process of separation when the two alleles at a locus act independently of alleles at other loci. 3. Recombination: the sorting of alleles into new combinations. It may create combinations the same as the parents or different combinations. 4. Linked Genes: Genes located close together on the same chromosome that belongs to the same linkage group. They travel together eventually arriving at the same location. They do not assort independently. C.

5.2 Linked Genes Segregate Together and Crossing Over Produces Recombination Between Them 1. Genes occasionally switch from one homologous chromosome to the other through the process of crossing over. a. Results in recombination’s and breaks up the association of genes that are close together on the same chromosome i. If two genes on different chromosomes, alleles assort independently into gametes ii. Linkage: keeps particular genes together 1. If two genes within close vicinity on same chromosome, allelic combinations assorted dependently (called LINKAGE) 2. Linkage can be broken due to recombination between allelic regions on homologous chromosome iii. Crossing Over: mixes them up 1. Breakage and reciprocal rejoining of homologous DNA double helices 2. Linkage only broken when recombination between nonsister chromatids 3. Occurs reciprocally between loci along homologous chromosomes 4. Recombination occurs during prophase of meiosis I  A single crossover meiosis yields two recombinant chromatids and two parental chromatids  Crossover is reciprocal: Parents Ab, aB -> Recombinant AB predicts ab 2. Crossing Over a. Frequency of crossover is the higher the further two loci are away from each other b. We can infer how far loci are from each other by counting the recombined offspring c. 1. Only parental combinations in F2: Complete linkage, loci very close to each other d. 2. Parental > Recombined : Linkage incomplete, on same chromosome, but loci further away from each other e. 3. Parental = Recombined: Different linkage groups (Either far away on same chromosome or on different chromosomes) 3. Complete Linkage Compared with Independent Assortment a. With complete linkage no crossing over occurs and only the gametes of the parents can be produced. No new combinations occur because the genes affecting the two traits are completely linked and are inherited together. b. Nonrecombinant gamete: or parental gametes contain only original combinations of alleles present in the parents c. Nonrecombinant progeny: display the orginial combinations of traits present in the P generation. i. New combinations can only arise if the physical connection between the two genes was broken. d. Recombinant gametes: gametes with new combinations of alleles i. With independent assortmen, nonrecombinant and recombinant gametes are produced in equal proportions. e. Recombinant progeny: the progeny with new combinations of traits formed from recombinant gametes. 4. Crossing Over with Linked Genes a. Genes that exhibit crossing over are independently linked

i.

5.

For closely linked genes, crossing over does not take place in every meiosis. In meiosis when there is no crossing over, only nonrecombinant gametes are produced. ii. In meiosis when there is a crossover only half of the gametes are recombinants and half are nonrecombinants. (Because it only affects two of the four chromatids) 1. Thus the frequency of recombinant gametes is always half the frequency of crossing over, and the maximum proportion is 50%. Calculating Recombination Frequency a. Recombination frequency: the percentage of recombinant progeny produced in a cross.

recombinant frequency = 6.

number of recombinant progeny × 100 % total number of progeny

Coupling and Repulsion a. p= wild type p+= normal phenotype b. Coupling or cis configuration: arrangement in which wild-type alleles are found on one chromosome and mutant alleles are found on the other chromosome.

b +¿ pb ¿ p ¿

+¿

c.

Repulsion or trans configuration: arrangement in which each chromosome contains one wild type and one mutant allele.

p b+¿ +¿b p ¿ ¿ d.

7.

It is essential to know the arrangement of the alleles on the chromosomes to accurately predict the outcome of crosses in which genes are linked. e. Interchromosomal recombination: between genes on different chromosomes. It arises from independent assortment f. Intrachromomsomal recombination: between genes located on the same chromosome. It arises from crossing over. Predicting the Outcome of Crosses with Linked Genes a. Knowing the arrangement of alleles on a chromosome allows us to predict the types of progeny that will result from a cross entailing linked genes and to determine which of these types will be the most numerous. b. To determine the proportions of the types of offspring you need to know the recombinant frequency. It tells you how often the alleles in the gametes appear in new combinations and allows us to predict the proportions of offspring phenotypes that will result from a specific cross with linked genes. i. Example: smooth fruit (t) is recessive to warty fruit (T) and glossy fruit (d) is recessive to dull fruit (D). These genes exhibit a recombination frequency of 16%

T D td × td td 1.

2.

Four types gametes will produced by the heterozygous parent. Two types of nonrecombinant genes (T D and t d) and two types of recombinant genes (T d and t D) The recombinant frequency tells us that 16% of the gametes produced by the heterozygote’s will be recombinants and because there are two types of recombinant gametes so the frequency will be

3.

16 % =8 % 2

or 0.08

All other gametes will be nonrecombinants so their frequency will be 100%- 16%= 84% Because there are two types of nonrecombinants they each should arise with a frequency of

84 % =42 % 2

or 0.42

4. 5.

6.

The other parent is homozygous and therefore only produces a single type of gamete (t d) with the frequency of 100% or 1.0 The progeny from the cross result from the union of two gametes producing four types of progeny. The expected proportion can be determined by using the multiplication rule and multiplying together the probability of each uniting gamete. Testcross progeny with warty and dull fruit will have the frequency 42% (.42

× 1.0) and the genotype TD td 8.

Testing for Independent Assortment a. In some crosses genes are obviously linked because there are clearly more nonrebombinant progeny than recombinant. b. Must answer the question. Is the inheritance of alleles at one locus independent of alleles at a second locus? It yes, genes are assorting independently. It no the genes are probably linked. c. A fast method for testing for independent in genotypes is with a chi square test of independence. (I skipped because he didn’t do it in the notes if interested on pages 116-119 9. Gene Mapping with Recombination Frequencies a. Genetic maps: chromosome maps calculated by using the genetic phenomenon of recombination b. Physical maps: chromosome maps calculated by using physical distances along the chromosome c. Map units: distances on genetic maps; one map unit equals 1% recombination i. Genetic distances are additive. | -------------------15 m.u. |---------------  | C---------10 m.u-------B-5 m.u.-A The most we can tell is which gene lies in the middle, but if we add an additional D gene we get the following frequencies Gene Pair Recombination Frequency (%) A and D 8 B and C 13 C and D 23 i. C and D exhibit the most recombination; therefore the must be the furthest apart, with genes A and B between them. ii. Using recombination frequencies and remembering 1 m.u. is 1% recombination we can add D to the map |-------------------------------------- 23 m.u.----------------------- | | | -------------------15 m.u. |---------------  | D-----8 m.u.-----C---------10 m.u-------B--5 m.u.--A iii. We cannot distinguish between genes on different chromosomes and genes located far apart on the same chromosome. If genes exhibit 50% recombination, the most that can be said about them is that they belong to different groups of linked genes (different linkage groups), either on different chromosomes or far apart on the same chromosome iv. A testcross for two genes that are relatively far apart on the same chromosome tends to underestimate the true physical distance; because the cross does not reveal double crossovers they might take place between the two genes. v. A double crossover arises when two separate crossover events take place between two loci. vi. A single crossover will switch the alleles on homologous chromosomes, but a second crossover will reverse the effects of the first, restoring the original parental combination of alleles and producing only nonrecombinant genotypes in the gametes, although parts of the chromosomes have recombined. 10. Constructing a Genetic Map with Two-Point Testcrosses a. Genetic maps can be constructed by conducting a series of testcrosses between pairs of genes and examining the recombination frequencies between them. b. Two-point testcross: a test cross between two genes (well that was easy!) Example: Suppose we carried out a series of two-point crosses for four genes a, b, c, and d and obtained the following frequencies. Gene loci in testcross Recombination frequency (%) a and b 50

a and c a and d b and c b and d c and d 1. 2. 3. 4.

5.

6.

D.

50 50 20 10 28

To begin constructing a map we see that a and b are 50 m.u. apart and therefore are on separate chromosomes or really far apart. Then we see that a and c are also 50 m.u. apart as well as a and d are 50 m.u. therefore a must be on separate linkage group. Next we see b and c are 20% and therefore separated by 20 map units. And b and d are 10% and 10 m.u. apart. Now to decide which side of b lies d: To decide this we must look at the d and c distance. If it was to the right it would only be 10 m.u. away from c, whereas if t lies to the right it would be 30 m.u. away. Because the recombination frequency of c and d is 28% it makes since for d to be to the left of b. This discrepancy arises from the double crossovers between the two outer genes that goes undetected. The genetic map would be as such i. Linkage group 1 a | ii. Linkage group 2 d b c | | | |--10 m.u.--|-----------20 m.u.---------| |--------------------30 m.u .-------------------|

5.3 A Three-Point Testcross Can Be Used to Map Three Linked Genes 1. A more efficient mapping technique is a three-point testcross. The order of the three genes can be established in a single set of progeny and some double crossovers can usually be detected, providing accurate map distance. 2. Constructing a Genetic Map with the Three-Point Testcross a. Three recessive mutations scarlet eyes (st) are recessive to red eyes (st+), ebony body color (e) is recessive to gray body color (e+), and spineless (ss) is recessive to normal bristles (ss+). They are all linked and located on the third chromosome. b. To map the genes, we need to determine their order on the chromosome and the genetic distances between them. 3. Determining the locations of crossovers a. We have already identified the nonrecombinants, the double crossovers, and that leaves four single crossover progeny. b. To determine where the crossovers to place in these progeny, compare the alleles found in the singlecrossover progeny with those found in the nonrecombinants. 4. Calculating the recombination frequencies a. Next we determine the map distances which are base on the frequencies of recombination. i. Add up all of the recombinant progeny and divide this number by the total number of progeny from the cross, and multiplying by 100%. To determine the map distances properly we must include all crossovers. 1. The crossover between (st) and (ss) consists of two single crossovers and two double crossovers. There are 755 progeny and the frequency between ss and st is

(50+ 52+5 + 3) ×100 %=14.6 % 755 2.

Therefore the distance between st and ss loci can be expressed as 14.6 m.u. The distance between (ss) and (e) can be calculated the same way.

(43+ 41+ 5 + 3) ×100 %=12.2 % 755

This the map distance between ss and e is 12.2 m.u. Finally the distance between the two outer loci st and e is simply the summation of the distances between st and ss and between ss and e (14.6 m.u. + 12.2 m.u. = 26.8 m.u.) We can now draw a map |-----------------------26.8 |m.u.----------------------- | st-------14.6 m.u-------ss-------12.2 m.u-------e Interference and coefficient of coincidence a. Map distances also give us information about the proportions of recombinant and nonrecombinant gametes that will be produced in a cross. b. Theoretically we should be able to calculate the double-recombinant gametes by using the multiplication 3.

5.

rule of probability, by multiplying ss and e, or 0.146

E.

F.

G.

×

0.122 = 0.0178. Multiplying thus by the total

number of progeny gives 13.4, but there was only 8 which is less than expected. c. This calculation assumes that each crossover event is independent, but crossovers are frequently not independent events: the occurrence of one crossover tends to inhibit additional crossovers in the same region of the chromosome, and so double crossovers are less frequent than expected. d. Interference: the degree to which one crossover interferes with additional crossover in the same region. Calculated by 1 minus the coefficient of coincidence e. Coefficient of coincidence: the ratio of observed double crossovers divided by the expected double crossovers f. Interference tells us of the double crossovers expected how much will not actually be observed. g. Sometimes a crossover increases the probability of another crossover taking place nearby and we see more double crossover progeny than expected. 6. Gene Mapping a. Frequency of recombinant gametes is 0-49.9%, if two loci are on the same chromosome = linked) b. Linkage group = genes on same chromosome that cosegregate more than predicted by independent assortment c. A chromosome may contain several linkage groups, especially if known genes on that chromosome are located in clusters d. Independent assortment: i. Two loci generate 50 percent of non-parental allele combinations ii. They may be located on non-homologous chromosomes iii. OR: they may be far apart on the same chromosome = unlinked iv. Additional loci that are linked to one of the unlinked loci >Are they also linked to the other locus? 7. Genetic maps a. Useful in understanding and experimenting with the genome of organisms b. Available for many organisms in the literature and at Web sites c. Maps based on recombination frequencies between phenotypic markers supplemented with molecular markers (e.g. DNA sequence differences) 8. Summary: Recombination a. Recombination between linked genes occurs with same frequency whether alleles are in cis or trans configuration b. Recombination frequency increases with increasing distances between genes c. Recombination frequency is specific for a particular pair of loci d. No matter how far apart two genes may be, the maximum detectable frequency of recombination between any two genes is 50 percent. e. Loci separated by more than 50cM behave as if they were on different chromosomes When does crossover occur? (during prophase of meiosis I pachytene) 1. CROSSOVERS DETECTABLE: a. 1. IN OFFSPRING BY NEW ASSORTMENT OF PHENOTYPES b. 2. IN MEIOTIC CELL BY MICROSCOPY after pachytene Genetic vs. Physical Distance 1. Map distances based on recombination frequencies are not a direct measurement of physical distance along a chromosome because frequencies are not constant for a given number of base pairs 2. Recombination “hot spots” result in an overestimate physical length 3. Low rates of recombination (e.g. in heterochromatin or at centromeres) underestimate actual physical length Problem

1. 2. 3.

Not all marker heterozygosities present in same strain because: Alleles are mutations Mutations arise spontaneously, in different individuals, are not all present on along same chromosome in same individual 4. We have to map multiple locus pairs, with one shared locus between two mappings H. Analysis of trihybrid testcross data 1. 1. Identify pairs of parental and recombinant offspring a. most abundant >> parental (noncrossover) b. intermediate abundance >> single crossovers c. least abundant >> double crossovers 2. 2. Double crossover occurs as product of single frequencies 3. 3. gene order: the parental gene order yields double crossovers by switching middle genes) 4. 4. Calculate RF for single crossovers, adding double crossovers each time 5. 5. Draw map Chapter 7: Outline A.

B.

C.

D.

Needed: Phenotypes = Nutritional needs 1. Wild type: Can produce biomolecules from anorganic nutrients (CHNOPS) (= prototrophic) 2. Mutant: Can’t produce one or several biomolecules (due to mutation in gene(s) that encode enzymes important for biosynthesis) (= auxotrophic) How can DNA be transferred into bacteria? 1. Conjugation: Some bacteria naturally transfer DNA among each other (“bacterial sex”) 2. Transduction: Bacterial virus erroneously package bacterial DNA when they make new virus in their host, take that piece along in next round of infection a. Bacterial virus replicates in host, erroneously packages bacterial DNA when new virus is made, transfers it into next host (remember Hershey experiment) 3. Transformation: Bacteria take up naked DNA from medium (added by a researcher) 4. Conjugation: DNA Transfer that occurs naturally between bacteria via cytoplasmic bridge 5. Transformation: Naked DNA (added to medium surrounding bacteria) is taken up by bacterial recipient (remember Avery experiment) a. Transfer detectable if wild-type DNA transferred into mutant Polyploids 1. Chromosome mutations: variations in the number and structure of chromosomes. They play an important role in evolution 2. Metacentric: The centromere is located approximately in the middle so the chromosome has two arms of equal length 3. Submetacentric: The centromere is displaced toward one end, creating a long arm and a short arm (short arm p, long arm q) 4. Acrocentric: the centromere is near one end, producing a long arm and a knob or satellite at the other end 5. Telocentric: The centromere is at or very near the end of the chromosome

7.2 Chromosome Rearrangement Alter Chromosome Structure 1. Chromosome rearrangement: mutations that change the structure of individual chromosomes 2. Chromsome duplication: a mutation in which part of the chromosome has been double a. AB CDEFG b. tandem duplication: the duplicated region is immediately adjacent to the original segment. i. AB CDEFEFG c. Displaced duplication: the duplicated segment is located some distance from the original segment, either on the same chromosome or on a different one. i. AB CDEFGEF d. Reverse duplication: an inverted duplication i. AB CDEFFEG e. An individual homozygous for a duplication carries the duplication (mutated sequence) on both homologous chromosomes f. An individual heterozygous for duplication has one unmutated chromosome and one chromosome with duplication. The problems arise in chromosome pairing at prophase I of meiosis, because the two chromosomes are not homologous in length.

3.

4.

5.

g. The only change is the presences of additional copies of normal sequences likely due...