Chapter 5: Chromosome Mapping in Eukaryotes Flashcards Preview

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Flashcards in Chapter 5: Chromosome Mapping in Eukaryotes Deck (52):
1

TRUE or FALSE?

Two genes that are separated by 10 map units show a recombination percentage of 10%.

True

One map unit is equal to 1% recombination between two genes; 10 map units would be equal to 10% recombination between the genes.

2

Genes X, Y, and Z are linked. Crossover gametes between genes X and Y are observed with a frequency of 25%, and crossover gametes between genes Y and Z are observed with a frequency of 5%. What is the expected frequency of double crossover gametes among these genes?

1.25%

The probability of a double crossover is the product of the probabilities of the single crossovers: 0.25 x 0.05 = 0.0125, or 1.25%.

3

Assume that the genes from the previous example are located along the chromosome in the order X, Y, and Z. What is the probability of recombination between genes X and Z?

30%

Recombination frequencies between linked genes along a chromosome are additive, so the recombination frequency between genes X and Z is 25 + 5 = 30.

4

The presence of wild-type F2 progeny indicates a dihybrid cross. The converse is also true: If a cross fails to produce wild-type progeny in the F2, it is a monohybrid cross, and the two lines are mutant in the same locus.

Put another way, there is no opportunity for independent assortment to produce a wild-type genotype in the F2 because the two lines fail to complement each other.

5

TRUE or FALSE?

Crossing over during prophase I of meiosis occurs between alleles on sister chromatids

FALSE

Crossing over during meiosis occurs between alleles on nonsister chromatids.

6

Meiosis Facts

Complete linkage results in the formation of only parental gametes.

Parental gametes contain the same combinations of linked genes as found in the parent cell.

Recombinant gametes contain combinations of alleles not found in the parent cell.

If crossing over occurs, half of the gametes formed are parental and the other half are recombinant.

7

If two genes on the same chromosome exhibit complete linkage, what is the expected F2 phenotypic ratio from a selfed heterozygote with the genotype a+b+ ⁄⁄ ab?

3:1

Each parent produces two types of gametes, a+b+ and ab, giving the simple Mendelian ratio of 3 a+b+ : 1 ab.

8

Assume that there are 12 map units between two loci in the mouse and that you are able to microscopically observe meiotic chromosomes in this organism. If you examined 200 primary oocytes, in how many would you expect to see a chiasma between the two loci mentioned above?

48 primary oocytes

9

TRUE or FALSE?

In a gene mapping cross, the term Frac NCO refers to the fraction of gametes that have the same genotypes as the parental gametes.

TRUE

Frac NCO refers to the fraction of gametes that have not undergone crossing over and thus their genotypes reflect those of the parental gametes.

10

the following statements about three linked genes that are spaced very close together along a chromosome are most likely to be true

Interference will be a significant factor in the number of crossovers observed.

Interference effects are more likely when crossovers are confined to a small region.

11

For linked genes A, B, and C, the fraction of single crossovers for genes A and B is 0.1, the fraction of single crossovers for genes B and C is 0.3, and the fraction of double crossovers is 0.03. Which of the following statements is true?

The distance between A and B is less than the distance between B and C.

The frequency of crossover events decreases as the distance between genes decreases.

12

In this case, the F1 is a dihybrid with AB on one chromosome and ab on the other chromosome. You can tell this by looking at the parents. The F1 hybrid got AB from one parent and ab from the other parent (AB/ab). A dihybrid that is AB/ab can produce four possible “output” gametes through meiosis. Output gametes that match one of the input gametes are nonrecombinant (AB or ab). Output gametes with new combinations of alleles are recombinant (Ab or aB).
Its important to realize that its also possible to have a dihybrid who is Ab/aB instead. In this case, one parent was AAbb and the other parent aaBB. A dihybrid who is Ab/aB can also produce four possible "output" gametes through meiosis, but in this case the nonrecombinant gametes are Ab or aB and the recombinant gametes are AB or ab. In Part C, be sure to check the parents of the F1 trihybrid to determine their configuration of alleles before answering the question. Which gametes are nonrecombinant versus recombinant depends how the alleles are organized in the F1 trihybrid.

In a testcross with a homozygous recessive line (tester), the tester will contribute only recessive alleles that do not affect the phenotypes of the testcross progeny. Therefore, such testcrosses allow you to determine the haploid genotype of the gametes produced by the F1 dihybrid.

Recombination occurs when combinations of alleles not found in either parent are placed into gametes during meiosis in the F1 generation. The F1 can be thought of as having been formed from two haploid “input” gametes.

13

TRUE or FALSE?

To construct a mapping cross of linked genes, it is important that the genotypes of some of the gametes produced by the heterozygote can be deduced by examining the phenotypes of the progeny.

To construct a mapping cross of linked genes, it is important that the genotypes of all of the gametes produced by the heterozygote can be deduced by examining the phenotypes of the progeny, taking into consideration that the homozygote produced only recessive gametes. Gametes and their genotypes can never be observed directly.

14

In a three‑point mapping experiment for the genes y‑w‑ec, the following percentages of events are observed: NCO events: 65%; SCO events between y and w: 15%; SCO events between y and ec: 17%; DCO events: 3% What is the map distance between y and ec?

20 map units

The map distance between any two genes is the sum of the percentages of all detectable recombination events between them, so 17 + 3 = 20.

15

For linked genes A, B, and C, the map distance A–B is 5 map units and the map distance B–C is 25 map units. If there are 10 double crossover events out of 1000 offspring, what is the interference?

0.2

The expected frequency of double crossovers would be 0.05 X 0.25 = 0.0125.

The observed double crossover frequency is 10/1000 = 0.010.

The coefficient of coincidence is 0.01/0.0125 = 0.8, so the interference is 1 – 0.8 = 0.2.

So, the interference is 1 - (0.01/0.0125) = 0.2

16

Suppose that 3 tomato genes from Part A did not assort independently, but instead were linked to one another on the same chromosome. Would you expect the phenotypic ratio in the offspring to change? If so, how?

All eight possible phenotypes could occur, but a greater proportion of the offspring would have the parental phenotypes.

Because all three genes are linked, it is more likely that the parental allele combinations would stay together rather than be recombined through a crossover event. That is why a greater proportion of the offspring would have parental phenotypes. Nevertheless, some crossing over would likely occur, which is why a small proportion of the offspring would have recombinant phenotypes.

17

The recombination frequencies between genes can be used to construct a linkage map, as you have just done. The closer two genes are to each other on the same chromosome, the less likely they are to be separated by a crossover event, resulting in a lower recombination frequency.

To calculate the distance between m and p, you would add the map distances between m and d and between d and p (12 cM + 5 cM = 17 cM). Noticed that the recombination frequency between m and p is slightly less than that sum. This is because of double crossover events--the times that crossovers occur both between m and d and between d and p. In a double-crossover event, the second crossover effectively “cancels out” the first, reducing the number of recombinants between m and p that are observed, while contributing to the number of recombinants between each of the other two pairs of genes. Therefore, adding the smaller map distances to calculate larger distances avoids inaccuracies due to double-crossover events that produce non-recombinant genotypes.

18

The results of the Stern experiment supported the general idea that _______.

genetic recombination is a result of physical exchange between homologous chromosomes

Whenever recombinant phenotypes occurred, the cytological markers indicated that a physical exchange between the X chromosomes had also occurred.

19

The steps described in this tutorial are a reliable way to map loci with a testcross:

Determine the input gametes that formed the F1 dihybrid, trihybrid, or other hybrid.
Examine the testcross progeny to determine the output gametes produced by the F1.
Determine if the output gametes are recombinant or nonrecombinant for all pairs of loci.
Calculate recombination frequencies between loci by calculating the proportion of the recombinant gametes for all pairs of loci.
Use the recombination frequencies to assemble a recombination (linkage) map. Add smaller map distances to calculate larger intervals, avoiding inaccuracies due to double-crossover events that produce nonrecombinant genotypes.

Note that this example had all 3 loci linked on one chromosome, but this may not always be the case. A 3-point testcross with only two linked loci would give 2 locus pairs with a recombination frequency of 50% and one pair with a frequency less than 50%. If all 3 loci were unlinked, all three frequencies would be 50%. In all cases, however, the steps outlined in this tutorial will work to map the loci.

20

The discernible difference in length between the two X chromosomes of the female fruit fly was important in Stern’s experiments because _______.

it allowed cytological detection of physical exchange between the chromosomes

The differences in structure between the two chromosomes allowed Stern to track the inheritance of recombinant and nonrecombinant X chromosomes.

21

Stern observed all of the following results in his experiment.

Recombinant phenotypes were associated with physically re-arranged X chromosomes. An X chromosome of normal length was the result of such a rearrangement.

the number of car, B+ male offspring was roughly equal to the number of car+, B male offspring
the number of males was roughly equal to the number of females
one of the recombinant phenotypes was associated with an X chromosome of normal length

22

If two lines have mutations in separate loci, then alleles of those loci can assort independently to produce wild-type progeny in the F2. Observing wild-type progeny in the F2 proves that the two lines are mutant in separate loci. Observing the F1 phenotypes alone is not reliable because dominant mutant alleles will produce a mutant F1 phenotype whether the cross involves a single locus or two loci.

Dihybrid crosses of unlinked loci do not always produce a 9:3:3:1 ratio. In some cases, the two loci interact and produce a modified 9:3:3:1 ratio in which certain genotypes share the same phenotype.
Modified dihybrid ratios can be identified in real data sets by multiplying the total number of F2 progeny by commonly found ratios (for example, 9/16 and 3/16) to check how well the data fit a hypothesized ratio.

23

Epistasis occurs when a genotype at one locus masks the phenotypic expression of a genotype at a second locus. In this cross, the doubly mutant genotype in the F2 cannot produce a plant that is both yellow and white as such plants will be one color or the other.
By looking at the F2 ratio, we see 3/16 of the progeny are yellow and 3/16 + 1/16 are white. This means that the doubly recessive class (1/16) has a white phenotype and that the recessive Victoria White genotype is epistatic to the Kansas Yellow mutant genotype. Because it is a homozygous recessive genotype that masks the expression of the second locus, this is an example of recessive epistasis.

Notice that the Victoria White and Kansas Yellow cross produces wild-type plants in the F2. What can you conclude from this result?

Recall that dominant epistasis produces a 12:3:1 ratio in the F2. Does that match the observed ratio? See Hint 1 if you need help with this question.

Genetic analyses of this type generally follow the same pattern, and they can be solved using the approach modeled in this tutorial. Examining the F2 of crosses between different mutant lines reveals whether they are mutant in the same locus (no wild-type progeny in the F2) or different loci (wild-type progeny in the F2). For crosses between unlinked loci, gene interactions can be deduced from modified 9:3:3:1 phenotypic ratios in the F2.

24

Independent assortment

Genes on different chromosomes segregate separately

25

But... What if two genes are on the same chromosome?

??

26

Non-sister chromatid crossover will shuffle alleles

Crossover must occur between 2 genes to get exchange

27

a single crossover between two nonsister chromatids and the gametes subsequently produced.

the exchange separates the alleles and results in recombinant gametes, which are detectable.

28

Morgan’s Fly experiments

Found Expected progeny & Unexpected progeny

Y chromosome does not contribute to somatic phenotype

No crossover occurs in male flies (simplifies experiments)

Crossover occurs on the X and the autosomes in females

Frequency of crossover? Very common

29

Morgan’s Fly experiments: Method

Create pure mutant strain female
Cross to white male, create heterozygotes females.
Cross heterozygote female to mutant male (crossing the F1s)
Observe offspring

The F1 and F2 results of crosses involving the yellow-body, white-eye mutations and the white-eye, miniature-wing mutations. In one cross, 1.3% of the F2 flies (males & females) demonstrate recombinant phenotypes, which express either white or yellow. In another cross, 37.2% of the flies (males & females) demonstrate recombinant phenotypes, which are either miniature or white mutants.

F1 cross: double heterozygote female x double mutant male >>> Non-recombinant offspring

Effect of crossover in the females: recombinant offspring
Recombinant F2 females..

30

Observations

Cross F1s

Observe wt and mutants in F2s

Also observe recombinant genotypes

Sturtevant (Morgan’s student) observed this rate of recombination could be correlated with distance between the genes

1% recombination=1 map unit (mu) = 1CM (Centimorgan)

31

Mapping Distance

Sturtevant observed recombination was correlated with distance, therefore could create a map based on recombination data

Compiled data from multiple crosses between y, w, and m that gave the following recombination percentages:
(1) yellow, white 0.5%
(2) white, miniature 34.5%
(3) yellow, miniature 35.4%

32

What is the highest percentage of crossover gametes that you will observe?

Only a single pair of non-sister chromatids are involved in the exchange

The other 2 chromatids are not involved and enter gamete unchanged

So….if crossover happens 100% of the time between linked genes in meiosis, what will be the OBSERVED recombinant gametes?

If crossover between these 2 loci happened in EVERY tetrad, you would only see 50% recombinant gametes

33

The consequences of a single exchange between two nonsister chromatids occurring in the tetrad stage.

Two noncrossover (parental) and two crossover (recombinant) gametes are produced

34

General Rule

The number of tetrads that are involved in exchange between 2 genes is twice as great as the number of recombinant gametes produced

35

Deducing genotypes from ratios

Genetic analysis works in two directions:
-predict genotypes in offspring
-determine genotypes of parents in cross

Specific expectations, e.g., 1:1:1:1 or 9:3:3:1 can be used to deduce genotypes

Cross example:
Phenotype # observed
A/– ; B/– 310
A/– ; b/b 295
a/a ; B/– 305
a/a ; b/b 290

The observed results are close to 1:1:1:1, allowing the deduction that the tested genotype was a dihybrid.

36

Chi square test pinpoints the probability that ratios are evidence of linkage

Transmission of gametes is based on chance events
-Deviations from 1:1:1:1 ratios can represent chance events OR linkage
-Ratios alone will never allow you to determine if observed data are significantly different from predicted values.
-The larger your sample, the closer your observed values are expected to match the predicted values.

Chi square test measures “goodness of fit” between observed and expected (predicted) results
-This accounts for sample size, or the size of the experimental population

37

Applying the chi square test:
Framing a hypothesis

Null hypothesis: observed values are not different from the expected values
-For linkage studies – no linkage is null hypothesis
-Expect a 1:1:1:1 ratio of gametes

Alternative hypothesis: observed values are different from expected values
-For linkage studies, genes are linked
-Expect significant deviation from 1:1:1:1 ratio

38

Null hypothesis

observed values are not different from the expected values

39

Alternative hypothesis

observed values are different from expected values

40

3 allele cross

Use to determine distance & order between the 3 genes

Used to build genetic maps of gene locations

41

Consequences of a double exchange occurring between two nonsister chromatids.

Because the exchanges involve only two chromatids, two noncrossover gametes and two double-crossover gametes are produced. The photograph illustrates several chiasmata found in a tetrad isolated during the first meiotic prophase stage.

42

double crossover

Two separate events of chromosome breakage and exchange occurring within the same tetrad.

43

Drosophila 3-point mapping cross

Follow 3 loci

Determine # of Non-crossover (NCO), Single crossover (SCO), and Double crossover (DCO)

Determine order of loci

Determine genetic distance between loci

44

3-point mapping cross

NCO: largest # of offspring
DCO: smallest # of offspring

Allele that changes position in DCO is in the middle

45

Remember, double crossovers are least frequent!

double crossovers = least number of offspring

46

Mapping sample problem
Interference

Crossover event in one region inhibits a second crossover event nearby.

Positive interference results in less DCO’s than expected

Negative interference results in more DCO’s than expected

47

_l_______________l_______l_
sc v s
33mu 10mu


DCO (expected) = (.33)(.10) = .033
0.033 x 1000 = 33
DCO (observed) = 24

C (Coefficient of coincidence) =
DCO observed/ DCO expected =
24/33 = .727

I(Interference) =
1-C = 1 - .727 =
.273 (positive interference)

48

Calculate gene distances from recombination frequency

Calculate expected double crossovers from single crossover data

Expected DCO = SCOy-w x SCOw-ec = (1.56% x 4.06%) x 10000= 6.33
Observed DCO = 6

49

Mapping Human Genes

Somatic Cell Hybridization (1960’s)

Mouse + Human ->-> Heterokaryon, Eventually loses all but one or a few human chromosomes ----> Synkaryon

FISH (fluorescent in situ hybridization) method
-Isolate mRNA
-Generate DNA using reverse transcriptase
-Fluorescently label ssDNA, hybridize with metaphase chromosome spread
-The physical location where fluorescent label is observed is location of gene

Lod Score method, pedigree analysis
-Determines linkage of a marker (RFLP) loci and a disease loci
-Provides probability that the 2 loci are linked to one another
-Logarithm of the Odds (Lod) Score, usually considered significant if >3.0 (1000 more likely linked than not linked)

50

Heterokaryon

A somatic cell containing nuclei from two different sources.

51

Synkaryon

The fusion of two gametic or somatic nuclei. Also,in somatic cell genetics,the product of nuclear fusion.

52

Synteny testing

Synteny testing = correlate the presence of particular chromosome with particular gene product

Use panel of somatic cell hybrid cell lines

Determines which chromosome contains particular gene of interest