Chapter 7 - Linkage, Recombo., and Eukaryotic Gene Mapping

Question 1

If a heterozygous parent (AB/ab) is test-crossed to a homozygous recessive (ab/ab) individual, what will the progeny phenotypic rate be if genes A and B are completely linked?

Answer 1

1:1 AB:ab

or 1AB:1ab

Question 2

For single crossovers, the frequency of recombinant gametes is half the frequency of crossing over because
A. Test cross between a homozygote and heterozygote produces 1/2 heterozygous and 1/2 homozygous progeny.
B. The frequency of recombination is always 50%.
C. Each crossover takes place between only two of the four chromatids of a homologous pair.
D. Crossovers occur in about 50% of meioses.

Answer 2

C. Each crossover takes place between only two of the four chromatids of a homologous pair.

Question 3

The % of meiotic cells with a single crossover is ___ the % recombinant gametes.
Given the above, the recombination frequency between the M and D genes is 12%. What % of meiotic cells had a single crossover between the M and D genes?

Answer 3

twice

24%

Question 4

What is the configuration of the alleles in the parental double heterozygote for the following progeny:
TD/td = 42, td/td = 42, Td/td = 8, tD/td = 9

Answer 4

Cis/coupling configuration

Question 5

From the cross AB/ab × ab/ab, what is the recombination frequency if the progeny numbers are 72 AB/ab, 68 ab/ab, 17 Ab/ab, and 21 aB/ab?

Answer 5

Recombinant = 17 + 21 = 38
Total progeny = 72 + 68 + 17 + 21 = 178
Recombo. freq. = (38/178)100 = 21.3%

Question 6

If the progeny numbers are 72 AB/ab, 68 ab/ab, 17 Ab/ab, and 21 aB/ab, are these genes of the parents in cis (coupled) or trans (repulsion)?

Answer 6

Cis (coupled)

Question 7

What is the most likely order of the linked genes R, S, and T if the distance between R and S is 22 m.u., the distance between S and T is 8 m.u., and the distance between R and T is 14 m.u.?

Answer 7

If R ➡️ S = 22; S ➡️ T = 8; R ➡️ T = 14, then

RTS or STR (longest flanking outlier)

Question 8

Given the following information, what genes are linked, and what is their order?
a and b = 50%
a and c = 50%
a and d = 50%
b and c = 20%
b and d = 10%
c and d = 28%

Answer 8

a is on one linkage group

cbd or dbc are on a linkage group together

Question 9

Consider the following three-point cross: ABC/abc x abc/abc. Which of the following is not a possible progeny
genotype?
A. ABC/abc
B. abc/aBC
C. AbC/abc
D. abC/Abc

Answer 9

D. abC/Abc (the homozygous recessive parent has only recessive alleles to give, so one of the gene combos has to be abc - can be on either side, although the WT is “usually” on top, but who knows how this mother fucker is going to ask).

Question 10

Is it possible for two different genes located on the same chromosome to assort independently?
A. No, if two genes are on the same chromosome, they will be linked and the recombination frequency will be less than 50%.
B. Yes, if the two genes are close enough to each other, there are a limited number of crossover events between them.
C. No, there will be a very high crossover interference such that the recombination frequency will be reduced significantly.
D. Yes, if the genes are far enough apart on the same chromosome, a crossover occurs between them in just about every meiotic event.
E. Yes, but only if the two genes are both homozygous.

Answer 10

D. Yes, if the genes are far enough apart on the same chromosome, a crossover occurs between them in just about every meiotic event.

Question 11

A situation where the coefficient of coincidence is greater than 1.0 would indicate that
A. the interference is high and one crossover suppresses the occurrence of a second one.
B. no double crossovers were found in the progeny of a testcross, even though some were expected based on probability.
C. double crossovers were found in the progeny of a testcross, but there were fewer of them than would be expected based uon probability.
D. there were more double crossovers in the progeny than would be expected based on probability
E. the genes involved were actually assorting independently.

Answer 11

D. there were more double crossovers in the progeny than would be expected based on probability

Question 12

Mendel’s 2 principles are the principle of ____ where diploid organisms have two alleles of each gene, and one allele segregates into each gamete; and the principle of ____ assortment where the segregation of alleles for one locus does NOT influence the segregation of allele for another locus.

Answer 12

segregation; independent

Question 13

Genes that are physically close are part of the same ___ ___, or just ____. Note: the ratio of progeny is ____ ____!

Answer 13

linkage group; linked

linked; NOT 9:3:3:1

Question 14

Express the configuration of alleles of genes Aa and Bb which are linked. How are they expressed if they are unlinked (or independent)?

Answer 14

AB/ab (cis or coupled) or Ab/aB (trans or repulsion)

AaBb

Question 15

NS/ns x ns/ns yields what progeny? How about NnSs x nnss?

Answer 15

NS/ns and ns/ns (50% each)

NnSs, nnSs, Nnss, nnss, (50% recombinant, 50% nonrecombinant)

Question 16

Why can’t recombination rates exceed 50%?

Answer 16

Because only two of the 4 chromatids cross over (during prophase 1 synapsis).

Question 17

If crossing over took place during every meiotic division, how would you determine if two genes are on different chromosomes?

Answer 17

You have no way of determining using two-point or three-point mapping.

Question 18

The percent of recombinant gametes is always ____ the percent of cells that underwent a single crossover.

Answer 18

half

Question 19

The percent of cells that underwent a single crossover is always ____ the percent of recombinant gametes.

Answer 19

double/twice

Question 20

If 4% of all cells in meiosis had a single cross over, ___% of gametes would be recombinant. The recombination rate would therefore be __%

Answer 20

2%; 2%

Question 21

If the recombination rate is 5%, what percentage of cells undergoing meiosis had a single cross over?

Answer 21

10%

Question 22

The percent of recombinant gametes is ___ the recombination frequency.

Answer 22

equal to

Question 23

Of 400 meiotic cells, 20 underwent a single crossover. What is the recombination rate? How many gametes are recombinant?

Answer 23

20/400*100 = 5% meiotic cells that had a crossover, so 2.5% recombination rate

20 had a single crossover, yielding 40 recombinant gametes.

Question 24

To determine whether alleles of genes are in the cis or trans positions, look at the positions of the ___ progeny (the most ___ progeny).

Answer 24

nonrecombinant; numerous

Question 25

Define interchromosomal recombination. Include the recombination frequency.

Define intrachromosomal recombination.

Answer 25

Interchromosomal recombination is recombining of different chromosomes (maternal and paternal) to create new genotypes (i.e. independent assortment /Mendelian/ card shuffle); the frequency is always 50%.
Intrachromosomal recombination is recombining of genes located on the same chromosome (crossing over); the recombination frequency is always <50%

Question 26

Crossing (A bl / a Bl) x (a Bl / a bl) you get (a bl / a bl). What kind of recombination occurred?

Answer 26

Intrachromosomal between the first parental to form an (a bl) gamete and an (A Bl) gamete. The (a bl) gamete is then fertilized by the non-recombinant gamete (a bl) of the second parent.

Question 27

One map unit (m.u.) is equal to?

Answer 27

1 cM (centiMorgan); or 1% recombination frequency

Question 28

Recombination between widely separated genes _____ the gene distance due to ____-crossovers (can be a problem for __-___ crosses). What might help to mitigate this problem?

Answer 28

underestimates; double; two-point

A three-point cross

Question 29

Recombination doesn’t occur in Drosophila ____.

Answer 29

males

Question 30

A three-point genetic cross will yield __ different phenotypes of progeny with the two most abundant being ___ or __-____ and the two least abundant resulting from a ___-____; and the four intermediate being ___-_____.

Answer 30

8; parental or non-recombinant; double-crossover; single-crossover.

Question 31

When mapping genes from a three-point cross, it is important to add what?

Answer 31

The single crossovers AND the double crossovers

Question 32

Describe the steps for obtaining recombination frequencies and mapping genes from a 3-pt cross.

Answer 32

Identify the 2 most abundant progeny (non-recombinant) and the 2 least abundant (double-crossover recomb.); compare the genotypes of the non-recombinant and double-crossovers to determine the middle gene (the one that’s different); rewrite/rearrange the genes in the right order; determine the recombination frequencies (single-cross + double-cross)/(total progeny)*100; map the loci.

Question 33

Describe the steps to determine the interference from double-crossovers.

Answer 33

First, determine the expected number of double-crossover progeny (recomb. rate of 1)(recomb. rate 2)(total progeny). Second determine the coeff. of coinc. (total double-crossover progeny)/(expected double-cross progeny). Last, calculate the interference: 1 - coeff. of coinc.

Question 34

Interference during crossing over occurs when…? What does it cause?

Answer 34

The first crossover interferes with the second crossover.

It causes an underestimation of recombination frequencies.

Question 35

Genetic mapping can be done with ____ frequencies, ____ markers, and ____ mapping.

Answer 35

Crossover; genetic markers; physical mapping

Question 36

Restriction Fragment Length Polymorphism (RFLP) is an example of what kind of mapping? What does it use? How does it work? What technology did it lead to?

Answer 36

Molecular marker mapping.
It uses type II restriction endonuclease (bacterial origin).
It works by cutting DNA in a predictable, reproducible way.
RFLP lead to recombinant DNA tech of the 70s.

Question 37

This type of molecular mapping uses short DNA sequences that are repeated in tandem. How are they distinguished from each other?

Answer 37

microsatelites.

Microsatellites have variations in the number of short tandem repeats.

Question 38

The highest resolution molecular marker technique for gene mapping is? What is it, exactly, and what’s a good analogy to describe them? How are they utilized?

Answer 38

Single nucleotide polymorphisms (SNPs) are variations at the nucleotide level (~0.1% or 4Mbp per individual).
They can be described as mile markers.
They’re utilized by comparing the SNPs of normal to diseased persons - there’s a nonrandom association between the disease and SNPs (diseased genes are close to the SNPs).

Question 39

Some drawbacks of crossover frequency mapping is it does not reveal the specific ___ of a ____ group and the distances are not ____ because recombination isn’t completely ____.

Answer 39

chromosome; linkage group; exact; random

Question 40

4 physical genetic mapping techniques include ___ mapping where chromosomes are stained to visually ID chromosomal ____; ___ cell ____ where two cells are fused together to create a “panel”; ___-___ hybridization (e.g. FISH) where a DNA or RNA ____ is tagged with a fluorescent marker; and direct DNA ____.

Answer 40

deletion mapping; deletions; somatic cell hybridization; in-situ; probe; sequencing

Question 41

Why is physical mapping more accurate than genetic mapping?

Answer 41

Both mapping systems will reveal the order of genes, but genetic mapping is less accurate due to recombination hot spots (causes less than equal probability) and to compressions/expansions of chromatids.

Question 42

How does deletion mapping reveal a gene’s location?

Answer 42

When a WT with a deletion is crossed to a homo. rec., the resulting progeny will be WT heterozygous (50%) and homo. rec. (50%) (instead 100% WT) due to the fact that the gene lies in the deleted area. The homo. rec. is actually hemizygous.

Question 43

In somatic cell hybridization, the hybrid cell that has ___ nuclei is called a ____.

Answer 43

two; heterokaryon

Question 44

The Drosophila genes for yellow body (y), white eyes (w) and cut wings (ct) lie on the X- chromosome with w in the middle. The y gene is 2.0 cM from w while the ct gene is 10 cM from w. Out of 1,000 progeny, only 1 was a double-crossovers. What is the interference?

Answer 44

Interference = 1 - Coefficient of interference
Coefficient of interference = actual double-x / expected; expected = (recombo. freq. 1)(recomb. freq 2)(total)
1/(.02)(.1)(1000) = 1/2 or .5
1 - 0.5 = 0.5, or 50% interference expected

Question 45

The basic triploid chromosome number in Chrysanthemum is 36. A new species of Chrysanthemum was found that has 48 chromosomes in its somatic cells. The ploidy of this new species is?

Answer 45

3n = 36; n = 12

48/12 = 4, or 4n

Tetraploid

Question 46

Given the following progeny, what is the gene order?  (write it out!)
bn sl px / bn sl px = 295
\+ + + / bn sl px = 270
bn + px / bn sl px = 10
\+ sl + / bn sl px = 15

Answer 46

bn–sl–px or px–sl–bn

Question 47

Which of the following is false concerning microsattelites (MSs)?
A. A single nucleotide change in the recognition sequence for a restriction endonuclease is sufficient to create an MSs.
B. Both the insertion or deletion of DNA sequences would create new MSs.
C. To be useful, MSs must be correlated with a known phenotype.
D. MSs can be used to create “DNA fingerprints” that are unique to each person (unless you are
comparing DNA from zygotic twins).
E. All the above are true.

Answer 47

C. To be useful, RFLPs must be correlated with a known phenotype.

CAUTION!! none of the techniques using molecular marker need phenotypic correlation (RFLP, microsattelite, SNP).

Question 48

Given the following, what are the order of the genes (1800 total progeny)? What are the map distances? What are the recombination frequencies? What is the interference?
ch + cn = 105
\+ + + = 750
\+ b cn = 40
\+ + cn = 4
ch b cn = 753
ch + + = 41 
\+ b + = 102 
ch b + = 5

Answer 48

ch-cn-b or b-cn-ch

ch→cn = 5; cn→b =12; ch→cn = 17
ch + cn = 5%
cn + b = 12%

expected = (.12)(.05)(1800) = 10.8; actual = 9; c of c = 9/10.8 = 0.83; Int = 1 - 0.83 = 0.17, or 17% interference expected