Lecture 11 Linkage Analysis I

Question 1

Sex is expensive - it comes with a 50% tax on reproduction….

So WHY DO ORGANISIMS HAVE SEX?

Answer 1

To generate NEW COMBINATIONS of genetic material, to improve prospects for survival and reproduction.

Question 2

How does SEX relate to RECOMBINATION

Answer 2

Sex is all about recombination

• literally making new genomes through the random assortment of pre-existing genetic material (with a few MUTATIONS FOR GOOD MEASURE)

And GENETICS is all about making sense of recombination.

Question 3

Dihybrid crosses

What is it?
What are the examples of it?

Answer 3

• (Remember: a monohybrid is a heterozygote for a single gene, e.g A/a)
  Notes: A and a = are 2 alleles
  (forms) of the same gene

DIHYBRID: double heterozygote (e.g. A/a;B/b)
- if the genes are on different chromosomes: A/a; B/b
-if the genes are on the same chromosomes AB/ab
- if the location of the two genes is not known: A/a; B/b

Question 4

1. Mendel - The monohybrid cross
   3 tall: 1 short
2. Dihybrid crosses

Answer 4

Monohybrid Crosses

P = R/R . y/y x r/r . Y/Y
(round, yellow). (wrinkled, yellow)
- Mendel did not know the location of units of inheritance

Gametes = R/y.Y/y
F1 (round, yellow)
- round and yellow are the dominant phenotypes, R and Y are the dominant alleles

Dihybrid Cross (from the previous example)

F1 x F1 <— “selfed” ( one of the major benefits of working with plants)

F2 =
315 round, yellow (9)
108 round, green (3)
101 wrinkled, yellow (3)
32 wrinkled, green (1)
————————————
Total = 556 seeds. (16)

• 9:3:3:1 phenotypic ratio including NEW PHENOTYPIC COMBINATIONS

Question 5

Dihybrid crosses: compound 3:1 ratios

Answer 5

Sum the number of individuals for each trait:
Seed shape:
R, round 315 + 108 = 423 3:1
r, wrinkled 101 + 32 = 133

Seed colour:
Y, yellow 315 + 101 = 416 3:1
y, green 108 + 32 = 140

The 9:3:3:1 ratio is made up of two different 3:1 ratios combined at random

3/4 of the F2 are round ->3/4 of these are yellow ->9/16 round, yellow3/4 of the F2 are round ->1/4 of these are green ->3/16 round, green.

1/4 of the F2 are wrinkled - > 3/4 of these are yellow ->3/16 wrinkled, yellow
1/4 of the F2 are wrinkled ->1/4 of these are green ->1/16 wrinkled, green

Question 6

WHAT ARE MENDEL’S 3 Laws?

Answer 6

1. Equal segregation: gene pairs segregate equally into male and female gametes
2. Law of dominance: one form of a gene masks the other (one allele is dominant, the other recessive)
3. Independent assortment: gene pairs (alleles) on different chromosome pairs segregate independently at meiosis

Is Law #3 always true? Do alleles at different loci always assort independently of one another?

Question 7

Review of Segregation and Recombination:

Answer 7

1. Principle of segregation – diploid organism, two alleles at a locus separate in meiosis, one into each gamete

• Independent assortment: this is the separation process – alleles at one locus act independently of alleles at other loci
• P: AABB x aabb, F1: AaBb What gametes does F1 produce?

New combinations of alleles that may differ from parent
Parental and recombinant gametes

Question 8

Why do Linked genes NOT assort INDEPENDENTLY?

Answer 8

1 * Genes located close together on the same chromosome are called linked genes and belong to the same linkage group

2 * Linked genes travel together in meiosis, arriving at the same destination (i.e. same gamete) and are not expected to assort independently

3 * Sweet peas – Bateson & Punnett early 1900s: purple flowers, long pollen X red flowers, round pollen

4 * All F1 progeny had purple flowers, and long pollen (purple dominant over red, long dominant over round)

5 * F2 not 9:3:3:1 – excess parental phenotypes

6 * Mendel was lucky: all characteristics he examined in peas did assort independently!

Question 9

Explain Crossing over at meiosis I can separate linked genes

Answer 9

1* Genes close together segregate as a unit and are inherited together.

2 * Genes occasionally switch from one homologous chromosome to the other
through crossing over.

3 * Results in the break up of genes that are close together.

4 * Linkage and crossing over result in opposite outcomes:

5. Linkage keeps genes together, crossing over mixes them up.

Question 10

Visualising linked genes & crossovers in Meiosis 1. = 6
process, reciprocal?

Answer 10

1. Chromosomes duplicate to form two sister chromatids.
2. Homologous chromosomes pair up.
3. A crossover swaps DNA strands between two non-sister chromatids of a homologous pair during meiosis I.
4. A crossover swaps DNA strands between two non-sister chromatids of a homologous pair during meiosis I.
5. The alleles will therefore swap positions.
6. A crossover is RECIPROCAL: for every Ab chromatid, there is also an aB chromatid produced.
   → The two classes of recombinants (Ab and aB) are going to be ~equal in number

Question 11

Notation for crosses involving gene linkage? = 6

Answer 11

1. In analysing crosses with linked genes, need to know the genotypes AND the arrangement of genes on chromosomes.

2 * New system of notation:
Consider a cross between an individual homozygous for dominant alleles at two linked loci and another individual homozygous for recessive alleles at
those loci (AABB x aabb)

3 * For linked genes need to write out specific alleles as arranged on homologous chromosome.

4 . One chr has two dominant alleles AB, the homologous chr has two recessive alleles, ab

5 * REMEMBER: two alleles at a given locus are always located on DIFFERENT homologous chromosomes

6 * A and a can never be on the same chr (this implies A and B are allelic, but in fact they are separate genes!)

Question 12

Detecting linkage: The Testcross = 5

Answer 12

1. Parents: AA BB × aa bb F1: Aa Bb (dihybrid)
   —– Now cross dihybrid (Aa Bb) with double homozygous recessive (aa bb)
2. A testcross allows you to consider gamete frequencies in one parent only (the dihybrid) by observing phenotypic ratios in progeny
3. We obtain two classes of gametes and progeny – recombinant (R) and non- recombinant (NR). In what proportions will these be?

4*****If A and B are unlinked, we observe a 1:1:1:1 ratio – reflecting equal frequency of gametesIf A and B are linked, the ratios of the various gametes, and therefore of the progeny, will change!

5*****If A and B are linked, the ratios of the various gametes, and therefore of the progeny, will change!

Think of the two classes of gametes and progeny – recombinant (R) and non-recombinant (NR). In what proportions will there be now?

Question 13

Explain: Complete Linkage vs Independent Assortment (4)

Answer 13

1. Leaves: normal vs mottled
2. Stature: tall vs dwarf
3. Conduct testcross (dihybrid x double recessive)
4. The ratio of phenotypes in progeny tells you about the linkage

Question 14

Progeny ratios from a dihybrid test cross

Answer 14

Learn the process/ equation
page 26 of LECTURE 11.

Question 15

In what proportion of meioses does a crossover take place?

Answer 15

This will determine how many recombinant gametes (and therefore progeny) OVERALL will be produced from the testcross.

Answer 16

1. Genes are linked when they are closely located on the same chromosome, so they do not assort independently (thus disobeying Mendel’s 3rd Law)
2. Testcross to look for linkage: AaBb X aabb If assorting independently, then:

– 50%gametes are parental,or non-recombinant (AB and ab)
– 50%gametes are recombinant (Ab and aB)
– 1:1:1:1(equal parental and recombinants)

If genes are linked, then crossing over occurs and the ratio depends on HOW FREQUENTLY crossing over occurs:

– >50% gametes are parental,or non- recombinant(AB and ab)
– <50% gametes are recombinant (Ab and aB)
– not1:1:1:1(excess of parentals)

3. Crossing over yields two reciprocal products, so two classes are equal in number on average
   (# recombinants Ab = # aB, and # parentals AB = #ab)

Question 17

KEY CONCEPT: Crossing over (CO) produces max. 50% recombinant gametes

Answer 17

1. In meiosis, single CO affects 2 of 4 chromatids, therefore 50% gametes are recombinant, 50% are NR
2. % recombinant gametes = 1⁄2 percentage of meioses in which CO occurs
3. Even if CO occurs every meiosis, only 50% of gametes will be recombinants, so:
   - Freq. of recombinant gametes is always 1⁄2 freq. of CO, max 50%
   - The two reciprocal recombinants are equal in number (half each)

Question 18

For single crossovers, the frequency of recombinant gametes is half the frequency of crossing over because:

Answer 18

For single crossovers, the frequency of recombinant gametes is half the frequency of crossing over because each crossover takes place between only two of the four chromatids of a homologous pair.

So the correct answer is “ each crossover takes place between only two of the four chromatids of a homologous pair.”

Question 19

Recombination frequency between two genes
What is RF and how to Calculate it?

Answer 19

1. Linked genes segregate together and crossing over produces recombination between them
2. Frequency of recombination can be determined from progeny of a testcross as follows:

RF = (number of recombinant progeny/ total number of progeny) x 100%

**What are we calculating here?
RF is directly proportional to the number of meioses that feature a crossover between the two loci of interest. Because meiosis with a single crossover results in 50% recombinant gametes, RF is half the frequency of crossovers.

RF can NEVER BE GREATER THAN 50%.
Otherwise crossing over would be happening in >100% of meioses!!

Question 20

EXAMPLE: But how is everything linked? Quantify RF = 8 steps

Answer 20

1. Gametes from MD/md dihybrid unite with
   gametes from homozygous recessive parent

55+53 = 108 are NR progeny (resemble parents)
15 have new combinations:
8 normal/dwarf
7 mottled/tall

2. If M and D were assorting independently, a testcross would produce 1:1:1:1 (25% each)
3. But, NR»R
4. Therefore genes are not assorting independently
5. When linked genes undergo crossover, >50% of progeny are NR and <50% are recombinant (because CO will happen in <100% meioses).
   RF
6. RF = (number of recombinant progeny/ total number of progeny) x 100%
   = (8+7)/(55+53+8+7) X 100%
   = 12.2%
7. Thus 12.2% of progeny exhibit a new combination of traits not observed in parents (they have recombined)
8. RF can be expressed as a decimal fraction i.e. 0.122

Question 21

In performing a dihybrid cross, you self a double heterozygote (Aa*Bb) and expect to see a 9:3:3:1 ratio in the resulting progeny. What is a good explanation if you do not see this ratio?

Answer 21

If you do not see the expected 9:3:3:1 ratio in the resulting progeny of a dihybrid cross of a double heterozygote (Aa*Bb), a good explanation could be that genes A and B are linked. This means that they are located close together on the same chromosome and tend to be inherited together.

So the correct answer is
** Genes A and B are linked**.

Question 22

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

Answer 22

1AB:1ab.

This is because the alleles of the two genes will always be inherited together, resulting in only two possible combinations: AB and ab.

Question 23

If genes A and B are linked, what is the maximum percentage of recombinant gametes that can be produced if a single crossover occurs during gametogenesis?

Answer 23

50%.

This is because a crossover event involves the exchange of genetic material between two of the four chromatids of a homologous pair. So, out of the four possible gametes produced, two will be recombinant and two will be non-recombinant.

Question 24

From the cross AB/ab X 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 24

In this case, the recombinant progeny are Ab/ab and aB/ab, which total to 17 + 21 = 38.

The total number of progeny is 72 + 68 + 17 + 21 = 178.

Therefore, the recombination frequency is 38/178 = 0.213.

So the correct answer is a. 0.213.