Lecture 12: Linkage Analysis II Flashcards
Last time in Linkage Analysis I…
1 * Linked genes do not assort independently at meiosis
2 * Recombination by crossing over separates linked genes
3 * Recombination frequency is maximum 50% if every meiosis produces a crossover, minimum 0% if very tightly linked
4 * Recombination frequency is most easily determined using testcrosses
5 * The greater the RF, the greater the number of recombinant progeny from a testcross
6 * RF is an important parameter for predicting outcomes of a cross, and for genetic mapping
Why do we want to know the recombination frequency? = 2
- To predict how many recombinants we might expect among progeny in future crosses, and therefore how many progeny we need to screen to recover a desired phenotype (today’s lecture) but more importantly…
- Mapping the location of genes relative to one another – next two lectures “Linkage & Mapping I & II”
The outcome of a testcross and allelic arrangement = 4
- repulsion? coupling?
1 * AB/ab and Ab/aB are phenotypically identical
2 * But the order of alleles on homologous chromosomes is different
3 * “In coupling”: dominant alleles are on the same homologous chr
4* “In repulsion”: dominant alleles are on opposite homologous chr
Predicting the outcome of a cross with linked genes = 3
1* Arrangement of genes tells us types of progeny
2 * Recombination frequency allows us to predict the proportions
of these types
3 * If we know how often alleles appear in new combinations in gametes we can predict the proportions of offspring phenotypes.
Predicting the outcome of a cross with linked genes - Freq. of progeny relates to recombination frequency
EXAMPLE: cucumber fruit (it’s a fruit, not just a vegetable) =14.
EXAMPLE: Example: cucumber fruit (it’s a fruit, not just a vegetable):
1. smooth fruit (t) recessive to warty fruit (T);
2. glossy fruit (d) is recessive to dull fruit (D)
- Cross warty/dull homozygote (TD/TD) with smooth, glossy homozygote (td/td) F1 is wild type warty/dull, double heterozygote TD/td
- If RF between T and D is 16%, how many individuals of each phenotype will result from a testcross? What about a selfed dihybrid?
Freq. of progeny relates to recombination frequency
5 * TD/td parent produces 4 gametes (2NR, 2R)
6 * RF tells us 16% of gametes will be R [i.e. Td and tD will be 16%]
7 * Each R gamete = 8% or 0.08
8 * NR gametes (TD and td) will be
100-16% = 84%
9 * Each NR gamete = 42% or 0.42
10 * Because testcross, other parent has only one type of gamete, td, at 100% or 1.00
11* Multiply probabilities to obtain frequency of progeny
12 * TD/td parent produces 4 gametes (2NR, 2R)
13 * RF tells us 16% of gametes will be R [i.e. Td and tD will be 16%]
14 * Each R gamete = 8% or 0.08
Testing for linkage (non-independent assortment)
- assumptions?
= 4
1 * With independent assortment we expect 1⁄4 of each genotype: 1⁄4 AaBb, 1⁄4 aabb, 1⁄4 Aabb, 1⁄4 aaBb
2 * Why? Because the chance of two independent events both occurring is the product of each event alone, i.e.
- If the probability of progeny being Aa = 1⁄2
- and probability of progeny being Bb = 1⁄2 Then - probability of progeny being AaBb = 1⁄4
- Probability of each single-locus genotype is 1⁄2 i.e. one or
other, neither is favoured or selected against - Genotypes at the two loci are inherited independently i.e. 1⁄2 x 1⁄2 = 1⁄4
Chi-square test:
P > 0.05 suggests chance may be responsible for the deviation between expected and observed values
P < 0.05 suggests that some causative factor other than chance is responsible – there is a statistically significant difference between exp. and obs.
Problems with the Goodness of fit Chi-square (χ2) test?
Better test: Chi-square Test for Independence
- Are these assumptions valid?
- Probability of each single-locus genotype is 1⁄2 i.e. one or other, neither is favoured or selected against
- Genotypes at the two loci are inherited independently
i.e. 1⁄2 x 1⁄2 = 1⁄4 (this is what we are testing!)
- These assumptions are what allow us to calculate the expected values for genotype distributions.
But how can we test for Assumption 2 if we’re not sure about Assumption 1? What if there is incomplete survival of a particular genotype?
The recombination frequency between two linked loci P and S is 20%. You allow a Ps/pS individual to self-fertilise. How many ps/ps individuals would you expect among a total of 500 progeny?
If the recombination frequency between two linked loci P and S is 20%, and a Ps/pS individual is allowed to self-fertilize, the expected proportion of ps/ps individuals among the progeny would be (0.2/2)^2 = 0.01.
This is because the recombination frequency of 20% means that 20% of the gametes produced by the Ps/pS individual will be recombinant (Ps or pS), and 80% will be non-recombinant (PS or ps).
Since ps/ps individuals can only result from the fusion of two ps gametes, the probability of obtaining a ps/ps individual is (0.8/2)^2 = 0.16.
Among a total of 500 progeny, we would therefore expect to see 500 * 0.01 = 5 ps/ps individuals.
So the correct answer is b. 5.
The recombination frequency between two linked loci P and S is 20%.
In a testcross between PS/ps and ps/ps, how many ps/ps individuals would you expect among a total of 500 progeny?
If the recombination frequency between two linked loci P and S is 20%, and a testcross is performed between PS/ps and ps/ps individuals, the expected proportion of ps/ps individuals among the progeny would be 0.5 * (1 - 0.2) = 0.4.
This is because the recombination frequency of 20% means that 20% of the gametes produced by the PS/ps individual will be recombinant (Ps or pS), and 80% will be non-recombinant (PS or ps).
Since all the gametes produced by the ps/ps individual will be ps, half of the progeny will receive a PS gamete from the PS/ps parent and a ps gamete from the ps/ps parent, resulting in a PS/ps genotype.
The other half of the progeny will receive a ps gamete from both parents, resulting in a ps/ps genotype. Among a total of 500 progeny, we would therefore expect to see 500 * 0.4 = 200 ps/ps individuals.
So the correct answer is c. 200.
The recombination frequency between two linked loci P and S is 20%. In a testcross between Ps/pS and ps/ps, how many ps/ps individuals would you expect among a total of 500 progeny?
If the recombination frequency between two linked loci P and S is 20%, and a testcross is performed between Ps/pS and ps/ps individuals, the expected proportion of ps/ps individuals among the progeny would be 0.5 * 0.2 = 0.1.
This is because the recombination frequency of 20% means that 20% of the gametes produced by the Ps/pS individual will be recombinant (PS or ps), and 80% will be non-recombinant (Ps or pS).
Since all the gametes produced by the ps/ps individual will be ps, half of the progeny will receive a Ps or pS gamete from the Ps/pS parent and a ps gamete from the ps/ps parent, resulting in a Ps/ps or pS/ps genotype.
The other half of the progeny will receive a PS or ps gamete from the Ps/pS parent and a ps gamete from the ps/ps parent, resulting in a PS/ps or ps/ps genotype.
Among a total of 500 progeny, we would therefore expect to see 500 * 0.1 = 50 ps/ps individuals.
In tomatoes, tall (D) is dominant over dwarf (d), and smooth fruit (P) is dominant over hairy fruit (p).
A farmer has two tall and smooth tomato plants, A and B. The farmer crosses plants A and B with the same dwarf and hairy plant and classifies the progeny.
page 26.
a) Are the loci that determine plant height and fruit hairiness linked?
b) What are the genotypes of plants A and B?
c) What is the RF between D and P?
d) Why are different proportions of progeny produced when A and B are crossed with the same dwarf Progeny of a hairy plant?
a) Are the loci that determine plant height and fruit hairiness linked?
*YES – not 1:1:1:1
b) What are the genotypes of plants A and B?
*DP/dp and dP/Dp
c) What is the RF between D and P?
*3.8%
d) Why are different proportions of progeny produced when A and B are crossed with the same dwarf Progeny of hairy plant?
*cis/trans
You perform a Chi-square test for genetic independence between genes A and B. You obtain a P-value close to 0.01. What does this signify?
If you perform a Chi-square test for genetic independence between genes A and B and obtain a P-value close to 0.01, this signifies that the observed distribution of progeny would be expected to occur by chance 1% of the time.
In other words, there is only a 1% probability that the observed deviation from the expected distribution could be due to random chance alone.
This suggests that genes A and B are not assorting independently and may be linked
** The observed distribution of progeny would be expected to occur by chance 1% of the time.
In tomatoes, tall (D) is dominant over dwarf (d), and smooth fruit (P) is dominant over hairy fruit (p).
A farmer has two tall and smooth tomato plants, A and B. The farmer crosses plants A and B with the same dwarf and hairy plant and classifies the progeny.
page 25.
a) Are the loci that determine plant height and fruit pubescence linked?
b) What are the genotypes of plants A and B?
c) What is the recombination frequency between D and P?
d) Why are different proportions of progeny produced when A and B are crossed with the same dwarf
pubescent plant?
a) Are the loci that determine plant height and fruit pubescence linked?
Yes, the loci that determine plant height and fruit pubescence appear to be linked. This is suggested by the fact that the proportions of progeny produced when plant A is crossed with the dwarf and hairy plant are very different from those produced when plant B is crossed with the same plant. If the loci were not linked, we would expect to see similar proportions of progeny in both crosses.
b) What are the genotypes of plants A and B?
Plant A is most likely DdPp, while plant B is most likely Ddpp. This is because when plant A is crossed with a dwarf and hairy plant (ddpp), approximately equal numbers of tall and smooth (DdPp), tall and hairy (Ddpp), dwarf and smooth (ddPp), and dwarf and hairy (ddpp) progeny are produced. This suggests that plant A is heterozygous for both traits. In contrast, when plant B is crossed with a dwarf and hairy plant (ddpp), most of the progeny are either tall and hairy (Ddpp) or dwarf and smooth (ddPp). This suggests that plant B is heterozygous for the height trait but homozygous recessive for the fruit pubescence trait.
c) What is the recombination frequency between D and P?
The recombination frequency between D and P cannot be accurately determined from the data provided because it appears that plants A and B have different genotypes.
d) Why are different proportions of progeny produced when A and B are crossed with the same dwarf pubescent plant?
Different proportions of progeny are produced when plants A and B are crossed with the same dwarf pubescent plant because plants A and B have different genotypes. Plant A appears to be heterozygous for both traits (DdPp), while plant B appears to be heterozygous for the height trait but homozygous recessive for the fruit pubescence trait (Ddpp).
