Download 1 - BrainMass

1. The Year of the Ram/Goat began February 1, 2003. In the Chinese calendar it is
Year 4701. Last summer in anticipation of the Chinese New Year, a new breed of
goats was developed. They had brown coats and were piebald (spotted). Males
from this new breed were crossed to true-breeding females that were dilute black.
The F1 males and females were solid black. They were testcrossed (to a triple
mutant) and produced the following 8 classes of male and female baby goats.
(a) Give the genotypes of the two parental goats, showing gene arrangement and gene
order.
(b) Give the % recombination between each pair of genes.
(c) Draw a map showing the positions of all genes and distances between genes, where
appropriate
(d) Give the formula and calculate the coefficient of coincidence
(e) Calculate interference.
2. Given the map a 10 b 20 d , and with no chromosome interference, how may
phenotypically a+b+d individuals would you expect in a progeny of 1000 of the following
cross:
a+ b d+ / a b+ d X a b d / a b d ?
1.
a) for a trihybrid testcross (which is what this is), where the genes are linked, the two
largest phenotypic classes correspond to the genotypes of the parentals.
In this case, d++/d++ and +bp/+bp.
For gene order, look at the two smallest phenotypic classes (+b+, d+p) (the double crossover phenotypes). The gene that is different between the parental genotypes and these
two classes is in the middle.
In this case – taking parental d++, compared to d+p – the p gene has changed
(recombined), taking parental +bp, compared to +b+, again, the p gene has changed,
therefore p is in the middle of the three genes, and the answer is:
Male +pb / +pb, Female d++ / d++
(And therefore the F1 genotype is +pb/d++)
b) For the % recombination (r.f. or recombination frequency) between d and p
we take the number of individuals who had a single cross-over between d and p only (dpb
+ +++), and add the double cross-overs (dp+, ++b) and the divide by the total number of
offspring (1000)
so, r.f. between d and p = (140+145+6+9)/1000 = 0.30 = 30%
Now we do the same for the r.f. between p and b
= (42 + 43 + 6 + 9)/1000 = 0.10 = 10%
For the r.f. between d and b (the “outside” pair of genes), we add all the single
crossovers, plus 2x the sum of the double crossovers.
Therefore r.f. between d and b = (140+145+42+43+2(6+9))/1000 = 0.40 = 40%
c) Map:
d ____________________p________b
30 m.u.
10 m.u.
(note, m.u. = map units = r.f., distance between d and b could also be included (40 m.u.),
also note that the map could be drawn in the other orientation, i.e. b to p to d, left to right)
d)coefficient of coincidence measures any bias towards having single cross-overs at the
expense of double cross-overs (i.e. of getting less double cross-overs than expected.)
formula c.o.c. = observed double crossovers / expected double cross-overs
the observed DxO in this case is 6+9 = 15. To calculate the expected DxO, we multiply
the probability of single crossovers between d and p and between p and b (0.30 and 0.10),
and then multiply by the total number of offspring (1000).
Therefore expected DxO = 30 (0.3 x 0.1 x 1000)
Therefore c.o.c = 15 / 30 = 0.50
e) Interference = 1 – c.o.c.
So in this example, I = 1 – 0.50 = 0.50
2. the cross is a+ b d+ / a b+ d x a b d / a b d
The question asks how many a+ b+ d individuals to expect out of 1000 offspring.
Comparing a+ b+ d to the parental chromosomal genetypes, note that the relationship
between a and b has changed (the mutant and wild-type alleles are on different
chromosomes in the parental genotype, but here they are both wild-type), whereas the
relationship between b and d has not changed (one wild-type and one mutant on the
parental chromosomes, same thing for a+ b+ d).
This means that a+ b+ d individuals have had a cross-over between a and b, but no crossover between b and d.
The probability of a cross-over between a and b is 10% = 0.10
The probability of NO cross-over between b and d is 100% - 30% = 70%
Therefore the probability of a cross-over between a and b AND none between b and d is
0.10 x 0.70 = 0.07
However, this 0.07 (7%) represents the total for a+ b+ d, and its reciprocal a b d+ which
will occur at the same frequency, so we need to divide by 2 to get the frequency of a+ b+
d alone (0.035)
Number of expected a+ b+ d is therefore 0.035 x 1000 = 35
3. Dominant alleles at two different loci affect the tail of mice. These genes are linked
and both are lethal in the embryo when homozygous. One of the tail genes is called
brachyury (T), and these mice have short tails. A third gene, histocompatibility-2 (H-2),
is linked to the tail genes and is concerned with tissue transplantation. Mice that are H2/+ will accept tissue grafts, unlike +/+. A cross was carried out between male T H-2/ +
+ and female + +/+ + mice, and produced the following baby mice:
From contingency tables, determine if the linkage between H-2 and T is significant using
the chi square test.
3. Note that because the mutant alleles are both dominant, crossing to a homozygous
wild-type mouse is effectively a test-cross.
Remember that for a dihybrid test cross, assuming independent assortment (no linkage),
we expect a 1:1:1:1 phenotypic ratio. (this is our null hypothesis for the Chi squared
test).
Therefore we expect 250/4 = 62.5 individuals in each of the 4 phenotypic classes of
offspring.
To calculate the experimental Chi-squared value, we use the following formula:
(#obs - #exp)2 / #exp for EACH of the four phenotypic classes, and then add the four
results together.
So,
Chi-squared = (105-62.5)2 / 62.5 + (115-62.5)2 / 62.5 + (16-62.5)2 / 62.5 + (14-62.5)2 /
62.5
= 143.7
To determine whether linkage is significant, we also need to determine the number of
degrees of freedom, which by rule is equal to the number of classes (4) minus one.
Therefore d.f. = 3 in this case.
To determine significance, you need to refer to a contingency table (there is probably one
in your text, look in the index under chi-sqared).
In this table are critical values for Chi-squared for varies probability (p) values, at various
degrees of freedom.
Usually, we assign significance at a 95% confidence level, which corresponds to a p
value of 0.05.
Find the critical value of Chi-squared for p = 0.05 and d.f. = 3. – the value is 7.815
Because our calculated value for chi-squared (143.7) is higher than the critical value, we
reject the null hypothesis that these genes assort independently, and therefore we can
conclude that the linkage between T and H-2 IS SIGNIFICANT