Answers to quiz 3:
1. Barr bodies are only seen in interphase nuclei and are correlated with the number of X
chromosomes present in the karyotype. If it were a normal male, it would be zero; if it were a
normal female, it would be one. There is also NO relationship between the number of Barr bodies
formed and Down Syndrome- Down Syndrome is an autosomal trisomy, and Barr bodies are a
result of X chromosome dosage compensation. Since the problem states we are examining Barr
bodies in a metaphase cell, the answer would be none. Ans: (a)
2. M(AP[L)E.SU]G4R
In this situation, since the breakage and fusions do not result in loss of gene activity at the break
points, the loss of genetic material would be most detrimental to phenotype. Among the
situations given, the first() is a paracentric inversion, the 2nd []is a pericentric inversion, and the
third is a deletion, which would be the most serious. Ans. (c).
3. This is true, and is depicted in your textbook. Please see figure 14.14.
4. The cross is
P: c bz wx sh d/c bz wx sh d x C Bz Wx Sh D/C Bz Wx Sh D
F1: C Bz Wx Sh D/c bz wx sh d
Backcross: C Bz Wx Sh D/c bz wx sh d x c bz wx sh d/c bz wx sh d
4. There are five genes that are segregating, so we would expect to see 25 or 32 different potential
phenotypic classes. Ans: (e).
The total number of progeny is ~1000. First classify the progeny as to where a crossover occurred
for each type. Then, total the number of crossovers between each pair of genes, and calculate the
observed map units.
Sh-D
5. Notice that a reduction of map units, or crossing-over, is seen in two intervals.
Results like this are suggestive of an inversion. The map distances between C and Bz and
Sh and D are similar to what would be expected. The inversion most
likely involves the Bz, Wx, and Sh genes. Further notice that all those instances in which
crossing-over occurred in the proposed inverted region involved a double crossover.
This is the expected pattern. Ans: none of the above (e).
A number of possible classes are missing: four single-crossover classes
resulting from crossing-over in the inverted region, eight double-crossover
classes involving the inverted region and the non-inverted region, and
triple crossovers and higher. The 10 classes detected were the only classes
that were viable. The viable classes involved a single crossover outside the inverted
region or a double crossover within the inverted region:
Class 1: parental; increased- owing to non-viability of some crossovers
Class 2: parental; increased- owing to non-viability of some crossovers
Class 3: crossing-over between C and Bz; approximately expected frequency
Class 4: crossing-over between C and Bz; approximately expected frequency
Class 5: crossing-over between Sh and D; approximately expected frequency
Class 6: crossing-over between Sh and D; approximately expected frequency
Class 7: double crossover between C and Bz and between Sh and D;approximately expected frequency
Class 8: double crossover between C and Bz and between A and D;approximately expected frequency
Class 9: double crossover between Bz and Wx and between Wx and Sh;approximately expected
frequency
Class 10: double crossover between Bz and Wx and between Wx and Sh;approximately expected
frequency
6. Micro-RNAs fulfill all these criteria- they are trans-acting, i.e. they are synthesized at one
locus and then bind to other molecules, they are processed into single stranded RNAs that interact
with RISC complexes, and some are derived from the introns of protein coding genes. Ans: (d)
7-8. To answer this problem, think about what would be expected in the situation where you had
a monohybrid individual with only two sets of alleles, relative to an individual who was trisomic
and would contain three sets of alleles. If we were carrying three copies of genes on the I
chromosome, we would NOT expect to see a 3:1 ratio for the genes that were segregating. For
example, suppose the Y gene were NOT on the I chromosome.
If not on I, the cross would be: Yy x Yy, or a 3:1 ratio of dominant to recessive phenotypes. This
is the ratio we observe for the Y/y alleles, so this allelic pair cannot reside on chromosome I.
Let us now examine allelic pair R/r. Suppose we had a trisomic plant that was RRr. In each
parent, the gametes that would give rise to diploids would have a 2/3 chance of containing a
dominant allele. We would expect the following ratios to be observed in diploid plants following
self-fertilization:
2/3 R
4/9 RR
1/3 r
2/9 Rr
2/3R
2/9 Rr
1/3 r
1/9 rr
2/3 R
1/3 r
7. Or about a 9:1 ratio of dominant to recessive phenotype in the diploid plants. This is what is
observed for trisomic I in the diploid Rr segregation pattern- therefore the R/r allelic pair must lie
on chromosome B. This is what we see for the R/r gene pair in the I chromosome trisomic, so R
lies on chromosome I. Ans: (d).
8. In examining the data set for the trisomic B parent, we see that all the segregation patterns,
with the exception of the pattern for L/l, show a 3:1 ratio characteristic of monohybrids. The L/l
pair shows a pattern characteristic of trisomy- the L/l allele pair is present on chromosome B.
Ans: AaDdLLlYyRr. (c)
9-10. This is a deletion mapping problem. A series of overlapping deletions will reveal (uncover)
a recessive phenotype if the recessive mutation on the homolog falls within the boundaries of the
deletion:
z+
x y
z
Pseudodominance:
phenotype of deletion
heterozygote is mutant
(-) for x and y; wild type
(+) for z
Start with deletions that uncover the smallest intervals, and build from these:
(d,f,h)
(c,e)
(f,g)
ace
(a,c)
(a,c,d)
dfh
dhf
hdf
dace
or
daceb
(a,b,c,e)
gfhdaceb
becadhfg
9. From comparing deletions 3 and 4, the c must be the middle gene in this three gene
array. Ans. True (a).
10. A similar argument can be made by comparing mutation 2 to these mutations; the d
gene must lie next to a to make dace. In comparing deletions 6 and 7 with this order,
neither f nor d could be the middle gene in the (d,f,h) cluster. The gene order must
therefore be: gfhdaceb or the inverse, becadhfg, since this procedure only places genes in
their relative order- it can’t order them relative to a chromosome structure like the
centromere. Ans. False (b).