Download Chromosome mapping and sex linked problems

A woman who is a carrier for hemophilia
marries a man who has hemophilia. Tell
the following:
a. Chances of having a child totally normal
b. Chances of having a boy with hemophilia
c. Chances of having a girl with hemophilia
d. Chances of having a girl who is a carrier
e. They have a girl. What are the chances
she is a carrier?
A woman who is a carrier for red-green
colorblindness marries a man with normal
vision. What are the chances they will:
a. Have a girl that is colorblind?
b. Have a boy that is colorblind?
c. Have a girl that is a carrier?
d. Have a girl that is normal?
e. Have a boy that has normal vision?
f. They have a boy. What are the chances he is
colorblind?
Chromosome mapping
Determine the sequence of the genes
below.
A – B = 34
C – B = 19
A – C = 15
Chromosome mapping
The following distances are determined for several
gene pairs.
a. Determine the sequence of these genes.
b. What is the distance between k and m?
j – k = 12
j–m=9
k–l=6
l – m = 15
The sequence is
k
l j
m or m j l k and the
distance between k and m is 21 map
units.
The following recombination frequencies
were determined. What is the order of
these genes on the chromosome?
a – c = 10; a – d = 30; a – e = 6;
b–c=4
b – d = 16; b – e = 20; c – d = 20
What is the distance between a and b?
The order is
e a c b d and the distance between a
and b is 14.
a. Determine the sequence of the genes
below.
b. What is the distance between c and d?
a – b = 8; a – c = 28; a – d = 25;
b – c = 20; b – d = 33
d a b c is the order. 53 map units apart.
 Determine the sequence of the following
genes:
P – Q = 17 S – T = 9 P – T = 12
P – S = 3 S – Q = 14
What is the distance between T and Q?
P S T Q distance between is T and Q = 5
1. A wild-type fruit fly (heterozygous for gray
body color and normal wings was mated
with a black fly with vestigial wings. The
offspring had the following phenotypic
distribution: wild type, 778; black-vestigial,
785; black-normal, 158; gray-vestigial,
162. What is the recombination frequency
between these genes for body color and
wing type.
2. In another cross, a wild-type fruit fly
(heterozygous for gray body color and red
eyes) was mated with a black fruit fly with
purple eyes. The offspring were as follows:
wild-type, 721; black-purple, 751; graypurple, 49; black-red, 45.
(a) What is the recombination frequency
between these genes for body color and
eye color?
 A space probe discovers a planet inhabited by creatures
who reproduce with the same hereditary patterns as
those in humans. Three phenotypic characters are
height (T = tall, t = dwarf), hearing appendages (A =
antennae, a = no antennae), and nose morphology (S =
upturned snout, s = downturned snout). Since the
creatures were not "intelligent" Earth scientists were able
to do some controlled breeding experiments, using
various heterozygotes in testcrosses.
 a. For a tall heterozygote with antennae, the offspring
were tall-antennae, 46; dwarf-antennae 7; dwarf-no
antennae 42; tall-no antennae 5. Calculate the
recombination frequencies for both experiment.
b. For a heterozygote with antennae and an
upturned snout, the offspring were
antennae-upturned snout 47; antennaedownturned snout, 2; no antennaedownturned snout, 48: no antennaeupturned snout 3. Calculate the
recombination frequencies for both
experiments.
Using the information from the problem
above, a further testcross was done using
a heterozygote for height and nose
morphology. The offspring were tallupturned nose, 40; dwarf-upturned nose,
9; dwarf-downturned nose, 42; talldownturned nose, 9. Calculate the
recombination frequency from these data;
then use your answer from the problem
above to determine the correct sequence
of the three linked genes.