Chapter 7
1
(1 point) From a cross between individuals with the genotypes Cc Dd Ee × cc dd
ee, 1000 offspring were produced. The class that was C- D- ee included 351 individuals.
Are the genes c, d, and e on the same or different chromosomes? Explain.
The genes are not on three different chromosomes. If that were true, eight possible
phenotypes would be produced in approximately equal numbers. Out of 1000 offspring,
there would be about 125 (1000 x 1/8) of each type. 351 is significantly more than 125.
2.
(2 points) Genes a and b are 20 cM apart. An a+ b+/a+ b+ individual was mated
with an a b/a b individual.
(a)
Diagram the cross and show the gametes produced by each parent and the
genotype of the F1.
(see work for answer)
(b)
What gametes can the F1 produce, and in what proportions?
A+B+/ab produces 4 types of gametes in the following proportions.
Recombinant 20%
A+b=10% ab+=10%
Non-recombinant gametes 80%
A+b+=40%
Ab=40%
(c)
If the F1 was crossed to a b/a b individuals, what offspring would be expected,
and in what proportions?
(see work for answer)
(d)
Is this an example of the coupling or repulsion linkage phase?
This is coupling linkage phase as the both dominant alleles of two genes on one
chromosome and the both recessive alleles on the other chromsomes.
(e)
If the F1 were intercrossed, what offspring would be expected and in what
proportions?
(see work for answer)
3.
(2 points) A phenotypically wild-type female fruit fly that was hetero-zygous for
genes controlling body color and wing length was crossed to a homozygous mutant male
with black body (allele b) and vestigial wings (allele vg). The cross produced the
following progeny: gray body, normal wings 126; gray body, vestigial wings 24; black
body, normal wings 26; black body, vestigial wings 124. Do these data indicate linkage
between the genes for body color and wing length? What is the frequency of
recombination? Diagram the cross, showing the arrangement of the genetic markers on
the chromosomes.
Yes the genes are linked as we do not see the normal ratio of progeny. The distance
between the genes can be calculated by determining the recombination frequency.
50/300x100= 16.6% the distance between the two genes is 16.6 m.u (see attached work)
No cross over occurs then only two phenotypes should be seen. That is 50% of the
offspring should be dominant to both traits. And the other 50 should be homozygous
recessive just as in the parents above.
class
phenotypes
(a) Frequency
interference
with
no
(b) Frequency
interference
1
scarlet,spineless
0.3784
0.37
2
ebony
0.3784
0.373
3
scarlet, ebony
0.0616
0.074
4
spineless
0.0616
0.075
5
scarlet,
ebony
0.0516
0.066
6
wild-type
0.0516
0.067
7
scarlet
0.0084
0
spineless,
with
complete
8
spineless, ebony
0.0084
0
4.
(3 points) In tomatoes, tall vine (D) is dominant over dwarf (d), and spherical fruit
shape (P) is dominant over pear shape (p). The genes for vine height and fruit shape are
linked with 20 percent recombination between them. One tall plant (I) with spherical fruit
was crossed with a dwarf, pear-fruited plant. The cross produced the following results:
tall, spherical 81; dwarf, pear 79; tall, pear 22; dwarf spherical 17. Another tall plant with
spherical fruit (II) was crossed with the dwarf, pear-fruited plant, and the following
results were obtained: tall, pear 21; dwarf, spherical 18; tall, spherical 5; dwarf, pear 4.
Diagram these two crosses, showing the genetic markers on the chromosomes. If the two
tall plants with spherical fruit were crossed with each other, that is, I × II, what
phenotypic classes would you expect from the cross, and in what proportions?
(see attached work)
5.
(3 points) The Drosophila genes vg (vestigial wings) and cn (cinnabar eyes) are
located at 67.0 and 57.0, respectively, on chromosome 2. A female from a homozygous
strain of vestigial flies was crossed with a male from a homozygous strain of cinnabar
flies. The F1 hybrids were phenotypically wild-type (long wings and dark red eyes).
(a)
How many different kinds of gametes could the F1 females produce, and in what
proportions?
Vg cn+/vg+ cn (long wings and dark red eyes) F1
Recombinant gametes: 10%
Vg+cn+=5%
Vgcn=5%
Non recombinant gametes 90%
Vg+cn=45%
Vgcn+=45%
(b)
If these females are mated with cinnabar, vestigial males, what kinds of progeny
would you expect, and in what proportions?
Vgcn+/vg+cn F1 x cinnabar,vegistal (vgcn/vgcn)- testcross
vg cn
vg+ cn+ = 5%
vg+ cn+ / vg cn = 5% ---long wings and dark red eyes
vg cn = 5%
vg cn/ vg cn = 5% ----vestigial wings and cinnabar eyes
vg+ cn = 45%
vg+ cn/ vg cn = 45% ---long wings and cinnabar eyes
vg cn+ = 45%
vg cn+/ vg cn = 45% ---vestigial wings and dark red eyes
6.
(3 points) In maize, the genes Pl for purple leaves (dominant over pl for green
leaves), sm for salmon silk (recessive to Sm for yellow silk), and py for pigmy plant
(recessive to Py for normal-size plant) are on chromosome 6, with map positions as
shown:
pl sm
py
45 55 65
Hybrids from the cross Pl sm py / Pl sm py × pl Sm Py / pl Sm Py were testcrossed with
pl sm py / pl sm py plants. Predict the phenotypes of the offspring and their frequencies
assuming (a) no interference and (b) complete interference.
(see attached work)
7.
(4 points) Female Drosophila heterozygous for three recessive mutations e (ebony
body), st (scarlet eyes), and ss (spineless bristles) were testcrossed, and the following
progeny were obtained:
Phenotype
Number
wild-type
67
ebony
8
ebony, scarlet
68
ebony, spineless
347
ebony, scarlet, spineless
78
scarlet
368
scarlet, spineless
10
spineless
54
(a) What indicates that the genes are linked?
Two of the classes (parentals) vastly outnumber the other six classes
(recombinant)
(b) What was the genotype of the original heterozygous females?
St++/+ss e
(c) What is the order of the genes?
St-ss-e
(d) What is the map distance between e and st?
[(145+122) x 1+ (18) x 2]
(e) Between e and ss?
122+18/1000=14.0cM
(f) What is the coefficient of coincidence?
(0.018)/(0.163 x 0.140)=0.789
(g)
Diagram the crosses in this experiment.
st + +/+ ss e females x st ss e/st ss e males
8.
(2 points) A normal woman with a color-blind father married a normal man, and
their first child, a boy, had hemophilia. Both color blindness and hemophilia are due to
X-linked recessive mutations, and the relevant genes are separated by 10 cM. This couple
plans to have a second child. What is the probability that it will have hemophilia? color
blindness? Both hemophilia and color blindness? Neither hemophilia nor color blindness?
chances of hemophilia disease to second child after first child having hemophilia is 50%
as the disease is due to X-linked recessive mutations and distance of separation of
hemophilia gene from color blindness gene is 10cM so recombination frequency is 10%
and non-recombinant frequency is 90%.
b) probability of having child with color blindness will be =90% of 50%=90/100 x
50/100=0.9 x 0.5=0.45.
c) probability of child having both color blindness and hemophilia is 10% of 50%
probability=10/100 x 50/100=0.1 x 0.5=0.05.
d) recombination frequency between color blindness gene and hemophilia gene on
chromosome locus =10cM=10%=0.1,probability of having child with neither color
blindness nor hemophilia will be =0.1-probability of having both color blindness and
hemophilia=0.1-0.05=0.05.