The first step is to write out the genotypes of the parents and
remember that there are two loci to keep tract of. This problem
then is basically a dihybrid cross with one of the loci on the
An ebony male is homozygous recessive for ebony and hemizygous
wildtype for eye color (X-+,Y:eb,eb). The white-eyed female is
homozygous recessive for white eyes and homozygous wildtype for
body color (X-w,X-w:+,+).
P generation X-+,Y:eb,eb x X-w,X-w:+,+
ebony male white-eyed female
the offspring in the F1 generation
X-w,X-+:+,eb wildtype female
X-w,Y:+,eb white-eyed male
The F1 by F1 cross is a sib-sib mating and produces in the F2 the
following genotypes and phenotypes.
1 X-w,X-w:+,+ 1 white-eyed female
2 X-w,X-w:+,eb 2 white-eyed female
1 X-w,X-w:eb,eb 1 white-eyed, ebony female
1 X-w,X-+:+,+ 1 wildtype female
1 X-w,Y:+,+ 1 white-eyed male
2 X-w,X-+:+,eb 2 wildtype female
2 X-w,Y:+,eb 2 white-eyed male
1 X-w,X-+:eb,eb 1 ebony female
1 X-w,Y:eb,eb 1 white-eyed, ebony male
1 X-+,Y:+,+ 1 wildtype male
2 X-+,Y:+,eb 2 wildtype male
1 X-+,Y:eb,eb 1 ebony male
In summary then
3 wildtype female 3 wildtype male
3 white-eyed female 3 white-eyed male
1 ebony female 1 ebony male
1 white-eyed, ebony female 1 white-eyed, ebony male
Note that the ratio of the X-linked trait is 1 wildtype to 1
for both sexes, just as you would predict for an X-linked trait
(see problem #1), while the ratio of wildtype to
ebony is 3:1, just as you would predict for a monohybrid cross.
The Punnett Square for the cross is given on the left.
P. 102, #26. For the pedigrees (a-c), determine the possible modes
of inheritance for each trait.
a) In this pedigree note that:
1) only males are affected, which indicates an X-linked trait
2) no affected individuals appear in the 4th generation, which
indicates a recessive trait.
Thus the most likely mode of inheritance is X-linked recessive but
an autosomal recessive mode of inheritance can not be ruled out.
b)In this pedigree note that:
1) no generations are skipped, which indicates a dominant more of
2) affected fathers seem to pass the trait only to there
daughters, but this could be due to chance and about half of an
affected mothers children are affected.
Thus, the likely mode of inheritance is X-linked dominant or
autosomal dominant. Autosomal recessive can not be ruled out,
c) In this pedigree note that
1) only males express the trait and that all the sons of an
affected father are affected but none of the daughters.
This is a Y-linked mode of inheritance.
P. 102, #27. A black and orange female cat is crossed with a black
male, and the progeny are:
females: 2 black, 3 orange and black
males: 2 black, 2 orange
Explain the results.
In cats, coat color is carried on the X-chromosome. The black and orange female cat is heterozygous for this trait (X-b,X-o), while the black male is hemizygous for this trait (X-b,Y). In the female progeny, the black females received a black allele from their father and a black allele from their mother, thus they are homozygous black. The orange and black females received an orange allele from their mother and a black allele from their father and they are thus heterozygous for the trait. In the male progeny they are all hemizygous in that they have only one X-chromosome from their mother and one Y-chromosome from their father. The black male offspring got the black allele from their mother and the orange male offspring got the orange allele from their father.
P. 102, #28. Based on the following Drosophila crosses, explain the genetic basis for each trait and determine the genotypes of all individuals:
white-eyed, dark-bodied female x red-eyed, tan-bodied male
F1: females: all red, tan
males: all white, tan
F2: 27 red, tan
24 white, tan
9 red, dark
7 white, dark
No differences between males and females in the F2 generation.
First we need to examine each locus individually. At the locus for eye color, we see a difference between the sexes in the F1 generation. This indicates that this locus is probably carried on the X-chromosome. As the females in the F1 are heterozygous and red-eyed, we can assume that red-eyed is the dominant trait. Thus in
the P generation
X-w,X-w x X-+,Y
white-eyed female x red-eyed male
In the F1 generation
X-w,X-+ red-eyed female
X-w,Y white-eyed female
In the F2 generation, the sib-sib mating of the F1s will yield a 1:1 ratio of white-eyed to red-eyed flies, which is about what we have in the F2 generation (31:36). For body color, there is no difference between the sexes in either the F1 or F2 generations, which indicates that this locus is carried on an autosome. As the F1s are all tan, then tan must be the dominant trait.
Thus in the P generation
tt x TT
dark female x tan male
In the F1 generation
Tt all tan
In the F2 generation the sib-sib mating of the F1s will yield a 3:1 ratio of tan to dark bodied flies, which is about what we see in the F2 generation (51:16).
Putting the two loci together, we have
white-eyed, dark-bodied female X-w,X-w:tt
red-eyed, tan-bodied male X-+,Y:TT
red-eyed, tan-bodied female X-+,X-w:Tt
white-eyed, tan-bodied male X-w,Y:Tt
red-eyed, tan-bodied X-+,X-w:T? or X-+,Y:T?
white-eyed, tan-bodied X-w,X-w:T? or X-w,Y:T?
red-eyed, dark-bodied X-+,X-w:tt or X-+,Y:tt
white-eyed, dark bodied X-w,X-w:tt or X-w,Y:tt
Last updated on 22 August 1996.
Provide comments to Dwight Moore at firstname.lastname@example.org.
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