PROBLEM SET
Chapter 5 -- Sex Determination, Sex Linkage, and Pedigree Analysis

P. 98, #1. In Drosophila, the lozenge phenotype, caused by a sex-linked recessive allele (lz), is of narrow eyes. Diagram to the F2 generation, a cross of a lozenge male and a homozygous normal female. Diagram the reciprocal cross.
A lozenge male is hemizygous in that he has one X-chromosome with the "lz" allele and one Y-chromosome (X-lz,Y). The homozygous normal female has two X-chromosomes with wildtype "+" alleles on each chromosome (X-+,X-+). Thus the parents are 
P generation            X-lz,Y     x      X-+,X-+
                     lozenge male       normal female

the offspring in the F1 generation 
                     X-+, X-lz   wildtype female
                     X-+, Y      wildtype male

The F1 by F1 cross is a sib-sib mating and produces in the F2 the
following genotypes and phenotypes.
         X-+, X-+   wildtype female
         X-+, Y     wildtype male
         X-lz, X-+  wildtype female
         X-lz, Y    lozenge male
Thus in the F2 generation, all of the females are wildtype and the males are 1 wildtype : 1 lozenge.
The reciprocal cross is between a wildtype male (X-+,Y) and a lozenge female (X-lz,X-lz). 
P generation            X-+,Y     x      X-lz,X-lz
                    wildtype male      lozenge female

the offspring in the F1 generation 
                     X-+, X-lz   wildtype female
                     X-lz, Y     lozenge male

The F1 by F1 cross is a sib-sib mating and produces in the F2 the
following genotypes and phenotypes.
         X-+, X-lz  wildtype female
         X-+, Y     wildtype male
         X-lz, X-lz lozenge female
         X-lz, Y    lozenge male
Thus in the F2 generation, the females are 1 wildtype : 1 lozenge, and the males are 1 wildtype : 1 lozenge.

P. 99, #5. The electrophoretic gel below shows activity for a particular enzyme. Lane 1 is a sample from a "fast" homozygote. Lane 2 is a sample from a "slow" homozygote. In lane 3 the blood from the first two was mixed. Lane 4 comes from one of the children of the two homozygotes. Can you guess the structure of the enzyme? If this were an X-chromosome trait, what pattern would you expect from a heterozygous female's a) whole blood b) individual cells? Can you guess the structure of the enzyme?
The structure of the enzyme is a dimer, in which two subunits that are coded by the same locus are combined to produce the functional enzyme. In a heterozygous individual both alleles are expressed and thus two slightly different subunits of the protein are produced, a "fast" form and a "slow" form. When the functional protein is assembled the subunits are assembled randomly. Thus sometimes two "fast" subunits are combined to produce a "fast" enzyme, sometimes two "slow" subunits are combined to produce a "slow" enzyme, and sometimes one "slow" subunit is combined with a "fast" subunit to produce an "intermediate" enzyme. This is why in lane 4 (the heterozygote) you see three bands.
a) whole blood?
If the locus was on the X-chromosome and the individual was a heterozygous female, the pattern that her whole blood would have would be that of lane 3. This is because only one allele is active in each cell and when the active enzyme is assembled from subunits only one type of subunit is available in each cell, thus no intermediate enzymes can be made.
b) individual cells?
As for individual cells, the pattern would be either lane 1 or lane 2, because within each cell only one allele is expressed and thus only one type of subunit is made.

P. 99, #8. How many Barr bodies would you see in the nuclei of perons with the following sex chromosomes? What would the sexes of these persons be? If these were the sex chromosomes of individual Drosophila that were diploid for all other chromosomes, what would their sex be?
In the table below, the human sex is given first and the second line is the sex of a fruit fly. 
  a. XO    0   female  no Y chromosome is present
               male    ratio of X's:sets of autosomes = 0.50

  b. XX    1   female  no Y chromosome is present
               female  ratio of X's:sets of autosomes = 1.00

  c. XY    0   male    Y chromosome is present
               male    ratio of X's:sets of autosomes = 0.50

  d. XXY   1   male    Y chromosome is present
               female  ratio of X's:sets of autosomes = 1.00

  e. XXX   2   female  no Y chromosome is present
               female  ratio of X's:sets of autosomes > 1.00

  f. XXXXX 4   female  no Y chromosome is present
               female  ratio of X's:sets of autosomes > 1.00

  g. XX/XY 1/0 male/female mosaic
               male/female mosaic
P. 99, #9. What are the possible modes of inheritance in the pedigrees below (a-c)? What modes of inheritance are not possible for a given pedigree?

a) In this pedigree note that:
1) only males are affected, thus a difference in expression between the sexes, which indicates a possible x-linked trait and
2) no individual expresses the trait in the 2nd generation, thus the expression of the trait skipped a generation, which indicates that the trait is recessive.
This is most likely an X-linked recessive trait, however, we can not rule out autosomal recessive trait. Because no individual expressed the trait in the 2nd generation we can rule out any dominant trait. Of course, we must assume that the trait is 100% penetrant.

b) In this pedigree note that:
1) both males and females express the trait in about equal numbers, which indicates an autosomal trait and
2) no generations are skipped, which indicates a dominant trait and
3) about half the offspring of an affected parent express the trait.
Thus the most likely choice is an autosomal dominant trait. That the daughter of an affected male does not express the trait rules out X-linked traits, if we assume that the trait is 100% penetrant. That the trait appears in every generation pretty well rules out any recessive trait, though autosomal recessive could be a possibility.

c) In this pedigree note that:
1) the only affected individuals come from unaffected individuals, which indicates a recessive trait and
2) this is further substantiated by the fact that the parents of the affected child are related to each other and
3) as only boys are affected, it could be an x-linked trait but because of the few numbers of offspring it is hard to draw this conclusion.
Thus the trait is a recessive trait on either an autosome or an X-chromosome. All dominant traits are ruled out.

P. 100, #10. In the pedigrees of the rare human traits (a-d, above), including twin production, determine which modes of inheritance are most probable, possible, or impossible.

a) In this pedigree note that:
1) note that individual IV-2 is a male that apparently expresses the trait of twin production as he is married to an unrelated female, which indicates that there is no difference in expression of the trait between the sexes and thus the trait can be either X-linked or on an autosome and
2) while it does not skip any generations, which indicates that is it is likely a dominant trait it could also be a recessive trait.
The most likely candidate is an autosomal dominant trait but all other modes are possible.

b) In this pedigree note that:
1) the trait is expressed in every generation, which indicates that it is probably a dominant trait and
2) II-2 is an affected male who has an affected son, which rules out the possibility of it being an X-linked trait.
Thus the trait is most likely an autosomal dominant trait, but autosomal recessive can not be ruled out. All X-linked traits are ruled out.

c) In this pedigree note that:
1) none of the offspring in the second generation express the trait, thus the trait is a recessive trait and
2) males are much more likely to be affected and that the affected females in the third generation are the result of an affected male and a potentially carrier female, which indicates that the trait is an X-linked recessive trait.
The most likely mode is X-linked recessive, though autosomal recessive is possible. All dominant modes of inheritance are ruled out.

d) In this pedigree note that:
1) no individuals express the trait in the 3rd and 4th generations, which indicates that this is a recessive trait and
2) males and females are equally affected and that V-14 is an affected female from unaffected parents, which indicates that the trait is carried on an autosome.
Thus the only possibility is autosomal recessive and all other modes of inheritance are ruled out.

P. 100, #21. A man with brown teeth mates with a woman with white teeth. They have four daughters, all with brown teeth, and three sons, all with white teeth. The sons all marry women with white teeth, and all their children have white teeth. One of the daughters (A) marries a man with white teeth (B), and they have two brown-toothed daughters, one white-toothed daughter, and one brown-toothed son, and one white-toothed son.
a) explain these observations.
b) Based on your answer to (a), what is the chance that the next child of the A-B couple will have brown teeth?
The first step is to draw out the pedigree.

Examination of the pedigree shows that
1) all of the daughters of a brown-toothed male have brown teeth but all of the sons have white teeth, which indicates that the trait is carried on the X-chromosome and
2) the trait is expressed in every generation, which indicates that it is a dominant trait.
Thus the best explanation is that the trait (brown teeth) is an X-linked dominant trait.
The chance that the next child of the A-B couple will have brown teeth is 0.50. This is because the brown-toothed mother is heterozygous for the trait and thus she has a 0.50 chance of passing the brown-toothed allele on to any of her offspring. As the birth of each child is an independent event, the outcome of previous births have no effect on the outcome of the next birth.

P. 101, #24. In Drosophila, white eye is an X-linked recessive trait and ebony body is an autosomal recessive trait. A homozygous white-eyed female is crossed with a homozygous ebony male. a) What phenotypic ratio do you expect in the F1 generation? b) What phenotypic ratio do you expect in the F2 generation? c) Suppose the initial cross was reversed: ebony female x white-eyed male. What phenotypic ratio would you expect in the F2 generation?
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 X-chromosome. 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 white-eyed 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 and
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 inheritance and
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, though.

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 
P generation
    white-eyed, dark-bodied female  X-w,X-w:tt
    red-eyed, tan-bodied male       X-+,Y:TT
F1 generation
    red-eyed, tan-bodied female     X-+,X-w:Tt
    white-eyed, tan-bodied male     X-w,Y:Tt
F2 generation
    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 mooredwi@esumail.emporia.edu.
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