Category: Ecology

Molecular Genetics Problem 10 10. An aneuploid person is obviously female, but her cells have two Barr bodies. What is the probable complement of sex chromosomes in this individual?

This individual probably is XXX.

The individual is a female. Nondisjunction of sex chromosomes produces a variety of aneuploid conditions in humans. Most of these conditions appear to upset genetic balance less than aneuploid conditions involving autosomes. Extra copies of the X chromosome are deactivated as Barr bodies in the somatic cells. Females with trisomy of the X chromosome (XXX), which occurs about once in approximately 1000 live births, are healthy and cannot be distinguished from XX females except by karyotype.

An Example of nondisjunction:

Klinefelter’s syndrome

49 ,XXXXY

This karyotype shows a variant of Klinefelter’s syndrome. Individuals with this syndrome are male, typically with the karyotype 47,XXY. Individuals with Klinefelter’s syndrome exhibit a characteristic phenotype including tall stature, infertility, gynecomastia and hypogonadism. Aneuploidy above one extra chromosome is usually fatal but because of X-inactivation, which “turns off” all but one X chromosome per cell, the effects of 3 extra chromosomes are reduced.

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Molecular Genetics Problem 12 12. About 5% of individuals with Downs syndrome are the result of chromosomal translocation. In most of these cases, one copy of chromosome 21 becomes attached to chromosome 14. How does this translocation lead to children with Down syndrome?

 

The case of having a chromosomal fragment joining to a nonhomologous chromosome is called translocation.

Children who have Down’s Syndrome will either have an extra chromosome 21, thus having a total of 47 chromosomes, or about 5% receive a combined 14-21 chromosome combination. (Chromosome 21 links to #14).

In meiosis, the combined 14-21 chromosome will actually behave as a single chromosome. Should this mutated gamete join a normal one during fertilization there will be three chromosome 21’s in the resulting zygote.

 

 

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Molecular Genetics Problem 13 13. Assume genes A and B are linked and are 50 map units apart. An individual heterozygous at both loci is crossed with an individual who is homozygous recessive at both loci. (a) What percentage of the offspring will show phenotypes resulting from crossovers? (b) If you did not know genes A and B were linked, how would you interpret the results of this cross?

 

a) 50 % because when the loci are at opposite ends of a chromosome they behave as if they were on different chromosomes.

The resulting cross gives results almost identical to a normal Mendelian dihybrid cross, which can be symbolized as follows:

expected genotypes

recombinant genotypes

 

b) You would assume that these genes were on separate chromosomes since they appear to segregate independently even though they are linked.

 

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Molecular Genetics Problem 14 14. In Drosophila, the gene for white eyes and the gene that produces “hairy” wings have both been mapped to the same chromosome and have a crossover frequency of 1.5%. A geneticist doing some crosses involving these two mutant characteristics noticed that in a particular stock of flies, these two genes assorted independently; that is they behaved as though they were on different chromosomes. What explanation can you offer for this observation?

One possibility is that a translocation has occurred where a segment of chromosome containing only one of the 2 genes has moved to a different chromosome.

 

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Molecular Genetics Problem 5 5. 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; gray-purple, 49; black-red, 45. (a) What is the recombination frequency between these genes for body color and eye color? (b) Following up on this problem and problem 4, what fruit flies (genotypes and phenotypes) would you mate to determine the sequence of the body color, wing shape, and eye color genes on the chromosomes?

 

To help see the process write out the genotypes. The genes can be symbolized as follows:

b+ = gray body b = black body pr+= red eyes pr = purple eyes

Genotype of the heterozygous wild type (gray body and red eyes) using Morgan’s method:

 

The genotype of the homozygous recessive individual (used in a testcross) would be:

 

Determine the recombination frequency:

Total flies = 1566

 

(a) The percent recombination is therefore 6%

 

(b) Example: cross a wild type fly heterozygous for body color and wing shape with a fly homozygous recessive for the same traits (black body/curly wings)

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