Heredity and Development Second Edition (1972) / Chapter Skim
Currently Skimming:
5 Morgan and Drosophila
5 Morgan and Drosophila
Pages 87-139
The Chapter Skim interface presents what we've algorithmically identified as the most significant single chunk of text within every page in the chapter.
Select key terms on the right to highlight them within pages of the chapter.
From page 87... ...
During this same decade Sutton had suggested that the chromosomes might provide the physical basis for inheritance but those biologists concerned with breeding experiments were unable to appreciate the force of his arguments and data. In 1910 the American geneticist, Thomas Hunt Morgan (1866–1945)
|
From page 88... ...
So Drosophila melanogaster was easy to collect, simple to maintain in the laboratory, and experimentation was most economical -- an important consideration at a period when very little money was available to support scientific work. In fact, Morgan used Drosophila because he was unable to obtain the funds to experiment with rabbits, which are far more expensive to maintain.
|
From page 89... ...
These results could not be explained by the usual Mendelian scheme. The peculiar relation of eye color to sex, with the absence of white females in the F2 of the first cross, suggested that sex chromosomes might be involved.
|
From page 90... ...
This scheme fits the experimental results, but Morgan pointed out that it is necessary to make one assumption about gamete formation in the F1 red males. In these males, which are heterozygous for white eyes, it is necessary to assume that the W gene always goes to the same pole of the spindle with the X and that the w gene never does during the chromosome movements of meiosis.
|
From page 91... ...
If such a male is crossed to a white female, the following results would be expected: Once again, the actual experiment yielded the expected results. Note, however, the assumption that W and X were always together in the same sperm and that no wX sperm could be formed by WwX males.
|
From page 92... ...
All of the females derived from this cross had red eyes and all of the males had white eyes. One might have expected that all of the F1 would have been red-eyed, since the male should have been of the genotype WWX.
|
From page 93... ...
If we assume that W and w are located on the X chromosomes, ‘W' should be interpreted as an X chromosome with the W allele. In the same manner, w will indicate an X chromosome with the w allele.
|
From page 94... ...
Those genes carried on any chromosome except the sex chromosomes are said to be autosomal genes. The inheritance of autosomal genes followed the usual Mendelian scheme.
|
From page 95... ...
The fact that the genetic results exactly paralleled the behavior of the X chromosome was strong evidence that the gene responsible for white eyes is part of the X chromosome. At least many biologists believed the data to be highly suggestive.
|
From page 96... ...
In this instance the problem was to interpret the white eye case in genetic and cytological terms.
|
From page 97... ...
The diploid cells of this species have four pairs of chromosomes. Let us assume that each of the first four mutant genes discovered are located on a different chromosome pair.
|
From page 98... ...
This would be true for all classes of gametes, since they would be combining with a gamete having both recessive genes. (These bl gametes, having only recessive genes, cannot alter the expression of genes in the gametes with which they combine.)
|
From page 99... ...
HEREDITY AND DEVELOPMENT: SECOND EDITION 99 When these results are compared with those of the first cross, the percentages for each phenotype are found to be different, but again we notice a preponderance of the parental types. Both crosses suggest an orderly, though non-Mendelian, mechanism of inheritance.
|
From page 100... ...
Morgan and his associates discovered many new mutants and used them in crosses. Some of the crosses involving two pairs of genes gave the independent assortment expected in the Mendelian scheme.
|
From page 101... ...
The following is an example of inheritance of two linked genes: In Drosophila gray body color (B) is dominant to black body color (b)
|
From page 102... ...
HEREDITY AND DEVELOPMENT: SECOND EDITION 102 This F1 individual will produce four types of gametes, two the result of crossing over and two non-crossovers. Since crossing over occurs in the tetrad stage we could diagram gamete formation as follows: The middle pair of chromatids will break and recombine.
|
From page 103... ...
Subsequent events have shown that Morgan's explanation satisfactorily accounts for the inheritance of genes located on the same chromosome. Many details were added, such as the occurrence of double or triple crossovers, and, for reasons still not understood, the absence of crossing over in Drosophila males.
|
From page 105... ...
Another interesting parallel between genetics and cytology was soon apparent. One deduction we could make from Sutton's hypothesis is this: If genes are on chromosomes and all chromosomes have genes, then the number of groups of linked genes would correspond to the number of pairs of homologous chromosomes.
|
From page 106... ...
The question arises, what determines the normal red eye color? The answer is that the wild type alleles of all of these 13 eye color genes, together with many undiscovered in 1915 when Morgan published his list, act together to produce the wild type red eye color.
|
From page 107... ...
. One of the X chromosomes was in two portions: one portion behaved as an independent chromosome and the other was attached to one of the tiny fourth chromosomes (Fig.
|
From page 108... ...
The other X, which had the piece of the Y attached to it, contained the wild type alleles, C and b, which when homozygous result in red eyes of normal shape. The essential point about this female is that she had two X chromosomes that could be distinguished from one another on both cytological and genetic grounds.
|
From page 109... ...
The carnation-normal flies, alone among the offspring, should have two normal-shaped X chromosomes. The red-bar flies, again alone among the offspring, should have one short X with a piece of Y chromosome attached and an X of normal proportions.
|
From page 110... ...
In this manner, the relative positions of the five genes were determined and a genetic map constructed. The y gene was taken as the reference point and the distances measured from it.
|
From page 111... ...
For this reason Sturtevant suggested that chromosome maps be based on crossover values of genes close to one another and not those far apart. Now the question arises, What is the relation of the chromosome map to the position of these genes on the chromosome?
|
From page 112... ...
The hypothesis that he sought to prove was that ‘sex-linked genes are located on the sex chromosomes.' Normal Inheritance of Sex Chromosomes. In order to understand Bridges' experiments, it is necessary to have clearly in mind the normal inheritance of sex chromosomes in Drosophila.
|
From page 113... ...
During meiosis a normal female produces only one class of ova so far as the sex chromosomes are concerned: those with a single X An XXY female would produce four types of ova during meiosis.
|
From page 114... ...
Bridges did this and found that approximately half of the daughters that he examined had two X chromosomes plus one Y (Fig.
|
From page 115... ...
The remaining females (classes 1 and 7) have two X chromosomes and a Y as shown in b (C.
|
From page 116... ...
The Y came to be looked upon as a nearly inert chromosome genetically in all organisms; a view strengthened by the discovery that in many animal species the males have a single X and no other sex chromosome. These data led to the concept that in species with a XX ♀ -XY ♂ sex chromosome constitution, the presence of a single X determined that the individual be a male and a pair of X chromosomes determined that the individual be a female.
|
From page 117... ...
These individuals were females. Bridges obtained flies with various combinations of autosomes and sex chromosomes (Fig.
|
From page 118... ...
This ratio is higher than that of a normal female, and the resulting imbalance in X chromosomes and autosomes produces a sterile and somewhat abnormal female. Bridges was able to obtain a value below the 0.5 ratio of normal males in individuals with a single X and three autosome sets.
|
From page 119... ...
Since the stock had been under observation for many generations, it is reasonable to assume that in one X chromosome the gene, which in the normal condition is involved in the production of red eyes, changed in such a way as to produce white eyes. One question this suggests is: If the original red eye gene can change to white, might it not change in another way to produce a still different result?
|
From page 120... ...
The distribution of these substances is as follows: BLOOD TYPE ANTIGEN IN CORPUS- ANTIBODY IN PLASMA CLE A A β B B α AB A and B none O none α and β The corpuscles are agglutinated (clumped) if those containing A antigen come in contact with α antibody, or if those containing B antigen come in contact with β antibody.
|
From page 121... ...
In the experiments attempting to produce mutations by physical or chemical means, mutations occurred but they also were very rare. Thus, if we expose flies to radium in an effort to produce mutations, and if a mutant form appears among the offspring or later descendants of the irradiated flies, we could not be sure whether radium was the cause or whether it ‘just happened.' Since new mutant genes appear infrequently and most of them are recessive, the mere detecting of them becomes a problem.
|
From page 122... ...
After considerable experimental manipulation, Muller developed the ClB strain of Drosophila. A ClB ♀ contains the C inversion on one of her X chromosomes, a recessive lethal gene l and a dominant mutant bar-eye B, both of the genes being within the inverted section of the chromosome.
|
From page 123... ...
With ClB flies it is possible to measure the mutation rate for lethal genes on the X of sperm. Once again it must be emphasized that this will not be a measure of rate for one locus, but of all the loci on the X that can mutate to a lethal condition.
|
From page 124... ...
Thus, if a lethal gene was present on the X of the original sperm, there will be no sons in the F2. This fact can be ascertained by a quick examination of the F2 flies.
|
From page 125... ...
Figure 5–9, from Painter's first paper on the subject, shows the appearance of the salivary chromosomes. The salivary glands are diploid, but instead of the expected eight chromosomes, Painter found only four.
|
From page 126... ...
(T. S.Painter, ‘A New Method for the Study of Chromosome Aberrations and the Plotting of Chromosome Maps in Drosophila melanogaster,' Genetics 19:175–88.
|
From page 127... ...
The concepts of chromosome structure previously based solely on the genetic data could be confirmed. Was there a close relation between the genetic map of a chromosome, which was nothing more than a way of describing linkage groups and crossover percentages, and the real chromosome?
|
From page 128... ...
HEREDITY AND DEVELOPMENT: SECOND EDITION 128 5–10 Inversions in Drosophila pseudoobscura. In a the standard band sequence in the third chromosome is shown.
|
From page 129... ...
It was done largely by means of inversions and translocations. Painter had studied the salivary gland chromosomes of normal flies so carefully that he was able to recognize the various regions.
|
From page 130... ...
Most deficiencies, except those that are very small, are lethal when homozygous. In the heterozygous condition they do not cause death, but 5–11 Diagrammatic representation of the experiment of Demerec and Hoover showing how the positions of genes on the salivary gland chromosomes can be determined.
|
From page 131... ...
were close to the end of the X chromosome. Demerec and Hoover made their crosses in such a way that a fly would receive one normal X chromosome carrying the recessive genes y, ac, and sc and another X carrying a deficiency but no mutant genes.
|
From page 132... ...
The genetic map is 5–12 Corresponding points in the salivary chromosome and linkage map for the tip of the second chromosome of Drosophila melanogaster. (Modified from C
|
From page 133... ...
Salivary gland chromosomes are of great importance in many genetic problems being studied today. One of their more spectacular applications has been in the field of evolution.
|
From page 134... ...
1921. ‘Triploid intersexes in Drosophila melanogaster.' Science 54: 252– 54.
|
From page 135... ...
, another autosomal character, is recessive to red eyes (S)
|
From page 136... ...
When you begin with the sex-linked gene remember these points: a. a female has two X chromosomes, which are distributed to both sons and daughters.
|
From page 137... ...
Type O is the homozygous recessive oo. These are autosomal genes.
|
From page 138... ...
27. Let us assume that an organism has a pair of autosomes and a pair of sex chromosomes (♀XX and ♂XY)
|
From page 139... ...
HEREDITY AND DEVELOPMENT: SECOND EDITION 139 gland techniques were known, Bridges examined the region of the X chromosome of normal flies and of bar-eyed flies where the bar locus was thought to be located. This is what he saw (modified from Bridges 1936; Science 83: 210)
|
Key Terms
This material may be derived from roughly machine-read images, and so is provided only to facilitate research.
More
information on Chapter Skim is available.