Chromosomal nondisjunction definition
During the criss-cross inheritance, the sons inherit the traits of the mother and daughter - a traits of the father. But scientists have noticed that sometimes - enough rarely, but there is a deviation from the scheme of criss-cross inheritance. From the crossing of white-eyed females and red-eyed males of fruit flies in the first generation of progeny, sometimes can be found the flies with parental phenotypes, which contradicts to the scheme of inheritance, when all females must be red-eyed and males white-eyed. Because the eye color trait of the fruit fly is linked to the X-chromosome, the scientists hypothesized, that the appearance of abnormal flies in the offspring is associated with the violation in the process of chromosome disjunction in meiosis. Sometimes can occur nondisjunction of chromosomes, and white-eyed female can produce gametes with two X-chromosomes and gametes without X-chromosome. If a gamete with two X-chromosomes merge with a normal gamete of father with Y-chromosome, then as result we have a female with two X-chromosomes and one Y-chromosome. And if the gamete without X-chromosomes are merged with a normal gamete of the father with X-chromosome, then as result we have a male, who has just one X-chromosome and no Y-chromosome. This assumption is fully confirmed when the researchers examined the chromosomes of these flies.
Nondisjunction of X-chromosomes in drosophila
Genetic calculation. Using the genetic calculator, we can demonstrate nondisjunction of chromosomes and model the described crossing. For criss-cross inheritance, we crossed white-eyed females - X(<w>)X(<w>) and red-eyed males X(<w+>)Y(-). As a result of this cross we have the phenotypes segregation with ratio: 1(50%) XX<w+> (red-eyed females) : 1(50%) XY<w> (white-eyed males). In order to demonstrate the possible chromosomal nondisjunction we should conclude in ampersand characters the notation of parents genotype, like this: &X(<w>)X(<w>)&. By choosing "Gametes for genotype 1" for the results of crossing you will see a combination of gametes, which may be formed with considering of the nondisjunction of chromosomes. This will be a normal gametes with one X-chromosome, as well as gametes with two X-chromosomes and without X-chromosome. Because the chromosomal nondisjunction occurs infrequently, the ratio of gametes in this case will not match reality, as well as the ratio of genotypes and phenotypes. To solve the problem, we must also conclude in ampersand characters the notation male genotype - &X(<w+>)Y(-)&. But from genotype of the male, we need only normal gametes. In order that the program has choose only normal gametes it's necessary to add percent characters in the end of the notation, like this - &X(<w+>)Y(-)%%&. Now, if you choose "Gametes for genotype 2", you will see the gametes, which will give us a male. In order to see the offspring from crosses for problems with considering of chromosomes nondisjunction, it is necessary to look at the results of genotypes rather than phenotypes. Since the frequency of nondisjunction of the chromosomes can not be determined, it is necessary to considering that the ratio in this case does not correspond to reality. Thus, the offspring will be mainly represented by red-eyed females with the genotype X(<w+>)X(<w>) and white-eyed males with genotype X(<w>)Y(-). However, in small amount in the offspring can be also white-eyed females with the genotype X(<w>)X(<w>)Y(-) and red-eyed males with genotype X(<w+>). Individuals with genotype Y(-) - probably lethal, and flies with the genotype X(<w+>)X(<w>)X(<w>) will be superfemales.
Gender determination in drosophila
It is necessary to give some explanation. The fact that, in the some animals, for example in the fruit fly, the gender is determined not by the presence of Y-chromosome, but the ratio of the number of X-chromosomes to the number of haploid sets of autosomes. The normal flies are diploid, ie they have two haploid sets of chromosomes. Thus, males who have one X-chromosome, the ratio is 1/2 = 0.5, and in females who have two X-chromosome is equal to 2/2 = 1. Therefore, flies with the genotype XXY is females, and with genotype XO - the males. Also, using certain procedures may be obtained abnormal flies with one X-chromosome and with three haploid sets of autosomes (1/3 = 0.33). Such flies are called supermales, because they have a very strong manifesting all male characters. Flies with normal - a diploid set of autosomes, but with three X-chromosomes are called superfemales (3/2 = 1.5), because they have a very strong manifesting all female characters. And the flies, which have two X-chromosomes and three haploid sets of chromosomes (2/3 = 0.67), are called intersexes, because they have intermediate characters between male and female. All these abnormal flies - supermales, superfemales, intersexes and males with genotype XO are sterile, it means, that they can not generate the offspring.
Genetic calculation. So, from all these abnormal flies the offspring can give us only flies with genotype X(<w>)X(<w>)Y(-). If cross these white-eyed females with red-eyed male, what progeny we can get? Let us write parents genotypes. Now we need to get only normal gametes, which can give us males and females, so we need add percent haracters to the end of notations like this &X(<w>)X(<w>)Y(-)%% & and &X(<w+> )Y(-)%%&. If you choose "Gametes for genotype" for first or for second genotype, you will see what kinds of gametes they can produce. The male gives the normal gametes and female will give the gametes with one X-chromosome, gametes with one X-chromosome and one Y-chromosome, as well as gametes with one Y-chromosome and gametes with two X-chromosomes. The formation of the last two sorts of gametes - is the result of secondary nondisjunction of chromosomes. The numerical ratio in this case is also will not match reality. The offspring will be mainly represented by red-eyed females with the genotypse X(<w+>)X(<w>) and X(<w+>)X(<w>)Y(-) and white-eyed males with genotype X(<w>)Y(-). In a small amount in the progany will be white-eyed females with genotype X(<w>)X(<w>)Y(-) and red-eyed males with genotype X(<w+>)Y(-) (as a result of secondary nondisjunction). But because of secondary nondisjunction these amount will be higher than in the first case. Individuals with two Y-chromosomes also probably lethal.
In some cases, when X-chromosome are physically bundled, the secondary nondisjunction can reach up to 100%. As an example, can cause the yellow coloration of flies body in the line of the double yellow. These abnormal flies have the genotype X(<y>)X(<y>)Y(-) (yellow body). From crosses of these females with normal males - X(<y+>)Y(-) (gray body) in the offspring all males was gray, and females was yellow. Females &X(<y>)X(<y>)Y(-)%%& could give gametes X, XY, XX, Y. But because the secondary nondisjunction in them equal to 100%, they produce only XX and Y gametes. Thus, ini the progeny will be only yellow females with the genotype X(<y>)X(<y>)Y(-) and gray males with genotype X(<y+>)Y(-).
Nondisjunction of chromosomes in humans
Changes in the number of chromosomes in humans and animals in most cases gives only a negative effect. One of the most common chromosomal mutation that occurs in humans as a result of nondisjunction of autosomes - is Down syndrome. This anomaly is caused by nondisjunction of 21-st chromosome. Among its symptoms include mental retardation, reduced resistance to disease, congenital heart anomalies, thick neck and short, stocky body, as well as the characteristic folds of skin above the eyes, which create a physical resemblance to the people of the Mongoloid race. Most often, children with Down syndrome and other similar anomalies are born to middle-aged women. It is likely that the reason for this is somehow connected with the age of eggs of the mother.
Nondisjunction of X-chromosomes in humans
In humans, it is also possible nondisjunction of X-chromosomes. Using a genetic calculator can show all the possible genotypes of offspring in violation of meiosis for X-chromosome. We know that women have the genotype "XX", and the men the genotype "XY". Let see an example, where the violation occurs during the formation of eggs.
Genetic calculation. Just as in the previous example to demonstrate the possible chromosomal nondisjunction we need each group of chromosomes (genes) in parents genotype notation conclude to ampersand characters like this: &XX&. Thus, from the female genotype will be generated all possible combinations of gametes, but from the male genotype, we need only normal gametes. For this, in the end of notation it is necessary to add percent characters like this : &XY%%&. If you choose "Gametes for genotype" for the first or second genotype, you can see what kind of gametes they can produce. A man will give a normal gamete and female give the gametes with one X-chromosome, as well as gametes with two X-chromosomes and without X-chromosome. The ratio of gametes in this case will not match reality, as well as the ratio of genotypes and phenotypes. Now let's see what kind of the genotypes will be in the offspring. But first we must say that in contrast to the fruit fly, at the human the male gender is largely determined by the presence of Y-chromosome. So, the individuals with genotype Y(-) - do not exist, probably because this mutation is lethal. Individuals with genotype "XY" is a normal man, and with the genotype "XX" - normal women. Individuals with genotype "XXY" is a man with Klinefelter syndrome, and with genotype "XXX" - woman with a Triple-X syndrome. Individuals with one X-chromosome is a woman with Turner syndrome.
As a result of nondisjunction may be also occur an increase of the number of chromosomes, which equal to haploid set, it means, the addition of whole haploid sets of chromosomes to the existing set of chromosomes. This phenomenon is called polyploidy. In the animal world polyploids are rarely observed, but a large number of plants are polyploids.
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