Reciprocal crosses

Fruit fly (Drosophila melanogaster) is the object of many genetic studies. Fruit fly has a short reproduction cycle, gives a sufficiently large offspring and can be easily kept in laboratory conditions. It also has a large number of easily identifiable traits that are inherited by the classical scheme of monohybrid crosses. But among the many traits that scientists have studied, have been discovered the traits, that inheritance was a different from the Mendelian scheme. Such differences was observed in reciprocal crosses. Reciprocal crosses - it is two crosses, differing by who from the parents - male or female bring in zygote dominant or recessive allele.

In the first case brick red-eyed females was crossed with white-eyed males, and in the second case is vice versa, white-eyed females was crossed with brick red-eyed males. In the first case all offspring in the first generation had a brick-red eyes, and in the second-generation was obtained the offspring with phenotype ratio - 3/4 (75%) of flies with brick-red eyes and 1/4 (25%) of flies with white eyes. It is shown that the brick-red eyes is dominant trait and white eyes - recessive. And the offspring ratio in the second generation - 3:1, is correspond to the classic monohybrid cross. But it was not quite true. The fact that only males had white eyes, and the ratio of male and female flies with brick-red eyes was 2:1. In the second case the offspring in the first generation was not the same. All females had brick-red eyes, and all the males white eyes. The ratio of males and females were equal - 1:1. Such inheritance was named criss-cross inheritance, when sons inherit the traits of the mother and daughter - the traits of the father. In the second generation of this cross was obtained the offspring with ratio 1:1 - 50% of flies with brick-red eyes and 50% of flies with white eyes, instead of 3:1, as we might expect for a classic monohybrid cross.

X - linkage definition

Only after comparing these results with characteristics of set of chromosomes of male and female fruit fly was able to give an explanation to obtained data. Females and males have the four pairs of chromosomes: three pairs of autosomes - the same chromosomes in the male and female, and another different pair of chromosomes. This pair of chromosomes responsible for the gender inheritance. In females, it is represented by two identical X-chromosomes, while males have one X-chromosome and one Y-chromosome. Random combination of these gametes provides an equal ratio of males and females in each generation. Y-chromosome differ from X-chromosome by form and usually Y-chromosome characterized by lack of genes that determine the traits of the organism. Thus, each crossing can be seen as test cross by gender trait.

If the genes, that are responsible for the manifestation of trait, located in the autosomes, the inheritance does not depend on what kind of parents - the female or male is a carrier of this trait. But the situation changes if the traits are determined by genes, which are located in the X - chromosome. Obviously, that eye color trait of the fruit fly is linked to X-chromosome and Y-chromosome does not contain this gene. Using the genetic calculator we can simulate this crossing.

X - linked inheritance examples

Eyes color of fruit fly (Drosophila melanogaster)

Let us introduce the necessary notation. Recessive allele of a gene of eye color we mark as "w" (white eyes), and the dominant allele of wild-type as "w +" (brick-red eyes). And since we mark the alleles with multiple letters, we need to conclud them into this characters "<" and ">" like this - "<w>" and "<w+>". The genotype of the female we mark as the "XX", and the genotype of the male as the "XY". In the genetic calculator to denote the linkage with X-chromosome we should enclose linked genes in the brackets "()" in the parental genotypes notation. Thus the genotype of the females with white eyes can be written as: X(<w>)X(<w>), and the genotype of the female with brick-red eyes as: X(<w+>)X(<w+>) or in the case of hybrid as: X(<w+>)X(<w>). In order to demonstrate the lack of genes in the Y-chromosome you need use the symbols "-" or "_" - for each missing gene. Then the genotype of the male with white eyes can be written like this: X(<w>)Y(-), and the genotype of a male with brick-red eyes as: X(<w+>)Y(-).

X(<w+>)X(<w+>) and X(<w>)Y(-) - it is our parents genotypes for the first crossing. As a result of this cross we get in the first generation the offspring with brick-red eyes. The genotype of the males will be X(<w+>)Y(-), and females X(<w+>)X(<w>). In the second generation from crosses of these males and females, we get the ratio of phenotypes in the offspring: 2 XX<w+> : 1 XY<w+> : 1 XY<w>. That means, 75% ( 2 XX<w+> and 1 XY<w+> ) of flies with brick-red eyes and 25% of flies with white eyes ( 1 XY<w> ), and only the males have white eyes, and the ratio of males and females in flies with brick-red eyes is 2:1.

X(<w>)X(<w>) and X(<w+>)Y(-) - - it is parents genotypes for the second crossing. As a result of this cross we get the phenotypes segregation in the first generation with the ratio: 1(50%) XX<w+> : 1(50%) XY<w>. And as you can see, all the females have brick-red eyes, and all the males white eyes (criss-cross inheritance). In the second generation from crosses of these males "X(<w>)Y(-)" and females "X(<w+>)X(<w>)" we get the phenotypes ratio in the offspring: 1 XX<w+> : : 1 XX<w> : 1 XY<w+> : 1 XY<w>. That means, 50% ( 1 XX<w+> and 1 XY<w+> ) of flies with brick-red eyes and 50% of flies with white eyes ( 1 XX<w> and 1 XY<w> ). Thus, the results is absolutely correct.

More clearly, these crosses can be demonstrated with Traits files. Using the rules of making genetic traits files we can create our file:

<w+>:brick red eyes
<w>:white eyes

Y:male
X:female

{
!gender:X
!simbol:(
!eyes color:<w+> <w>
!simbol:)
!gender:X Y
!simbol:(
!eyes color:<w+> <w> -
!simbol:)
}

Genetic calculation: You can open this file ( X-Linked traits 1.txt ) and calculate results for Traits phenotypes. On the tab "Find" you can choose each phenotype and see from what kind of genotypes it is consist.

Tortie coloration in cats

Another interesting example of X - linked traits inheritance is inheritance of tortie coloration in cats. Fact is that only females cats have a tortie coloration. The males cats with tortie coloration does not exist (with very rare exceptions). This fact could not be explained, until it became known that the inheritance of this trait is linked with X - chromosome. The black coloration of cats we designated by dominant allele "B", and red coloration by recessive allele "b". But if the cat has both these alleles in genotype, then it will have the tortie coloration. Obviously, that in this case we have a dealing with the allelic genes interaction - codominance. Therefore, we can't solve this problem without writing the traits file. Using the rules of making genetic traits files we can create our file:

B b:tortie cat
B:black cat
b:red cat

Y:male
X:female

{
!gender:X
!simbol:(
!color:B b
!simbol:)
!gender:X Y
!simbol:(
!color:B b -
!simbol:)
}

The gene is localized in the X-chromosome, but lack in the Y-chromosome. Therefore, we can write all possible parental genotypes for cats as follows:
For males -
X(B)Y(-) - black
X(b)Y(-) - red
For females -
X(B)X(B) - black
X(b)X(b) - red
X(B)X(b) - tortie

Genetic calculation: Open traits file ( X-Linked traits 2.txt ) and calculate results for Traits phenotypes. You can use any of these genotypes for crossing, but you can't get in the offspring the males with a tortie color. Males may be only black or red, but can't have the turtle coloration, because it is only possible, when both alleles - the dominant "B" and the recessive "b" represented simultaneously in the genotype.

Hemophilia inheritance

In humans, also known the traits, that are linked with X-chromosome. These include, for example, blindness or a serious hereditary disease of hemophilia. People who are sick with hemophilia have the poor blood clotting and even a small scratch can cause severe bleeding. This disease, with rare exceptions, can be found only in men. It was established that hemophilia is caused by a recessive gene, located in the X-chromosome, and therefore woman's, which heterozygous by this gene have a normal blood clotting.

Let see, what offspring can be obtained from women, which heterozygous by hemophilia trait and normal men. Trait of the normal blood clotting mark as dominant allele "H", and trait of hemophilia as recessive allele - "h". Female genotype, we can write as X(H)X(h), and male genotype as X(H)Y(-). From this marriage, we can get the offspring with ratio: 2(50%) XXH : 1(25%) XYH : 1(25%) XYh. Thus, all the girls will be healthy, and half the boys will be sick with hemophilia (XYh). Now choose a different type of result - Genotypes. The ratio of genotypes in the offspring will be: 1(25%) XXHH : 1(25%) XXHh : 1(25%) XYH : 1(25%) XYh. As you can see all the daughters will always have a normal blood clotting, but half of them will be heterozygous carriers of the disease ( XXHh ). More clearly, these crosses can be demonstrated with Traits files. Using the rules of making genetic traits files we can create our file:

H:normal
h:hemophilia

Y:male
X:female

{
!gender:X
!simbol:(
!blood clotting:H h
!simbol:)
!gender:X Y
!simbol:(
!blood clotting:H h -
!simbol:)
}

Genetic calculation: You can open traits file ( X-Linked traits 3.txt ) and see results for Traits phenotypes in this cross.

Traits dependent from gender

There are traits, which are controlled by genes is not linked with X - chromosome, but gender of animal influences on expression of this trait. These signs are called dependent from gender.

Absence of horns in sheeps

An interesting example may be the presence or absence of horns in sheeps. Allele "H" determines the presence of horns, and the allele "h" - the absence. But allele "H" is dominant in males and recessive in females, and allele "h" - on the contrary - recessive in males and dominant in females. In this case we have a deal with the interaction of these gene with the gender trait. Therefore, this genetic problem can't be solved without using of traits file. Using the rules of making genetic traits files we can create our file:

H h Y:horned
H h:hornless
H:horned
h:hornless

Y:male
X:female

{
!gender:X
!simbol:(
!horn:H h
!simbol:)
!gender:X Y
!simbol:(
!horn:H h -
!simbol:)
}

We can write all possible parental genotypes of sheep in this way:
For males -
HHXY- horned
HhXY - horned
hhXY - hornless
For females -
HHXX- horned
HhXX - hornless
hhXX - hornless

Genetic calculation: Open traits file ( Gender Linked traits 1.txt ) and calculate results for Traits phenotypes. You can use any of these genotypes for crosses. Animals with the genotype "HH" will be horned, and with the genotype "hh" - hornless. But the animals with the genotype "Hh" will be horned if it's males are hornless if it's females.

Hair loss in humans

In humans, also known the trait that dependent from gender, it's gene responsible for hereditary of hair loss. It is dominant in males and recessive in females. Bald men are more common be found than females, because for this is necessary the presence both of this alleles in woman's genotype, but for men is enough the presence only one from these alleles. Using the rules of making genetic traits files we can create our file:

H h Y:hairless
H h:hairy
H:hairless
h:hairy

Y:male
X:female

{
!gender:X
!simbol:(
!hair:H h
!simbol:)
!gender:X Y
!simbol:(
!hair:H h -
!simbol:)
}

We can write all possible parental genotypes are as follows:
For men -
HHXY- hairless
HhXY - hairless
hhXY - hairy
For womans -
HHXX- hairless
HhXX - hairy
hhXX - hairy

Genetic calculation: Open traits file ( Gender Linked traits 2.txt ) and calculate results for Traits phenotypes. You can use any of these genotypes for crosses. People with genotype "HH" will be bald, and with the genotype "hh" - hairy with no matter is this male or female. But people with genotype "Hh" would be bald if it's man, and hairy if it's woman.

Want more interesting articles ?

• Punnett Square Calculator
• How to use genetic calculator
• How to write parents genotypes
• How to work with genetic traits files
• Punnett square: practice and examples
• How to solve genetic linkage problems
• How to solve incomplete and codominance problems
• How to solve dominant and recessive epistasis problems
• How to solve polygenic inheritance problems
• How to solve chromosomal nondisjunction problems
• How to solve polyploidy problems