Recessive and dominant alleles.
It is easy to get confused about terms of recessive and dominant alleles. The common misunderstanding is that dominant alleles somehow "win against" recessive. It is not true!
Let's clarify what this means
- C(WT) is dominant over a recessive allele C(LOF), because a heterozygote still produces at least 50% of the normal amount of functioning protein, which is sufficient to produce a wild type phenotype. The extra non-functioning protein from the C(LOF) allele is quickly degraded and has no effect.
- In this case, C(GOF) is a dominant allele, because it codes for a protein which is visible on phenotype and masks a wild type protein.
A good example of haplosufficiency is a genetic disease Phenylketonuria (PKU). It is inherited as a Mendelian recessive disorder. This disease is caused by a mutated allele that encodes the liver enzyme phenylalanine hydroxylase (PAH). A wild type PAH coverts phenylalanine from food into the amino acid tyrosine:
A mutated gene which encodes for a PAH enzyme can alter amino acid sequence of an enzyme resulting in deformed substrate binding site of PAH. Hence, PAH cannot bind to phenylalanine and convert it into tyrosine. In this situation, phenylalanine starts to build up in the body and is converted instead into phenylpyruvic acid:
Phenylpyruvic acid interferes with the nervous system development and leads to mental retardation.
However, as mentioned above PKU disease is recessive. The common denomination is P = wild type allele and p = mutated allele. If a person inherits one P allele and one p allele (P/p), he would not be sick, because one P allele produces enough amount of PAH protein to breakdown phenylalanine into tyrosine leading to no accumulation of phenylpyruvic acid. That illustrates complete dominance, or haplosufficiency. People with two p alleles (p/p) would not produce any PAH and that would result in no breaking down of phenylalanine and accumulation of phenylpyruvic acid that would cause a mental retardation.
There are other cases, however, when one wild type allele is not enough to achieve normal levels of function, creating what is known as haploinsufficiency.
This inheritance pattern is most common for genes encoding proteins that are used in large numbers, such as structural proteins, energy metabolism enzymes and some pigments. Let's assume that cell requires 15 units of protein to be produced each hour to function properly. Given constraints on DNA transcription and translation speed, dominant allele X can produce 10 protein units per hour, and a recessive allele x produces none. Hence, homozygote dominant organism would produce 10 + 10 = 20 protein units. It is more than required and a phenotype appears normal. In that case, when only one X allele is present just 10 protein units would be produced per hour. This amount of a protein is not enough for normal chemistry and a phenotype would be altered, but milder than a phenotype having xx alleles, when no protein is being produced at all. To summarize, the homozygous recessive phenotype is more severe (in disease case) than the heterozygous phenotype. In contrast to complete dominance, heterozygous dominant would not have the same phenotype as homozygous dominant, but - intermediate phenotype.
Haploinsufficiency can be well illustrated with a famous example of incomplete dominance - flowers called snapdragons (Antirrhinum majus) that show intermediate pink phenotype, when heterozygotes.
When a pure-breeding wild type white flower (C(W)C(W)) is crossed with a pure-breeding wild type red flower (C(R)C(R)) all offspring will be of intermediate phenotype - pink (C(R)C(W)). It is incomplete dominance of the C(R) allele, which encodes for pigment proteins. Two copies of this allele produce the most copies of transcript, hence producing the greatest amount of pigment protein resulting in red petals of flowers. A single C(R) allele can only produce half the amount of pigment, leading to pink petals. Homozygote for C(W) allele appears to be white, because no pigment is expressed.
Co-dominance is a third inheritance pattern that is subtly different from incomplete dominance. Here, too, heterozygotes have clearly identified phenotypes, but these are generated by expressing the phenotypes of both alleles at once. This is unlike the case with incomplete dominance, where heterozygotes exhibit an intermediate phenotype, which differs from both the homozygous phenotypes. An example should help clarify the distinction, so let us take a look at ABO blood groups:
- Alleles code for red blood cell surface antigens
- Three alleles of one gene: I(A), I(B) and i
- An organism can inherit only two out of three alleles
- 4 different blood groups
The alleles I(A) and I(B) determine different antigens and an allele i codes for no antigens. Because allele i produces no proteins, it has no effect on the phenotype of heterozygotes, and therefore alleles I(A) and I(B) are fully dominant over an allele i. BUT! Alleles I(A) and I(B) both encode for proteins, and cells are able to produce the necessary dose of the antigens even from a single copy of each allele. When genotype is I(A)/I(B) both types of antigens are present because each of the alleles expresses its own determined cell surface antigens. I(A) and I(B) alleles are co-dominant!
Most traits are actually under the control of multiple genes, and here things become more complicated. The agouti mouse is a good example of illustrating multiple alleles for a gene responsible for mouse fur colour.
We know that an allele:
A - expresses agouti fur pigment;
a - black pigment
a(y) - yellow pigment in belly zone and black pigment in back zone
Crossing pure breeding mice allows determining, which allele is dominant over which allele. That can be illustrated by the figure:
When a mouse has a heterozygote genotype Aa(Y) its phenotype looks like a homozygote dominant AA. That is because only one copy of an allele A (on a paternal or maternal chromosome) is enough to produce agouti colour pigment and masks colour pigment expressed by the allele a(Y). That means that the allele a(Y) is recessive to A. However, it is a relationship between those two alleles, not an intrinsic property of the yellow belly allele!
Agouti allele is also dominant over an allele a coding for a black fur pigment. Black pigment is masked by agouti pigment and phenotype appears to be agouti mouse.
Thus, the allele A dominates over both alleles. What about a(Y) and a alleles relationships? As you can see from the diagram crossing of an a(Y)a(Y) mouse with an aa one, their heterozygote offspring would have yellow belly and black back. In other words, the allele a(Y) expresses enough pigment protein for a heterozygote to appear as homozygote, hence in this case the allele a(Y) is considered dominant over the allele a.
First point to note is that if an allele is lethal, it means that a gene plays an essential role in terms of embryo development and vital functions. Such alleles are usually recessive, so can be maintained in a population in heterozygotes. In some cases dominant lethal alleles exist, where any individual carrying the allele even in single copy does not survive. Clearly, these arise through mutation and cannot be maintained in natural populations. Despite being very rare in the wild, lethal alleles are often used as markers in work with model organisms.
One example of human lethal mutations is brachidactyly that means "short digits" in Greek. An allele is lethal in the homozygous state. A single lethal allele causes phenotype alterations. In the case of brachydactyly type A1, one dose of lethal allele causes forelimbs alterations such as short 2nd, 4th and 5th middle phalanges. A mutation occurs in IHH gene which encodes proteins responsible for bone growth and differentiation that are essential in development. Having a single mutated copy of the allele does not result in death, because one dose of functional IHH allele is almost enough to produce a required amount of a protein essential for a skeletal formation. As a result, a phenotype has just few deformations of skeletal bones such as abnormally short fingers. If an organism inherits two mutated copies of IHH allele no protein essential for skeletal bones formation is produced and development of embryo cannot be continued - the embryo dies.
Let's say that an allele a is recessive and codes for a completely dysfunctional form of a protein essential for bone growth, and A is a dominant wild type allele. If heterozygotes for these alleles procreate, then:
The table shows that homozygotes aa are never born (crossed out). Hence, the F1 offspring phenotypic ratio would be 2:1 (altered bone structure phenotype and normal respectively).
Resources and further reading:
Campbell N. A. and Reece J. B. (2002). Biology. (6th edition), chapter 14 - University level textbook.
Griffiths A. J. F., Wesslee S. R., Carroll S. B., Doebley J. (2012). Introduction to Genetic Analysis.(10th edition), chapter 6 - University level textbook.
Kirkpatrick T. J., Au K. S., Mastrobattista J. M., McCready M. E., Bulman D. E., Northrup H. (2003). Identification of a mutation in the Indian Hedgehog (IHH) gene causing brachydactyly type A1 and evidence for a third locus. doi:10.1136/jmg.40.1.42 - research paper.
Wilkie A. OM (2006). Dominance and recessivity. doi: 10.1038/npg.els.0005475 - review article.