Lect. 5 Non-Mendel's Laws PDF

Lect. 5: Non-Mendel's Laws Dr./ Ahmed Mahdy ❑ Non-Mendelian Inheritance (Variations on Mendel’s Laws) Incomplete dominance Codominance Lethal alleles Multiple alleles 2  Extensions to Mendel When Mendel’s laws/results...

Lect. 5: Non-Mendel's Laws Dr./ Ahmed Mahdy ❑ Non-Mendelian Inheritance (Variations on Mendel’s Laws) Incomplete dominance Codominance Lethal alleles Multiple alleles 2  Extensions to Mendel When Mendel’s laws/results may not be observed Genetic Occurrence Definition Examples 4 genes are involved in More than one gene can affect a single Polygenic inheritance determining eye color. trait Human height A pleiotropic allele dominant for yellow fur in A single gene can affect more than one Pleiotropy mice is recessive for lethal trait developmental defect. Sickle cell anemia Multiple alleles for one gene Genes may have more than two alleles ABO blood types in humans In incomplete dominance the heterozygote is intermediate. Dominance is not always In co-dominance no single allele is Human blood groups complete dominant, and the heterozygote shows some aspect of both homozygotes. Genes may be affected by the Environmental factors Siamese cats environment. Products of genes can interact to alter The production of a purple Gene interaction genetic ratios. pigment in corn ❑ Incomplete dominance Neither allele is dominant over the other Expression of both alleles is observed as an intermediate phenotype in the heterozygous individual Incomplete dominance results in intermediate phenotypes. ❑ Incomplete dominance: A condition in which neither allele is completely dominant and heterozygotes (F1 hybrids) phenotype is intermediate between the two parental homozygotes. Examples: 1- When red snapdragons are crossed with white ones, All F1 hybrids will be pink flowers. When F1 is self-pollinated, the genotypic and phenotypic ratios in F2 will be 1:2:1. 2- When a homozygous Andalusian fowl with black feathers is crossed with a homozygous fowl with splashed white feathers the F1 hybrids are found to contain blue feathers. P generation Red White RR rr Gametes R r F1 generation Pink Rr 1 1 – Gametes –2 R 2 r Sperm R r 1 – 1 2 – 2 1 – F2 generation 2 R RR rR Eggs RR 1 – Rr rr 2 r Incomplete dominance in snapdragon color 7 8 9 3- Palomino coat in horses: 10 ❑ Codominance. When both the alleles (dominant and recessive) are expressed in the hybrid, the phenomenon is known as codominance. Examples; in Animals: when red and white cattle are crossed, they produce a roan offspring which possess both red and white hairs on the skin. 13 14  Lethal gene or Allele (2 : 1) The genes which cause the death of the individual carrying it, is called lethal factor. Recessive lethal allele are expressed only when they are in homozygous state and the heterozygotes remain unaffected. e.g., Gene producing albino seedlings in barley  Many gene products are essential to an organism’s survival. Mutations resulting in the synthesis of a gene product that is nonfunctional can often be tolerated in the heterozygous state. However, such a mutation behaves as a recessive lethal allele, and homozygous recessive individuals will not survive. For example, a mutation that causes a yellow coat in mice. The yellow coat varies from the normal agouti (wild-type) coat phenotype. Crosses between the various combinations of the two strains yield unusual results. These results are explained on the basis of a single pair of alleles. With regard to coat color, the mutant yellow allele (AY) is dominant to the wild-type agouti allele (A), so heterozygous mice will have yellow coats. However, the yellow allele is also a homozygous recessive lethal. When present in two copies, the mice die before birth. Thus, there are no homozygous yellow mice. In other cases, a mutation may behave as a dominant lethal allele. In such cases, the presence of just one copy of the allele results in the death of the individual. 16 17 ❑ Multiple alleles ABO blood grouping. Three or more alternative forms or versions of a gene (alleles) that can occupy the same locus. However, only two of the alleles can be present in a single organism. 18 Phenotype Genotype AA or AO BB or Bo AB OO 19 Inheritance of ABO blood grouping 20 Inheritance of Rhesus (Rh) factor 22 Rhesus (Rh) factor; Genotype or phenotype 23 Only for reading 24 DEVIATION FROM MENDEL’S DIHYBRID PHENOTYPIC RATIO The Mendelian dihybrid phenotypic ratio of 9 : 3 : 3 : 1 is obtained only when the alleles at both gene loci display dominant and recessive relationship. Combinations of Two Gene Pairs with two modes of Inheritance modify the 9:3:3:1 Ratio. If one or both gene loci have incompletely dominant, or codominant or lethal alleles, the dihybrid ratio becomes modified variously. ❑ 3 : 6 : 3 : 1 : 2 : 1 Ratio When dihybrid parent have dominant and recessive alleles at one gene locus and co-dominant or incomplete alleles at second gene locus, (3 : 1) x (1 : 2 : 1) = 3 : 6 : 3 : 1 : 2 : 1 Example. In cattle, hornless or polled (P) is dominant to horned (p) trait, and white (W) coat is codominant to red (w) coat. ❑ 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1 Ratio Each parent of has incompletely dominant alleles at both gene loci. Formula: (1 : 2 : 1) x (1 : 2 : 1 ) = 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1. Example. In snapdragons, red flower color (R) and broadness of leaf (B) are incompletely dominant to white flower color (r) and narrowness of leaf (b). ❑ 6 : 3 : 2 : 1 Ratio When the F1 dihybrids have dominant-recessive alleles at one gene locus and recessive lethal alleles at second gene locus, the F2 offsprings manifest the phenotypic ratio of 6 : 3 : 2 : 1 Formula: (3 : 1) (2 : 1) = 6 : 3 : 2 : 1 ❑ 2 : 4 : 2 : 1 : 2 : 1 Ratio When the F1 dihybrids contain codominant alleles at first gene locus and recessive-lethal alleles at second gene locus, then their F2 progeny display the phenotypic ratio 2 : 4 : 2 : 1 : 2 : 1. Formula: (2 : 1) (1 : 2 : 1) = 2 : 4 : 2 : 1 : 2 : 1 ❑ 4 : 2 : 2 : 1 Ratio This phenotypic ratio is obtained when at both gene loci of F1 dihybrids occur the recessive lethal alleles. Formula: (2 : 1) (2 : 1) = 4 : 2 : 2 : 1 31