Quantitative heritability refers to the expression of a trait that depends on the contribution of each of the alleles involved in the interaction. More than two genes interact to express a phenotype. 1) 140 mm / 2) 0.2344
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Quantitative heritability is the transmission of a phenotypic trait in which expression depends on the additive effect of a series of genes.
The interaction of more than one gene determines the expression of the trait. And these genes can also have more than two alleles.
The action of many genes and alleles can cause many different combinations that are the reason for genotypic graduation.
Quantitative traits can be measured, such as longitude, weight, eggs laid per female, among others.
These characters do not group individuals by any precise and clear categories. Instead, they group individuals in many different categories that depend on how the genes were intercrossed and distributed during meiosis.
The result depends on the magnitude in which each allele contributes to the final phenotype and genotype. When they interact, they create a gradation in phenotypes, according to the level of contribution.
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In the exposed example, we know that
→ three diallelic genes control the length of wheat leaves L/l, W/w, H/h.
→ each allele contributes equally to the length of the plant leaves
ll ww hh → leaves of 100 mm in lengthLL WW HH → leaves of 220 mm in length
Knowing that each allele contributes equally to the length, we can assume there are 7 possible phenotypes
LLWWHH ⇒ 220 mmLlWWHH, LLWwHH, LLWWHhLlWwHH, LlWWHh, LLWwHh, llWWHH, LLwwHH, LLWWhhLlWwHhllWwHh, LlwwHh, LlWwhh, llwwHH, llWWhh, LLwwhhllwwHh, llWwhh, Llwwhhllwwhh ⇒ 100 mm
The differences in leaf length between the two h0m0zyg0us strains of wheat is 220 - 100 mm = 120 mm
We know that
The minimal length of leaves is 100 mm and correspond to the h0m0zyg0us recessive genotype, llwwhh.Each dominant allele contributes equally to the leaves legth. The difference between the minimal length and the maximum one is 120 mm. There are six dominant alleles.
The contribution of each dominant allele is
120mm / 6 alleles = 20 mm/allele.
Gentoype Dominant Length Phenotype
alleles contibution
llwwhh 0 0 100 mm
llwwHh, llWwhh, Llwwhh 1 20 mm 120mm
llWwHh, LlwwHh, LlWwhh,
llwwHH, llWWhh, LLwwhh 2 40 mm 140 mm
LlWwHh, LLWwhh, LLwwHh,
llWWHh, llWwHH, LlWWhh 3 60 mm 160 mm
LlwwHH
LlWwHH, LlWWHh, LLWwHh,
llWWHH, LLwwHH, LLWWhh 4 80 mm 180 mm
LlWWHH, LLWwHH, LLWWHh 5 100 mm 200 mm
LLWWHH 6 120 mm 220 mm
Now, let us analyze the crosses.
1st cross:
Parentals) LL WW HH x ll ww hh
F1) Ll Ww Hh
2nd cross: F1 selfed-cross
Parentals) Ll Ww Hh x Ll Ww Hh
F2) Genotypes
1/64 LLWWHH 2/64 LLWWHh 1/64 LLWWhh 2/64 LLWwHH 4/64 LLWwHh 2/64 LLWwhh 1/64 LLwwHH 2/64 LLwwHh 1/64 LLwwhh 2/64 LlWWHH 4/64 LlWWHh 2/64 LlWWhh 4/64 LlWwHH8/64 LlWwHh4/64 LlWwhh 2/64 LlwwHH 4/64 LlwwHh 2/64 Llwwhh 1/64 llWWHH 2/64 llWWHh 1/64 llWWhh 2/64 llWwHH 4/64 llWwHh 2/64 llWwhh 1/64 llwwHH 2/64 llwwHh 1/64 llwwhh
Phenotypes
1/64 ⇒ 220mm6/64 ⇒ 200 mm15/64 ⇒ 180 mm20/64 ⇒ 160 mm15/64 ⇒ 140 mm6/64 ⇒ 120 mm1/64 ⇒ 10 mmLeaves of the llwwHH genotype plant will be 140 mmThe proportion of the F2 progeny with the same phenotype as the llwwHH genotype will be 15/64 = 0.2344
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