12. You are studying a completely new, uncharacterized mutation in maze that cau

Question

12. You are studying a completely new, uncharacterized mutation in maze that causes low sweetness in the kernels. You ascertain that it is an autosomal recessive genetic mutation. You cross a pure breeding low kernel sweetness strain that is homozygous for all markers with a carrier strain that is heterozygous for markers. You get genotypes and phenotypes from 500 kernels from the cross. (a) What is the implied gametic phase? (b) What order do the markers and mutation appear to be in? (c) What is the estimated map unit distance between the mutation and the markers? (d) Under what circumstances would you expect to get the result from part c? D1 D1 E1 E1: Normal Sweetness: 65 D1 D1 E1 E2: Normal Sweetness: 62 D1 D2 E1 E1: Normal Sweetness: 63 D1 D2 E1 E2: Normal Sweetness: 61 Low Sweetness: 60 Low Sweetness: 63 Low Sweetness: 62 Low Sweetness: 64

Explanation / Answer

The uncharacterized mutation mentioned in the question above has been shown to contribute to low sweetness in the kernels of maize. It has also been discovered that this uncharacterized mutation is an autosomal recessive one. This implies that the strains of maize in which this mutation is getting manifested phenotypically (that is, the lowering of the sweetness of maize kernel is taking place), must be homozygous for the recessive allele.

Let us denote this genotype as D1D1 (where D1 is the allele recessive to D2).

a) It is stated in the question that a pure-breeding low sweetness kernel strain, which is homozygous for all markers has been crossed with a carrier strain heterozygous for markers. The marker gene are denoted by E1 and E2 each representing one allele type.

Based on these, the pure -breeding strain homozygous for markers can be denoted by D1D1 / E1E1

(since it is homozygous for both the sweetness gene and the marker)

On the other hand, genotype of the carrier strain which is heterozygous for markers can be denoted by D1D2 / E1E2.

(since it is a carrier of the low sweetness allele, and hence have the heterozygous dominant genotype. It is also heterozygous for the marker).

Therefore the parental genotypes of the cross, or the gametic phase (the allelic combination obtained from the parents) are : D1D1 / E1E1 and D1D2 / E1E2.

b) Since the mutation has been stated to be an autosomal recessive mutation, therefore manifestation of the phenotype corresponding to an autosomal recessive mutation would require a homozygous recessive genotype characteristic of the mutation. But the wild type carrier strain is heterozygous in nature, that is, it already had one recessive allele. Thus, one mutation would be enough to bring about the homozygous recessive condition in the maize kernel. Therefore, the mutation appears to be of the first order.

c) These two strains are being crossed to yield 500 offspring kernels.

As per the recombination results provided in the question, the cross betweent the two parental varieties yielded the following genotypes in the following frequencies :

D1D1 / E1E1 : (65 + 60) = 125 (Parental genotype)

D1D1 / E1E2 : (62 + 63) = 125 (Parental genotyoe)

D1D2 / E1E1 : (63 + 62) =125 (Recombinant genotype)

D1D2 /E1E2 : (61 + 64) = 125 (Recombinant genotype)

Map distance between the mutation and the marker can be calculated by dividing the number of recombinant gametes by the number of total gametes analyzed, multiplied by 100.

Here the number of total gametes = 500

Number of recombinant gametes = 250

Therefore, map distance (between D and E genes) = [ 250/500 ] * 100 = 50

The distance between the mutation gene (D) and the marker gene (E) is 50 cM.

d) As we can see, the number of recombinant gametes obtained in the experiment is 50% of the total numbe rof gametes. The rest 50% gametes retained the parental genotype. Equal numbe rof parental and recombinant gametes signify non-linkage between the genes under consideration. Or in other words, the mutation gene and the marker gene is located quite distant from each other, such that they are free to recombine with homologous chromosomes, and do not get inherited together in a linked manner.

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