If you cross dominant genotype PV/PV male Drosophila with double recessive female and obtain F 1 hybrid. Now you cross F 1 male with double recessive female. i) What type of linkage is seen? ii) Draw a cross with correct genotype. iii) What is the possible genotype in F 2 generation?

i) Again if you cross F 1 male with double recessive female a complete linkage can be seen and it occurs due to the absence of chiasma or crossing over in between the linked genes present on homologous chromosomes. Crossing over will not present in male drosophila. ii) Correct cross over is shown in the below diagram and In F1 generation it shows only two gametes due to lack of crossing over iii) F 2 hybrid exhibits only parental genotype due to complete linkage and Test cross failed to produce 1:1:1:1 ratio & instead produced 7:1:1:7. PV Pv PV pv PV PPVV PPVv PPVV PpVv Pv PPVv PPvv PPVv Ppvv PV PPVV PPVv PPVV PpVv pv PpVv Ppvv PpVv ppv

S.No. Gamete types Number of progenies 1 ABC 349 2 Abc 114 3 abC 124 4 AbC 5 5 aBc 4 6 aBC 116 7 ABc 128 8 abc 360 i) What is the name of this test cross? ii) How will you construct gene mapping from the above given data? iii) Find the correct order of genes.

i) The name of the test cross given above is three point test cross, and it refers to analysing the inheritance patterns of three alleles by test crossing a triple recessive heterozygous with a triple recessive homozygous. It enables to determine the distance between the three alleles and the order in which they are located on the chromosome. Double cross-overs can be detected which will provide more accurate map distances. ii) Construction of gene map: To construct the gene map, the recombinant frequency (RF) of the alleles has to be calculated. From the given data it is clear that ABC and abc are parental (P) types and the others (Abc, abC, AbC, aBc, ABc) are recombinant (R) type. Recombinant for the alleles Gametes No. Of progenies. A and B A and C B and C ABC 349 Abc 114 R R abC 124 R R AbC 5 R R aBc 4 R R aBC 116 R R ABc 128 R R abc 360 Let's analyse the loci of two alleles at a time starting with A and B. Since the genes AB and ab are parental type, the recombinants will be Ab and aB.Therefore, Recombinant frequency of alleles Ab and aB = Recombinant frequency for the loci B and C. The parent form are Bc and bC and the recombinant are Bc and bC. Recombinant frequency of alleles Bc and bC = Since the recombinant frequency of the alleles A and C shown the highest frequency, they must be the farthest apart and alleles B must lie in between A and C. So the gene map can be constructed as follows: iii) Gene orders are the permutation of genome arrangement. From the above calculations the correct gene order is ABC/abc.

E M B I B E

Biology Botany Standard 12>Genetics>Chapter 3 - Chromosomal Basis of Inheritance>Evaluation

Evaluation

Author:Tamil Nadu Board

Tamil Nadu Board Biology Botany Solutions for Exercise - Evaluation

Simple step-by-step solutions to Evaluation questions of Chromosomal Basis of Inheritance from Biology Botany Standard 12. Also get 3D topic explainers, cheat sheets, and unlimited doubts solving on EMBIBE.

Questions from Evaluation with Hints & Solutions

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12th Tamil Nadu Board

IMPORTANT

Biology Botany Standard 12>Chapter 3 - Chromosomal Basis of Inheritance>Evaluation>Q 14

If you cross dominant genotype PV/PV male Drosophila with double recessive female and obtain F₁hybrid. Now you cross F₁male with double recessive female.

i) What type of linkage is seen?

ii) Draw a cross with correct genotype.

iii) What is the possible genotype in F₂generation?