Question

A heterozygous trihybrid corn plant is test crossed and the following progeny are oduced. The mutant traits-booster (b), silkless (s), liguleless (1) - are recessive to er wild-type (+) alternatives. (Hint: The wild-type alternatives (+) for each of the three mutant traits can be named B, S, and L, respectively.) (6 points) Note: This is a three point cross. Show all your work to get full credit. You will get partial credit for any work that is correct even if you do not get the entire answer correct. Yes, you can use a calculator! 291 Three Possible Phenotypes Progeny booster booster silkless 137 booster silkless liguleless liguleless 142 silkless liguleless 3 silkless booster liguleless 34 339 982 = Total Offspring 30 a) (2 pt) In the left margin next to the phenotypes above, identify the parental types (P), the phenotypes resulting from double crossovers (D), and the phenotypes resulting from single crossovers (S) b) (2 pt) Which gene (booster, silkless, liguleless) is between the other two on the chromosome? How can you tell? c) (2 pt) Calculate distances and draw a linkage map showing the gene order and the distances between the genes on the chromosome in centi Morgans. Show your work.

Answer 1

Abbreviations - (i) booster, (ii) silkless - s and liguleless - l

In a three point cross, you must always remember that in a dataset like this, 2 data sets represent parent phenotype, 2 datasets represent single cross over between gene A and gene B (here gene s and gene b), 2 datasets represent single cross over between gene B and gene C ( here gene b and gene l) and 2 datasets represent double cross over.

(a) The simplest way to identify parental genotype is to see the maximum number of offsprings with that phenotype. In this data, 291 offsprings have genotype 'b', 's' and 'l' and 339 offsprings have genotype +,+ and + (OR B,S and L). So these represent the parental genotype

The simplest way to determine the double cross over genotype is to see the lowest number of offspring given in the data. Here 3 (+,s,l)and 6 (b,+,+) are the lowest number of offspring and hence they represent the double cross over.

The remaining number of offsprings represent the single cross over, i.e., (b,s,+), (+,+,l), (+,s,+) and (b,+,l)

(b) The sequence of the genes can be determined as follows:

• compare parental phenotypes and double cross overs to detemine the gene present in betwee

Parental genotype	b s l	+ + +
Double cross overs	+ s l	b + +

In double cross overs only the centre gene is switched, and when comparing the parental and double cross over sets, only the b gene has switched, hence we can say that the sequence of the gene on chromosome is s-b-l, and b gene is in between s and l

(c) Distance between genes can be calculated by the formula :

distance in cM = offsprings with single cross over + offsprings with double cross over/ total numer of offsprings*100

• Distance between gene s and gene b = 30+34+6+3/982*100 = 7.43 cM
• Distance between gene b and gene l = 137+142+6+3/982*100 = 29.32 cM

Linkage map: the genes will be as shown below in the chromosome.

s b - 7.43 29.52

PART -II

(a) Not consistent with X- linked recessive inheritance

(b)Not consistent with X- linked recessive inheritance

(c) Consistent with X- linked recessive inheritance