genetic biology (4) In chickens, the gene for red color (R), shaggy feathers (Sg
ID: 148527 • Letter: G
Question
genetic biology
(4) In chickens, the gene for red color (R), shaggy feathers (Sgf) and black tail (Bt) appear to be on the same autosome and have the following phenotypes and dominant/recessive relationships (8 points total) R-red body feathers, dominant r- white body feathers, recessive Sgfstraight feathers, dominant sgf - shaggy feathers, recessive Bt - black tail feathers, dominant bt white tail feathers, recessive A random red body, straight feathered, black tail rooster from a pen containing chickens with a mix of phenotypes of red or white body feathers, straight or shaggy feathers, and black or white tails, was crossed to hens from a pen that contained pure-breeding white body, shaggy, white tailed chickens The progeny had the following phenotypes and counts red body, straight feather, black tail white body, straight feather, black tail white body, shaggy feather, black tail white body, shaggy feather, white tail red body, shaggy feather, black tail red body, shaggy feather white tail red body, straight feather, white tail white body, straight feather, white tail 250 310 46 264 37 331 42 34Explanation / Answer
The data provided is re-written in the following way considering only gene symbols,
R Sgf Bt - 250
r Sgf Bt - 310
r sgf Bt - 46
r sgf bt - 264
R sgf Bt - 37
R sgf bt - 331
R Sgf bt - 42
r Sgf bt - 34
From the above data, genotype found most frequently are parental genotype.
Thus,
r Sgf Bt - 310 and R sgf bt - 331 are parental phenotype.
Double crossover gametes are always in the lowest frequency.
So,
R sgf Bt - 37 and r Sgf bt - 34 are double cross overs.
Next, double cross over events move the middle allele from one sister chromatis to other. on comparing parental genotypes with double cross over genotype, you can find that Bt/bt allele has been moved from one sister chromatid to another. Hence Bt/bt is the middle allele.
The genes are linked and the order can be written as
R bt sgf and r Bt Sgf.
The answer for the question D is,
Option D). Bt (middle gene)
The map distance(linkage distance) can be calculated by the following formula
= (Number of recombinant offspring/Total number of offspring) * 100
The distance between R and bt is,
=(37 + 34 + 250 + 264)/1314 * 100
=44.52 cM (cM stands for centiMorgan, unit of measurement of genetic linkage)
The distance between bt and sgf is,
=(37 + 34 + 46 + 42)/1314 *100
=12.11 cM
Thus the map distance between the genes farthest based on the sum of the distance between each of those genes and the gene in the middle is
=44.52 cM + 12.11 cM
=56.6 cM
Hence the answer for the question E is
Option A). 56.6
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