Answer 1. RARG QTQC x RARA QCQC
JUSTIFICATION : The parents are the ones who are crossed to generate progeny. Therefore, the parent genotypes are the same as the genotypes mentioned in the question
Answer 2. 47.7cM
Justification : the genotypes with crossing over are 242 and 235.
Sum = 242+235 = 477
Linkage distance = (477/ 1000) *100 = 47.7 cM
Answer 3. No, the hypothesis of no Linkage can not be rejected.
Justification :
If the genes are far apart on the chromosome a cross over will occur every time that pairing occurs and an equal number of parental and recombinant chromosomes will be produced. Test cross data will then generate a 1:1:1:1 ratio.
In the given case, a similar 1:1:1:1 ratio is observed that indicates that the genes are widely apart and there is no linkage between the two genes.
9. You have been asked to determine the map distances between two SNP loci R and Q in wheat. The following cross was undertaken: RARG QcQrx RARA QcQc and yielded progeny as shown below. RARA QrQc RAR...