Please provide clear, step by step answers. Thank you!!! 1. Saint Bernard dogs a

Question

Please provide clear, step by step answers. Thank you!!!

1. Saint Bernard dogs are known to exhibit certain traits, including excessive drooling (D-), large and gentle personality (G-). These traits are dominant to no drooling (dd), small head head (H-) and gentle personality (G-). These traits are dominant to no drooling (dd), small head (hh) and aggressive personality (gg) and all are located on the same chromosome. Two pure- breeding strains of Saint Bernard dogs are crossed to produce F1 dogs that are heterozygous with alleles D d, H h, G g. Data from a three-point testcross are shown here. (10 points) Genotype D-, H-, G Phenotype Normal Saint Bernard dog Small head Aggressive personality No drooling Small head, aggressive personality No drooling, small head No drooling, aggressive personality No drooling, small head, aggressive personality Number 52 15 106 350 375 122 10 70 dd, hh, gg A. What is the genotype of the F1 triply heterozygous parent? Please be as specific as possible, with the correct alleles on each chromosome. (4 points) B. What is the order of these three traits in the Saint Bernard linkage group? (3 points) C. What is the distance in map units between these three genes? (3 points)

Explanation / Answer

Hi,
This is a typical 3 -point cross. Here 3 genes are considered for linkage determination. The F1 triple gene heterozygote is crossed to a triple recessive homozygous variety. The resulting progeny are counted, their phenotypes are analyzed and gene order is mapped.
In the given example,the F1 is a triple heterozygous , so its genotype = DHG/dhg or DdHhGg.
The recessive test cross parent = dhg/dhg or ddhhgg

In order to find the order of the genes,we need to find out the parental and double cross over mutants. The parental types are ones with highest number of occurrence. So 350 and 375 are parental. The double cross over (DCO) are the ones with lowest number of progeny. So, 15 and 10 are the DCO. Now, compare one of parental (dd,H-,G-) with a DCO (dd,H-,gg). two genes have same allels and only one is different i.e dd and H- are in same position and G- and gg are changed. So G is in middle.
So gene order is = DGH

Map distance is calculated by determining recombination frequency (RF).
RF between D and G = no. of progeny with recombination between D and G + DCO / total progeny
total= 1100
RF between D and G = 122+70+25 / 1100 = 0.1972 =19.72% =19.72mU
RF between G and H = 106+52+25 / 1100 = 0.1663 = 16.63 mU

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.