1. In guinea pig, fur color is determined by one gene, and fur length by anotherindependently segregating gene. The black fur allele (B) is dominant to brown furallele (b) and the short allele (S) is dominant to long fur allele (s). From progenylisted in the following table, provide the probable genotype for the parents of eachcross.Phenotype of ProgenyBlack short black long brown shortParentsBlack short XBlack shortBlack short XBrown longBlack long XBlack long9017032163529140brown long8152. A mouse with the genotype BbccEeFf is mated with another mouse withBBCCEEFF genotype. What is the probability of obtaining the followinggenotype in the progeny?a. BBCcEeFFb. BbccEEFfC.BbccEeFF123. Determine the type of gametes that are formed from the following individualswith the given genotype?A. AaBbCcB. AABBCcC. AABbCe

Name: 1. In guinea pig, fur color is determined by one gene, and fur length
Uploaded: 2024-03-20T06:43:04.000Z
Channel: Bartleby
Description: 1. In guinea pig, fur color is determined by one gene, and fur length by anotherindependently segregating gene. The black fur allele (B) is dominant to brown furallele (b) and the short allele (S) is dominant to long fur allele (s). From progenylist

Dominant traits are always expressed even if only one copy of the dominant trait exists. Recessive traits are only expressed in homozygous conditions. DISCLAIMER - Since you have asked multiple questions, we will solve the first question for you. If you want any specific question to be solved then please specify the question number or post only that question. Case 1 - Black short * Black short Progenies formed - Black short Black long Brown short Brown long This shows that both the parents are heterozygous as the progenies formed show both the fur colours (black and brown) and both the fur lengths (short and long). So parental genotype will be - BbSs * BbSs Gametes formed will be - BS, Bs, bS, bs BS Bs bS bs BS BBSS(Black short) BBSsBlack short) BbSS(Black short) BbSs(Black short) Bs BBSs(Black short) BBss(Black long) BbSs(Black short) Bbss(Black long) bS BbSS(Black short) BbSs(Black short) bbSS(Brown short) bbSs(Brown short) bs BbSs(Black short) Bbss(Black long) bbSs(Brown short) bbss(Brown long) Case 2 - Black short * Brown long Progenies formed are - Black short Black long Brown short Brown long Genotype of the parents would be - BbSs (black short) * bbss (Brown long) Gametes formed - BS, Bs, bS, bs and bs bs BS BbSs(Black short) Bs Bbss(Black long) bS bbSs(Brown short) bs bbss( Brown long) Case 3 - Black long * Black long Progenies formed - Black long Brown long Parental genotype - Bbss * Bbss Gametes formed - Bs, bs Bs bs Bs BBss(Black long) Bbss(Black long) bs Bbss(Black long) bbss(Brown long)