**Bio3110 Genetics: Additional Problems (Chapter 2)**1. **In guinea pigs, fur color is determined by one gene, and fur length by another independently segregating gene.** The black fur allele (B) is dominant to brown fur allele (b) and the short allele (S) is dominant to long fur allele (s). From progeny listed in the following table, provide the probable genotype for the parents of each cross. **Phenotype of Progeny** | **Parents** | **Black Short** | **Black Long** | **Brown Short** | **Brown Long** | |----------------------------|----------------|---------------|----------------|---------------| | Black short x Black short | 90 | 32 | 29 | 8 | | Black short x Brown long | 17 | 16 | 14 | 15 | | Black long x Brown long | 0 | 35 | 0 | 12 |2. **A mouse with the genotype BbccEeFf is mated with another mouse with BBCcEEFF genotype.** What is the probability of obtaining the following genotype in the progeny? a. BBCcEeFF b. BbccEeFF c. BbccEEFF 3. **Determine the type of gametes that are formed from the following individuals with the given genotype:** A. AaBbCc B. AABBcc C. AABbCc

Name: **Bio3110 Genetics: Additional Problems (Chapter 2)**1. **In guinea p
Uploaded: 2024-03-20T06:42:55.000Z
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Description: **Bio3110 Genetics: Additional Problems (Chapter 2)**1. **In guinea pigs, fur color is determined by one gene, and fur length by another independently segregating gene.** The black fur allele (B) is dominant to brown fur allele (b) and the short all

1. Given: The black fur allele (B) , brown fur allele (b), short allele (S) and long fur (s) The phenotype of parents is given as follows, find genotype a) Black short X Black short Genotype = BbSs X BbSs b) Black short X Brown long Genotype =BBSs X bbSs c) Black long X Black long Genotype = Bbss X Bbss2. The probability of obtaining following genotype in a population when BBccEeFf is mated with BBCcEEFF The probability of B is 1. (calculate after drawing punnet square of BB x Bb, The chances of B is 100%) The probability of C is 1/2 (half the offspring is Cc and other cc) The probability of E is 1 ( Ee x EE, 100% of E in first filial generation and 1/2 is EE and 1/2 is Ee) The probability of F is 1 ( Ff x FF, 100% of F in first filial generation, 1/2 is FF, 1/2 is Ff) So keeping this in mind calculate the probability for the following genotypes a) BBCcEeFF = 1/2.1/2.1/2.1/2= 1/16 b) BbccEEFf = 1/2.1/2.1/2.1/2 =1/16 c)BbccEeFF = 1/2.1/2.1/2.1/2 =1/16 3. The possible gametes of the given individuals a) AaBbCc = ABC, ABc, AbC, Abc, aBC, abc, abC, aBc b) AABBCc = ABC, ABc c) AABbCc =ABC, Abc, AbC, Abc