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Answers-to-Lecture-Q_40308 - Answers to Lecture Questions...

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Unformatted text preview: Answers to Lecture Questions from 2110 Lecture (Question numbers below match those in the VNET document, "1362 CH 1216 Questions to be covered during lecture") 8. Somatic cells of a camel have 70 chromosomes. Somatic = diploid, thus 70 is the diploid #; gametes of this camel would have a haploid # of 35. A. How many chromatids are present during Anaphase I? To answer this, remember Anaphase I is part of meiosis I during which the cell would still have: (a) the diploid # of chromosomes (70), & (b) these chromosomes would be in the replicated form, each having 2 chromatids. Thus 70 chromosomes x 2 chromatids = 140 chromatids total. B. How many chromatids during Metaphase II? In meiosis II, cells now possess the haploid # (i.e., 35). During Metaphase II the chromosomes are in their replicated form (2 chromatids each). So, 35 chromosomes X 2 chromatids each = 70 chromatids. C. How many tetrads during Prophase II? None; tetrad formation followed by crossing over of nonsister chromatids only occurs once, during Prophase I. D. How many chromosomes in a camel skin cell? skin cell = somatic cell = 70 chromosomes E. How many chromatids during Metaphase? This is mitosis & metaphase has the diploid # of 70; 70 replicated chromosomes x 2 chromatids each = 140 chromatids. F. Haploid gametes can undergo meiosis. T or F? False; there would be no homologous pairs in a haploid cell to undergo meiosis. 11. Test your knowledge of the Law of Segregation: A. How many loci in AABbCC genotype? B. How many different gamete types would be formed? AABbCC represents genotype of diploid individual; 2 alleles for each gene. Thus there are 3 gene loci = genes A, B, C; gene B represented by one dominant allele & one recessive allele, A & C genes each represented by 2 dominant alleles. Different gamete genotypes?: Law of Segregation says allele pairs segregate during meiosis such that each gamete receives one allele per pair for each gene. For the genotype AABbCC, each gamete will get from: gene A, 1 allele A allele gene B, 1 allele either B or b allele gene C, 1 allele C allele Different types = 2, either ABC genotype, or AbC genotype. How this looks during meiosis: Haploid gametes containing only one allele of each allele pair 14. Provide a genetic model explaining the data in the following experiment: A. Cross 1: Redeyed mouse x Whiteeyed mouse gives the following F1 offspring: all redeyed mice Conclude: red eye trait dominant to white eye trait; dealing with a single gene for eye color & 2 alleles (red eye allele, white eye allele). Use R, r as alleles. Since cross 1 yielded all redeyed mice in F1 above, parents of F1's must be homozygous, i.e., RR (red) x rr (white); F1's are Rr (heterozygous) B. Cross 2: Redeyed F1 x Redeyed F1 gives the following F2 offspring: 36 redeyed mice & 13 whiteeyed mice Conclude: F1's are heterozygous: Rr x Rr; this cross yields: 15. Provide a genetic model explaining the data in the following experiment: A. Cross 1: Longeared mouse x Shorteared mouse gives the following F1 offspring: 12 longeared mice and 10 shorteared mice Dealing with a single gene for ear length & 2 alleles (longear allele, shortear allele). Call it gene E, and use E, e as alleles. Conclude: not sure which is dominant trait since both traits appear in approx equal amounts in F1. So, look at Cross 2 where the F1 longeared mice are crossed with each other and see what phenotypes are obtained: B. Cross 2: Longeared F1 x Longeared F1 gives the following F2 offspring: 34 longeared mice and 14 shorteared mice Conclude: this 34:14 ratio is close to the 3:1 phenotypic ratio obtained when crossing heterozygotes. This indicates that the longeared trait is dominant to short ears, thus these F2 mice are heterozygous, Ee. Crossing Ee x Ee gives 1 EE, 2 Ee, 1 ee; longeared, shorteared (3:1 phenotypic ratio). This is also consistent with cross 1 above: the longeared mouse is heterozygous (Ee), the shorteared mouse is homozygous (ee). This cross yields: 16. Achondroplasia is a form of dwarfism controlled by one gene with 2 alleles. 2 achondroplastic dwarfs marry and have a dwarf child and later have a child which is of normal size. Based on this: A. Is achondroplasia a recessive or dominant phenotype? B. What are the genotypes of the 2 parents? Answer: Achondroplasia must be dominant trait for this reason: if it were recessive, then the dwarf parents would both have the recessive genotype of aa. Any children they have (aa x aa) could only be dwarfs (aa); no children of normal height could be produced. Since they eventually had a normal height child, achondroplasia must be a dominant trait. Thus, the dwarf parents would each be heterozygous, Aa. Such a cross, Aa x Aa generates AA, Aa, and aa genotypes; AA, Aa are dwarf phenotypes, aa are normal height. 17. In the pedigree to the right, is the trait expressed by the shaded individuals inherited as a dominant or recessive trait? (or are both possibilities?) Cannot be dominant trait; if it were, affected offspring B would have inherited dominant trait from a parent; neither of his parents are affected (i.e., shaded); thus it must be recessive. Genotypes: A: Aa B: aa C: Aa; can only be this genotype (her son in 3rd generation is affected, i.e., aa) D: Aa ...
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