Answers-to-Lecture-Q_40308

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

This note was uploaded on 04/23/2010 for the course BIOL 1361 taught by Professor Knapp during the Spring '05 term at University of Houston.

Ask a homework question - tutors are online