Answers to Practice Items

Transcription

1 nswers to Practice Items Question 1 TEKS 6E In this sequence, two extra G bases appear in the middle of the sequence (after the fifth base of the original). This represents an insertion. In this sequence, two bases have been deleted after the fourth base of the original sequence. This represents a deletion. This sequence is the exact same as the original. There is no mutation in this sequence. In this sequence, the fifth base has been chaged from to. This represents a point mutation and is the correct answer. Question 2 TEKS 6E Reading the codon chart, UU originally coded for leucine. U also codes for leucine and would be a silent mutation. U also codes for leucine and would be a silent mutation. UG also codes for leucine and would be a silent mutation. U codes for histidine. This point mutation would result in a different amino acid sequence. This is the correct answer. Question 3 TEKS 6E Insertion of N bases causes the chromosome to become longer with extra information and can lead to different proteins being produced. Gametes that have extra chromosomes are caused by nondisjunction during meiosis. Insertion of N bases could indeed change the sequence of amino acids in a protein. This is the correct answer. Insertion of N bases into a gene would have no effect on other chromosomal fragments. The structure of the ribose sugar does not change under any circumstance. Question 4 TEKS 6E In this diagram, the fragment of one gene is transferred to another chromosome. This represents translocation. This is the correct answer. uplication would result in the repeat of a sequence of the gene in the chromosome. eletion would result in the removal of a sequence of the gene from the chromosome. Inversion would result in the reversal of a sequence of the gene within the chromosome.

2 Question 5 TEKS 6E Mutations can only be passed on to offspring if they occur in the cells which create offspring, the gametes, sperm and egg cells. This is the correct answer. ancerous liver cells may contain mutations, and cancerous cells may spread to other parts of the body. However, only gametes are passed to offspring and mutations must occur there in order to be spread to children. rain cells are not passed on to offspring, and mutations there will not be passed to future generations. Mutated skin cells are not passed on to offspring, and mutations there will not be passed to future generations. Question 6 TEKS 6E Translocation would result in parts of a gene appearing on a different chromosome, not the appearance of an extra chromosome. Inversions are reversals within a gene. These are not likely to be detected on a karyotype. Nondisjuction occurs when chromosomes fail to separate during meiosis. This results in the appearance of extra chromosomes in the karyotypes of offspring. This is the correct answer. The deletion of part of chromosome would simply appear as an abnormally shaped chromosome, not as the appearance of an extra chromosome. Question 7 TEKS 6F If one of the parents was homozygous dominant () and the other parent was heterozygous (b), or if both parents were homozygous dominant, then there would be 0% chance of one of the children being blue-eyed. The probability of one child being blue-eyed (bb) from the mating of two heterozygous dominant (b) parents is 25%. This is the correct answer. 50% probability of having blue-eyed (bb) children would result from the mating of a heterozygous dominant (b) parent and a homozygous recessive (bb) parent. There is no possible scenario that would result in a 75% probability of having blue-eyed children.

3 Question 8 TEKS 6F 0% probability of normal hearing would result from the mating of two homozygous recessive (dd) parents. There is no possible scenario that would result in 25% of offspring having normal hearing. 50% probability of normal hearing would result from the mating of a heterozygous dominant (d) parent and a homozygous recessive parent (dd). This is the correct answer. 100% probability of normal hearing would result from the mating of a homozygous dominant () parent and a heterozygous dominant (d) parent or form the mating of two homozygous dominant () parents. Question 9 TEKS 6F ecause one parent is homozygous dominant for green feathers and the other is at least heterozygous dominant for green feathers, it can be assured that all offspring of these birds will have green feathers. This is the correct answer. For yellow feathers to be present in all offspring, both parents would have had to have yellow feathers (gg). Since the male is heterozygous dominant for a long beak (Ll), the mother would have had to have been homozygous dominant for the trait (LL) in order for all offspring to have long beaks. Short beaks would have been present in all offspring if both parents were heterozygous recessive for the trait. Question 10 TEKS 6F This probability would have shown up only from a cross of yyrr YyRr. See choice for the explanation of how to get the answer. This probability would have shown up only from a cross of YyRr YyRr. See chouce for the explanation of how to get the answer. To solve this problem easily, calculate the probabilities of each trait separately and then multiply the probabilities together. The probability of a yellow pea appearing from a cross of heterozygous dominant traits is 3/4. The probability of a wrinkled pea showing up from a cross of a heterozygous plant and a homozygous recessive plant is 1/2. 3/4 1/2 is 3/8. This is the correct answer. This probability would have shown up from a cross of YYrr YYRr or from YYrr YyRr or from Yyrr YYRr. See choice for the explanation of how to get the answer.

4 Question 11 TEKS 6F Incomplete dominance shows up as an even mixing of the two traits. In this case, red (RR) white (WW) would have offspring who are all RW. This shows up as pink if the trait is incompletely dominant. This is the correct answer. This is not possible. Offspring must contain half the genetic material of each parent. While the offspring plants will be hybrid (RW), the incomplete dominance of the trait means the flowers will be pink, not red. Red and white striped flowers would show up if the trait were co-dominant, not incompletely dominant. Question 12 TEKS 6F Since both parents are hybrid (RW), there will be a 25% chance that the offspring will have a WW genotype. Since both parents are hybrid (RW), there will be a 25% chance that the offspring will have a WW genotype. This is the correct answer. 50% chance of a WW genotype in offspring would result from the cross of a RW cow and a WW cow. There is no possible combination of parents that would produce a 75% chance of a WW genotype in offspring. Question 13 TEKS 6F will be heterozygous with each allele. It is not possible for the children to be homozygous for the type blood trait (I I ). will be heterozygous with each allele. This is the correct answer. will be heterozygous with each allele. It is not possible for the children to have type O blood (ii). will be heterozygous with each allele. It is not possible for the children to be homozygous for the type blood trait (I I ).

5 Question 14 TEKS 6F The dominant trait has been passed on to the child of individual 1 and his wife. Since the wife is recessive, the dominant trait had to have come from individual 1. t least one allele must be dominant, so it doesn t matter what the other allele is. Individual 1 does not necessarily have to be EE only. If individual 1 were ee only, then he and his wife could only have children who have attached earlobes. This cannot be the answer. The dominant trait has been passed on to the child of individual 1 and his wife. Since the wife is recessive, the dominant trait had to have come from individual 1. t least one allele must be dominant, so it doesn t matter what the other allele is. Individual 1 does not necessarily have to be Ee only. The dominant trait has been passed on to the child of individual 1 and his wife. Since the wife is recessive, the dominant trait had to have come from individual 1. t least one allele must be dominant, so it doesn t matter what the other allele is. Individual 1 could be either EE or Ee. This is the correct answer. Question 15 TEKS 6G Meiosis results in the independent assortment of genes with each cycle. The distribution of heterozygous traits becomes randomized with the production of gametes, so no two are exactly alike with respect to the content of their genes. Gametes necessarily indicate sexual reproduction. sexual reproduction is simply mitosis. Through the independent assortment of genes with each meiotic cycle that results in the randomized distribution of traits in gametes, meiosis is the process by which genetic variation is introduced into the species. This is the correct answer. ctually, because three polar bodies result from the production of one egg cell and the production of male gametes does not result in the production of polar bodies, only one egg is produced per meiotic division, whereas four sperm are produced.

6 Question 16 TEKS 6G Normal cells are diploid and contain 23 chromosomal pairs for 46 total chromosomes. ells only have 92 chromosomes at the end of the prophase step of mitosis or meiosis I. Normal cells are diploid and contain 23 chromosomal pairs for 46 total chromosomes. The result of meiosis is a group of haploid cells, each only containing one copy of each of our 23 chromosomes. This is the correct answer. Human cells never have less than some multiple of 23 chromosomes. Question 17 TEKS 6G ell division does not result in the decrease of the number of cells in an organism. Meiosis creates haploid cells, which have only one copy of each of the 23 chromosomes, as opposed to normal cells which are diploid and contain two copies. This is the correct answer. fertilized egg cell only results at the end of the sexual reproduction process when the sperm and egg combine to form the new offspring. Genetically identical individuals only result from asexual reproductive processes, from cloning, or the creation of identical twins by fission of the zygote after fertilization. Question 18 TEKS 6H Inversion of chromosome would not likely be visible on a karyotype. In this karyotype, you can plainly see that this individual has three copies of chromosome 21, which is called trisomy 21. This is the correct answer. Translocation on chromosome 1 might show up as a slight variation between the chromosomes on the karyotype. This is not seen here. eletion on chromosome 5 might show up as a slight variation between the chromosomes on the karyotype. This is not seen here. Question 19 TEKS 6H ecause one gene is cut from one organisms genome and the combined with the genome of a different organism, this technique is called recombinant N. This is the correct answer. Messenger RN is produced during the normal transcription process in a cell s nucleus. Multiple alleles are just multiple copies of genes. This is a normal occurrence in many organisms. Gene mutations are changes of the genome within an organism, and not the actual insertion of foreign N from another organism.

7 Question 20 TEKS 6H Species only shares one band with the unknown organisms, and Species shares none. There are other choices that share more bands. Species shares all four bands with the unknown organism, and Species shares all three of its band with the organism. These are the most closely related organisms in this analysis. This is the correct answer. While Species shares all of its bands, Species only shared one. While Specied shares all of its bands, Species shares none.