Genetics Answer Note #2

Question 1 choose abcd Which is a correct Punnett square for this scenario, and what is the probability for that one of their children will also be a carrier for the disorder? Answer According to Mendelian inheritance principles, the probability of a child inheriting a gene from their parents is 50%. However, this probability can vary depending on the type of inheritance pattern and the specific genes involved. In the case of a carrier for a genetic disorder, the probability can also be affected by the inheritance pattern and the gene in question. In order to determine the probability of a child being a carrier for a genetic disorder, we must first understand the concept of carrier and how it relates to genetics. A carrier is an individual who carries one copy of a gene for a genetic disorder, but does not display any symptoms or traits. This is because the individual also has a second copy of the gene that is functioning properly. When two carriers of a genetic disorder have children, there is a possibility that the child can inherit one copy of the mutated gene from each parent, resulting in the child being affected by the disorder. Now, let us consider the scenario of a couple, both of whom are carriers for a genetic disorder. In this case, the Punnett square would look like this: |A|A| ----------------------------a | AA | Aa | ----------------------------a | Aa | aa | In this Punnett square, "A" represents the dominant, functioning gene and "a" represents the recessive, mutated gene. As carriers, the parents both have one copy of the mutated gene, represented by the lowercase "a". In this scenario, there are four possible outcomes for their children. 25% will inherit two copies of the dominant gene (AA) and will not be carriers for the disorder. 50% will inherit one copy of the dominant gene and one copy of the mutated gene (Aa) and will also be carriers for the disorder. Finally, 25% will inherit two copies of the mutated gene (aa) and will be affected by the disorder. Therefore, the probability of one of their children being a carrier for the disorder would be 50%. It is important to note that this is a theoretical calculation and may not always accurately represent the actual probabilities in real-life situations. Factors such as mutations, genetic recombination, and other genetic variations can affect the inheritance patterns and the resulting probabilities. Question 2 DNA analysis to discover if fetus is diseased or not Answer DNA analysis, also known as genetic testing, is a highly advanced and accurate method of diagnosing genetic disorders in a fetus. It involves analyzing the genetic material of the fetus to identify any abnormalities or mutations that could lead to a variety of diseases. This process has revolutionized the field of prenatal diagnosis and has provided expectant parents with valuable information about their baby's health. The first step in DNA analysis is the collection of a sample from the fetus, which can be obtained through amniocentesis or chorionic villus sampling. These procedures involve extracting a small amount of amniotic fluid or tissue from the placenta, respectively. The cells in this sample contain the genetic material of the fetus, which can then be analyzed in a laboratory. The next step is to isolate the DNA from the cells and amplify it using techniques such as polymerase chain reaction (PCR). This produces multiple copies of the DNA for easier and more accurate analysis. The amplified DNA is then subjected to various tests, such as karyotyping, to examine the structure and number of chromosomes, and fluorescence in situ hybridization (FISH), which can detect specific genetic abnormalities. Another important aspect of DNA analysis is the use of specialized software and databases to interpret the results. Comparison with a reference genome can help identify any variations or mutations in the genetic code of the fetus that could lead to a disease. Some of the most common genetic disorders that can be detected through DNA analysis include Down syndrome, cystic fibrosis, and sickle cell anemia. This information allows parents to make informed decisions about their baby's health, including prenatal treatment options and planning for the future. Question 3 The child's father has blood group AB (IV), the mother has blood group 0 (I). What blood group and genotype are possible for the child? Select one answer: a. AB(IV) ; IАIВ b. A (II) ; IAIA c. A (II) ;IАI0 d. 0 ( I) ; I0I0 Answer According to basic Mendelian genetics, the phenotype (observable characteristics) of an individual is determined by his or her genotype (genetic makeup). Blood type is determined by the presence or absence of certain antigens on the surface of red blood cells. The ABO blood group system is based on the presence of two antigens, labeled A and B, and the absence of both antigens gives the O blood type. Meanwhile, the genotype of an individual is the combination of genes inherited from both parents. In this scenario, the father has the blood group AB, which means he carries both A and B antigens. Additionally, the genotype for AB blood type is IAIB, with each gene representing one of the blood type antigens. On the other hand, the mother has blood group O and therefore carries neither A nor B antigen. The genotype for O blood type is ii, with two lowercase "i" representing the absence of both antigens. Based on these parental genotypes, the possible blood types and genotypes for the child are as follows: 1. A (II) ; IAIA: In this scenario, the child inherits one A antigen from the father (genotype IA) and one from the mother (genotype IA). Therefore, the child would have blood group A and a genotype of IAIA. This option is incorrect as the father has the AB blood group and cannot pass down a single A antigen to the child. 2. A (II) ; IАI0: Similarly, in this scenario, the child would inherit the A antigen from the father (genotype IA) and the absence of A antigen from the mother (genotype i (lowercase i)). Thus, the child will have blood group A and a genotype of IAIi. Again, this option is not possible as it does not account for the mother's O blood type. 3. AB (IV) ; IАIВ: This option is correct and the most likely scenario. The child would inherit one A antigen from the father (IA) and one B antigen from the mother (IB), resulting in the AB blood type and a genotype of IAIB. 4. 0 ( I) ; I0I0: In this scenario, the child inherits the absence of A and B antigens from both parents (genotype ii). Therefore, the child would have blood group O and a genotype of ii. While this option is possible, it does not account for the father's AB blood group. To summarize, based on the blood types of the parents and their genotypes, the most likely blood group and genotype for the child would be AB (IV) ; IАIВ. This means that the child would have both A and B antigens, just like their father, and their genotype would consist of one A gene and one B gene inherited from their parents. Question 4 Biology test Personal Area My courses EORs Institute of Linguistics and Intercultural Communication Biology test General 02 Test. Genetics. Question 1 No answer yet Score: 1 Not markedFlag question Question text The child's father has blood group AB (IV), the mother has blood group 0 (I). What blood group and genotype are possible for the child? Select one answer: a. AB(IV) ; IАIВ b. A (II) ; IAIA c. A (II) ;IАI0 d. 0 ( I) ; I0I0 Clear my selection Question 2 No answer yet Score: 1 Not markedFlag question Question text Features of mutational variability - Select one answer: a. reversible b. individual c. adaptive d. group Clear my selection Question 3 No answer yet Score: 1 Not markedFlag question Question text Mendel's third law is called Select one answer: a. independent combination of features b. gamete purity c. uniformity of first generation hybrids d. splitting of first generation hybrids Clear my selection Question 4 No answer yet Score: 1 Not markedFlag question Question text How many types of gametes does an organism with the genotype AaBBCC produce? Select one answer: a. 4 b. 2 p. 1 d. 3 Clear my selection Question 5 No answer yet Score: 1 Not markedFlag question Question text When implementing which law is a 3:1 phenotype split observed? Select one answer: a. 3 Mendel's law b. Morgan's Law c. 2 Mendel's law d. 1 Mendel's law Clear my selection Question 6 No answer yet Score: 1 Not markedFlag question Question text How many sex chromosomes does a human have: Select one answer: a. 2 b. 46 p. 23 d. 24 Clear my selection Question 7 No answer yet Score: 1 Not markedFlag question Question text An organism with the genotype AaBbCc can form gametes - Select one answer: a. Vv b. AaB c. ABC d. Aa Clear my selection Question 8 No answer yet Score: 1 Not markedFlag question Question text The dihomozygote has the genotype - Select one answer: a. aaBv b. AABB c. AaBB d. AaBb Clear my selection Question 9 No answer yet Score: 1 Not markedFlag question Question text What amino acid corresponds to the codon AUG? Select one answer: a. Proline b. Phenylalanine c. Lysine d. Methionine Clear my selection Question 10 No answer yet Score: 1 Not markedFlag question Question text Which tRNA anticodon is complementary to the UGC triplet? Select one answer: a. UGC b. UAC c. CSU d. ACG Clear my selection Question 11 No answer yet Score: 1 Not markedFlag question Question text An individual with the AABB genotype Select one answer: a. forms two types of gametes b. forms one type of gametes c. heterozygous d. homozygous for a recessive trait Clear my selection Question 12 No answer yet Score: 1 Not markedFlag question Question text From a marriage between a man with II and a woman with I blood groups, the children may have the following genotype(s) Select one answer: a. IАIВ b. I0I0 c. IВI0 d. IАIА Answer The possible blood group and genotype for the child of a father with AB (IV) and a mother with 0 (I) are determined by Mendelian genetics. According to Mendel's law of inheritance, blood type A is dominant over O. Therefore, the possible blood group for the child is A (II). The genotype of the child can either be IAIA or IAI0. This is because the father can pass on either the A or B allele, while the mother can only pass on the O allele. This results in the child's genotype being either IAIA (homozygous for A) or IAI0 (heterozygous for A). The possible combinations for the child's blood group and genotype are as follows: a. AB (IV); IАIВ (not possible as both parents need to pass on the AB allele, which is not present in the mother's blood group 0) b. A (II); IAIA (possible) c. A (II); IAI0 (possible) d. 0 (I); I0I0 (not possible as the child would have to inherit both O alleles from the mother, but the father can only pass on the A or B allele) Notes: - Blood type inheritance follows the principles of Mendelian genetics, with the A and B alleles being dominant over the O allele. - Both parents' blood types and genotypes must be considered to determine the possible combinations for the child's blood group and genotype. - The child's blood group will be determined by the dominant allele they inherit from each parent, while the genotype will depend on the combination of alleles they receive from both parents. Question 5 Two dogs give 46 grey dogs and 49 black dogs. Those 46 grey dogs give (f1 generation) 64 grey dogs and 19 black dogs. a) What’s the type of hereditary for the two colours. b) What are the genotypes? Answer Based on the information given, the type of hereditary for the two colors is likely to be codominance. This means that both colors, grey and black, are able to express themselves fully in the offspring. This is in contrast to incomplete dominance where there would be an intermediary color. B) The genotypes for the two colors can be represented as follows: Grey color: Gg Black color: Bb This represents codominance, meaning the dominant allele for grey color (G) and the dominant allele for black color (B) are both expressed in the offspring. The lowercase letter represents the recessive allele.