Lab On X-Linked Genetics

LABORATORY 16 Human Genetics (Patterns of Genetic Inheritance) X-Linked Disorders Does a color-blind male give his son a recessive-bearing X or a Y that is blank for the recessive allele? Yes 1. a. What is the genotype for a color-blind female? Xb Xb How many recessive alleles does a female inherit to be color blind? two b. What is the genotype for a color-blind male? XbY How many recessive alleles does a male inherit to be color blind? One 2. a. With reference to Figure 16.5a, if the mother is a carrier ( XB Xb ) and the father has normal vision (XBY), what are the chances that a daughter will be color blind? None b. A daughter will be a carrier? 50% c. A son will be color blind? 50% 3. a. With reference to Figure 16.5b, if the mother has normal vision ( XBXB ) and the father is color blind (XbY) what are the chances that a daughter will be color blind? None b. A daughter will be a carrier? 100% c. A son will be color blind? None X-Linked Genetics Problems 1. A woman with normal color vision ( XBXb ) whose father was color blind ( XbY ) marries 1 a man with normal color vision ( XBY ) What genotypes could occur among their offspring? Their children could be XBXB, XBXb, XBY, or XbY What genotypes could occur if it was the normal-visioned man’s father who was color blind? This means his wife is not a carrier and since both parents are normal, the children could be only XBX or XBY. 2. Antonio’s father is color blind. ( XbY)) but his mother is not color blind ( XBXb or XBXB) Is Antonio necessarily color blind? Explain. No. Even if his mother is XBXb he could inherit the XB Could he be color blind? Explain. Yes. If his mother is XBXb he could inherit the Xb. 3. Make up a cross involving hemophilia that could be answered by a Punnett square. What is the answer to your genetics problem? For example, a normal man reproduces with a carrier female. What are the chances that a son will have hemophilia? For this cross, the answer is 50%. Multiple Alleles Blood Typing Problems 1. A man with type A blood reproduces with a woman who has type B blood. Their child has blood type O. Using IA, IB, and i give the genotype of all persons involved. Man: IA i Woman: IBi Child: ii 2. If a child has type AB blood and the father has type B blood, what could the genotype of the mother be? IAIA or IAi 3. If both mother and father have type AB blood, they cannot be the parents of a child who has what blood type? Type O blood 4. What blood types are possible among the children if the parents are IAi  IBi? Types A, B, AB and O. Experimental Procedure: Using Blood Type to Help Determine Paternity 2 Complete Table 16.2Blood Types of Involved Persons Mother Child Father? Wanda Sophia #1 #2 #3 Blood AO+ AB+ O+ Atype 4. If a person had AB+ blood, which wells would show agglutination? Conclusion 1. Which man is Sophia’s father based on the results of your blood typing experiment? Explain your reply based on what you’ve learned about the inheritance of alleles for blood type and Rh. 2. (not in book) What's Wanda's full blood genotype? 16.3 Genetic Counseling Determining Chromosomal Inheritance Observation: Sex Chromosome Anomalies Label the karyotype in Figure 16.6 as one of the syndromes just discussed. 1. Poly X there are 3 x’s 3. Kingefelter there are two x’s and a y 2. Turner there is a sing X 4. Jacob there is a single and two y’s Determining the Pedigree Pedigree Analyses 1. a. Notice that neither of the original parents is affected, but several children are affected. This could only happen if the trait were … Autosomal recessive b. What is the genotype of the following individuals? Generation I, individual 1: Aa. This individual has to be heterozygous because some of the children are affected Generation II, individual 1: aa. This individual has to be homozygous recessive because he is affected Generation III, individual 8: Aa. This has to be the case because the mother is homozygous recessive, and 3 the individual has to inherit at least one of her recessive alleles. 2. a. Notice that only males are affected. This could only happen if the trait were … X-linked recessive b. What is the genotype of the following individuals? Generation I, individual 1: XAX?. This female parent must be carrier because she has an affected son. Generation II, individual 8: XAX? Unable to determine whether this female is a carrier or not because she had no children. Generation III, individual 1: XAY This male is unaffected; therefore, he must have received a dominant allele. Construction of a Pedigree 2. What is your key for this trait? a = normal eyelashes A= double row of eyelashes 3. Construct two pedigrees with symbols only for … Student shows the four generations once on right and once on left. Henry’s grandparents, Henry’s parents, Henry and his wife, Henry’s daughter, Polly. 4 4. Which pattern is correct? Autosomal dominant 5. Use correct genotypes to show a cross between Henry and Isabella and from experience with crosses, state the expected phenotypic ratio among the offspring: Cross Henry Isabella Aa × aa Phenotypic ratio: 1:1 5. What are the percentage chances of Henry and Isabella will have a child with double eyelashes? 50% LABORATORY REVIEW 16 1. Homozygous dominant phenotype. Like AA genotype 2. Alleles possessed by an individual is described by genotype 3. 22 pairs of autosomes a human has 4. Autosomal dominant. Since the presence of at least one copy of W is giving widow peak it is dominant. The question does not talk about only males or only females getting widow peak. SO it is autosomal. 5. Pedigree can be used to determine the genetic inheritance if we know the genotype of parents. 6. If an individual carries one copy of recessive allele, he or she is called a Carrier 7. Huntington disease is autosomal dominant. The individual with recessive genotype will not develop the disease. So it is “hh” genotype. 8. Color blindness is a sex-linked (X-lined) disease. A male child receives the X chromosome only from his mother and father gives Y. To get the disease the mother must pass on the disease to him 9. Haemophilia is a sex-linked recessive disorder. Genotype of disease = xhxh or xhY. The genotype of the carrier will be xHxh. a carrier is always a female. 10. AB blood type arises when 2 dominant allele A and B are present together. 11. A child with AB blood group will have received the allele A and B from both parents. Hence a person with O group can’t be a parent of the child. 5 12. The presence of three chromosomes is called Trisomy. Human disease is Down syndrome. 13. When chromosomes are paired in size and shape, homologues chromosomes are created. 14. Pedigree is used to determine the inheritance within the family. Thought Questions 15. Three allele are present in determining the blood types. IA, IB & i. The allele "i" is recessive to both IA & IB whereas IA & IB are codominant. Father (A type blood) genotype = IAi Mother (O type blood) genotype = ii The genotype of the child with type A blood = IAi The genotype of the child with type O blood = i IA i (father) x (mother) ii  Parents 16. XC = Normal Xc = Colorblind XcY (colorblind man) x (normal woman) XCXC----Parents 6 Normal son = 50% Color-blind son = 50% The chances that a color-blind man will have a color-blind grandson = 50% 17. do not answer this question 7