Inheritance (Mendelian Genetics)

Inheritance Mendelian Genetics Describe the methodology and conclusions drawn from Mendel’s pea plant experiment Method: Mendel crossed pure breeding pea plants (P generation), then crossed large numbers of the offspring (F1) ………………………………………………………………………………………………………………………………………………….......... He did this for a variety of different characteristics (e.g. flower colour, plant height, etc.) ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... Conclusions: Mendel concluded that organisms have discrete factors that control inheritance of traits (i.e. genes) ………………………………………………………………………………………………………………………………………………….......... There are different versions of these factors (i.e. alleles) and each parent passes on one copy to offspring ………………………………………………………………………………………………………………………………………………….......... Only one version gets expressed (i.e. dominance / recessiveness) ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... Explain how sexual reproduction results in diploid zygotes with two alleles for each characteristic Parents each pass on one copy of every chromosome to offspring via haploid gametes ………………………………………………………………………………………………………………………………………………….......... The resulting zygote is diploid and therefore possesses two copies of every gene ………………………………………………………………………………………………………………………………………………….......... These alternative copies of a gene are called alleles ………………………………………………………………………………………………………………………………………………….......... Differentiate between homozygous, heterozygous and hemizygous Homozygous – Alleles are the same for a given characteristic ………………………………………………………………………………………………………………………………………………….......... Heterozygous – Alleles are different for a given characteristic ………………………………………………………………………………………………………………………………………………….......... Hemizygous - There is only one allele (occurs in males for sex chromosome traits) ………………………………………………………………………………………………………………………………………………….......... Modes of Inheritance Distinguish between complete dominance and co-dominance Complete Dominance – One allele (dominant) masks expression of other allele (recessive) in heterozygotes ………………………………………………………………………………………………………………………………………………….......... Codominance – Both alleles are expressed within a heterozygous phenotype ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... Solve the following questions (HINT: use Punnett squares) 1. Albinism is a recessive trait. An albino man marries a normal woman who had an albino parent. How likely is it that their first child will be albino? 50% (mother is heterozygous) …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………... 2. In guinea pigs, curly hair (H) is dominant over straight hair (h). Work out F1 and F2 generations of a cross between a homozygous curly hair guinea pig and a homozygous straight hair guinea pig. 100% curly hair (100% Hh) F1: ……………………………………………………………………………………………………………………………………………. 75% curly hair, 25% straight hair (1 HH : 2 Hh : 1 hh) F2: ……………………………………………………………………………………………………………………………………………. 3. Red coat colour (R) is dominant to silver-grey coat (r). Work out genotypes and phenotypes for: a) Homozygous red × silver grey: 100% Red (100% Rr) ……………………………………………………………………………………………... 50% Red (Rr) ; 50% silver grey (rr) b) Heterozygous red × silver grey: ……………………………………………………………………………………………… 100% Red (2 RR : 2 Rr) c) Homozygous red × heterozygous red: ……………………………………………………………………………………. 4. In hamsters, black hair is dominant over white hair and curly hair is dominant over straight hair. How would you tell if a black and curly hamster was homozygous or heterozygous for each trait? Perform a test cross (i.e. cross with a known homozygous recessive hamster) …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………... 5. In humans, widow's peak (W) is dominant over a continuous hairline (w), and short fingers (F) are dominant over long fingers (f). Two individuals with widow's peak and short fingers have a child with continuous hairline and long fingers. Determine the genotype of the parents. Parents must be heterozygous (don’t have recessive trait but both pass it on to offspring) …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………... 6. Achondroplasia (dwarfism) is caused by a dominant allele that is embryonically lethal when in a homozygous state. Calculate the likelihood of two dwarf parents having a normal sized offspring. Two thirds (two heterozygotes normally have 3:1 ratio, but homozygous dominant genotype is lethal) …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………... List the possible genotypes for the following blood groups Phenotype Genotype Type A A I I A and I A i Type B B I I B Type AB B and I i A I I B Type O ii Identify which blood group is the universal donor and which is the universal acceptor O is the universal donor ………………………………………………………………………………………………………………………………………………….......... AB is the universal acceptor ………………………………………………………………………………………………………………………………………………….......... Define sex linkage and identify two examples of sex-linked conditions Sex linkage refers to any gene / trait that is located on the sex chromosomes (X or Y) ………………………………………………………………………………………………………………………………………………….......... Two examples of X-linked recessive conditions are haemophilia and red-green colour blindness ………………………………………………………………………………………………………………………………………………….......... Outline why X-linked recessive disorders are more common in males Males have only one X chromosome and so cannot be carriers for recessive conditions ………………………………………………………………………………………………………………………………………………….......... Describe the inheritance and cause of the following genetic diseases Cystic Fibrosis: Autosomal recessive disease ………………………………………………………………………………………………………………………………………………….......... Caused by a mutation to the CFTR gene on chromosome 7 ………………………………………………………………………………………………………………………………………………….......... Causes the production of sticky mucus leading to obstructed airways and digestive issues ………………………………………………………………………………………………………………………………………………….......... Huntington’s Disease: Autosomal dominant disease ………………………………………………………………………………………………………………………………………………….......... Caused by a mutation to the HTT gene on chromosome 4 ………………………………………………………………………………………………………………………………………………….......... Abnormal production of the Huntingtin protein leads to neurodegeneration and dementia ………………………………………………………………………………………………………………………………………………….......... Compare the health consequences associated with the Hiroshima bombing and Chernobyl meltdown Chernobyl meltdown released more fissionable material with longer half lives ………………………………………………………………………………………………………………………………………………….......... Hiroshima bomb was detonated above ground and radiation was dispersed ………………………………………………………………………………………………………………………………………………….......... Hence, Chernobyl responsible for more severe long-term health effects (cancer, etc.) ………………………………………………………………………………………………………………………………………………….......... Solve the following questions (HINT: use Punnett squares) 1. A young man has type AB blood and his sister has type O blood. What are the genotypes and phenotypes of the parents? A I i B and I i …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………... 2. A woman with blood type A marries a man with type O. They have six children, all with type A blood. What are the probable parental genotypes of the parents and could they possibly have a child with type O blood? Woman is PROBABLY homozygous (because all offspring have type A blood) …………………………………………………………………………………………………………………………………………………... However, the woman COULD be heterozygous (meaning there would be a 50% chance of type O offspring) …………………………………………………………………………………………………………………………………………………... 3. A young woman sues a man for parental support for her illegitimate child. The man’s blood type is B, while the mother and child are both blood type O. Based on this information, the judge rules in the man’s favour. Was this wise? No. It is possible the man could be the father if he was heterozygous. More testing required …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………... 4. A woman with haemophilia marries a normal man. What are the probabilities of them having children with haemophilia (include genders)? 100% sons have haemophilia, 0% daughters have haemophilia (but all are carriers) …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………... 5. Can a colour-blind woman have a son with normal vision? Explain No. The son must inherit the X chromosome from their mother (they receive a Y from the father) …………………………………………………………………………………………………………………………………………………... As mother is homozygous recessive, son must receive the recessive trait from the mother …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………... 6. Baldness is X-linked recessive. Identify parental genotypes that could produce bald daughters Father MUST be bald …………………………………………………………………………………………………………………………………………………... Mother may either be bald (homozygous recessive) or a carrier of the bald allele (heterozygous) …………………………………………………………………………………………………………………………………………………... Analyse the data in the following pedigree chart to draw appropriate conclusions Fatal familial insomnia is a rare autosomal condition characterised by a progressive inability to sleep (insomnia), eventually leading to death 1 3 4 9 2 5 6 10 7 11 8 12 13 15 14 16 17 Is the condition caused by a dominant or recessive allele? Explain Dominant – two affected parents have an unaffected offspring ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... Work out (where possible) the genotypes of each of the individuals Hh 1. ……………………… hh 6. ……………………… H_ 11. ……………………… Hh 16. ……………………… hh 2. ……………………… Hh 7. ……………………… hh 12. ……………………… hh 17. ……………………… hh 3. ……………………… Hh 8. ……………………… Hh 13. ……………………… Hh 4. ……………………… hh 9. ……………………… hh 14. ……………………… hh 5. ……………………… hh 10. ……………………… Hh 15. ……………………… What is the probability that individual 11 is homozygous? One in three …………………………………………………………………………………………………………………………………………………..........