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topic 3.4 answers, Schemes and Mind Maps of Genetics

University of Bahrain (UOB)Genetics

3.4 Inheritance. Mendelian Genetics. Describe the methodology and conclusions drawn from Mendel's pea plant experiment. Method:.

Typology: Schemes and Mind Maps

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3.4 Inheritance-
Mendelian-Genetics-
Describe(the(methodology(and(conclusions(drawn(from(Mendel’s(pea(plant(experiment(
Method:(
…………………………………………………………………………………………………………………………………………………..........(
…………………………………………………………………………………………………………………………………………………..........(
…………………………………………………………………………………………………………………………………………………..........(
Conclusions:(
…………………………………………………………………………………………………………………………………………………..........(
…………………………………………………………………………………………………………………………………………………..........(
…………………………………………………………………………………………………………………………………………………..........(
…………………………………………………………………………………………………………………………………………………..........(
(
Explain(how(sexual(reproduction(results(in(diploid(zygotes(with(two(alleles(for(each(characteristic(
…………………………………………………………………………………………………………………………………………………..........(
…………………………………………………………………………………………………………………………………………………..........(
…………………………………………………………………………………………………………………………………………………..........(
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Differentiate(between(homozygous,(heterozygous(and(hemizygous(
…………………………………………………………………………………………………………………………………………………..........(
…………………………………………………………………………………………………………………………………………………..........(
…………………………………………………………………………………………………………………………………………………..........(
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Modes-of-Inheritance-
Distinguish(between(complete(dominance(and(co-dominance(
…………………………………………………………………………………………………………………………………………………..........(
…………………………………………………………………………………………………………………………………………………..........(
…………………………………………………………………………………………………………………………………………………..........(
(
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.)
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)
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
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)
Complete Dominance – One allele (dominant) masks expression of other allele (recessive) in heterozygotes
Codominance – Both alleles are expressed within a heterozygous phenotype
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3.4 Inheritance

Mendelian Genetics Describe the methodology and conclusions drawn from Mendel’s pea plant experiment Method: ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... Conclusions: ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... Explain how sexual reproduction results in diploid zygotes with two alleles for each characteristic ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... Differentiate between homozygous, heterozygous and hemizygous ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... Modes of Inheritance Distinguish between complete dominance and co-dominance ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... 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.) 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) 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 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) 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? …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………...
  2. In guinea pigs, curly hair (H) is dominant over straight hair (h). Work out F 1 and F 2 generations of a cross between a homozygous curly hair guinea pig and a homozygous straight hair guinea pig. F 1 : ……………………………………………………………………………………………………………………………………………. F 2 : …………………………………………………………………………………………………………………………………………….
  3. Red coat colour (R) is dominant to silver-grey coat (r). Work out genotypes and phenotypes for: a) Homozygous red × silver grey: ……………………………………………………………………………………………... b) Heterozygous red × silver grey: ……………………………………………………………………………………………… 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? …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………...
  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. …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………...
  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. …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………... 50% (mother is heterozygous) 100% curly hair (100% Hh) 75% curly hair, 25% straight hair (1 HH : 2 Hh : 1 hh) 100% Red (100% Rr) 50% Red (Rr) ; 50% silver grey (rr) 100% Red (2 RR : 2 Rr) Perform a test cross (i.e. cross with a known homozygous recessive hamster) Parents must be heterozygous (don’t have recessive trait but both pass it on to offspring) Two thirds (two heterozygotes normally have 3:1 ratio, but homozygous dominant genotype is lethal)

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? …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………...
  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? …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………...
  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? …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………...
  4. A woman with haemophilia marries a normal man. What are the probabilities of them having children with haemophilia (include genders)? …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………...
  5. Can a colour-blind woman have a son with normal vision? Explain …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………...
  6. Baldness is X-linked recessive. Identify parental genotypes that could produce bald daughters …………………………………………………………………………………………………………………………………………………... …………………………………………………………………………………………………………………………………………………... I i and I i A B 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) No. It is possible the man could be the father if he was heterozygous. More testing required 100% sons have haemophilia, 0% daughters have haemophilia (but all are carriers) 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 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 Is the condition caused by a dominant or recessive allele? Explain ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... Work out (where possible) the genotypes of each of the individuals

  1. ………………………
  2. ………………………
  3. ………………………
  4. ………………………
  5. ………………………

What is the probability that individual 11 is homozygous? …………………………………………………………………………………………………………………………………………………..........

Dominant – two affected parents have an unaffected offspring Hh hh hh Hh hh hh Hh Hh hh hh H_ hh Hh hh Hh Hh hh One in three