review genetics lecture 1

CLASSICAL GENETICS I

Review for Lecture P101

most important concept:

Genetic inheritance is based on probabilities of specific chromosomes and genes being present in the gametes which produce a zygote, the cell which multiplies to produce a new individual.

Chapter 9 explains how genes get distributed to gametes

Meiosis:

• fig 9.2

• CD activity 9.1.

Meiotic errors: karyotypes used for diagnosis p. 178, and CD activity 9.2

Classic (Mendelian) P-1 Cross

• Two different pure-strain individuals, • example: one is round and the other wrinkled.

• Pure strain or true breeding means that the genetic trait in question (round or wrinkled) has been the same in all known ancestors of these individuals.

• The two P-1 individuals are crossed (mated). sex.

Classic P-1 Cross

• Phenotypes = what you see; results of experiments (what you can count or measure)

• genotype = hypothesis, the cause of the phenotype;

• you can’t see genotypes in classical genetics; genotypes are hypothetical

Classic P-1 Cross

• The two parents have a homozygous genotype (hypothesis)

– One is homozygous dominant AA– The other homozygous recessive aa

Recessive Alleles

• Sometimes “skip” a generation

• Definition: a version (allele) of the gene which can be hidden by another version of the gene on the other homologous chromosome.

CLASSIC P-1 CROSS

• How do we find out whether round or wrinkled is recessive?

• We observe the phenotype of the F-1 generation -- the children of the P-1’s

• (F-1’s all will have the Aa genotype;WHY???)

F-1 generation in a P-1 crossall will have the Aa genotype

So… which allele was recessive?

The Aa Genotype

• (or Rr or Hh or Ff or AbAc)

• is heterozygous

• is sometimes called hybrid or monohybrid

• the genes from the two parents are two different alleles.

four possible types of crosses

• P-1 or homozygous cross (AA x aa)

• Hybrid cross (Hh x Hh)

• Test cross (Tt x tt) (Failed test cross) (Ff x FF)

•Hh cross

All possible problem types are based on two basic questions:

• Given these parents, what is the probability that an offspring will have __ genotype or ___ phenotype?

• Given this ratio of offspring, what are the most likely parental genotypes or phenotypes?

four possible types of crosses

• P-1 or homozygous cross (AA x aa)

• Hybrid cross (Hh x Hh)

• Test cross (Tt x tt) (Failed test cross) (Ff x FF)

most important concept:

Genetic inheritance is based on probabilities of specific chromosomes and genes being present in the gametes which produce a zygote, the cell which multiplies to produce a new individual.

PKU PROBLEMS(phenylketonuria: see the label on a diet drink)

• It’s recessive; to have the disease your genotype must be– a. Rr– b. RR– c. rr– d. either “a” or “b” above

PKU PROBLEMS(phenylketonuria: see the label on a diet drink)

• It’s recessive; to have the disease your genotype must be– a. Heterozygous– b. homozygous dominant– c. Homozygous recessive– d. either “a” or “b” above

PKU PROBLEMS(phenylketonuria: see the label on a diet drink)

• Carriers are heterozygous; they don’t have the disorder, but they may pass it to their descendents.

• When two carriers marry, what is the probability that their first child will have PKU?

PKU PROBLEMS(phenylketonuria: see the label on a diet drink)

• Carriers are heterozygous; they don’t have the disorder, but they may pass it to their descendents.

• When two carriers marry, what is the probability that their second child will have PKU?

PKU PROBLEMS(phenylketonuria: see the label on a diet drink)

• Carriers are heterozygous; they don’t have the disorder, but they may pass it to their descendents.

• Two carriers marry, and they plan to have two children. What is the probability that both children will have PKU?

PKU PROBLEMS(phenylketonuria: see the label on a diet drink)

• Carriers are heterozygous; they don’t have the disorder, but they may pass it to their descendents.

• Two carriers marry, and they plan to have two children. What is the probability that both children will have PKU and that both children will be boys?

PKU PROBLEMS(phenylketonuria: see the label on a diet drink)

• When a phenylketonuriac marries a person known to be a carrier, what is the probability that their first child will have PKU?

PKU PROBLEMS(phenylketonuria: see the label on a diet drink)

• When a phenylketonuriac marries a person who does not have PKU and has been tested and found not to be a carrier, what is the probability that their first child will have PKU?

“RULES FOR WORKING GENETICS PROBLEMS”

1. For labs and predicted results on quizzes, the phenotype ratios are the results (or observations). Genotypes are hypotheses and conclusions (inferences and interpretations) and explanations for the phenotypes. Genotype includes words like dominant, recessive, heterozygous, sex-linked, etc. Genotype words cannot be in the “results” sections or in the predictions. Genotypes are only in the hypothesis and conclusion parts.

“RULES FOR WORKING GENETICS PROBLEMS”

2. Do a SIMPLE Punnett square for every problem. If it’s a dihybrid type problem, do two Punnett squares and multiply the fractions. If the problem is more complex, do more Punnett squares.

“RULES FOR WORKING GENETICS PROBLEMS”

3. For almost every problem, begin by writing down the phenotypes and genotypes like this example:

– Sickle-cell aa

– Normal Aa or AA

– (Carrier Aa)

“RULES FOR WORKING GENETICS PROBLEMS”

• 4. For almost every problem, symbolize or reiterate the question. For example, if it asks how many of the children will be carriers, write down something like this:

– ? Aa

• This is just a good habit to make sure you’re aiming toward the answer and not wasting time calculating how many are not carriers or something else not needed for answering the question.

“RULES” FOR WORKING GENETICS PROBLEMS

1. Results = observations = phenotypes, not….

2. Keep Punnett squares simple.

3. Write genotype/phenotype chart first

4. Make sure you’re answering the question.

5. Don’t even think about ….

Dihybrid cross

Dihybrid = 2 kinds of genes

• Color + texture. Do 2 Punnett Squares.

• Then apply the product principle of probability (multiply)

Bad high school method.

• Don’t do this

• It’s good for explaining, like theprevious slide

• but not goodfor working problems

Cystic fibrosis is recessive. Sickle cell anemia is recessive

• If Frances and Francis are both heterozygous for both cystic fibrosis and sickle cell anemia, – what is the probability that their first child will

have cystic fibrosis?– what is the probability that their first child will

have sickle cell anemia?

Cystic fibrosis is recessive. Sickle cell anemia is recessive

• If Frances and Francis are both heterozygous for both cystic fibrosis and sickle cell anemia, – what is the probability that their first child will

have both diseases?– What is the probability that their second child

will have both diseases?

Cystic fibrosis is recessive. Sickle cell anemia is recessive

• If Frances and Francis are both heterozygous for both

cystic fibrosis and sickle cell anemia, – what is the probability that their first child will

be a carrier of both diseases?– What is the probability that their second child

will be homozygous dominant for both diseases?

Polydactyly is dominant

• Two 6-toed dogs mate and have 7 puppies, 6 with 6 toes but one with normal toes. What is the most likely genotype of the parent dogs?

Polydactyly is dominant

• A 6-toed dog, Fluffy, mates with a normal-toed dog, Ralph. They have 14 puppies, all with 6 toes. What are the most likely genotypes of the parent dogs?

Polydactyly is dominant and albinism is recessive

• An albino dog with normal toes, Archie, mates with a yellow 7-toed dog, Precious. Precious was sired by a normal-toed albino, Fido. What is the probability that Archie and Precious could have a yellow normal-toed puppy?

Polydactyly is dominant and albinism is recessive

• An albino dog with normal toes, Archie, mates with a yellow 7-toed dog, Precious. Precious was sired by a

normal-toed albino, Fido. If Archie and Precious have 12 puppies, how many are likely to be yellow and normal-toed?

Polydactyly is dominant and albinism is recessive

• An albino dog with normal toes, Archie, mates with a yellow 7-toed dog, Precious. Precious was sired by a

normal-toed albino, Fido. If Archie and Precious have 12 puppies, how many are likely to be yellow and normal-toed and female?

“RULES” FOR WORKING GENETICS PROBLEMS

1. Results = observations = phenotypes, not….

2. Keep Punnett squares simple.

3. Write genotype/phenotype chart first

4. Make sure you’re answering the question.

5. Don’t even think about ….