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"Experience is what you get when you don't get what you want." --Tori Filler.

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Presentation on theme: ""Experience is what you get when you don't get what you want." --Tori Filler."— Presentation transcript:

1 "Experience is what you get when you don't get what you want." --Tori Filler


3 Chapter 14 RQ 1.What are the alternative versions of a gene called? 2.An unknown genotype breeding with a recessive is known as a what? 3._________ is when one gene can have multiple phenotypic effects. 4.A pink offspring from a red parent and a white parent is demonstrating what type of inheritance? 5.Blood types exhibit which type of inheritance pattern?

4 1. Define the terms character, trait, true- breeding, hybridization, monohybrid cross, P generation, F 1 generation, and F 2­ generation. Character  a heritable feature Trait  each variant of a character (ex: colors) True-breeding  self-pollinated plants Hybridization  the crossing of two varieties Monohybrid cross  tracks the inheritance of a single character (flower color) P generation  the true-breeding parents F 1 generation  the 1 st set of hybrid offspring F 2 generation  the 2 nd generation from allowing the F 1 set to self-pollinate


6 2. List four components of Mendel’s hypothesis that led him to deduce the law of segregation. 1.Alternative forms of genes are responsible for variations in inherited characters (alleles) 2.For each character, an organism inherits 2 alleles, one from each parent 3.If the 2 alleles differ, one is fully expressed (dominant) and the other is completely masked (recessive) 4.The 2 alleles for each character segregate during gamete production (allele pairs separate during gamete production)


8 3. Do a Punnett Square to predict the results of a monohybrid cross between pure white and pure purple flowers and state the phenotypic and genotypic ratios of the F2 generation. P = purple: p = white P  PP x ppPp F 1  Pp F 2  Pp x PpP Phenotypic: 3:1 Genotypic: 1:2:1 p PP (purple) Pp (purple) Pp (purple) pp (white)


10 4. Distinguish between genotype and phenotype; heterozygous and homozygous; dominant and recessive. Genotype  an organism’s genetic makeup Phenotype  an organism’s expressed traits Homozygous  2 identical alleles for a trait (RR) Heterozygous  2 different alleles for a trait (Rr) Dominant  the gene expressed in the phenotype Recessive  the allele not represented in the phenotype (masked)

11 5. How can a testcross be used to determine if a dominant phenotype is homozygous or heterozygous? Testcross  the breeding of an organism of unknown genotype with a homozygous recessive PP x ppor Pp x pp all Pp 2Pp : 2pp


13 6. Define random event, and explain why it is significant that allele segregation during meiosis and fusion of gametes at fertilization are random events. The outcome of random events are unaffected by the outcome of previous such events If we know the genotypes of the parents, we can predict the most likely genotypes of their offspring using the laws of probability Probability scale  0 to 1 (0 won’t happen) - probabilities for all possible outcomes of an event must add up to 1

14 7. What is the law of independent assortment? The independent segregation of each pair of alleles during gamete formation (meiosis)


16 8. How would you use the rule of multiplication to calculate the probability that a particular F2 individual will be homozygous recessive or dominant? How would you use the rule of addition to calculate the probability that a particular F2 individual will be heterozygous? Rule of multiplication  probability that independent events will occur simultaneously is the product of their individual probabilities Ex: Pp x Pp – what is the probability it will be homozygous recessive (pp)? egg – F 1 (Pp) will get p allele = ½ sperm – F 1 (Pp) will get p allele = ½ ½ x ½ = ¼ chance of (pp)

17 Addition… Rule of addition  probability of an event that can occur in 2 or more independent ways is the SUM of the separate probabilities of the different ways. Ex: in a cross of 2 heterozygotes (Pp x Pp) what is the probability that the offspring will be heterozygous? egg – recessive allele – ½ sperm – dominant allele – ½ = 1/4  1/2 egg – dominant allele – ½ sperm – recessive allele – ½ = 1/4

18 9. Use a Punnett Square to predict the results of a sample dihybrid cross and state the phenotypic and genotypic ratios of the F2 generation. Samples on board

19 Using the laws of probability, predict from a trihybrid cross between two individuals that are heterozygous for all three traits, what expected proportion of the offspring would be: a) homozygous for the three dominant traits PpRrYy x PpRrYy  PPRRYY Treat as 3 separate monohybrid crosses Pp x Pp – prob. for PP = ¼ Rr x Rr – prob. for RR = ¼ Yy x Yy – prob. for YY = ¼ Rule of multiplication (1/4 x 1/4 x 1/4 ) = 1/64 chance of ‘PPRRYY’

20 b) heterozygous for all three traits PpRrYy x PpRrYy  PpRrYy Treat as 3 separate monohybrid crosses Pp x Pp – prob. for Pp = ½ Rr x Rr – prob. for Rr = ½ Yy x Yy – prob. for Yy = ½ Rule of multiplication (1/2 x ½ x ½) = 1/8 chance of ‘PpRrYy’

21 c) homozygous recessive for two specific traits and heterozygous for the third List the genotype possibilities that are homozygous recessive for 2 traits and heterozygous for 1 trait. Use the rule of multiplication to get the probability that one genotype shows Use the rule of addition to calculate probability pprrYy – ¼ x ¼ x ½ = 1/32 ppRryy – ¼ x ½ x ¼ = 1/32  3/32 chance Pprryy – ½ x ¼ x ¼ = 1/32

22 11. Give an example of incomplete dominance and explain why it is not evidence for the blending theory of inheritance. It is the pattern of inheritance in which the dominant phenotype is not fully expressed in the heterozygote, resulting in a phenotype intermediate between homozygous dominant and homozygous recessive Ex: red (RR) and white (rr) = pink (Rr) Alleles maintain their integrity in the heterozygote and will separate during gamete formation


24 12. Explain how the phenotypic expression of the heterozygote is affected by complete dominance, incomplete dominance, and codominance. Complete ------Incomplete ------Codominance In complete dominance, the AA and Aa look the same In incomplete dominance, the A is incompletely dominant and Aa is an intermediate phenotype In codominance, there is no dominant allele and the alleles are expressed equally

25 13. Describe the inheritance of the ABO blood system and explain why the I A and I B alleles are said to be codominant. Blood types A, B, AB, and O Both A and B antigens are expressed in heterozygotes

26 14. Define and give examples of pleiotropy. Pleiotropy  the ability of a single gene to have multiple phenotypic effects Ex: hereditary diseases – a single defective gene causes complex sets of symptoms - tigers and siamese cats have a gene for abnormal coloration which also affects the eyes, which end up with a cross-eyed look

27 15. Explain how epistasis affects the phenotypic ratio for a dihybrid cross. Epistasis  the interaction between two nonallelic genes in which one modifies the phenotypic expression of the other The dihybrid phenotypic ratio will deviate from the usual 9:3:3:1 to 9:3:4 Ex: BbCc x BbCc  9 Black B _ C _ 3 Brown bbC _ 4 Albino _ _ cc


29 16. Describe a simple model for polygenic inheritance, and explain why most polygenic characters are described in quantitative terms. It is the model of inheritance in which the additive effect of two or more genes determines a single phenotypic character Ex: skin color - 3 genes with dark allele A, B, C contribute one “unit” of darkness to the phenotype * AABBCC  very dark * alleles have a cumulative * aabbcc  very light effect so ‘units’ are used *


31 17. Describe how environmental conditions can influence the phenotypic expression of a character. A single gene may produce a range of phenotypes in various environments Ex: blood cell count varies on altitude, activity levels, and infections Ex: skin color is multifactorial – depends on many factors (genotype, environmental influences..)


33 18. Given these family pedigrees, deduce the genotypes for some of the family members.

34 19. Describe the inheritance and expression of cystic fibrosis, Tay-Sachs disease, and sickle-cell disease. All are recessively inherited disorders Cystic fibrosis  4% of Caucasians are carriers - need 2 recessive alleles to occur - 1 in 2500 Caucasians inherit the disease Tay-Sachs  primarily found among Jewish descent - it is a lipid accumulation in the brain - 1 in 3600 inherit the disease Sickle-cell  primarily found in African-American descent - 1 in 400 inherit disease - it is the result of a single amino acid substitution in hemoglobin - 1 in 10 is heterozygous, which provides an enhanced resistance to malaria


36 20. Explain how a lethal recessive gene can be maintained in a population. Heterozygotes act as carriers in the population who do not show the lethal recessive

37 21. Explain why consanguinity increases the probability of homozygosity in offspring. Shared ancestry contributes to the same alleles being passed In other words, you are more likely to inherit the same recessive alleles from related parents than from unrelated parents

38 22. Explain why lethal dominant genes are much more rare that lethal recessive genes. Because heterozygous dominant individuals exhibit the trait Homozygous dominants usually spontaneous abort Homozygous recessives have the normal phenotype They are rarer because they are always expressed (effects are not masked)

39 23. Give an example of a late-acting lethal dominant in humans and explain how it can escape elimination. If the disorder does not appear until an advanced age after afflicted individuals may have transmitted the lethal gene to their children Ex: Huntington’s disease  degenerative disease of the nervous system - children of afflicted parent have a 50% chance of also getting the disease later in life

40 24. Explain how carrier recognition, fetal testing and newborn screening can be used in genetic screening and counseling. Carrier recognition  tests are available to check for heterozygous carriers – can make informed decisions about having children Fetal testing  - amniocentesis – remove amniotic fluid at 14 – 16 weeks - chorionic villus sampling (off placenta) – can test at 8 – 14 weeks, can know results in 24 hours - ultrasound– create image - fetoscopy – fiberoptic scope in uterus (used when problems are high) Newborn screening  screen for problems right away - ex: phenylketonuria - 1 in 15,000 inherit this disease - cannot break down the amino acid phenylalanine



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