Mendel’s cross:



Chapter 8 – GeneticsGregor Mendal – proved particulate genetics - that inheritance is due genes that remained separate and intact when gametes fused. What he studied: mostly the English Pea PlantNormally, these plants self fertilize, but Mendel was able to easily cross them by manually removing the stigma before it matured. Why pea plants?Easy to maintainMake many offspringShort life cycleMany easy to measure traits in which there are only two forms and it is clear which one is dominantMendel’s cross:P generation - both parents are true-breeding for their trait (homozygous). The P generation is manually crossed. F1 – First filial generation are planted using the seeds formed by that that initial cross. Data is recorded. They exhibit the dominant trait. When the flowers form, they are isolated (place a bag over each flower) to make sure that the F1 generation self pollinatesF2 – Second filial generation seeds are planted. Data is again recorded. These offspring exhibit the dominant trait in a 3:1 ratio to the recessive trait. In other words the recessive trait reappearedTerms to remember/review:HomozygousHeterozygousPhenotypeGenotypeTest CrossMendel’s First Law – The law of segregationWhen any individual produces gametes, the two copies of a gene separate, so that each gamete only gets one copy of the geneMendels second law: Law of independent assortment. – Alleles of different genes assort independently of one another during gamete formation. Punnett Square:On your own you can try this page if you ‘re needing a really good basic Punnett review: Cross: Purple corn is dominant to yellow corn. Starchy corn is dominant to sweet corn. If two plants that are heterozygous for both traits are crossed, what is the phenotype of the offspring?A faster way to do this: use probabilityThe product rule is a useful principle of probability. Here’s the principle: the probability of independent events occurring at the same time is the product of each of their probabilities.If you flip a penny and a nickel at the same time, the probability of both coins coming up heads is ? X ? = ?If you flipped a penny, a nickel and a dime, the probability that they would all come up heads is ? X ? X ? = 1/8Use probability and the product rule to solve the following problems1. In sweet peas purple flowers (P) are dominant over red flowers (p) and short seeds (S) are dominant over long seeds (s). A sweet pea plant that is heterozygous for flower color and seed length is crossed with a plant with red flowers and long seedsa. write the cross : ______________________________X____________________________What fraction of the offspring from this cross with have the genotype Ppss?d. What fraction of the offspring from this cross will have the genotype PpSS? 2. Two individuals with the genotype AaBbDd are crossed.a. what fraction of the resulting offspring will have the genotype aabbdd? what fraction of the resulting offspring will have the genotype will have at least one dominant trait? What fraction of the resulting offspring will have the at least two dominant traits? A plant with genotype DdEdFfGgHh reproduces by self-pollination.What fraction of its offspring will have the genotype DDEeffGGHh What fraction of its offspring will have the genotype be homozygous recessive for at least one trait?PedigreesBecause planning crosses and counting many offspring is impossible with humans, geneticists rely on pedigrees which are family trees that show the occurrence of inherited phenotypes in several generations of related individualsCreate a pedigree that traces the inheritance of cystic fibrosis for the following family:Susan and Thomas are both healthy but Thomas’s brother died of cystic fibrosis. Susan and Thomas marry have have 6 kids; three girls followed by three boys. One of them, a boy, dies of cystic fibrosis at 10. The oldest daughter marries and has a girl with cystic fibrosis. The next oldest daughter marries but has no children. The youngest daughter marries and has 3 daughters and 4 grandchildren and none have cystic fibrosis. One of the two surviving boys of Susan and Thomas marries and has 3 sons. None of them have cystic fibrosis. The last surviving son of Susan and Thomas marries and has 4 children: A daughter who has cystic fibrosis, a son who does not, a daughter who does not have c.f. and a son who does. Is CF recessive or dominant? Non – Mendelian GeneticsMultiple Alleles – more than 2 alleles for a given trait. Example: fur color in rabbits. Here are the possibilities: C - dark gray - encodes for tyrosinase which is an enzyme that makes pigment cchd –chinchilla - encodes for a mutated enzyme that makes the rabbit light gray ch – Himalayan – pigment is restricted to extremities (ears, tips of feet, nose)c – no pigmentC is dominant over the other 3, cchd is dominant over ch and c and ch is dominant over c.Sample problem: A dark gray rabbit is crossed with a Himalayan. The F1 consists of a ratio of 2 black to 2 Chinchilla. Can you determine the genotypes of the parents?Sample problem 2: A second cross was done between a black rabbit and a Chinchilla. The F1 contained a ratio of 2 black to 1 Chinchilla to 1 Himalayan. Can you determine the genotypes of the parents of this cross? Incomplete dominance – the alleles are neither dominant nor recessive to one another. Instead, the heterozygotes have an intermediate. See page 153. Codominance – two alleles of a gene both produce their phenotypes when present in a heterozygote. Example: blood type- I = codes for an enzyme that attaches sugars (carbohydrates) to the surface of red blood cellsIA - attaches A sugarIB – attaches B sugarIo – attaches no sugarBoth IA and IB are dominant over IoCodominance: IA and IB will result in a person that has both A and B sugars attached.As an aside here: people with type A blood can receive type A blood and type O blood but will have a serious immune reaction to type B or ABpeople with type B blood can receive type B and type O but will have a serious immune reaction to type A or type ABPeople with type AB can receive AB, A, B and O and people with O can only receive O. Type O is called the universal donor, type AB is called the universal recipient.Sample problem: Before the use of DNA testing, blood typing has often been used as evidence in paternity cases, when the blood type of mother and child may indicate that a man alleged to be the father could not possibly have fathered a child. (Think of this as old fashioned Jerry Spinger Show). For the following mother and child combinations, indicate which blood groups of potential fathers would be exonerated* *In the past, this problem has resulted in a number of students asking me what ‘exonerated’ means. You possess (or sit near someone that possess) a handheld device that is capable of accessing the entirety of information known to man. You use it to look at pictures of parrots on roller skates and for posting selfies with a latte saying “Getting my starbucks on!!” but you might try using it to look up ‘exonerated’ before asking your tired, overworked and underpaid teacher. Blood Group of MotherBlood Group of ChildMan Exonerated if He Belongs to Blood Group(s)ABAa.OBb.AABc.OOd.BAe.EpistasisOccurs when the phenotypic expression of one gene is affected by another gene. Example: Coat color in laboradorsB black pigment is dominant to b brownE pigment deposition in hair is dominant to e no deposition, so hair is yellow.GenotypePhenotypeEEBBBlackEeBBEEBbEeBbEEbbBrownEebbeeBByelloweeBbeebbExample. Cross two Black labs that are heterozygous for both coat color genes. What are the expected phenotypic ratios??Genes and Environment The phenotype of an individual does not result from its genotype alone. Genotype and environment interact to determine the phenotype.Two terms you need to understand (page154):Penetrance – whether it is always expressedExpressivity – how it is expressed or the degree which it is expressedLinked genesSome genes do not sort independently. They are said to be linked. Ex: Red hair and freckles are linked in that they are very close to each other on the same chromosome and do not separate due to crossing over. Ex. Sex linkage – genes on the X or Y chromosomes won’t sort independent of gender. Any mutation on the X chromosome is sex linked. The sex linked recessive mutations have these characteristicsThe phenotype appears more often in males than femalesA male with the mutation passes it on only to his daughters. Daughter’s who receive one X-linked mutation are heterozygous carriers. They are phenotypically normal but they can pass the mutant allele to their sons or daughtersEx: Color blindness - the result of a mutation on the X chromosome and is always expressed in males when they carry that allele.A woman who is a heterozygous carrier for color blindness marries a man who is not color blind. They have a baby boy. What is the chance he is color blind?A few years later they have a baby girl. What is the chance she is color blind?Not all genes are nuclearIn humans we have about 24,000 genes in the nuclear genome. We have 37 in the mitochondria. All of your mitochondrial DNA came from your mother since the sperm only contributes nuclear DNA to the egg. This portion of your DNA that came from your mother’s mitochondria is called cytoplasmic inheritance. Prokaryotes – All prokaryotes lack a nucleus and reproduces asexually. So how do the evolve? What accounts for the variation we see from one strain of bacteria to another? Mutation –more common in bacteria because they lack a protective nuclear membraneTransduction – DNA enters from invading virusesTransformation – small circular pieces of DNA (called plasmids) slip into the cell Conjugation – bacterial sex. You read that correctly. Let’s take a look at a picture that prokaryote would categorize as pornographic: PlasmidsA plasmid is a small, circular, self-replicating DNA molecule separate from the bacterial chromosome. Some plasmids can exist on their own in a cell or can be taken up into the cell’s chromosome. Plasmids that can do both are called episomesPlasmids are generally helpful however a cell can exist and reproduce without plasmidsThe role of plasmids in bacterial conjugationAn F factor (fertility factor) exists as a plasmid (F plasmid) F factors consist of about 25 genes that cause a bacterium to make a sex pilus. A sex pilus is the connection that forms between 2 bacteria. Bacteria that have the F plasmid can make the sex pilus and are called “male”. F- bacteria cannot make a pilus and only receive genetic information.Antibiotic ResistanceBacteria use R plasmids to pass along resistance to antibiotics. R plasmids can contain as many as 10 genes for resistance to antibiotics making the bacteria with this plasmid resistant to more than one type of antibioticGenetic PracticeFrom your book complete “Apply the Concept” on page 153. There is a misprint so you really only have to answer 1 and 5. What looks like questions 2, 3 and 4 is really just data you use to answer #5 From your book answer “Do you understand the concept 8.2 all 4 questions From your book answer “Apply the concept” on page 157. Explain your answer From your book answer’Do you Understand Concept 8.3” on page 159. Answer just the 1st and 3rd bullet point, skipping the middle one.For the following crosses, determine the probability of obtaining the indicated genotype in an offspring:CrossOffspringProbabilityAAbb X AaBbAAbbAaBB X AaBbaaBBb.AABbcc X aabbCCAaBbCcc.AaBbCc X AaBbccAabbccd. In rabbits, the homozygous CC is normal, Cc results in rabbits with deformed legs, and cc is lethal. For a gene for coat color, the genotype BB produces black, Bb brown and bb a white coat. Give the phenotypic ratio of offspring from a cross of a deformed-leg, brown rabbit with a deformed-leg, white rabbit. Polydactyly (extra fingers and toes) is due to a dominant gene. A father is polydactyl, the mother has a normal phenotype and they have had one normal child. What is the genotype of the father? Of the mother? What is the probability that a second child with have the normal number of digits? In dogs, black (B) is dominant to chestnut (b), and solid color (S) is dominant to spotted (s). What are the genotpes of the parents that would produce a cross with 3/8 black solid, 3/8 black spotted, 1/8 chestnut solid and 1/8 chestnut spotted puppies? In Labrador retriever dogs, the dominant gene B determines black coat and bb produces brown. A separate gene E, however, shows dominant epistasis over the B and b alleles, resulting in a golden coat color. The recessive e allows expression of B and b. A breeder wants to know the genotypes of her three dogs, so she breeds them and makes note of the offspring of several litters. Determine the genotypes of the three dogs. Golden female (Dog 1) X golden male (Dog 2). Offspring: 7 golden, 1 black, 1 brown.Black female (Dog 3) X golden male (Dog 2). Offspring: 8 golden, 5 black, 2 brown The ability to taste PTC is controlled in humans by a single dominant allele (T). A woman nontaster married a man taster and they had three children, two boy tasters and a girl nontaster. All the grandparents were tasters. Create a pedigree for this family for this trait. Indicate, where possible, whether tasters are TT or Tt. Two true-breeding varieties of garden peas are crossed. One parent had red, axial flowers, and the other had white, terminal flowers. All F1 individuals had red, terminal flowers. If 100 F2 offspring were counted, how many of them would you expect to have red, axial flowers? True breeding tall red-flowered plants are crossed with dwarf white-flowered plants. The resulting F1 generation consists of all tall pink-flowered plants. Assuming that height and flower color are each determined by a single gene locus on different chromosomes, predict the results of an F1 cross of the TtRr plants. List the phenotypes and predicted ratios for the F2 generation. In Coleus, some plants have shallowly scalloped edges and others have deeply scalloped edges. A cross is made between homozygous deep and shallow individuals. The shallow trait is dominantusing S and s to symbolize the genes for this trait., give the phenotypic and genotypic ratios for the F1 If self-pollination is allowed, what is the phenotypic ratio for the F2 generation?14. a. In a pea plant that breeds true for all tall, what possible gametes can be produced? Use the symbol D for tall and d for dwarf,b. in a pea plant that breeds true for dwarf, what possible gametes will be produced?c. what will be the genotype of the F1 offspring from a cross between these two types?d. Assuming that the allele for tall is dominant, what will be the phenotype of the F1 offspring from a cross between these two typese. What will be the probable distribution of traits in the F2 generation? 15 The ability to taste a biter chemical, phenylthiocarbamide (PTC) is due to a dominant gene. Use T and t to symbolize the two alleles of this gene.a. What is the genotype of a nontaster?b. What are the possible genotypes of the taster?c. Could a person with two tasters as parents be a nontaster? How?16 Albinism, the total lack of pigment, is due to a recessive gene. A man and woman plan to marry and wish to know the probability of their having any albino children. What are the probabilities if:a. both are normally pigmented, but each has one albino parentb. the man is an albino, the girl is normal but her father is an albinoc. the man is an albino and the girl’s family includes no albinos for at least three generations17. Two short-haired female cats are mated to the same long-haired male. Several litters are produced. Female No. 1 produced eight short-haired and six long-hair kittens. Female No. 2 produced 24 short-haired ones and no long-haired. From these observations, what deductions can be made concerning hair-length inheritance in these animals? Assuming the allelic pair S and s, give the likely genotypes of the two female cats and the male.18. In peas, a gene for tall plants (T) is dominant over its allele for short plants (t). The gene for smooth peas (S) is dominant over its allele for wrinkled peas (s). Determine the probability of obtaining the offspring for each cross:CrossOffspringProbabilityOffspringProbabilityTtSs X TtSsTTSSa.Tall plant, wrinkled pease.Ttss X ttssTtSsb.Tall plant, wrinkled peasf.ttSs X TtssttSsc.Tall plant, wrinkled peasg.Ttss X TtssTTssd.Short plant, wrinkled peash.19. In watermelons, the genes for green color and for short length are dominant over their alleles for striped color and for long length. Suppose a plant with long striped fruit is crossed with a plant heterozygous for both of these characters. What phenotypes would this cross produce and in what ratios?20 Mr. and Mrs. Fecundity, both having blood type B, have 12 children. ? of whom are type B and ? of whom are O, what are the genotypes of the parents?21. A family of six includes four children, each of whom has a different blood type: A, B, AB, and O. What are the genotypes for the parents for this trait?22. Samantha and Jennie were roommates at Breckenridge Hospital and both had daughters at about the same time. After Samantha took baby Tabitha home, she became convinced that the babies had been switched. Blood tests were performed with the following results: Samantha and her husband: Both type AB; Jeannie and her husband are both type A; Tabitha is type A and Jeannie’s baby (Sabrina) is type O. Had a switch occurred? Explain23. Mortimer has type B blood. His wife Molly is unsure of her blood type. If their first child, Paris, is type B, their second child Nicole is AB and the twins MaryKate and Ashley are A, can you determine the blood types of Mortimer and Molly?24. Cystic fibrosis is caused by a recessive lethal gene and can be detected by an excess concentration of chloride in sweat. If a sweat test reveals a man to be heterozygous and his wife to be homozygous normal, what are the chances that their children will have the disease? ................
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