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Non-Mendelian Genetics. Inheritance patterns that don’t follow the rules. A. Linkage. 1. Image of Bateson and Punnett 2. tried to replicate work of Mendel using other traits of garden pea 3. traits investigated -flower color P (purple) and p (red flowers)
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Non-Mendelian Genetics Inheritance patterns that don’t follow the rules
A. Linkage • 1. Image of Bateson and Punnett • 2. tried to replicate work of Mendel using other traits of garden pea • 3. traits investigated • -flower color P (purple) and p (red flowers) • -pollen size L (long) and l (short pollen)
4. Parental cross • a. pure breeding purple flowers with long pollen grains were crossed with pure breeding red flowered plants which possessed short pollen • b. write down the genotypes of the parental generation • c. write down the genotypes of the F-1 generation • d. the F-1 generation were then test crossed • e. expected outcome? • f. actual outcome 7P_L_:1ppL_:1P_ll:7ppll
5. Explanation • a. results far from anticipated 1:1:1:1 • b. showed that alleles for purple color and long pollen traveled together and that red color and short pollen traveled together usually • c. create punnett square for the test cross • d. which parent determined the outcome of the offspring • e. expected that equal numbers of each gamete type would be produced • f. appears that alleles traveled as a unit most of the time-why
If crossing over happens, four different gamete types are produced, but in very different numbers.
Sample Problem • In Drosophila melanogaster there is a dominant gene for gray body color and another dominant gene for normal wings. • The recessive alleles of these two genes result in black body color and vestigial wings respectively. Flies homozygous for gray body and normal wings were crossed with flies that had black bodies and vestigial wings. The F1 progeny were then test-crossed, with the following results: • Gray body, normal wings 236 • Black body, vestigial wings 253 • Gray body, vestigial wings 50 • Black body, normal wings 61 • Would you say that these two genes are linked? If so, how • many units apart are they on the chromosome?
B. Mapping • 1. in the example above crossing over occurred 2/16 times or 12.5% of the time • 2. if the above two genes had been found further apart on the chromosomes, would they have been more or less likely to rearrange themselves by crossing over? • 3. further apart two genes are on a chromosome, the more likely they are to recombine by the process of crossing over • 4. by convention, a 1% crossing over frequencey occurs when the two genes are 1 map unit apart from each other
5. Sample problem-determine the order of the following genes • a. the frequency of crossing over between the following genes is listed below • b. A and B 40% B andC 20% C and D 10% C and A 20% D and B 10%
6. Other points about mapping • a map unit previously was called a Dalton • the physical length of a map unit is not constant • some regions of a chromosome experience crossing over more frequently than others • more money was spent on fruit fly genetics determining order of genes on their chromosomes than was getting men to the moon
C. Multiple allele inheritance-blood type • 1. definitions of antigens and antibodies • 2. agglutination • 3. three alleles involved in blood type • IA produces antigen A on the surface of the RBC • IB produces antigen B on the surface of the RBC • i produces neither of the antigens A nor B • notice that the first two alleles are codominant • An individual who doesn’t possess the antigen does possess a circulating antibody against the antigen
6. Practice genetic’s problem • If a man with blood type B, one of whose parents had blood type O, marries a woman with blood type AB, what will be the theoretical percentage of their children with blood type B?
D. Gene Interactions-don’t appear as typical 9:3:3:1 ratios • Collaboration • Two dominant alleles produce a phenotype that neither can produce alone • Mutually dependent upon one another to produce the new phenotypes • Comb types in chickens good example • Single comb shown here
e. Other comb types • R produces rose combs • P produces pea combs • R_P_ produces walnut
f. RRpp (Wyandotte chicken) X rrPP (Brahma rooster) • g. list the F1 • h. free range breeding to produce the F2
Sample Problem • A dominant gene, A, causes yellow color in rats. • The dominant allele of another independent gene, R, produces black coat color. • When the two dominants occur together (A/-R/-), they interact to produce gray. • Rats of the genotype a/a r/r are cream-colored. • If a gray male and a yellow female, when mated, produce offspring approximately 3/8 of which are yellow, 3/8 gray, 1/8 cream, and 1/8 black, • What are the genotypes of the two parents?
2. Epistasis • a. the dominant allele of one gene overshadows the expression of the dominant allele of a second gene • b. C is the color gene in fowl • c. cc produces a white fowl • d. I is epistatic over the C allele of the color gene • e. any bird who possesses “I” in his/her genotype is going to be white • f. ii will be colored if the dominant allele C is present
g. IICC (white leghorn chicken) X iicc (white Wyandotte rooster) • h. show the F1 and F2 results
Sample Problem • In Leghorn chickens colored feathers are due to a dominant gene, C; • White feathers are due to its recessive allele, c. • Another dominant gene, I, inhibits expression of color in birds with genotypes CC or Cc. Consequently both C-I- and cc-- are white. • A colored cock is mated with a white hen and produces many offspring, all colored. • Give the genotypes of both parents and offspring.
3. Multiple gene inheritance a. so far been discussing traits that are governed only by one gene b. far from the truth c. most phenotypes that we are aware of are distributed in a bell-shaped curve like human height d. often multiple genes affect such traits e. height in plants might be affected by three genes each possessing two alleles f. the dominant allele of each gene might add 1 cm to basic height of plant g. the recessive allele of each gene would not affect the basic 10 cm height h. aabbcc X AABBCC i. F1 generation selfs itself j. 1/64 6/64 15/64 20/64 15/64 6/64 1/64 k. the more genes affecting a trait, the smoother is the bell curve l. the environment also affects phenotype smoothing off the curve even more
E. Sex Determination in humans 1. human condition there are 23 pairs of homologous chromosomes • 2. 22 pairs of autosomes-nonsex chromosomes • 3. one pair of sex chromosomes that exist in two different configurations
4. X and Y • XX female and XY is male • 5. Concept of Barr Body and determination of genetic sex • 6. Females are mosaics when it comes to traits carried on the X chromosome as either one or the other X chromosome can be active
7. XO Turner’s syndrome -a. appear as a female phenotypically -b. the female pattern is the default pattern set to develop as female unless something triggers a change- -c. but doesn’t develop into the adult pattern -d. hips and breasts remain immature -e. no Barr bodies -f. mental function is not impaired -g. 1/5000 affected -h. approximately 15-20% of aborted fetuses
F. Sex Linkage • 1. genes that are carried on the X chromosome are said to be sex-linked • 2. since the male only has one X chromosome, he only has one allele present for that trait • 3. if he has the allele, it is expressed no matter what • 4. males are genetically weaker than females for this reason • 5. the female has a back up copy of the allele if the first one is defective • 6. another interesting point is that a male inherits his sex-linked traits from his mother ie male pattern baldness • 7. example is hemophilia • a. the trait for hemophilia is inherited as a sex-lined recessive trait • b. express the alleles as • XH for normal and Xh as the allele that represents the recessive state • c. example
Practice Problems • Red-green color blindness is inherited as a sex-linked recessive. If a color-blind woman marries a man who has normal vision, what would be the expected phenotypes of their children with reference to this character? • Suppose that gene b is sex-linked, recessive, and lethal. A man marries a woman who is heterozygous for this gene. If this couple had many normal children, what would be the predicted sex ratio of these children?