I.New Explanations of Inheritance A. Understanding Gene Function 1. Prokaryote gene regulation a. genes that code for enzymes

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2 I.New Explanations of Inheritance A. Understanding Gene Function 1. Prokaryote gene regulation a. genes that code for enzymes using lactose are expressed only when lactose is present b. Expression of this gene depend on RNA polymerase binding at the promoter c. Then pass through the operon

3 d. Operon set of genes & its control seq.

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5 e. Repressor protein can bind & block RNA poly. from transcribing the genes f. When lactose (inducer) present repressor protein change shape & moves off DNA

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7 g. RNA poly can transcribe the genes encoding the needed enzymes

8 2. Eukaryotic Gene Regulation a. hormones & environmental factors can influence if & when a gene will be expressed b. Hormone enters through p.m. & binds to receptor protein c. Hormone-receptor complex enters nucleus & acts as a transcription factor

9 d. Binds to regulatory DNA seq. called response element e. Nearby gene is activated

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11 3. Cytoplasmic Inheritance a. mitochondrial DNA (mtdna) mitochondria has several copies of mtdna that is circular and double stranded 1. copied independently of nuclear DNA 2. new mito made by fission

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13 b. Maternal Inheritance 1. many mito in egg cell to meet energy and growth needs 2. mito in sperm tail = energy for movement. 3. when sperm enter egg tail disintegrates 4. zygote only inherits mothers mito! p Mother of all of Us?

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15 6. Genomic Imprinting activation or inactivation of certain genes depends on the location on chromosome & parental origin. a. Huntington's Disease inherited by dominant allele: -- if child inherited allele from father symptoms develop earlier -- inherit from mother symptoms develop later

16 b. myotonic muscular dystrophy is more severe if allele inherited from mother c. Every individual has a maternal and paternal copy of a gene d. Maternal gene (imprint) can affect the paternal gene and vice a versa e. Genomic imprinting affects ~ 100 genes

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21 more are inactivated than activated f. Gene imprinting occurs in meiosis

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25 X-inactivation is removed during meiosis a The de novo inactivation model requires many rounds of inactivation and reactivation: the paternal germline initiates meiotic sexchromosome inactivation, but the X chromosome is completely reactivated after meiosis. The zygote inherits two fully active X chromosomes and begins re-inactivation of the paternal X chromosome (X P ) at the 4- to 8- cell stage. In the trophectoderm (extraembryonic cells, shown in blue), X P silencing is maintained, therefore accounting for the imprinted form of X-chromosome inactivation. By contrast, in the epiblast (green cells), yet another round of reactivation takes place in preparation for a final round of inactivation in the form of random X-chromosome inactivation. b In the pre-inactivation model, the female zygote inherits a partially silent X P and maintains the silent state throughout preimplantation development. Silencing becomes globalized and complete in extra-embryonic tissues. This accounts for the imprinted form of X-chromosome inactivation. By contrast, the epiblast cells of the inner cell mass (ICM) undergo a single round of reactivation followed by a random form of X-chromosome inactivation.

26 Epistasis Epistasis is a form of gene interaction in which one gene masks the phenotypic expression of another. There are no new phenotypes produced by this type of gene interaction.

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28 Mutation in Gene 1 interferes with Gene 2. Gene 2 is active, but lack of enzyme 1 prevents enzyme 2 from producing molecule C

29 Epistatic versus Hypostatic The alleles that are masking the effect are called epistatic alleles The alleles whose effect is being masked are called the hypostatic alleles.

30 How do we solve epistatic problems? We are dealing with polygenic traits as in the previous section, however we now have one pair of alleles masking the other. This means we will still be using dihybrid crosses!

31 Recessive or Dominant? Epistasis can be described as either recessive epistasis or dominant epistasis. Let s look at an example of recessive epistasis.

32 Labrador Retrievers Fur color in Labrador Retrievers is controlled by two separate genes. Fur color is a polygenic trait! Gene 1: Represented by B : Controls color Gene 2: Represented by E : Controls expression of B

33 If a Labrador retriever has a dominant B allele, they will have black fur. If they have two recessive alleles (bb) they will have brown fur.

34 If a retriever receives at least one dominant E allele, they will remain the color that the B allele coded for. Either black of brown However, if a dog receives a pair of homozygous recessive e alleles, they will be golden regardless of their B alleles!

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36 BBEE and BbEe --> Black retrievers bbee and bbee --> Brown retrievers BBee, Bbee, or bbee --> Golden retrievers

37 Try this cross You have decided to cross your golden retriever (bbee) with the neighbor s chocolate retriever (bbee). What color pups will they have?

38 bbee x bbee FOIL: be FOIL: be or be Genotypes of F1 generation: bbee and bbee Pups phenotypes: Brown and golden

39 Dominant Epistasis Let s have a look at dominant epistasis Squash fruit color is controlled by two genes. Gene 1 is represented by a W Gene 2 is represented by a G

40 Squash Fruit Color Genotypes and Phenotypes: W-/G- white W-/gg white ww/g- green ww/gg yellow

41 Squash Fruit Color Which allele is epistatic in squash color? The dominant W allele is epistasis How do you know? Because every time a dominant W allele shows up in a squash genotype, the squash fruit color is white.

42 Try this cross. Cross a green squash (wwgg) with a white squash (Wwgg). What color are the offspring?

43 Wwgg x wwgg FOIL: Wg or wg FOIL: wg or wg F1 generation genotypes: Phenotypes:

44 4. Epistasis gene cannot exert its phenotypic effect unless a 2 nd gene is also expressed. a. Ex:) dominant B = black coat homo rec. bb = brown coat b. 2 nd gene = E/e affects the deposition of dark pigment hair c. Dominant E = expression of B/b d. Homo rec. ee = pigment defective & little pigment in hair = light yellow

45 e. light yellow no matter which alleles B/b they carry f. E/e is epistatic to B/b g. e alleles masks the expression of the B/b gene that encodes black or brown

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47 Epistasis often involves B = melanin gene; where a gene at one locus modifies the phenotype (C = deposits pigment) of gene at a second locus (Black or brown) and results in an altered dihybrid ratio. 9 B_C_ 3 bbc_ 3 B_cc 1 bbcc black brown white 9:3:4

48 5. Genetic Anticipation symptoms show up earlier in age over each new generation a. Ex:) Huntington's Disease dominant disorder b. Trinucleotide repeat expansion # of codon repeats increase steadily with each generation c. CAG on chromosome 4 has more than repeats = HD d. Under 39 repeats normal

49 e. Greater # of repeats earlier symptoms appear

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53 Samples of coronal and sagittal magnetic resonance imaging from a patient with Huntington's disease (top row) and a normal control (bottom row) showing the outlines of caudate and putamen (left), cerebral (center) and cerebellar volumes (right).

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56 7. Transposable Elements (AKA: Jumping Genes) pieces of genetic material moved from one chromosome to another (transposon). a. Discovered by: Barbara McClintock 1950 b. took more than 30 yrs to find enough evidence to support hypothesis c. Move in and out of genes that involved pigment

57 d. Observed in Zea mays corn with spotty or streaky coloring

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60 The following illustration shows how grain color in Indian corn may be affected by transposons. The different cards represent a linear sequence of genes on a chromosome. The ace of spades represents a transposon that moves to different positions on the chromosome. The jack of diamonds represents the gene for purple pigmentation in the corn grain. When the transposon (ace of spades) moves to a position adjacent to the gene for pigmentation (jack of diamonds), the pigmentation gene is blocked and no purple is synthesized (white area): When the transposon (ace of spades) moves away from the gene for pigment production (jack of diamonds), the production of purple pigment is resumed (continuous purple area). In this example the gene for pigment production (jack of diamonds) is not adjacent to a transposon (ace of spades):

61 When a transposon moves to different positions within cells of the corn kernel, the coloration gene is "turned on" or "turned off" depending on whether it lands in a position adjacent to the pigmentation gene.

62 8. X Inactivation 1 of female X chromosomes is inactive (turned off) a. random inactivation of maternal or paternal X b. may vary from cell to cell c. 1 cell maternal inactive, another cell paternal might be inactive d. Ex:) Calico Cat females ONLY! patchy colors

63 Calico Cat Is there a human calico? 2001 Lee Bardwell Lionization in a Calico cat. The random pattern of orange and black fur is due to the random inactivation of the X chromosomes Heterozygous for black and orange hair: X B X b (X B =black) (X b =orange) randomly inactivated during embryo development

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66 Lionization in a human female. The pink areas represent patches of skin that have no sweat glands, a condition known as dermal displasia. The gene controlling dermal displasia is located on the X chromosome

67 Lyon hypothesis 1. One X chromosome in each cell becomes inactivated early in development a. The one that is deactivated is random b. All cells derived from cell with deactivated chromosome will have that chromosome deactivated c. This results in genetic mosaics 1. Examples a. Calico cats b. Sweat gland distribution in females heterozygous for extodermal dysplasia