Bio 30 4th Exam Reviewer

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BIO 30 4TH EXAM REVIEWER

Merlyn S. Mendioro

Delayed and Extra-chromosomal Inheritance
1.) Genetic factors that are located outside the chromosome: plasmagenes, plasmons, cytogens, plasmids. 2.) Plasmid inheritance implies: perpetuation through DNA Replication. 3.) Killer gene particulate material in Paramecium aurelia: kappa 4.) Mirabilis jalapa shows extrachromosomal inheritance in the ___________. When a pale male parent is crossed with a green female the result usually is __________. : chloroplast, green 5.) Hereditary mitochondrial diseases are transmitted only through the ____________ line since ______________ can hardly contain mitochondria. : maternal, spermatozoa 6.) Gradual loss of the ability to control eye movement: Progressive External Ophthalmoplegia 7.) Occurs during childhood characterized by a combination of anemia, reduction of all blood cells, dysfunction of the pancreas, liver and kidneys: Pearson Syndrome 8.) Cytoplasmic nuclear male sterility:

9.) Mothers transmit virus-like particles called sigma. What is this? Infective Heredity 10.) Extrachromosomal particles/plasmids free of the host organism or integral of the organism chromosome that are infective. For example E. Coli’s fertility trait. Episomes 11.) What are the different criteria for extrachromosomal inheritance?: Difference in reciprocal cross results, Maternal Inheritance, Non-mappability, Non-segregation, Non-Mendelian Segregation, Indifference to nuclear substitution, Infection-like transmission 11a.) Difference in reciprocal crosses – if the normal cross is equal to the reciprocal cross Maternal inheritance – if the traits are mainly from the mothers because of the cytoplasm contributed. Non-mappability – the extra-chromosomal gene cannot be mapped. Non-segregation – failure to show segregation merits extrachromosomal heredity Non-Mendelian Segregation – does not follow Mendelian proportions. Indifference to nuclear substitution – when characteristic persists in presence of nuclear transmission. Extrachromosomal inheritance comes into play. Infection-like transmission – transmitted without nuclear transmission, IT IS Extrachromosomal. Quantitative Genetics

1.) A quantitative trait is _________. Quantitative effects are __________ if they can be added to produce phenotypes, the sum total of the negative and positive effects of individual ____________: polygenic, additive, polygenes 2.) The several basic assumptions for polygene hypothesis: Gene determining quantitative traits = Gene determining qualitative traits only that the former has NO INDIVIDUALLY RECOGNIZED PHENOTYPIC EFFECT

Series of genes independent of one another governs a quantitative trait Genes have cumulative effect
Dominance is ordinarily absent
The F1 appears intermediate of the parents
There is an appreciable influence of the environment on the expression of the trait The only adequate system of classification is through measurement of the trait

3.) Contributory effect per allele = large phenotype-smallest phenotype2n

4.) Finding the frequency of each combination: Use binomial distribution or Pascal Triangle 5.) (a+b)2n , where n is the number of gene pairs, 2n is the number of alleles. 6.) Problem Solving I: The gene AA controls for the phenotype of length of corn. Three genes are responsible for the length of corn. The longest measurement of corn in a sample of 1000 is 130cm while the shortest is 24 cm. Find all the possible progenies of the parent corn and their respective measurements. Include their frequencies.

130 cm – 24 cm = 106 cm
106cm / 6 alleles = 17.67 cm per allele

Gene| aaaaaa| Aaaaaa| AAaaaa| AAAaaa| AAAAaa| AAAAAa| AAAAAA| Length| 24cm| 41.67cm| 59.34cm| 77.01cm| 94.68cm| 112.35cm| 130.02cm| Frequency| 1/64| 6/64| 15/64| 20/64| 15/64| 6/64| 1/64|

7.) Finding the gene pairs:

8.) The tendency of the offspring of extreme...
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