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Medelian Genetics

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卡片总数: 21内容版本: v4公开卡包更新时间: 8/1/2026

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#1
正面 (问题)

Blending theory of inheritance (before Mendel)

背面 (解答)

Traits blend evenly in offspring through mixing of parents blood Explains intermediate traits

#2
正面 (问题)

Mendel garden pea

背面 (解答)

• removed stamens from purple flower and transferred pollen to white flower • control self fertilization and cross fertil. • P was homozygous • F1 was all purple, not pure breed. Heterozygous • F2 both purple and white

#3
正面 (问题)

Law of segregation

背面 (解答)

• inheritance of single character • genes exist in alternative versions called alleles • for each characteristic, organism inherits 2 alleles one from each parent • alleles differ, dominant determines organisms appearance • allele pairs separate during gamete production

#4
正面 (问题)

Recessive alleles for albino coloration in animals

背面 (解答)

• expressed when both are recessive homozygous - Aa brown mice can pass a allele so they carriers

#5
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Testcross

背面 (解答)

To determine unknown genotypes • mating between individual of unknown genotype and homozygous recessive Individual • will show whether unknown genotype includes recessive allele • identify dominat phenotype genotype

#6
正面 (问题)

Independent assortment (2 principle of Mendel)

背面 (解答)

Alleles at different loci (on different chromosomes) are inherited independently -experiment with 2 or more different traits

#7
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Chi-square

背面 (解答)

-if calculated value is smaller than tabulated value at the .5 probability than independent assortment is supported

#8
正面 (问题)

Polygenic inheritance (exception to Mendel one gene codes for one trait)

背面 (解答)

Single character many genes

#9
正面 (问题)

Incomplete dominance (exception to Mendel 1 gene codes for 1 trait)

背面 (解答)

Allele is not completely dominant, the other allele will have effect Ex: snapdragons: pink F1 generation, heterozygote phenotype is different from either parent -1 allele produces functioning protein. Other non-functional

#10
正面 (问题)

Co-dominance (exception to Mendel 1 gene codes 1 trait)

背面 (解答)

Both alleles produce protein that creates trait

#11
正面 (问题)

Pleiotropy (exception to Mendel 1 gene codes 1 trait)

背面 (解答)

Ex:sickle cell -affects type of hemoglobin produced and shape RB -causes anemia and organ damage Single gene many phenotypes!!! -Marfans syndrome: dominant allele ( responsible for tall. Curved spine. Elongated fingers)

#12
正面 (问题)

Epistasis (exception to Mendel 1 gene codes 1 trait)

背面 (解答)

• gene at one locus alters expression of a gene at another • regulatory genes may give rise to epigenetics: turning on and off genes • mice, black is B dominant. b is brown recessive. • presence of second allele C determines pigment is expressed • C is color expression. c is color inhibition. CC and Cc have black or brown. cc are white.

#13
正面 (问题)

Autosomal recessive disorder

背面 (解答)

• sickle cell. Cystic fibrosis. Albinism. Methemoglobinemia • problem with recessive allele • even though not expressed. Parents can be carriers.

#14
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Autosomal dominant disorder

背面 (解答)

• one faulty dominant allele causes disorder • Huntingtons, polydactyl, brittle bone • person with disorder can be homo or hetero

#15
正面 (问题)

X-linked disorder

背面 (解答)

Faulty allele on X chromosome • faulty recessive allele • mother to son and daughter • father to daughter • color blind, hemophilia, duchenne • male with faulty allele x have no other allele on Y to mask effect

#16
正面 (问题)

Faulty chromosome number

背面 (解答)

Chromosome pair fail to separate during meiosis. -monosomy: chromosome with no pair -trisomy: 3 copies of chromosome Non disjunction: in meiosis 1 100% abnormal. Meiosis 2 50%

#17
正面 (问题)

Aneuploidy

背面 (解答)

Gametes with too many or too few chromosomes

#18
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Monosomy

背面 (解答)

Turners syndrome • non disjunction in 23 pair • only X chromosome present • 95% die

#19
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Trisomy

背面 (解答)

Down syndrome -impotency in male -21 non disjunction XXX super female XYY super male -delayed walking emotional difficulties XXY: Klinefelter syndrome

#20
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Linkage

背面 (解答)

• result of two loci being located close together on same chromosome. Causes departure from independent assortment • causes certain combo of genes to be over represented in indivdiaul gametes • F1 wild type, • no linkage expected 25%

#21
正面 (问题)

Linkage mapping

背面 (解答)

• tell how far apart loci are by proportion of F2 from a test cross that are recominants • take number of recombinants and divide by total • recombinants are F2 that do not resemble grandparents • from r get distance between loci by multiplying r by 100 in map units • farther away, higher r, more recominants