本期的內(nèi)容是獨(dú)立分配。本文集的這一部分是遺傳、進(jìn)化與生態(tài)學(xué) Genetics, Evolution, and Ecology. 這門課理論上建議在閱讀完文集的第一部分的內(nèi)容之后再開始學(xué)習(xí)，但基礎(chǔ)不足的朋友也可以嘗試閱讀喔~

這一部分的主要內(nèi)容均來自 Prof. Angela J. Roles 的 BIOL 200 課程，因此本文集的這一部分均不會(huì)標(biāo)記為原創(chuàng)。但由于文本來源不清晰，UP主還是一個(gè)字一個(gè)字碼出來的文章，本文禁止非授權(quán)的轉(zhuǎn)載，謝謝！

Lesson 14: Independent Assortment

[1] Principle of Independent Assortment

?Dihybrid cross = breeding of parents who are both heterozygous for 2 different genes whose proteins influence different phenotypes (e.g., seed color and shape)

?????Seed shape alleles: R = round, r = wrinkly

?????Seed color alleles: Y = yellow, y = green

?Expect to see phenotypes:

?????9 round, yellow

?????3 round, green

?????3 wrinkled, yellow

?????1 wrinkled, green

Observed phenotypic pattern

?????Independent inheritance of seed color and shape;

?????Visible in dihybrid crosses as 9:3:3:1 phenotypic ratio (with dominance).

?Actual genetic mechanism

?????Genes for different traits are located on different chromosomes;

?????During meiosis, different chromosomes, carrying genes for different phenotypes, move independently of each other.

[2] Non-Independent Assortment

Contexts where phenotypes for different traits are NOT inherited separately:

?????Pleiotropy: a single gene may influence multiple phenotypes, thus variation in these phenotypes is not inherited independently;

?????Epistasis: phenotype can’t be predicted from genotype for only one gene;

?????Linkage: alleles of genes that are physically near to each other on chromosomes may be inherited together (they are only separated by crossing-over events).

?

?When assortment isn’t independent:

? ? - Phenotypes from a dihybrid cross (GgWw × GgWw) will not show the expected 9:3:3:1 ratio;

????- Phenotypes from a test cross (GgWw × ggww) will not show the expected 1:1:1:1 ratio.

?

[3] Test crosses: determining an unknown genotype

?A test cross is when an organism with a known phenotype (but uncertain genotype; homozygous or heterozygous) is crossed with an organism of known phenotype and genotype.

?For example, Y (yellow) is dominant to y (green) for seed color.

????- We have a plant with yellow seeds, call it Unk; its genotype could be either homozygous (YY ) or heterozygous (Yy);

????- We can write Unk’s genotype as: Y / ? or Y / ?

????- We cross yellow-seeded Unk with a green seeded plant (genotype must be yy):

Y / ? ?× ?yy

????Note that all of yy’s gametes will have the y allele for this seed color gene.

?

?We cross yellow-seeded Unk with a green seeded plant (genotype must be yy):

Y / ? ?× ?yy

?If UnkGeno is homozygous (YY ), what will be the phenotypes and genotypes of the offspring? In what ratios?

????- All of Unk’s gametes will get a Y from it.

????- Thus, for the cross YY × yy, all offspring will have the genotype Yy and the phenotype yellow seeds.

?If UnkGeno is heterozygous (Yy), what will be the phenotypes and genotypes of the offspring? In what ratios?

????- Half of Unk’s gametes will carry Y for this gene and half will carry y.

????- Thus, for the cross Yy × yy, half the offspring will have genotype Yy with phenotype yellow seeds and half will have genotype yy with phenotype green seeds.

?

When to use a test cross

(1)?When you know that the phenotype is inherited with a dominant/recessive pattern for 2 alleles of a single gene.

????- You must be able to use a homozygous recessive as one parent so that you can tell which alleles are inherited from the other parent.

(2)?To figure out the genotype of an individual that could be either homozygous or heterozygous, you can’t tell from the phenotype.

(3)?To determine whether two genes assort independently of each other.

????- Parents must be a double homozygous recessive and a double heterozygote: AaBb × aabb;

????- If the A gene is assorting independently of the B gene, then the AaBb parent will produce equal proportions of gametes with AB, Ab, aB, and ab.

????- If assortment is NOT independent, then the AaBb parent will produce mostly gametes that are AB and ab with very few Ab or aB gametes.

?

[4] Pleiotropy: phenotype in chickens

?Gene F in chickens (frizzle gene)

?Frizzle allele exhibits multiple phenotypes:

????- Curled feathers

????- High metabolism

????- High blood flow

????- High body temperature

????- Higher digestive capacity

????- Fewer eggs laid

?It’s an α-keratin gene ---?forms fibrous structural proteins.

?Pleiotropic changes may result from loss of body heat.

?

[5] Epistasis: phenotype depends on genotype at more than one locus

?Three squash shapes: 9 Disk, 6 Sphere,?1 Long

?At least 2 genes each with 2 alleles

?How is squash shape determined?

????- Long shape only aabb

????- Sphere when either aa or bb AND a capital allele at the other locus (ex: Aabb or aaBB) – Disk when at least 1 A and 1 B

?Molecular genetic basis unknown.

?

[6] Linkage: genes on the same chromosome

?Genes that are physically near to each other on the same chromosome are called linked.

????- When genes are physically linked, the phenotypes encoded by their alleles can be inherited together

?Sex-linked is a different kind of linkage:

????- 1 of 10 men is colorblind but only 1 of 200 women is colorblind

????- Colorblindness is sex-linked—not the same as physical linkage

????- “Sex-linked” indicates that XX and XY individuals differ in the frequency of a phenotype because the causal gene is located on a sex chromosome.

?

Autosomal linkage is physical

- Parental gametotypes

????Only chromosomes with the haplotypes gC or gC are observed, as inherited from parents (no crossing-over).

- Resulting haploid gametes

????? If the Gee and Cee genes are independently assorting, we expect to see the following gametotype proportions:25% gC, 25% gC, 25% gC, 25% gC;

????? Instead, we see: 50% gC and 50% gC.

?

Linkage causes an association between traits

????Plant height and pod shape are physically closely linked --- probably not independently assorted.

If plant height and pod shape are linked...

????Then maybe some phenotypes are commonly found together, like tall and constricted, while others are rare.

Crossing-over reduces linkage

?Here we see 3 separate points of exchange beween homologous chromosomes;

?Most chromosomes experience more than 1 chiasma per meiosis;

?Genes located farther apart are MORE likely to be separated by crossing-over →?high recombination rates:

????-?Parental gametotypes: XYZ and xyz

????-?Recombinant gametotypes: Xyz, XyZ, xYz, and xYZ

?

Test cross to detect linkage

?To check for linkage, we do a test cross: YyWw × yyww (black bodied,?red eyed fly?crossed with yellow-bodied, white-eyed fly)

?What phenotypic ratio do we expect if the?genes are unlinked?

????- Parental?types:

????????1 black body, red eyes YyWw parental types

????????1 yellow body, white eyes yyww

????- Recombinants:

????????1 black body, white eyes Yyww

????????1 yellow body, red eyes yyWw

?What if genes are tightly linked? Then recombinants will be?under-represented.

?

?To show our hypothesis of linkage, we can re-write our cross:?YW /yw × yw/yw

?We actually observed:

????- Parental?types:

????????542 black body, red eyes YW /yw

????????441 yellow body, white eyes yw/yw

????- Recombinants:

????????9 black body, white eyes Yw/yw

????????8 yellow body, red eyes yW /yw

?We observed 1.7% recombinants. These genes are linked, they are 1.7?cM apart.

?

Building linkage maps from recombination rates

?The rate of recombination?= the rate at which?recombinant phenotypes?are observed.

?Unlinked genes will assort?independently, yielding a?50% rate of recombination.

?Distance of 1.5 between yellow body gene and white eyes gene means?1.5% recombination rate, 1.5 centimorgans (cM)

????- Caveat: linkage maps are relative?--- can’t convert cM into base pairs

?Use a test cross to check for linkage between a pair of genes.

?

Constructing the linkage map

?To get the distances from the Yellow body?gene, we would have?observed in test crosses:

????- Distance to white eye gene: Yw or yW occurred 1.5% of the time;

????- Distance to vermilion eye gene: Yv or yV occurred 30.7% of the time;

????- Distance to miniature wings gene: Ym or yM occurred 33.7% of the time;

????- Distance to rudimentary wings gene: Yr or yR occurred 50% of the time?(unlinked).

?