Problems in Genetics and Their Solution

2021-05-12 01:44:35
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How would you advise the parents about the probability of their child being a deaf boy, a deaf girl, a normal boy, or a normal girl? Be sure to state any assumptions that you make. (4 points)

The probability of having a deaf boy is 1/4 since this recessive gene runs in the family and the boy will get his X-Chromosome from the mother. This is a case of consangunity

The probability of having a deaf girl is is 1/4 since the mother is a carriar and the father has the recessive gene, and the disorder can only appear in homozygous recessive girls.

The probability of having a normal boy is since the son can only get an X-chromosome from the mother who is a carrier. Meaning on of her X-chromosomes is normal.

The probabilty of having a normal girl is the mother has one normal X-Chromosome since she is a career and the father is affected.

2. The pedigree below shows the inheritance of a rare human disease. Is the pattern best explained as being caused by an X-linked recessive allele or by an autosomal dominant allele with expression limited to males? Please state the reasons why you selected either choice. (3 points)

This is a case of X-linked recessive inheritance. This is because

This allele has appears most frequently in males, whereas in autosomal dominant inheritance, the allele is expected to be distributed evenly in both males and females.

In autosomal dominant inheritance, unaffected parents cannot transmit the allele to the children. This contrary to what is seen it this pedigree which shows unaffected parents transmitting the allele to the children as seen in X-linked recessive inheritance.

In X-linked recessive inheritance affected fathers transmit the mutant allele to their daughters who become carriers and give birth to affected sons. On the contrary, in autosomal dominant inheritance there is no carrier state.

3. The pedigree below is for a rare human disease called spastic paraplegia, nervous disorder in which there is an inability to coordinate voluntary movements. What mode of inheritance is suggested by this pedigree? Which family members must be heterozygous according to your model? (4 points)

This is an X-linked recessive mode of inheritance.

The males must be heterozygous to present with the condition.

4. Draw a four-generation pedigree, following color blindness, using the following information: In generation I, neither parent is color blind. In generation II, a son (II-1) is color blind, a daughter (II-2) is not. II-2 and a man with normal color vision (II-3) have a daughter (III-1) who is not color blind. III-1 and a man with normal color vision (III-2) have a daughter (IV-1) who is not color blind. Write in all the genotypes that are known. Use symbols Xc and Y to follow the sex chromosomes and sex-linked genes. (4 points)

211455028575 X+Y XcX

31242013619500

F1 XcY X+? X+Y

X+?

F2 X+Y

F3 X+?

5. In corn, purple endosperm (P) is dominant to white (p) and full (F) is dominant to shrunken (f). A plant that is homozygous for purple and full is crossed with a plant that is homozygous for white and shrunken. The resulting F1 plants are then testcrossed to the homozygous recessive parent. From the testcross, the following progeny are obtained: 1265 purple & full, 356 purple & shrunken, 341 white & full, and 1154 white & shrunken. Are these two traits sorting independently? Conduct a chi-square test of independence to determine if these two traits are linked. Draw a genetic map that shows the distance between the two genes that control these traits. (6 points)

PpppRow total

Ff1265 341 1606

ff365 1154 1519

Collumn total 1630 1495 3125

Grand total

Observed Expected=Row total x Collumn totalGrand totalPpFf1265 837.69

Ppff365 792.31

ppFf341 768.31

ppff1154 726.69

X2=(Observed-expected)2expected

(1265-837.69)2837.69+(365-792.31)2792.31+341-768.312768.31+1154-6. A series of two-point crosses among fruit flies is carried out between genes for brown eyes (bw), arc wings (a), vestigial wings (vg), ebony body color (e), and curved wings (cv). The following number of nonrecombinant and recombinant progeny were obtained from each cross. (8 points)

Genes in cross Progeny (NR) Progeny (R)

a, bw1110 1114

a, cv 2299 133

a, e 2672 528

a, vg 3775 3776

bw, cv 3160 3160

bw, e 4150 4150

bw, vg 3003 1272

cv, e 4923 1390

cv, vg 1203 1004

vg, e 4559 4559

Using these data from two-point crosses, create a genetic map for these genes, providing the order of the genes and the best estimates of the distances separating them.

7. In corn, colorless endosperm (c) is recessive to colored (C), waxy endosperm (wx) is recessive to normal endosperm (Wx), and shrunken endosperm (sh) is recessive to full (Sh). Plants with colored, normal & full endosperm that are heterozygous are testcrossed to plants with colorless, waxy, & shrunken endosperm. The following numbers of progeny were obtained from these crosses.

Colored, normal, full13

Colored, normal, shrunken2420

Colored, waxy, full171

Colored, waxy, shrunken514

Colorless, normal, full527

Colorless, normal, shrunken142

Colorless, waxy, full2365

Colorless, waxy, shrunken14

Determine if colorless is linked to waxy endosperm, waxy linked to shrunken endosperm, and colorless linked to shrunken endosperm. Draw a genetic map that shows the distance between the three genes that control these traits. Calculate the coefficient of coincidence and interference. (12 points).

 

 

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