Question

#1.

In peas, flower location and plant height are controlled by genes that follow Mendel’a law. One allele at each locus is donminat to the other at that locus: (i) AXIAL FLOWERS (A), terminal flowers (a), and (ii) TALL PLANT (T), short plant (t). A pure breeding plant for the dominant traits is mated with pure breeding plant for the recessive traits. What possible gametes can F1 individuals that interbreed make?

HWS, Check canvas for due date 20 pts NAME LW2 NOTE: Liwsir your anwers te 1. I allele at locus is dominant to the other at that locus: () AXIAL FLOWERS (A), terminal flo ower location and plant height are controlled by genes that follow Mendelrs laws. one a PLANT (T) short plant (). A pure breeding plant for the dominant (a), and (ii) TALL P terminal fnowers (a) traits. What raits is mated with pure breeding plant for the recessive possible gametes can F1 individuals that interbreed make ? (Cirele I answer, 2 pS) TTAA, ttar TT, aa, Ta, LA TT, aa Ta, tA TA. ta TA, Ta, tA, ta 2.1 Red-green colorblindness is an X-linked recessive trait. A normal woman, but a carrier, has children with a normal man. What is the chance that their first child will be red-green colorblind ? ( pts) A, 0% B, 12.5% C.25% D.50% E, 75% F. 100% in the pace at right noathy setap d show your Use standard notation to indicate X-linkage and define your ymbols for full credir (2 pts) 2.2 If the couple in 2.1 has had 3 sons (no twins), what is the chance that they all will be red-green colorblind? a pas) A. 50% B, 25% C, 12.5% D. about 6% E. about 3% F. about 1.5% 2.3 Ifa normal woman whose father was red-green colorblind has children is the chance that they will have red-green colorblind daughter ? l a red-green colorblind son? A, 0%//0% B. 0% // 50% C50% // 50% D50%/y 100%. E. 100%"50% with a red-green colorblind man, what (2 pts) F. 100 % // 100 % Word Problem: In cats, two of many genes that affect coat color are the tabby locus and the orange locus The tabby locus is autosomal The dominant allele M produces the mackerel stripe pattern (Fig. A), while the recessive allele m produces blotched or whorled stripes (Fig. B). The orange locus is X-linked with 2 alleles: O, which produces an orange- based pigment, and o, which is a non-functional allele that does not influence pigment color. The far of cats without an O allele will have a black-grey color base, and other genes, like the mackerel locus can still affect the stripe pattern (as in Fig, A&B). When the O allele is present and active, the fur's color base will be orange/grey. Hence, O exhibits spistasis to other color genes. For example, the cat Garfield from the comie strip, who is an orange mackerel tabby, must carry the O allele and at least one M allele In Reno near the Wild Orchid, a black-grey whorled stripe tabby tomcat mates with a dihybrid female. a) Describe &explain the mother's phenotype (Hint: see Ch 12 class notes, p. 2) b2) Female genotype bl) Male genotype cl) Male gametes: d) Neatly set up the mating using a Punnett square. Use gametes consistent with (c1) and (c2) above. c2) Female gametes: e) Show the expected phenotypic ratio of the offspring,including their sex.

Answer 1

Ans 1. The gametes formed by the F1 individuals will be :- TA, Ta, tA, ta.

Explanation:- It os because of the fact that the genotype of the parent is TTAA and ttaa. The F1 plants that is produced by them is TtAa. So it will produces the gametes like TA, Ta, tA and ta.

Ans 2.1 Option D is the correct option.

Explanation:- There are 50% chances that the first child will be green red colourblindness. Because its gene is present on the X chromosome . So girl child will be always carrier. And the chances are only 50% for colourblind to first child.

Ans 2.2. Option C is the correct option.

Explanation:- There are 12.5% chance that the all 3 son will be colourblind. It os because for a son to be colourblind the probability is 50%.

Ans 2.3 Option A is the correct option.

Explanation:- It is because the normal women has genotype XX and his husband has the genoype X^CY. When they mates , the daughter will be carrier only , so probability is only 0% . And son will have probability upto 0% , it is because mother carry no gene for colourblind. It has normal X chromosome which mate with Y to produce the normal sons.