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In human beings, the absence of molars is inherited as a dominant tat if two 5)

ID: 143024 • Letter: I

Question

In human beings, the absence of molars is inherited as a dominant tat if two 5) beterozy potes have four children what is the probability that a) all will have no molars? b.) three will have no molars and one will have molars? c) the first two will have molars and the second two will have no molars? 6) Galactosemia is inherited as a recessive trait a) one of four children will be affected? children, what is the chince that b.) three children will be borm in this order normal boy, affected girl, affected boy? 7) Mendel self-fertilized pea plants with round and yellow peas. In the next generation he recovered the following numbers of peas 315 round and yellow pras 108 round and green peas 101 wrinkled and yellow peas 32 wrinikled and sreen peas a) What is your hypothesis about the genetic control of the phenotype? Do the data support this hypochesis? 8) Two curly-winged flies, when mated produce sixty-one curly and thirty-Eve straight-winged progeny. Use a chi-square test to determine whether these numbers fit a 3:1 ratio A short winged, dark-bodied fly is crossed with a long-winged, tan-bodied fly. All the Fi progeny are long-winged and tan-bodied F, flies are crossed among themselves to yield 84 long-winged, tan-bodied flies: 27 long-winged dark-bodied flies: 35 short- winged, tan-bodied flies, and 14 short- winged, dark-bodied flies 9) What ratio do you expect in the progeny? Use the chi-square test to evaluate your hypothesis. Is the observed ratio within the expected range? a) b)

Explanation / Answer

5) the trait for absence of molar present on x chromosome and is dominant (xd)

therefore the genotypes of the two heterozygotes will be xdx (female) and xdy(male)

we can assess the progeny by using punnette square

so here the progeny are

xdxd - no molar teeth

xdx - no molar teeth

xdy - no molar teeth

xy - molar teeth

so the answer is b)three will have no molar teeth and one will have

6)same as in 5th question

consider trait for galactosemia present on x chromosome and is recessesive (xg)

the two normal heterozygotes are xgy(male) xgx(female)

using punnette square

progeny are

xgxg - galactosemia

as the trait is recessesive the remaining progeny is normal

so the answer is a)one of the children is affected

7)mendel self fertilised RrYy(round and yellow)

by the dihybrid cross ,we will get the phenotypic ratio as 9:3:3:1

9 - round and yellow

3 - round and green

3 - wrinkled and yellow

1 - wrinkled and green

the given data doesnot suppoet this dihybrid cross hypothesis

=>the remaining questions comes under different subject because it require chi square test and this test doesnot belong to the biology.it comes under statistics.u can place this question there

gametes xd x xd xdxd xdx y xdy xy