Question

1. (2pts) Which of the following are features of heterochromatin (choose all that apply)

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Increased histone acetylation

Decreased histone acetylation

Densely packed nucleosomes

Loosely packed nucleosomes

Increased DNA methylation

Decreased DNA methylation

Questions 2-3: The Igf2 gene encodes insulin-like growth factor 2, a protein that is important for normal growth. In mice, no expression of the Igf2 gene results in a growth deficiency and a small mouse phenotype. The Igf2 gene is maternally imprinted and autosomal.

2. (2pts) A female mouse is homozygous wildtype for the Igf2 gene and mates with a male mouse that is heterozygous for a null allele of the Igf2 gene (the other allele is wildtype). What are the expected phenotype proportions of their progeny?

All small progeny

All wildtype progeny

25% small progeny and 75% wildtype progeny

50% small progeny and 50% wildtype progeny

75% small progeny and 25% wildtype progeny

All small males and all wildtype females

All wildtype males and all small females

3. (2pts) A female mouse is homozygous for null alleles of the Igf2 gene and mates with a male mouse that is homozygous wildtype for the Igf2 gene. What are the expected phenotype proportions of their progeny?

All small progeny

All wildtype progeny

25% small progeny and 75% wildtype progeny

50% small progeny and 50% wildtype progeny

75% small progeny and 25% wildtype progeny

all small males and all wildtype females

all wildtype males and all small females

Answer 1

Answer. 1 Gene silencing result in formation of heterochromatin regions. Gene silencing is done by DNA methylation, Decreasing acetylation, and heterochromatin regions are densely packed.

So correct option are

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Decreased histone acetylation

Densely packed nucleosomes increased DNA methylation

Answer 2 maternal imprinting means when allele inherited from mother that will be silenced.

So here maternal allele will not be considered, then male is heterozygous for a null allele then half of gamete will have null allele and half have normal,

So 50% small progeny and 50% wildtype progeny will be produced.

Answer 3 All wildtype progeny because male is homozygous for wild allele.